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What metal is this?
Does anyone know what metal this is? (right click -> "Open image in new tab" for larger image)

It has to be one of the following: magnesium, aluminum, zinc, tin, antimony, lead.
This metal was found in a laboratory where the students were performing a calorimetry experiment to determine specific heats of a few metals. I don't think it is lead (because of the color and that it is too toxic for students to use) and I don't think it is antimony (because that may be too expensive, but I could be wrong).

I'd say there's a fair chance that the students were doing calorimetry experiments for the oxidation of magnesium. However, it doesn't quite look like "typical" magnesium shavings and doesn't appear to have much of the gray oxide coating that you would expect after a while. How was it stored, and for how long?

Looks like tin to me.

Metallic lead is not that toxic.

I would guess that it is tin. The coloration and cut look right compared to an image of tin turnings (eBay listing for tin metal turnings). I'm more used to seeing aluminum and magnesium as strips rather than turnings.
Beyond that, I think it will probably take more than just the image to positively identify the metal. You could test the material with something akin to part 3 of this experiment (Linus Chem Department .pdf file).
Alternatively, you could also rerun the calorimetry experiment on the metal and see if the specific heat is consistent with any of these.

It is probably tin judging from its appearance. The other metals are unlikely because:

zinc, aluminium, magnesium will have an oxide layer (no shine) unless stored without exposure to air.
antimony is very unlikely for calorimetry.
although lead is toxic, many schools still continue to use compounds of lead, like lead acetate and lead nitrate, for salt analysis (being a common use). So using lead is not unusual.

It HAS to be one of the following: Magnesium, Aluminum, Zinc, Tin, Antimony, Lead

If you have narrowed it down to that group it should be trivial to narrow it down even further. I assume you didn't take high school chemistry.
Lead is both very heavy and very soft. Is the pile of shavings unusually heavy? Pick up a larger piece with pliers. Squeeze. if it deforms easily, it's lead. Lead also melts at very low temperatures. Heat a piece with a gas torch. If it's heavy, soft and melts almost immediately, it's lead.
Both magnesium and aluminum are very light. Is the pile of shavings unusually light? Pick up a piece, put it on a metal pan and heat it with a gas torch. If it ignites and burns with an intense, blue-white flame, it's magnesium (aluminum is much harder to ignite).
Zinc, tin and antimony are a bit harder to separate, so do the easy ones first.

Tin is just about as pliable as lead, and melts even easier. True, it feels somewhat less heavy, but I wouldn't rely on that feeling alone to tell them apart.

If only everyone was as smart as you Peter...


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What does the number mean in Sylgard 170, Sylgard 184, etc?
My lab uses a lot of PDMS, a silicon-based organic polymer, to make microfluidic devices. PDMS is available in many different formulations, and we usually use Dow Corning's Sylgard 184. Recently we got a batch of Sylgard 170 (which is supposed to cure to a solid much faster than 184), and apparently there's a whole bunch of different Sylgard types with numbers ranging from 160 to 577.
My question is, what does that number in the name of each different Sylgard type mean?
I can't seem to find a clear answer online. I know it's not directly related to curing speed, since 182 cures slower than 184. My guess is that it refers to the number of monomer residues in the polymer. Does anyone actually know?

Perhaps molecular weight?

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How do I calculate the equilibrium constant of a reaction from two related reactions' constants?

At a certain temperature the following reactions have the constants shown:
$$\begin{align}\ce{S(s) + O2(g) &<=> SO2(g)} && K'_\mathrm{c} = 4.2\times 10^{52}\\
\ce{2S(s) + 3 O2(g) &<=> 2 SO3(g)} && K''_\mathrm{c} = 9.8\times 10^{128}\end{align}$$
Calculate the equilibrium constant $K_\mathrm c$ for the following reaction at that temperature:
$$\ce{2SO2(g) + O2(g) <=> 2 SO3(g)}$$

I already have the answer (which is given in the back of the textbook): $$K_\mathrm{c}=5.6\times10^{23}$$
But whenever I complete the problem (I've done it thrice) I get $2.3\times10^{76}$.
My process for solving this is as follows. I reverse the first equation and double the coefficients. Since I've done this, I use the inverse of $K'_\mathrm{c}$, $(K_\mathrm{c}')^{-1}$. I then multiply both $K$ values and the result is $2.3\times10^{76}$. Where is my mistake?

when you double the coefficients you don't get double the $K_{eq}$ but the square of $K_{eq}$.

Equilibrium constants depend upon the choice of reaction equations. I have noted the extended way to calculate $K''_\mathrm{c}$ below:
$$\begin{align}\ce{2S + 3 O2 &<=> 2 SO3}\tag{2}\\[0.4em]
K''_\mathrm{c} &= \frac{[\ce{SO3}]^2}{[\ce{O2}]^3}\end{align}$$
(Sulfur does not take part in the equilibrium constant because its activity, as a solid, is $a(\ce{S}) = 1$.)
Having this, we can approach the question. First, we need to add particles in an equivalent way (i.e. the same to reactant and product side) to get from equation $(3)$ to something that can be expressed as a linear combination of $(1)$ and $(2)$ ($a(1) + b(2)$):
\begin{align}\ce{2SO2 + O2 &<=> 2SO3}\tag{3}\\[0.4em]
\ce{2SO2 + 3 O2 &<=> 2 SO3 + 2 O2}\tag{$3'$}\\[0.4em]
\ce{2 SO2 + 3 O2 + 2 S &<=> 2 SO3 + 2 O2 + 2 S}\tag{$3''$}\end{align}

 $\displaystyle (3'') = -2 \times (1) + 1 \times (2)$

With that in hand, we can determine what the equilibrium constant of $(3'')$ would look like and how that can be traced back to the other equilibrium constants:

 $$\displaystyle K_\mathrm{c} = \frac{[\ce{SO3}]^2[\ce{O2}]^2}{[\ce{SO2}]^2[\ce{O2}]^3}\tag{4}$$

 $$\displaystyle K_\mathrm{c} = \frac{[\ce{O2}]^2}{[\ce{SO2}]^2} \times \frac{[\ce{SO3}]^2}{[\ce{O2}]^3} = (K'_\mathrm{c})^{-2} \times K''_\mathrm{c} \tag{5}$$

And thus, your final solution is:

 $$(4.2\times10^{52})^{-2} \times (9.8\times10^{128}) = 5.6\times10^{23}\tag{6}$$

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Are there compounds more conductive than Silver (Ag)?
Silver is the most electrically conductive element on the Periodic Table. However, I am interested in knowing if there are any compounds (combinations of elements) that would be more conductive than Silver (without Superconductor cooling). The extreme electrical conductivity of specific chemicals has many real world applications, from simply calling someone miles away, to processing enormous amounts of data in servers.

Graphene supposedly has higher conductance.

The [...] resistivity of graphene sheets would be 10−6 Ω⋅cm. This is less than the resistivity of silver, the lowest otherwise known at room temperature.

but it is not a compound - it is a homonuclear molecule.

Sorry, but, "graphene" isn't even mentioned on that page.

@jjm sure was in 2017. Just go through the edit history.

So, nowadays it is expected from users to rely on historical archives of third party web sites... Damn, what happened to SE supposingly being self-contained?

@JJMDriessen I was hoping you would understand that upon seeing something is in effect a dead link, you would take it upon yourself to edit in the proper details. SE is supposed to be a shared responsibility. I could be dead, or long gone. It is not my sole responsibility to mind the quality of my old answers, it is the collective's.

Well, actually, I just came here looking for an answer. Since it became apparent this answer was inadequate, I simply pointed that out, as part of my "shared responsibility". I never said it was your responsibility to edit, but you could have since you replied. Then, I was hoping that you would understand this answer is not self-contained in the first place, dead link or not. In fact, it does not explain anything, not even that graphene is not actually a compound. Now that I know this, I can edit it in, essentially changing the answer, although such practices are usually not appreciated...

@JJMDriessen Fine, you have aggravated me just enough to update the answer. And you insist on the value of "pointing things out" as if that is a skill worth having on the internet. It is not in short supply, I am afraid. The difference between a homonuclear molecule and a compound is not very practical to make for the reasoning to this question - but then again, it is also easily avoided. It is common to intercalate or insert interstitial or epitaxial additions to graphene to exploit its properties. That would make them compounds.

My apologies for the aggravation. I suppose both of us were starting to become a bit snarky about... not much. You know more about this subject than me, and I appreciate the update. +1

@jjm I wasn't at my best behavior either. Bygones. The link is restored, answer improved. The system works.

I need to be perfectly honest here. This whole comment thread serves as point in case for why link-only answers are discouraged. Stuff on the internet changes, and while there is probably most of the times a trace one could follow, especially with Wikipedia that provides such a history, it's always better to aim for completeness of a question. I'm sure the conversation could have been nicer, but I guess we all learned something valuable here.

The short answer is "no".  There are materials, both actual and theoretical, which would have a higher conductivity, but not as a simple "material".  They only make better conductors under very precise conditions.  Since you specifically ruled out superconductive cooling I think we can also safely rule out materials that are a single atom thick as a prerequisite for great conductivity.
Graphene, for example, is known as a two dimensional or "2D" material.  Its conductive properties only materialize when it is very, very thin.  A single atom thick.  You're not going to go out and buy a spool of graphene wire and expect it to be the best conductor ever.
Stanene is a relatively new discovery.  So new, in fact, it hasn't even been produced or tested yet, though experiments are currently underway.  It, also, is a 2D material made from a single layer of tin atoms.  Tin is a "topological insulator", meaning that its interior structure isn't a great conductor, but its surface is a very good conductor.  Electrons all flow on the outer surface.  In electronics dealing with alternating current this is know as the "Skin Effect".  So, theoretically, if it were a single atom thick it would be all surface and no interior, making it a better conductor than even graphene.  Researchers are actually claiming 100% efficiency, a very bold claim.
So no, there are no "materials" which are better conductors, at least that I can find.  There are, however, "atomic structures" which are better conductors.  But since you're asking this question I assume you are not a microchip manufacturer, making these materials pretty much useless to you, personally.  If you are a microchip manufacturer then, yeah, there's some exciting stuff out there.  Otherwise, though, there is still nothing better than silver.

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How to determine the order of elution in silica gel column chromatography?

Benzofuranoids ustusoranes A (1), B (2), and E (3) were isolated from Aspergillus ustus 094102. If these compounds were separated by column chromatography on silica gel, what would be the elution order? (top — comes off the column first, bottom — comes off last).

As I get it, column chromatography is like TLC where in a silica gel the nonpolar compound comes out first.

The comparison with TLC is valid, and your judging of polarity, based on the interaction of the $\ce{OH}$ groups with the silica sounds good. So, yes, I would come to the same conclusion. If you would perform the separation in the lab, you would nevertheless keep all three fractions and determine the structures by independent methods (NMR).

Your conclusion is absolutely correct. Column chromatography can be thought of as three-dimensional version of TLC (and vice-versa). So the most polar compound which interacts with silica gel most elutes slowest and the least polar compound leaves the column first.
I also support your suggested order as an initial starting point. However, As Klaus mentioned in a comment, unless one has performed the same separation multiple times one should never rely on paper-chemical polarity analysis alone and always confirm structures independently — typically that means ‘measure an NMR of all fractions’.

Why three-dimensional?

@orthocresol Diffusion in a column can go $x$ and $y$, elution is $-z$. In a TLC, elution is also $z$, but diffusion can only happen in $x$ direction.

I always thought 2D TLC referred to the thing where you develop along $z$ direction, then rotate it by $90^\circ$, stick it in a different solvent system, and develop along $x$ direction.

@orthocresol Yes, that’s the method of two-dimensional TLC evolution. But even in standard, single evolution TLCs you get a certain diffusion perpendicular to the elution direction and spot broadening.

Ah, okay; I wasn't meaning to nitpick, I was just curious as to what you meant.

@orthocresol I thought that. But once you mentioned 2D TLC (which I regularly forget), I thought it would be helpful to edit anyways =)


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What is fused calcium chloride?
As far as I have researched, fused calcium chloride is the hydrated form of calcium chloride. If that is true, then why is it called 'fused', why not simply hydrated?

Typically, "fused" refers to a technique to produce synthetic crystals (https://en.wikipedia.org/wiki/Verneuil_process)

That's a wholly different kind of "fused".

Fused means it contains fused water of crystallization i.e it is in hydrated form/melted form. http://www.chemicalforums.com/index.php?topic=77259.0

The term "fused" in chemistry generally refers

to become physically joined together

as by melting, adherence or other means of consolidation [1].
At an industrial scale solutions containing $\ce{CaCl2}$ are obtained either as by-products in the production of soda (by the ammonia method) and Berthelot's salt, or by the adding $\ce{HCl}$ to limewater. Solutions are then evaporated and $\ce{CaCl2}$ is isolated as a white or greyish porous mass, which is called fused calcium chloride and is generally water-free or contains trace amounts of water.
To make it clear, in this case "fused" ≠ "melted", as it's not about melting $\ce{CaCl2}$ (m.p. = $\pu{772^\circ C}$), but rather a physical adhesion as a result of a set of steps applied to calcium chloride solution (or its deliquescent form) of gradually increasing the temperature from $\pu{70 .. 80^\circ C}$ to approx. $\pu{300^\circ C}$. This results in formation of hydrated forms ($\ce{CaCl2 * x H2O}$) with subsequent complete dehydration and solidifying above $\pu{250^\circ C}$ alongside with minor partial hydrolysis to form $\ce{CaO}$ and $\ce{HCl}$.
Fused refers to the fact that this porous mass is granulated, but not dispersed, in order to increase the surface area and improve its dehumidifying ability and at the same time to prevent the salt from deliquescing swiftly.
Reference

fused. (n.d.) American Heritage® Dictionary of the English Language, Fifth Edition. 2011.

Basically, fused $\ce{CaCl2}$ is an anhydrous salt. It is called fused because it is ionic and solid crystal and is formed by electrolysis and fusing. It is highly water absorbing in nature.


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Molybdenum/Tungsten alloy armor piercing round
As we all know, the military's use of depleted uranium in armor piercing rounds and shells is very dangerous as far as health concerns, and an alternative is needed. Tungsten is already a very dense metal, but molybdenum is widely used to harden metals, and also has a very high density. Would a molybdenum and tungsten alloy work for armor piercing capabilities? If so, why not?

There's a good reason to stay with depleted uranium: It is cheap!

Molybdenum's density is much lower. Tungsten is OK, and indeed is used in similar applications, but I think that depleted uranium is cheaper. Its radioactivity is not a dig deal. Anyway, I would not care much for the health of somebody I shoot at.

https://www.youtube.com/watch?v=UDWD4VM-31E tungsten. It's a video of people shooting with it btw.

Depleted uranium is also self-sharpening and flammable: https://en.wikipedia.org/wiki/Depleted_uranium#Ammunition.

I really like this sentence "use of depleted uranium in  armor piercing rounds and shells is very dangerous as far as health concerns". We are talking about munition - the use of guns, especially caliber 30 mm,  is always bound to some "health concerns". Just ask anybody who stand once on wrong side of A10  Thunderbolt.

Probably not. Molybdenum and Tungsten are both hardening alloys to steel, they perform the same function, though differently. I suspect a Mo - W alloy would have the fragility of both molybdenum and tungsten while not hardening each other especially much nor increasing density.
The answer is: Sure, it would be hard, heavy and likely armor piercing, but very inefficient use of both Mo and W. Just use W alone, or alloy it with elements that are lighter, but melt within reasonable ranges. A quick glance at wikipedia confirms that Fe, Ni and Co are usual alloying elements. Tungsten carbide is also used.
Creating this Mo-W alloy would also be very difficult, usually you solvate the high mp material in the liquid low mp material but these both do not melt within reasonable limits so you would have to use some seriously high temperature equipment, we are talking beyond bright white heat - or rely on very, very slow processes (diffusion - and these are large atoms...).


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Can i use deionized water instead of distilled water
My question is really simple : can i use deionized water instead of distilled in experience ?
Because i have a air dehumidifier at home and it produce a lot of deionized water
I know that deionized water ($\ce{diH2O}$) is :

Water that has had almost all of its mineral ions removed, such as cations like sodium, calcium, iron, and copper, and anions such as chloride and sulfate.

And distilled water ($\ce{dH2O}$) is :

Water boiled and then condensed into a clean container, leaving solid contaminants behind

So basically the difference between $\ce{diH2O}$ and $\ce{dH2O}$ is that $\ce{dH2O}$ keep his ions mineral right ?
Can i use deionized water anyway ? Or not ?

No, distilled H2O does not keep its mineral ions. Whether or not you can use deionized water depends on your application.

@Ivan Neretin I need water to wash products (like crystal), and to use as reactant

Water from a dehumidifier is similar to distilled, in that it contains little in the way of dissolved materials.   It condensed on (typically) anodized aluminum,
and dripped through polymer collection trays into a bucket.
How clean is the bucket?  There may be residue from dust/lint etc.
It's water.   You can moisten plants, fish can swim in it, and it's pure enough
to drink or wash your windows (in fact, it's softer than most tap water, it
makes a good final rinse for windows).  I'd stick to deionized or distilled
for lab use, though, because those didn't come from a dirty bucket.


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Why is the hydbridization of the N atom (rightmost) in diazomethane sp and not sp2?
Why is the hydbridization of the $N$ atom (rightmost) in diazomethane $sp$ and not $sp^2$ ?
Most books claim that the negative charge (electron pair) of nitrogen can participate in resonance and is placed in a pure $p$ orbital. However, according to the general rule the hydribization number (number of sigma bonds + number of lone pairs) of the $N$ comes out to be $3$ implying ($sp^2$). However, according to several books the hybrdization is $sp$ and not $sp^2$. Why is it so?

As always in these questions, draw the resonance structure.

@Zhe I don't agree with you. We can draw a resonance structure only if we know beforehand that the electron pair is in pure p orbital. But, unless we know that the hybridization is $sp$ we cannot do so. If both the electron pairs were in hybrid sp2 orbitals then no resonance would have taken place.

You don't have to agree with me, but it does. That's why diazomethane behaves like a carbene and is nucleophilic at the carbon.
In addition, the logic here is backwards. You should be thinking about resonance and whether the electron should go into a $p$ orbital to begin with, not the other way around.


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Why are alkyl halides used as solvents for relatively non polar compounds and not polar compounds?
It is written in my textbook that :

Alkyl halides are used as solvents for relatively non polar compounds.

Alkyl halides are themselves polar with carbon having delta positive charge and halide having delta negative charge. If polar compounds are dissolved in them, then there will be more interactions between the compound and alkyl halide. The polar compound should dissolve in it.
What do they mean "relatively non polar compounds are soluble in alkyl halides" ? How will non polar compounds become soluble and why the textbook does not mention solubility of polar compounds?

Alkyl halides are less polar than they seem.

But "still" they are polar. How will non polar compounds be soluble in them at all?

OK, they are even less polar than you seem to think now.

I thought they were polar based on their electronegativity diffrence which is around 0.5 in Pauling scale.

Like I said before, electronegativity is hardly a precise measure of anything.

Then how else do we decide?

Practice.$\mathstrut$

Polarity is relative. Dipole moment (macro scale) and dielectric constant of the solvent (molecular scale)  are probably the most decisive. Water has ε~80, ethanol 25 and DCM is "only" 9. By comparison toluene, benzene or tertachlormethane have ε~2. The last example also illustrates why differences in electronegativity are not enough. Being symmetric (Td) tetrachloromethane has dipole moment of zero, even if there is a big difference between electronegativity of carbon on chlorine.

I understand that more is the dielectric constant, more is the force applied by the solvent molecules on the charged particles.  But I don't understand how dielectric constant affects the process of dissolution?


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How to increase the concentration of HCl
Today I have purchased some $\ce{HCl(aq)}$ on my local market with a concentration of 23%.
However, I need to obtain a concentration of 37% in order to perform an experiment. Can I use distillation process to increase the acid concentration?
If yes, can I use standard chemistry glassware to process it? (In an isolated system to prevent the $\ce{HCl(g)}$ from escaping of course.)
Distillation unit

If that experience is you feeling a lot of pain, this would do it. Depending on how much acid you have casually bought, this could be really dangerous. HCl is a gas, if you heat up your little experiment, it will go wherever it pleases, most likely heavily react with your body, you will also breathe it in and it could get into your eyes, making you blind and helpless.  Please stop whatever you are doing before people get hurt.

Yeah maybe you right. Distillation of HCl doesn't seem to be the best idea. So is there any other way to increase HCl concentration ? Or does I have to buy concentrated HCl from chemistry supplier ?

Why not adjust the other reactants in your reaction in order to reach the same concentrations throughout the process? This seems a lot safer.
Would you mind sharing with us your procedure?

What experiment are you attempting? 23% HCl strikes me as concentrated enough to do most reactions that call for conc. HCl.

As sternly warned by users above, there are serious safety issues at play here, so please do your due diligence about precautions and consult an expert if need be.

you mention using an isolated system, closed glassware apparatuses are seriously dangerous when performing chemical reactions! Your usually heating up something, or the reaction might produce gas by itself. This means your apparatus has to withstand the significant pressure this causes, and it won't do that. The result will be that you're picking out the glass shrapnel from your face while you're lungs are burned by the gaseous HCl you set free. There is no way to perform this reaction safely without a fume hood and expertise.

I don't want to perfom this experiment, I dont want to ruin my life for something that will probably won't work anyway. Instead of increase my HCl concentration I'll balance my others products to keep the proportion for my experiment. Anyway thank for the answers !

Add Calcium Chloride to destroy the azeotrope.  Then distill and capture the vapor phase.
See the second method "Extractive Distillation" here:
http://www.jacobs.com/uploadedFiles/wwwjacobscom/20_Learn_About_Us/25_Products/252_Chemetics/Hydrochloric%20Acid%20Concentration.pdf
(Obviously, proper equipment and safety precautions are essential.)

Irrespective of the potential hazards to an inexperienced chemist attempting distillation of hydrochloric acid, especially absence of adequate safety measures such as a fume hood, in this case, distillation will not result in a higher concentration of acid. Hydrogen chloride and water form a constant-boiling mixture (azeotrope) at ~20% HCl, which is what you are already starting with. I suggest you read the Wikipedia page on Hydrochloric Acid for starters.

$HCl/H_2O$ forms an azeotrope at atmospheric pressure at ~108.6°C with an $HCl$ concentration of ~20.2%
If you try to distill your 23% mixture, you will outgass a bit of HCl as the mixture heats, thereafter you'll get constant boiling azeotrope. If you bought a pool acid or other commercial HCl mix, distilling this might be a good first step, though. They sometimes put "stabelizer" or a colourant in it and there might be other impurities as well. Also, if you could distill, cooling water would not work to condense HCl in the setup shown in your drawing, as it boils at around -85°C.
You can use $CaCl_2$ or $MgCl_2$  (hygroscopic) to capture some of the water and "break" the azeotrope that way.
Take half of your dilute $HCl$ (or as much as you need) and chill it (ice bath), add $CaCl_2$ to the non-chilled half (or however much you need) heat it gently and bubble the captured gas/vapour phase through the chilled dilute $HCl$. Avoid suck-back of the chilled $HCl$ (use a reverse funnel or appropriate setup). You should be able to get up to 37% $HCl$
Please note the safety precautions needed to handle HCl gas. The solution may "foam" when the HCl is bubbled through. HCl has 42% solubility in water at atmospheric pressure and 20°C, but around 37% it will start fuming in moist air.
The second way (wouldn't recommend it): If you have the facilities to do it safely, you could produce $HCl$ gas (as per the old Manheim process) from $NaCl$ and $H_2SO_4$ and then bubble the captured gas through your dilute $HCl$ solution (or through distilled water if you want to be sute of your final product concentration and make sure your final product doesn't have any other contaminants) to make your concentrated $HCl$ (Note: be careful you don't get suck-back of dilute HCl into your hot reagents when doing this, especially if you heat it enough to get the second reaction.)
This is a two-step process that produces HCl at both steps. First an exothermic reaction that will supposedly occur at room temperature, but in practice needs some heat and moisture to get going:
$NaCl + H_2SO_4 → HCl + NaHSO_4$
The second step is endothermic:
$NaCl + NaHSO_4 → HCl + Na_2SO_4$
You have to heat your reagents to high temperatures for the second step.


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Can anomers be destinguished based on Fischer projection?
I would like to ask if it is possible to tell an anomer using Fischer projection (e.g. glucose). In high school we learned only Haworth and Tollens are hepful in this case.
However, I'm not exactly sure since normal or wedged bonds used in drawn Fischer projection should give the idea. If the double bonded oxygen is on the right that is where poloacetal hydroxyl group will be formed, should stay above the ring after cyclisation and based on normal/wedged bond connecting $\ce{-CH2OH}$ goup with the rest of the molecule one should be able to tell the anomer.
I feel like I am missing a very basic principle that causes any attempt of my logical explanation to fail.

Considering that Emil Fischer was the one who named the sugars, his projection is probably able to tell them apart... :)

The problem with Fischer is that it is designed for linear compounds. An extension for cyclic ones does exist (specifically for anomers, in fact) but I forgot it.

For a mnemonic of pyranose anomers in the Fischer, Haworth and  chair conformations,  look here: http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/carbprojection.pdf


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Problem in Colour in Cerium (IV)Sulphate due to charge transfer
My text says that $\ce{Ce(SO4)2}$ is coloured due to charge transfer spectra.
It also gives other examples such as $\ce{KMnO_4}$ and $\ce{K_2Cr_2O_7}$, to which I agree but when I read the conditions for charge transfer spectra to take place it says

+6;+7;+8 show it more
if the central metal atom is diamagnetic but still if compound is showing colour.

I see $\ce{Ce}$ in its +4 oxidation state and configuration
$\ce{[Xe] 4f^1 5d^1 6s^0}$
doesn't show 1st  the conditions then where am I going wrong?

ohh! no it should have been $Ce(SO_4)_2$ ; But still it doesn't hold the first condition true why?

Related: http://chemistry.stackexchange.com/questions/34733/why-does-a-mixture-of-siloxene-and-ceriumiv-sulfate-luminesce

I want to know why both conditions in my text are not true simultaneously?  why?

Cerium(IV) is not $\ce{[Xe] 4f^1 5d^1 6s^0}$. With $58-4=54$ electrons it must be $\ce{[Xe]}$ only, so diamagnetic and yet colored.

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text:
Reduction of silicon dioxide
Silicon dioxide, in the form of quartz, is in abundance around my local area and I would be interested in creating silicon as a separate element.
My problem is that I don't know what to react silicon dioxide with to reduce it to silicon. I have read that it can be reduced with carbon but I couldn't find any information about the temperatures this takes place at. I am also aware that a thermic reaction can take place between aluminium and silicon when mixed with sulfur but I don't have any aluminium powder.
Thank you in advance.

https://en.wikipedia.org/wiki/Silicon#Metallurgical_grade

@hex can you build an answer around that reference?  The given answer is rejected as not an answer, and for my part I am not convinced it would work.

@OscarLanzi When I post it a year ago the link was different as it is today. However if you read the wikipedia page and the references you'll find what you want. I have no time to read for others and as I am not a specialist of the question ... :-)


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Why is double bond stronger than single bond?
If a pi-bond is weak due to its sideways overlapping, then how can we say that breaking a double bond is a bit difficult? Is it because of the slight increase in energy as compared to a single bond to break it?

A double bond is composed of a π-bond and a σ-bond. In order to break a bond such as this you must provide the molecule with the energy needed to break the σ-bond alone plus the one needed to break the π-bond alone. In other words, you must provide the sum of those two amounts of energy. Thus, it goes without saying that the sum is a greater number than the two energies alone.

True but you may want to add that the portion the pi bond adds is typically less than the portion the sigma bond adds.

Double bonds are stronger than single bonds and they are characterized by the sharing of four or six electrons between atoms, respectively.
Double bonds are comprised of sigma bonds between hybridized orbitals, and pi bonds between unhybridized p orbitals. Double bonds offer added stability to compounds, and restrict any rotation around the bond axis.

Double bond is more stronger than single bond because,

Energy required to break double bond is 614 J while in breaking single bond is 349 J, thus the energy to break double bond is more than single bond so it is stronger than single bond.
Double bond is restricted to rotation while single bond rotate freely so, double bond is stronger.
Bond length in double bond is 1.34 angstrom while in single bond is 1.54 angstrom  where bond length is inversely proportional to bond energy so less bond length, more bond energy is required to break thus it is more stronger bond.

Are you talking about benzene? I don't think the question asks specifically about benzene, so most of your answer is irrelevant to this question. (except point 2)


text:
Is home synthesis of cellulose acetate possible, and why?
Is it remotely possible to synthesize detectable quantities of cellulose acetate from materials available in common department stores (such as cotton balls, vinegar, and battery acid) and using tools available in a home kitchen (such as ovens, stoves, freezers, glass jars, and metal pots)?  If so, how would the reagents need to be pre-processed, and what would the procedure be?  If not, where do the insurmountable difficulties lie?
On a webpage titled "From Cotton to Cellulose Acetate" it is indicated that, in general, cellulose acetate is manufactured from specially prepared cellulose in a solution of 100% acetic acid with acetic anhydride.  That same site goes on to specifically state that immersing cotton balls in vinegar will only result in "cotton balls that smell like pickles".  Other sources add concentrated sulfuric acid into the mix.  And an answer to another question on the Chemistry SE indicates that household vinegar could possibly be concentrated up to 58% using fractional freezing in a kitchen freezer.
It's difficult to tell from the sources available whether making homemade cellulose acetate is completely impossible, somewhat inconvenient, surprisingly hazardous, or merely commercially unviable.

SAFETY - In general using a kitchen for chemistry experiments is a really bad idea. Glacial (100%) acetic acid is not just a stronger vinegar. Glacial  acetic acid should be used with proper safety equipment and a chemical hood.

I think you should give up trying to make cellulose acetate by simple treatment of cotton wool with acetic acid. You might want to consider a rayon synthesis. This is an experiment you could do in a school chemistry lab inside the fume hood.
If you get some copper sulphate solution and add a little ammonia to it then you form a pale blue copper(II) hydroxide percipitate. Then add more ammonia to form a deep blue solution of tetraaminecopper(II) sulfate. This solution will dissolve cellulose. Leave in a sealed container cotton wool or bloting paper with the copper sulfate / ammonia mixture.
When it has reacted then remove the cellulose by straining or filtration (do not use filter paper). Then add the solution to acetic acid. You will be rewarded with regenerated cellulose. This is a form of rayon. I know it works becuase I did it in my early teens. Just be careful not to inhale too much ammonia. Do the experiment in a well ventillated placed.

The bottom line is that this would be a very bad idea to try due to safety reasons alone, and is not realistically feasible using only household materials regardless.
Glacial (100%) acetic acid is dangerous if not handled properly and requires an appropriate fume hood at a minimum. You also are not going to get to glacial acetic acid from vinegar, and water will kill this reaction.
Acetic anhydride is a reagent that you just won't be able to obtain or synthesize yourself. Even if you did, it would be hydrolyzed by the water in your impure acetic acid.
Sulfuric acid is commonly used as a catalyst, and while you could obtain this, it should absolutely not be handled by someone without appropriate equipment and training in it's safe usage.
So again, this is just not a feasible project and there would be serious dangers along the way if you tried. I like your creative thinking process and I hope you will pursue some other, safer, home science project.


text:
How to connect structures in Avogadro at certain angles?
I am working with this carbon nano-structure in Avogadro and the junction between my two nano-structures I try to join never seems to be aligned well. It always seems to be off kilter, which is most likely due to how I move the junction into position (manually dragging).

I posted a picture of my issue above. Where the two white arrows are is what is bothering me. I would like those to be as vertical as possible but I do not know of ways to do this aside from manually adjusting the atoms/bonds, which I assume would not get them to align exactly.
UPDATE:
As per khaverim's advice, I followed his first instructions, using MMFF94 instead of UFF (forgot to change it and my structure is large so it takes a while to optmize, will try with UFF later), and got the following structure:

I am pretty sure they are all equal angled, at least they look to be so, so thanks.

If you don't mind, please share a link to the .xyz file of your system.

I'm afraid I don't think this is at all possible with Avogardo. I personally have little to no experience with the program itself, as I find it especially tedious to build a molecule with symmetric features there. As far as I know, it only has drag and drop tools to manipulate the geometry. If you open up your question to suggest alternative programs, you probably get a workable solution. I'd think it would be quite easy with ChemCraft which is unfortunately not free.

I do not have access to it right now, but the build tool of Turbomole (as bundled with TmoleX) is free for academic or personal use and cross-platform. I shall endeavour to post a picture of its application later, given that the only available answer thus far resorts to outside programs.

@Martin-マーチン    How can I edit the bounty to allow for your response. Also, would the trial version of ChemCraft allow me to do this?

@khaverim        I would need to ask my advisor for permission to link the .xyz file sorry.

@Martin-マーチン Avogadro has (very fast) geometry optimization algorithms, as mentioned in the first part of my answer. Making symmetric molecules is much easier than you may have imagined. (I added an example in my answer)
Also -- ChemCraft is free for the first 150 days, currently. Just installed it last week though I'm not familiar with using it.

You cannot edit the bounty message directly, but you can amend the question and edit the title. For example you could add a sentence where you ask for alternatives if Avogadro cannot do it. Since the current answer already does that, I see no problem.

Can you set a bounty greater than your reputation???!!!

@AaronJohnSabu          At the time I set the bounty, it subtracted the 150 reputation from the total I had.

@khaverim        Sorry if this warrants another question, but how could I determine if my final structure I get post optimization is in fact symmetrical, as you spoke about with Martin in the comments above.

@RoSiv probably a good separate question at this point, but I would 1) make sure the system is centered about 0,0,0 (e.g. by some program. I think Avogadro does it automatically when you save an .xyz but I wrote a shell script for it too: https://github.com/khavernathy/useful_computational_chem/blob/master/center_it.sh); 2) check if desired atoms are coplanar (have the same cartesian coordinate for the desired dimension). i.e. I don't think there is an easy way apart from just doing the math.

Did you try a (Universal Force-Field) geometry optimization? I would:

Select all atoms
Deselect (hold Ctrl + click) the 4 (6? 8?) carbons of interest
Click Extensions -> Molecular Mechanics -> Ignore Selection
Click Extensions -> Optimize Geometry

Alternatively, you could optimize a few more atoms around/in the bridge while the others are constrained. The G.O. algorithm is pretty robust and handles many systems quite well. A simple example of hand-drawn cyclohexane:

I got a result like this by making 2 nanotubes with chiral index (4,4), drawing 6 bonds and optimizing the bridge:

If this is ineffective, I would resort to:

Change the 4 carbons to different elements e.g.

$\ce{O        H}$
$\ce{N        S}$
(like the arrangement you have in the picture, so you can identify them later)

Click File -> Save As... and save the file as an .xyz
Calculate the O-N and H-S distance by $r_{ij} = \sqrt{\Delta x^2 + \Delta y^2 + \Delta z^2}$ from the coordinates in the .xyz file.
Ideally, Avogadro saved the .xyz coordinates aligned in z (which is default when you draw molecules without rotating the initial view).
(Even if not,) re-write the coordinates of the atoms aligned in x and z, e.g. $\ce{O}$ at $(x, y, z)$; $\ce{N}$ at $(x, y-r, z)$.
Or (yikes) write a quick program to generate possible systems with the fixed $r$ you calculated, name them system-000.xyz, system-001.xyz, system-002.xyz, etc., and view the "trajectory" frames until you find one pleasing. (Viewing can be made much easier with VMD, e.g. vmd system-*.xyz).

Avoiding Avogadro: If you use Molden and load your system (as an .xyz, .car, .pdb etc.), you can use the ZMAT-editor to manually change bond angles and dihedrals:

Finally, if all else fails, it may be time to consider a full quantum-mechanical treatment of the system (if you're familiar with High Performance Computing) with a free (for academia!) program like Orca (where one can constrain certain atoms during the optimization). An example input would be:
! hf 6-31g* opt  # or copt, for cartesian optimization

%pal
nprocs 128   # num. processors
end

%scf
maxiter 2000 # max self-consistent field iterations
end

%geom
Constraints # atoms to keep fixed starting at index 0
{C 61 C}
{C 62 C}
{C 290 C}
{C 291 C}
{C 292 C}
end
end

*xyz 0 1  # charge=0; mult=2S+1 = 1 (closed shell)
  C      -1.306743      4.170005      1.012072
  C      -0.042414      3.718873      0.686759
  C       0.210647      2.367809      0.785537
  C      -2.267754      3.271952      1.435683
  C      -2.920909      0.889649      1.948837
  C      -4.153277      1.187877      2.498723
  C      -4.989183      0.155126      2.896328
  C      -4.571713     -1.152339      2.735382
  C      -3.332221     -1.380404      2.172864
....... etc.
*

Even though I like Molden a lot, it often has problems finding a good z-matrix to manipulate for complex systems. Changing one angle can easily ruin the whole molecule depending on how the relative connections are defined. However, I kind of agree that this task needs a different program than Avogadro.

@Martin-マーチン  Or "simple" systems with high symmetry.

@LordStryker Indeed. So far I only know that ChemCraft handles symmetry well.

@khaverim        In your latest picture you made a bridge, even though I wanted pentagons to compose the bridge only, and it appears you have squares, which molecules did you select before optimizing the bridge? Only the ones within the bridge? And then did you ignore the rest of the selection before optimizing? Also, as an aside why use UFF as opposed to other optimizations? Thanks.

@RoSiv to clarify: you want 5 bonds linking the two tubes? It looks like an even number in your picture, like 4 maybe. I did the procedure in the first part of my answer: select all, deselect bonding atoms (which is 6 in my example), ignore selection, optimize. I say UFF because sometimes it is necessary if you have a large system. Otherwise Avogadro defaults to a better force-field. (The program usually gives you the option if the system is too big to handle with non-UFF)

I don't know if you can do it with Avogadro, since I have no extended experience with the program. As far as I can see, there is no possibility to edit bond  lengths/ angles/ dihedral angles directly. It provides a cartesian coordinate editor, but that's as tedious to manipulate your structure with as it is by doing it by hand.
I assume your problem is quite simple to be solved with ChemCraft. This program provides the tool to merge and align two structures from two different windows. This allows you to easily combine pre-computed fragments without loosing symmetry and not doing any calculations by hand.
Since I have no access to any of your structures, I am going to run you through it with an example. I am going to cup a $\ce{C70}$ nano-tube fragment with a $\ce{C40}$ buckminster fullerene, twice. Both fragments have (at least) $C_5$ symmentry, but this is no necessity. At the end we will produce a Fullerene with $D_\mathrm{5h}$ symmetry. For reasons of spacing this post, I'll include the coordinates at the end. (I am using build 485 currently, but the updated versions should still be able to perform this.)
You start with opening two instances of ChemCraft and loading each fragment into one. You want the parts to be aligned facing each other. In my layout I am modifying the $\ce{C70}$ fragment on the left, while the window on the right provides me with the fragments I use (not that it matters in this context). After this is done, you select the atoms you want to align in each window. The more atoms you select, the more input you provide for the program to align structures. Your display should look somewhat like below:

In the left window you choose "Tools > Structure combiner > Merge two structures from two Chemcraft windows by selected subunits". Because of the high symmetry the algorithm might get the orientation wrong. It is then best to try selecting (one or two) fewer atoms.

The right window should now look like below. Obviously the atoms you have taken to align are double in the mix. In some structures you can easily select them and delete them, when they are matching too well, you can switch between displays.

If you decide to "show erasen atoms" they will be displayed as dummies; after you click apply those will disappear and the program will re-bond you molecule.
Now one side is successfully capped, and I'll proceed with adding the second cup. We have created a nice little nano-pill. From the menu, you can choose to symmetrise the molecule. Ideally in this case it should already be very, very close to $D_\mathrm{5h}$. Cartesian space will obviously introduce some artefacts. Coose "Edit > Set point group ... > Auto" to let the program determine it for you.

Indeed there is a slight variation for the point group. Fortunately the program can also correct for it.

You don't need to stop there. ChameCraft comes with a xyz/ distance/ angle/ dihedral editor on the lower left corner of the window, dpending on how many atoms you select. You can choose whether you want to average the structure or only move the last selected atom. If the atoms are not connected with visible bonds, it will manipulate the whole fragment accordingly. With this feature you can fine tune everything, but then you might need your pen, paper and calculator again.

As I don't know the context of your research, it should suffice to say that these are techniques for creating suitable input geometries for more advanced calculation programs, of which there are quite a few. A well designed geometry might take crucial optimisation time off your hands. I am not sure how well suited force field methods are for nanotubes, and while Avogadro provides methodology for simple organic molecules, projects like this might just be too complex for it to be reasonable. As far as I know, Avogadro has no symmetry recognition.
Appendix
The $\ce{C70}$ fragment has the following coordinates (in angstrom):

70
symmetry c1
C        1.385307752     -3.115487343     -5.540304632
C        2.534920901     -2.280244499     -5.540304632
C        3.391088177      0.354767430     -5.540304632
C        2.951975027      1.706218740     -5.540304632
C        0.710500000      3.334745672     -5.540304632
C       -0.710500000      3.334745672     -5.540304632
C       -2.951975027      1.706218740     -5.540304632
C       -3.391088176      0.354767430     -5.540304632
C       -2.534920901     -2.280244499     -5.540304632
C       -1.385307752     -3.115487343     -5.540304632
C       -0.710500000     -3.334745672     -4.309125825
C        0.710500000     -3.334745672     -4.309125825
C        1.385307752     -3.115487343     -3.077947018
C        2.534920901     -2.280244499     -3.077947018
C        2.951975027     -1.706218740     -4.309125825
C        3.391088177     -0.354767430     -4.309125825
C        3.391088177      0.354767430     -3.077947018
C        2.951975027      1.706218740     -3.077947018
C        2.534920901      2.280244499     -4.309125825
C        1.385307752      3.115487343     -4.309125825
C        0.710500000      3.334745672     -3.077947018
C       -0.710500000      3.334745672     -3.077947018
C       -1.385307752      3.115487343     -4.309125825
C       -2.534920901      2.280244499     -4.309125825
C       -2.951975027      1.706218740     -3.077947018
C       -3.391088176      0.354767430     -3.077947018
C       -3.391088176     -0.354767430     -4.309125825
C       -2.951975027     -1.706218740     -4.309125825
C       -2.534920901     -2.280244499     -3.077947018
C       -1.385307752     -3.115487343     -3.077947018
C       -0.710500000     -3.334745672     -1.846768211
C        0.710500000     -3.334745672     -1.846768211
C        1.385307752     -3.115487343     -0.615589404
C        2.534920901     -2.280244499     -0.615589404
C        2.951975027     -1.706218740     -1.846768211
C        3.391088176     -0.354767430     -1.846768211
C        3.391088176      0.354767430     -0.615589404
C        2.951975027      1.706218740     -0.615589404
C        2.534920901      2.280244499     -1.846768211
C        1.385307752      3.115487343     -1.846768211
C        0.710500000      3.334745672     -0.615589404
C       -0.710500000      3.334745672     -0.615589404
C       -1.385307752      3.115487343     -1.846768211
C       -2.534920901      2.280244499     -1.846768211
C       -2.951975027      1.706218740     -0.615589404
C       -3.391088176      0.354767430     -0.615589404
C       -3.391088176     -0.354767430     -1.846768211
C       -2.951975027     -1.706218740     -1.846768211
C       -2.534920901     -2.280244499     -0.615589404
C       -1.385307752     -3.115487343     -0.615589404
C       -0.710500000     -3.334745672      0.615589404
C        0.710500000     -3.334745672      0.615589404
C        1.385307752     -3.115487343      1.846768211
C        2.534920901     -2.280244499      1.846768211
C        2.951975027     -1.706218740      0.615589404
C        3.391088176     -0.354767430      0.615589404
C        3.391088176      0.354767430      1.846768211
C        2.951975027      1.706218740      1.846768211
C        2.534920901      2.280244499      0.615589404
C        1.385307752      3.115487343      0.615589404
C        0.710500000      3.334745672      1.846768211
C       -0.710500000      3.334745672      1.846768211
C       -1.385307752      3.115487343      0.615589404
C       -2.534920901      2.280244499      0.615589404
C       -2.951975027      1.706218740      1.846768211
C       -3.391088176      0.354767430      1.846768211
C       -3.391088176     -0.354767430      0.615589404
C       -2.951975027     -1.706218740      0.615589404
C       -2.534920901     -2.280244499      1.846768211
C       -1.385307752     -3.115487343      1.846768211

The $\ce{C40}$ fragment has the following coordinates:

40
symmetry c1
C       -0.004840185     -3.437248260     -1.231937277
C       -1.267748660     -3.000481413     -1.692390182
C       -2.250219438     -2.561103110     -0.763208689
C       -3.030739803     -1.417632648     -1.047784233
C       -2.833456162     -0.706703611     -2.263242764
C       -2.833456162      0.706703611     -2.263242764
C       -3.030739803      1.417632648     -1.047784233
C       -2.250219438      2.561103110     -0.763208689
C       -1.267748660      3.000481413     -1.692390182
C       -0.004840185      3.437248260     -1.231937277
C        0.283148522      3.437248269      0.160450194
C        1.546056995      3.000481414      0.620903095
C        2.528527766      2.561103104     -0.308278399
C        3.309048125      1.417632641     -0.023702855
C        3.111764498      0.706703610      1.191755677
C        3.111764498     -0.706703610      1.191755677
C        3.309048125     -1.417632641     -0.023702855
C        2.528527766     -2.561103104     -0.308278399
C        1.546056995     -3.000481414      0.620903095
C        0.283148522     -3.437248269      0.160450194
C       -0.693482914     -3.000481414      1.084109073
C       -1.963942392     -2.561103104      0.620903094
C       -2.567533811     -1.417632641      1.191755675
C       -3.226850870     -0.710929034      0.160450192
C       -3.226850870      0.710929034      0.160450192
C       -2.567533811      1.417632641      1.191755675
C       -1.963942392      2.561103104      0.620903094
C       -0.693482914      3.000481414      1.084109073
C       -0.034165862      2.293777797      2.115414553
C        1.349945920      2.293777797      1.829137515
C        2.135133060      1.143470461      2.115414554
C        1.531541637      0.000000000      2.686267134
C        2.135133060     -1.143470461      2.115414554
C        1.349945920     -2.293777797      1.829137515
C       -0.034165862     -2.293777797      2.115414553
C       -0.641366192     -1.143470461      2.689680299
C       -1.904274667     -0.706703610      2.229227401
C       -1.904274667      0.706703610      2.229227401
C       -0.641366192      1.143470461      2.689680299
C        0.139154169      0.000000000      2.974255840

Final $\ce{C130}$ nano-pill:

130
symmetry d5h
C        3.334745673     -0.710500000     -2.462357615
C        3.334745673      0.710500000     -2.462357615
C        3.115487344      1.385307752     -1.231178807
C        2.280244500      2.534920901     -1.231178807
C        1.706218741      2.951975027     -2.462357615
C        0.354767431      3.391088176     -2.462357615
C       -0.354767429      3.391088176     -1.231178807
C       -1.706218739      2.951975027     -1.231178807
C       -2.280244498      2.534920901     -2.462357615
C       -3.115487342      1.385307752     -2.462357615
C       -3.334745671      0.710500000     -1.231178807
C       -3.334745671     -0.710500000     -1.231178807
C       -3.115487342     -1.385307752     -2.462357615
C       -2.280244498     -2.534920901     -2.462357615
C       -1.706218739     -2.951975027     -1.231178807
C       -0.354767429     -3.391088176     -1.231178807
C        0.354767431     -3.391088176     -2.462357615
C        1.706218741     -2.951975027     -2.462357615
C        2.280244500     -2.534920901     -1.231178807
C        3.115487344     -1.385307752     -1.231178807
C        3.334745673     -0.710500000      0.000000000
C        3.334745673      0.710500000      0.000000000
C        3.115487344      1.385307752      1.231178807
C        2.280244500      2.534920901      1.231178807
C        1.706218741      2.951975027      0.000000000
C        0.354767431      3.391088176      0.000000000
C       -0.354767429      3.391088176      1.231178807
C       -1.706218739      2.951975027      1.231178807
C       -2.280244498      2.534920901      0.000000000
C       -3.115487342      1.385307752      0.000000000
C       -3.334745671      0.710500000      1.231178807
C       -3.334745671     -0.710500000      1.231178807
C       -3.115487342     -1.385307752      0.000000000
C       -2.280244498     -2.534920901      0.000000000
C       -1.706218739     -2.951975027      1.231178807
C       -0.354767429     -3.391088176      1.231178807
C        0.354767431     -3.391088176      0.000000000
C        1.706218741     -2.951975027      0.000000000
C        2.280244500     -2.534920901      1.231178807
C        3.115487344     -1.385307752      1.231178807
C        3.334745673     -0.710500000      2.462357615
C        3.334745673      0.710500000      2.462357615
C        1.706218741      2.951975027      2.462357615
C        0.354767431      3.391088176      2.462357615
C       -2.280244498      2.534920901      2.462357615
C       -3.115487342      1.385307752      2.462357615
C       -3.115487342     -1.385307752      2.462357615
C       -2.280244498     -2.534920901      2.462357615
C        0.354767431     -3.391088176      2.462357615
C        1.706218741     -2.951975027      2.462357615
C       -1.717973404      3.000456514      3.693536421
C       -0.373754410      3.437219745      3.693536421
C        0.373754411      3.437219749      4.903041136
C        1.717973407      3.000456525      4.903041139
C        2.322720745      2.561081857      3.693536424
C        3.153493774      1.417620883      3.693536424
C        3.384486708      0.706697748      4.903041140
C        3.384486708     -0.706697748      4.903041140
C        3.153493774     -1.417620883      3.693536424
C        2.322720745     -2.561081857      3.693536424
C        1.717973407     -3.000456525      4.903041139
C        0.373754411     -3.437219749      4.903041136
C       -0.373754410     -3.437219745      3.693536421
C       -1.717973404     -3.000456514      3.693536421
C       -2.322720749     -2.561081859      4.903041134
C       -3.153493775     -1.417620884      4.903041132
C       -3.384486697     -0.706697747      3.693536419
C       -3.384486697      0.706697747      3.693536419
C       -3.153493775      1.417620884      4.903041132
C       -2.322720749      2.561081859      4.903041134
C       -1.579654762      2.561081856      6.105357127
C       -0.227398602      3.000456514      6.105357127
C        0.745287399      2.293758766      6.848429277
C        1.947593421      2.293758772      6.105357132
C        2.783333689      1.143460975      6.105357132
C        2.411800691      0.000000000      6.848429279
C        2.783333689     -1.143460975      6.105357132
C        1.947593421     -2.293758772      6.105357132
C        0.745287399     -2.293758766      6.848429277
C       -0.227398602     -3.000456514      6.105357127
C       -1.579654762     -2.561081857      6.105357127
C       -1.951187752     -1.417620880      6.848429274
C       -2.923873754     -0.710923134      6.105357124
C       -2.923873754      0.710923134      6.105357124
C       -1.951187752      1.417620880      6.848429274
C       -0.972686005      0.706697746      7.595944297
C        0.371532989      1.143460974      7.595944302
C        1.202306016      0.000000000      7.595944306
C        0.371532989     -1.143460974      7.595944302
C       -0.972686005     -0.706697746      7.595944297
C       -1.717973404     -3.000456514     -3.693536421
C       -0.373754410     -3.437219745     -3.693536421
C        0.373754411     -3.437219749     -4.903041136
C        1.717973407     -3.000456525     -4.903041139
C        2.322720745     -2.561081857     -3.693536424
C        3.153493774     -1.417620883     -3.693536424
C        3.384486708     -0.706697748     -4.903041140
C        3.384486708      0.706697748     -4.903041140
C        3.153493774      1.417620883     -3.693536424
C        2.322720745      2.561081857     -3.693536424
C        1.717973407      3.000456525     -4.903041139
C        0.373754411      3.437219749     -4.903041136
C       -0.373754410      3.437219745     -3.693536421
C       -1.717973404      3.000456514     -3.693536421
C       -2.322720749      2.561081859     -4.903041134
C       -3.153493775      1.417620884     -4.903041132
C       -3.384486697      0.706697747     -3.693536419
C       -3.384486697     -0.706697747     -3.693536419
C       -3.153493775     -1.417620884     -4.903041132
C       -2.322720749     -2.561081859     -4.903041134
C       -1.579654762     -2.561081857     -6.105357127
C       -0.227398602     -3.000456514     -6.105357127
C        0.745287399     -2.293758766     -6.848429277
C        1.947593421     -2.293758772     -6.105357132
C        2.783333689     -1.143460975     -6.105357132
C        2.411800691      0.000000000     -6.848429279
C        2.783333689      1.143460975     -6.105357132
C        1.947593421      2.293758772     -6.105357132
C        0.745287399      2.293758766     -6.848429277
C       -0.227398602      3.000456514     -6.105357127
C       -1.579654762      2.561081856     -6.105357127
C       -1.951187752      1.417620880     -6.848429274
C       -2.923873754      0.710923134     -6.105357124
C       -2.923873754     -0.710923134     -6.105357124
C       -1.951187752     -1.417620880     -6.848429274
C       -0.972686005     -0.706697746     -7.595944297
C        0.371532989     -1.143460974     -7.595944302
C        1.202306016      0.000000000     -7.595944306
C        0.371532989      1.143460974     -7.595944302
C       -0.972686005      0.706697746     -7.595944297

This method actually worked better than the first answer, as I got weird angles on one part of the joining nanotube when just using Avogadro.  Is there a way to split the bounty between you and khaverim?

@RoSiv No, there is no way to split the bounty, but that is okay. I am happy to help and I really don't need more reputation. After all, your question gave me some nice opportunity to play with my toys :D


text:
Why don't minerals evaporate from seawater along with the water?
When water evaporates from the ocean, of course the minerals such as salt remain in the ocean-- that's why seawater is so saline. However, I'm very puzzled: why don't minerals evaporate along with the water? Could you give me a somewhat simple answer for this? (I never studied chemistry, so I really have no clues as to the possible answer.)

The definition of evaporation is "the process whereby atoms or molecules in a liquid state  gain sufficient energy to enter the gaseous state." Does that mean minerals can evaporate also if they are imparted enough energy?

You don't want to breathe deeply over an open container of molten iron... Now, it is somewhat unlikely that a mineral (normally a fairly complex crystal with multiple elements) would go in to the gas state as a single molecule mimicking the mineral composition.

Related: What happens to the molecules of a liquid when it evaporates?

In order to get a measure of how much a compound is prone to evaporation one can check boiling points. Boiling point of water is 100 C in atmospheric pressure, for molecular NaCl, sodium sulfate and Magnesium chloride (main minerals in seawater) boiling point is above 1400 C. For ionic salts dissolved in water there is a different story since they are ionized. The ions are surrounded by water molecules in liquid phase, while if vaporized they would not have a chance of having water molecules closeby so they prefer to stay in the liquid phase. All of that being said, one can envision a certain temperature, pressure and concentration (extremely high for three) at which ions would somehow get out to the vapor phase. Since sea level temperature is quite low, mineral or salt evaporation is practically zero.
Side note on why would water evaporate at 4 C instead of 100 C:
Although it take 100 C of kinetic energy (molecule's velocity) to have "all" water molecules to escape from the intermolecular attractive forces, still "some" of them end up having the energy to escape at 4 C (sea temperature average). It is because kinetic energy has distribution over all the molecules, and you will always find some that have the necessary energy. At the same time "some" of the water molecules in the air end up not having enough kinetic energy to remain in the air, and so they end up being attracted back to the liquid. These two numbers (The molecules of water in air that condense back and the water molecules in the ocean that evaporate) are equal so we have thermodynamic equilibrium. The net number of molecules that escape the sea depends on kinetic energy distribution which is exemplified in temperature.

I think the part prior to the side note makes a good answer. The side note is a bit shaky , i.e. things like "100 C of kinetic energy, etc., and really isn't about what the OP asked. In short, I get the point of the side note, but I think it takes away from the quality of the answer. My 2 cents.

Very good answer. Could you give me a website where I can find the melting/boiling points of various minerals? This website gives me the states of the elements at any temperature; however, it will not let me search for compound states at various temperatures. I'd really appreciate being directed to a website that will give me this information.

NaCl dissolved in water is ionized; why does the NaCl molecule have to be the vapor composition?

Whit3rd is right. It is not much relevant to think about NaCl molecule being in the vapor phase in normal conditions. I should amend my answer.The ions are in lower potential when surrounded by close water neighbors. In gas phase they are not surrounded by close water molecules, so they stay in the liquid. I am trying to put up a case for when you see ions in the vapor. It maybe possible at very high temperature and pressures. Maybe at very high temperatures they get back to elemental form or even molecular form when there is less water around.


text:
Is histidine drawn wrong in these mechanisms for chymotrypsin?
While trying to understand the mechanism of chypotrypsin, I have been looking at a few mechanisms, which are found in my lecture notes, and one found in my textbook. In both cases, it looks like the histidine has the wrong connectivity with the rest of the amino acid chain. Is there a reason for this, or is it an oversight on behalf of the publishers?

Under aqueous conditions you can tautomerise the double bond, so it doesn't matter. The two nitrogens are, for all intents and purposes, equivalent.


text:
Ozonated oil Mixed with water.
I wish to emulsify ozonated oil with water and other ingredients to make a cream. Will mixing the ozone oil and water together with an emulsifier lessen the potency of the ozone oil as an oxidant? what if i were to ozonate the water and oil together and add an emulsifier at the end and blend them? with emulsifying wax for example.  THankyou!

LOL - Instead of ozonated oil use snake oil

I've tried this with a sota ozonator and it doesn't work out that well using water.  First when you ozonate oil by itself it gets really thick which clogs up your aerator for use with anything else so if you do this you have to use a separate aerator for oil and water.
I'm looking at doing something similar with bees wax and colloidal silver and placing the contents into a chap stick like container, when I figure something out I'll let you know.  Water can be mixed with oil with out using an emulsifier.  Dr Gerald Pollack from the University of Washington goes into what he calls the The Fourth Phase of Water "EZ water" and this type of water is known to mix with oil based substances. https://youtu.be/p9UC0chfXcg
His work can be found here
http://faculty.washington.edu/ghp/
Oil and water aren’t suppose to mix but they can see interview below
https://www.youtube.com/watch?v=4KQwh6j-uOg&feature=youtu.be&t=28m17s


text:
Understanding Avogrado's hypothesis
I have read that gases under identical conditions of temperature, pressure, and volume contain the same number of molecules, but every gas has different types of atoms so their force of attraction is also different then how can this be true?

LOL - Avacado's number https://en.wikipedia.org/wiki/Avogadro_constant

The assumption of ideal gas says that the attractions are zero. To first approximation at low pressure, this is mostly true.

If we consider **Ideal gas equation **
and ignore the type of interaction which are present in the real gases then ;
$PV=nRT $
wherevP= pressure ; v=vol. ; n= moles; R = universal gas constant ; T= temp.
$PV/RT=constant $
so $n=constant$
so if$ n$ is constant then if we multiply it with avogadro's constant $N_A $ it would give you number of molecules .
Again that's when ideal gas .

You are right to question Avogadro's hypothesis and your intuition is correct that it would not be true for gases with different attractive and repulsive forces. However, Avogadro's hypothesis defines an ideal gas: a gas where the particles themselves have no volume (no repulsive force) and the particles have no attractive forces. A gas that meets these conditions would perfectly follow Avogadro's hypothesis. While no gas is truly ideal, many behave almost ideally under the right conditions, such as very low pressures or high temperatures.


text:
Friedel Craft Alkylation
Why do we get isoproyl benzene on treating benzene with 1-chloropropane instead of n-propyl benzene?
Is it because of Markovnikov's rule?

First AlCl3 converts 1-chloropropane into propane with a positive charge on terminal carbon
Since secondary carbocation is more stable than primary carbocation it undergoes rearrangement and positive charge shifts to secondary carbonation atom.
Hence we get isopropyl benzene

Markovnikov's rule is just the beginning..It's better to understand the mechanism of how carbocations rearrange themselves to form stable compounds. Then markovnikov's rule will become obvious.

The Lewis acid will pull off the chlorine creating a primary carbocation, primary carbocations are very unstable and are almost always unfavorable. The propane will preform a hydride shift bringing the carbocation to the 2' carbon creating a secondary carbocation. Side note because this is an intermolecular reaction it takes place significantly faster than the actual alkylation. Then the Pi electrons from one of the double bonds in the benzene will attack the carbocation and you clearly know the rest.


text:
What are the common products of the reaction: 2-bromobutane and NaOH
I am self-studying organic chemistry to better my understanding, I cannot formally enroll at a university due to being active duty.
From my understanding there are 2 possible mechanisms when 2-bromobutane reacts with NaOH or any strong hydroxyl base: The E2 and SN2. The carbon attached to bromine is secondary. In the E2 reaction, the oxygen from the hydroxide attacks the "beta" hydrogen, deprotinating it and allowing the electron to travel and form a double bond. Bromide is a leaving group. We get 2-butene as well as water and the bromide ion.
But we have two products from this reaction: trans-2-butene and cis-2-butene.
We also have (2,S)-butanol.
My question is which of these are more favored and why? I know that cis-2-butene would not be favored due to sterics. If possible, I would also like to know when E2 reactions are favored over SN2.

A chemical reaction is the result of competition; it is a race that is won by the fastest runner. A collection of molecules tend to do, by and large, what is easiest for them.
An alkyl halide with β-hydrogen atoms when reacted with a base or a nucleophile has two competing routes: substitution ($\mathrm{S_N1}$ and $\mathrm{S_N2}$) and elimination.
Which route will be taken up depends upon the nature of alkyl halide, strength and size of base/nucleophile and reaction conditions.
Thus, a bulkier nucleophile will prefer to act as a base and abstracts a proton rather than approach a tetravalent carbon atom (steric reasons) and vice versa.
Similarly, a primary alkyl halide will prefer a $\mathrm{S_N2}$ reaction, a secondary halide – $\mathrm{S_N2}$ or elimination depending upon the strength of base/nucleophile and a tertiary halide – $\mathrm{S_N1}$ or elimination depending upon the stability of carbocation or the more substituted alkene.
So, I think preferably in aqueous NaOH or KOH there should be formation of alcohol but that too by $\mathrm{S_N1}$ if there's a tertiary; and, if primary halide, then it would follow $\mathrm{S_N2}$. It's bit tricky for secondary; again it depends on the reagents and solvents used.


text:
Why is it that only ozone absorbs UV radiation?
I know the mechanism of how ozone absorbs UV but why is it that no other molecules do this?
Is there any special reason?
PS: If I'm mistaken and there are other molecules that can absorb UV please mention them.

Do you just mean in the atmosphere or in general? In general, there are lots molecules which absorb UV light. One everyday example is sunscreen, which, depending on the formulation, contain compounds that absorb UVA and UVB light. http://www.compoundchem.com/2014/06/05/sunscreenchemicals/

In atmosphere. And also ur comment gave me another question. If ozone absorbs UV radiation...Why do we need UV protection using sunscreen? Is it that not all UV gets absorbed?

I think a good answer to this is under another question here:http://chemistry.stackexchange.com/questions/35734/what-property-allows-ozone-to-absorb-uv-light                                                                                The first answer gives an explanation of how both ozone and oxygen absorb UV. In regards to your second question, there isn't enough ozone (or other UV absorbing compounds) to perfectly screen the Earth's surface. In addition, ozone absorbs only specific wavelengths of UV light, so other UV wavelengths could get through.

Although you ask about ozone and UV light, the process is completely general is the same for any molecule, chlorophyll in plants, the dye in your T shirt, ink on paper etc. In the molecule, if the photon has the right amount of energy, it excites an electron to a higher molecular orbital and so forms an excited state. The photon is absorbed and so destroyed in the process. We see this as a colour (in transmitted or reflected light) if it occurs in the visible region of the spectrum. If it occurs in the UV we can't detect it by eye but can with photodetectors.

OK .. I think you guys misunderstood me let me rephrase. It is widely accepted that life on Earth wouldn't be possible without ozone layer, so why is it so? Even if ozone layer didn't exist why are there no other gases in the atmosphere that can absorb this radiation?

You should edit your question to ask that then.

Ozone is produced by the reaction of oxygen with UV light. But ozone absorbs more wavelengths of UV than oxygen and adds extra protection to living organisms which no longer need to protect from higher energy UV (though, if ozone didn't absorb these wavelengths, they would probably evolve such protection). Ozone's protective effects are a side effect of the production of oxygen by photosynthesis. We might consider this a fortuitous accident.

The ISO 21348 Definitions of Solar Irradiance Spectral Categories defines UV radiation as that portion of the electromagnetic spectrum having wavelengths from 10 to 400 nm.  Different atmospheric compounds absorb more or less strongly than others throughout the UV portion of the spectrum.
The following table was reproduced from this Wikipedia article:

Although not fully quantitative, this table indicates that nitrogen, oxygen and ozone all absorb different portions of the UV spectrum.  Water vapor is also an important UV absorber that is not discussed in the above table.
Also, regarding the UV absorption of (liquid) water and your comment that "It is widely accepted that life on Earth wouldn't be possible without ozone layer", it was in large part the protection from UV radiation afforded by liquid water that allowed aquatic organisms to survive prior to the existence of the ozone layer. If the ozone layer were to vanish today, although the nature of Earth's biosphere would change drastically, I suspect many aquatic species, particularly microbes, would adapt and continue to survive in the absence of an ozone layer. It would be bad news for you and me though ;)


text:
Themochemistry-unknown metal

Pellets of an unknown metal were heated to 95.0 °C and then transferred to a calorimeter containing 175 g of water at 20.0 °C. The highest temperature recorded was 23.6 °C and 96.0 g of the metal were used.
a) Calculate the specific heat capacity of the unknown metal.

(Metal)Hg=Hl(Water)
$$95\ \mathrm g\times c\times \Delta T=175\ \mathrm g\times4.19\ \mathrm{J/(g\ ^\circ C)}\times(23.6\ \mathrm{^\circ C}-20\ \mathrm{^\circ C})$$
I just wanna know how to get the change in temperature for the metal.

I think you are right so far ;
heat given by metal = heat take by water
so neglecting losses of energy ;
$\delta T_{metal} = 95- 23.6$
$\delta T_{water} = 23.6-20 $
and by substituting values with correct units in :
$ m_{metal} c' \delta T_{metal} = m_{water} c \delta T_{water} $
you should get the desired result.


text:
Separation techniques for different organic compounds

Select the best (most appropriate, easiest, fastest) separation technique for each pair of compounds shown below:

Separation techniques:

evaporation under reduced pressure
column chromatography
water-diethyl ether extraction
acid-base extraction

This requires knowledge of what compounds are separated most efficiently/appropriately by what separation technique. Since I did the labs separately I can only attempt to try to meld the concepts together.
Pair 1 is a carboxylc acid and chlorodiene. Both are polar.
Pair 2 is an ether and hydrocarbon. One is polar the other nonpolar.
Pair 3 looks to be an ester and cyclodiene. One is polar and the other slightly polar.
My attempt:
Based on the acid I'd use acid-base on pair 1.
Pair 2, since one is polar and the other nonpolar, seems like it needs liquid-liquid extraction (water-diethyl ether extraction)
Pair 3 are polar. Unsure of this one. They'd be slow on a column. Evaporation is based on a low boiling point.

You have correctly identified the carboxylic acid in pair 1. The second conpound is 9-chloroanthracene. Calling it a chlorodiene isn't correct. It's an aromatic compound. Anyway, your conclusion to use acid-base extraction is reasonable.
In pair 2, you seem to overestimate the polarity of the first compound. But there is another, more striking difference between the dipentyl ether and n-hexane.

 Dipentylether, $\ce{C5H11-O-C5H11}$, has a much higher molar mass than n-hexane, $\ce{C6H12}$. As a rule of thumb, you can assume that the compound with the higer molar mass will have a much higher boiling point. (There will be exceptions when strong interactions between molecules come into play, such as in the case of hydrogen bonds. But this isn't the case here!) If you look up the boiling points (bp) for both compounds, you'll see that our rule of thumb is supported here. The bp of n-hexane is 68 °C, that of the much heavier ether is > 180 °C. Consequently, distillation (under reduced pressure) is the way to go here.

In pair 3, you seem to misinterpret the nature of the second compound. It's not a diene, but an aromatic hydrocarbon named naphthaline. You might want to read about the properties of these compound and/or the comcept of aromaticity some other time. But for now, let's concentrate on another detail in the structure of the first compound in this pair.

 There is an $\ce{OH}$ group in the the first molecule. Here, the term polar is justified! Think about the typical setup of column chromatography. The stationary phase is usually silica (or aluminium oxide), while the mobile phase typically consists of mixtures of not-so-polar organic solvents. Compound 1 can strongly interact with the silica through hydrogen bonding, while compound 2 is an unsubstituted aromatic hydrocarbon. Separation by colunm chromatography is a very promising method for this pair.

Evaporation under reduced pressure will dip n-hexane's bp?

@user41987 At reduced pressure, the boiling points of both compounds will be lower.


text:
Which enolate of butanone is more stable?
Consider butanone (A) and its two possible enolates B & C.
Which of the two enolates will be more stable? B because of the more stable primary carbanion, or C because of more stable enolate (better hyperconjugation)?

Very much related: http://chemistry.stackexchange.com/questions/17103

This is a classic example of forming enolates under thermodynamic or kinetic control.

in my opinion it should be C.


text:
Phase Diagrams and Equilibrium
In this link
http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch14/phase.php
It says (in the latter part ) that for all combinations of Pressure and Temperature along line BC
the rate of boiling of liquid to form a gas= the rate of condensation of gas to form a liquid.
And C is the Critical Point.  Similar thing has been said  for other lines also
Now 100℃ (or boiling point of the liquid) will lie somewhere on BC
So according to the link rate of boiling of water to form gas is the same as the rate of condensation of gas to form water.
But I have read that at boiling point all the liquid changes to gas or there is no equilibrium at that point.
So why then it says- the rate of boiling of liquid to form a gas= the rate of condensation of gas to form a liquid.
I am not good at chemistry so this might not be a good question. But I was trying to understand phase Diagrams for the first time !

The phase diagram shown is an imaginary diagram. It is not for water, and there are no temperatures or pressures given on the axises so you can't pull 100 ℃ out of thin air.

@MaxW  No I don't refer to 100℃. I meant that he says that along line BC rate of boiling = rate of condensation. But I have read it can be rate of evaporation which can be equal to rate of condensation. At boiling point there is no equilibrium . Is there ?

The boiling point of a liquid is the temperature at which 1 atmosphere of vapor pressure is created. If you look at the lower diagram on the weblink you can see that there is noting particularity unique about the boiling point. The lower diagram shows that the substance has a vapor pressure  in equilibrium with the liquid phase below the boiling point and above the boiling point (albeit at a higher pressure than 1 atmosphere).

I got persnickety because carbon dioxide, for example, doesn't have a "boiling point."  The critical pressure is above one atmosphere.

In the phase diagram of a pure substance there are regions of a single phase e.g. solid, liquid or vapour. The boundary lines you refer to are where the substance exits as $2$ two phases, say solid and vapour in equilibrium with one another.
The Gibbs phase rule states that the number of degrees of freedom, F,  which is the number of (intensive) variables that can be varied is given by
$$F = C-P+2$$
where C is the number of components, $1$ for a pure substance, and P the number of phases, which is $1$ anywhere in the diagram except on a boundary line. Therefore the number of degrees of freedom is $2$, (except on a boundary line) which means that pressure and temperature can both be varied independently.
The boundary lines indicates the condition of P and T where there are two phases in equilibrium, say solid and vapour. As there are $2$ phases there is now only $1$ degree of freedom so that temperature and pressure are no longer able to be varied independently. Thus on the liquid /vapour curve, if the pressure is increased some vapour is condensed (heat is given out) and the temperature increases to a new value but one that that is still on the boundary line, i.e. vapour is still in equilibrium with the liquid. Along any boundary curve the two phases are in equilibrium and so, e.g., the rate of melting is equal to rate of freezing. (Any point along the liquid/vapour boundary line can be considered to be boiling, but we define the normal boiling point to be the boiling temperature at $1$ atm pressure.)
At the triple point, where the three boundary curves meet, there are no degrees of freedom and this point is fixed. At and above the critical point the density of the liquid and vapour are the same and the liquid's meniscus  disappears. The super-critical fluid looks very much like the swirling foam that one sees for a short while on pouring a bottle of Guinness into a glass.
The boundary lines can be calculated for a substance using the Clausius and Clausius-Clapeyron equations.

This seema to be better. Just confirm what I have understood is right or not. The boiling I am referring to , the normal one at say 100℃ is not an equilibrium ( IS THIS MUCH RIGHT) and it is not in equilibrium because Pressure is constant but temperature is rising ( but at 100℃ temperature is constant till boiling happens or non equilibrium exist.). But here both P & T are varying and hence equilibrium exist. That means I would have got the ' normal boiling point ( out of equilibrium ) , if I would have searched it inside the region of line BC towards the liquid side , not on the line BCitself

RE: "Along any boundary curve the two phases are in equilibrium and so, e.g., the rate of melting is equal to rate of freezing." // Melting-freezing  is only true for solid/liquid curve, not any curve on phase diagram.

Re: "Any point along the liquid/vapour boundary line can be considered to be boiling..." // Not really "boiling." Boiling doesn't really have any meaning in phase diagram other than there is a temperature which is the "boiling point." The phase diagram is about adiabatic equilibrium. You can't have "boiling" without the addition of heat.

@MaxW So what I said is that wrong ?

@MaxW  Can you if you wish write an answer as to why the link says that rate of boiling is equal to rate of condensation along BC. I have no problem with lime BC but the usage " equal". Because I have read that at boiling equilibrium cannot exist so rates cannot be equal. That is why it is boiling. If you think there is any fault in the link please point it too.

The boiling point of water at 100℃ is an equilibrium. It is the temperature  at which water has a vapor pressure of 1 atmosphere. There is nothing special about the 100℃ really. What is special is 1 atmosphere which is the air pressure at sea level on earth. So if you live in Denver Co., then water boils at a lower temperature than 100 ℃ because there is less air pressure than at sea level.

@MaxW  Do by equilibrium you mean that at 100℃ for water the rate of boiling is equal to rate of condensation of steam. If that would be  true water wouldn't turn to steam completely . AM I wrong . Please correct me because I am not good at this.

A liquid boils when its equilibrium vapour pressure is equal to the overall pressure. The liquid-vapour curve shows how the vapour pressure above the liquid, and in equilibrium with it,  varies with temperature. The boundary line thus defines the vapour pressure.  The boiling temperature is found by determining when the equilibrium partial vapour pressure is equal to the total  pressure. So my use of 'boiling' was a bit loose.

@porphyrin  So what I said in my 1st comment is right. Please clarify that at 1atm normal boiling of water at 100℃  is equilibrium or not. I think it cannot be an equilibrium process because more water turning to to vapour , hence boiling. Equilibrium can be at some temperature at a pressure ( say 1atm ) when both rates are equal. As temperature increases and pressure stays constant one process increases than the other

yes at equilibrium at exactly boiling point. You don't get this in a kettle of course as steam escapes. In thermodynamics equilibrium always applies. It is just the same for ice in water at exactly the freezing point. Both exist together indefinitely. Hardly ever seen in practice, as  fractionally above this temp all ice melts, fractionally below all water freezes. That the rate of A equals the rate of B is just another way of saying that equilibrium exists.  Thermodynamics says nothing about the rates of things, the rate statements come from a chemical kinetic approach to equilibrium.

An excerpt from the link you provided:

The figure below shows an example of a phase diagram, which summarizes the effect of temperature and pressure on a substance in a closed container...

When heating a liquid in a rigid, close container, no boiling* will occur. The pressure (and therefore density) of the vapour will rise as the temperature is rised. If you carry this on for long enough, the temperature on the liquid surface will reach what is called a critical temperature, $T_{c}$. At $T_{c}$ the pressure of the system is called critical pressure, $p_{c}$. At this point the container will be filled with a supercritical fluid.
*  Boiling, in this case, refers to free vapourization throughout the liquid.

Ok I understand Boiling here means the normal vapourization. So I was right that at boiling there is no equilibrium. Right ? And one thing more , I was thinking that even the pressure varies. At normal boiling pressure is constant. Even this might be a difference.

And one thing more. Throughout the line BC the liquid and gas coexist in equilibrium. Is saying equilibrium right ? But then what's special about the Critical Point .

First question: With an open container, the pressure of the vapour equals the external pressure. When two phases are at an equilibrium, the system  temperature is not freely variable if the pressure is set, and vice versa.
Second: Yes, it is correct. Throughout the BC line the liquid has some vapour pressure, which corresponds to some temperature.

Ok I got it. So in the closed box the temperature and pressure are dependent on each other just like you said and hence equilibrium . But what is special about critical temperature. Is it the maximum temperature at which they coexist ?

At the critical temperature the interface between the gas and the liquid disappears, and a supercritical liquid is formed.

But can I ask one thing. During the line BC why does the equilibrium physically exist ?

@Bdrs - You've used an odd definition of "boiling." // The text did note that point C was the critical point.  You're right about the critical temperature and critical pressure, but the supercritical fluid phase wasn't shown on the diagram.

@MaxW This is how Atkins's Physical Chemistry defines it while considering phase diagrams. I'll double check and revise my answer when I get home.

@MaxW  So the normal boiling does not occur because the container is closed but dor some other reason ?


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Why opposite of electrophilic is not Protophilic
In organic chemistry , considering definitions of  electrophilic and nucleophilic:

Electrophilic:having an affinity for electrons : being an electron acceptor.
nucleophilic : having an affinity for atomic neuclei (also can we say proton?) : being an electron donor.

So both being opposite to each, why dont we say nucleophilic as protophilic?

Because not all "nucleophilic" agents react equally well with all types of relatively exposed nuclei.  Phosphines are very unwilling to extract a proton but quite nucleophilic if the available electronic vacancy is in carbon or a transition metal.

Nucleophile-  nucleus loving .
but it means the one which donates electron or has a negative charge on itself.
The name/ term is given as nucleophile because it would try to give away its electron pair to molecule which has less electron density possibly an electrophile.
like in nucleophillic addition reactions the nucleophile attacks that part for example say in : carbonyl compounds on the carbon which has a less density of electron as the $\pi e^-$ land on oxygen .
So was that a proton site or nucleus site  or a site where electron density is less .


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Why is ruthenium(III) intensely colored in water
I was preparing a solution of ruthenium(III) chloride (aqueous) when I added some (nearly black) $\ce{RuCl3 . H2O}$ to the water. I was surprised that a tiny amount could dye the water so well, I'd say you can even compare this to $\ce{KMnO4}$.
The question is why is the color so intense?
Ruthenium(III) is a $\mathrm{d^5}$-system so the transitions should be spin-forbidden if we are talking about $\ce{d\bond{->}d}$ transitions. If this is the case we can consider a CT-transition. But if I look at the chemistry of manganese(II) it is slightly pink and doesn't dye water at all although it has the same ligands (chlorido and aqua).
And if we compare the colors of manganese(VII), chromium(VI) and vanadium(V) as well as manganese(VII) to renium(VII) with the same ligands, in this case oxido-ligands we see that the more stable states (vanadium(V) and renium(VII)) do not pull electrons that strong anymore and the color shifts further into something invisible to the human eye.
So ruthenium(III) should be even less colored than manganese(II) if we ignore the fact that it has 3 chlorido ligands for a second. So the only option I have left would be the π-donor ability of the chlorido-ligands.
I have read some time ago, that manganese is one of the only elements that form in as their aqueous $\ce{MCl2}$ a cis-dichlorido-coordinated species while all the other (3)d-metals form trans-coordinated ones. And I have already asked about the cis to trans difference in another post so I know now that it does indeed make a difference.
I'd like to compare this to the ruthenium(III) now but ruthenium(III) in water is somewhat complicated as a lot of possible compounds can form especially as the trihydrate will start to exchange chlorido-ligands for aqua ligands at a certain period of time and the trihydrate salt is said to be yellow-orange like iron(III) while I start with a monohydrate which is black.
I'm not sure whether this effect is now related to the rutheium only or if this is just the case because of some symmetry effect but I can't find a good answer on this anywhere.

Ru(III) is d5, but is it high- or low-spin?

Might be. You mean like with Co(III) where nearly every complex is low spin? Might be. They even have a diagonal relationship...

It's not about a diagonal relationship - which barely exist in the d-block - more like 4d/5d metals are nearly always low spin because of their large ligand-field splittings $\Delta$.

Ah good to know. I always thougt this was only the case for the 5d metals. Ok question solved then, thank you.

Perhaps the other relevant thing is that there's also greater vibronic coupling in 4d/5d metals, which helps to relax the Laporte selection rule. So, 4d/5d typically have more intense absorption than 3d.

As you said, in $\ce{RuCl3}$, the charge of the Ruthenium ion is +3, meaning d5. In water, the RuCl3 becomes $\ce{[Ru(H2O)6]^3+}$, an octahedral complex. The electron configuration of the Ruthenium ion can either be $\mathrm{t_{2g}{}^5\,e_g{}^0}$ (low spin), or $\mathrm{t_{2g}{}^3\,e_g{}^2}$ (high spin), as shown in the pictures below:

You also need to know about CFSE (crystal field stabilization energy), which is the difference in the energy of the free metal ion, and the metal ion in the complex.
According to the electron configuration diagrams above, the CFSE for each configuration is:
$$\text{Octahedral }\mathrm{d^5}\text{ — low spin}\\
\text{CFSE} = -0.4\Delta_\mathrm{O} \times 5 + 2P = 2P - 2 \Delta_\mathrm{O}\\[1.5em]
\text{Octahedral }\mathrm{d^5}\text{ — high spin}\\
\text{CFSE} = -0.4\Delta_\mathrm{O} \times 3 + 0.6\Delta_\mathrm{O} \times 3 = 1.2\Delta_\mathrm{O} - 1.2\Delta_\mathrm{O} = 0$$
Where $P$ is the mean pairing energy, the energy required to overcome the repulsion between the electrons and pair them together. $\Delta_\mathrm{O}$ is the octahedral splitting parameter, which I assume you already know.
As you can see, a high spin $\mathrm{d^5}$ octahedral complex will not lead to a change in energy compared to a free ion. Therefore, the spin is determined by the low spin configuration. If the pairing energy is bigger than the octahedral splitting parameter, the metal ion will gain energy, and will rather be high spin. If the pairing energy is smaller than the octahedral splitting parameter, the metal ion will lose energy and will rather be low spin.
The octahedral splitting parameter is mostly affected by the ligand type (sigma, pi acceptor, pi donor). The pairing energy is mostly affected by the size of the d orbitals — the bigger they are, the smaller it is (because the electrons have more room, the repulsion between them is smaller).
4d and 5d transition metals have large d orbitals, so that the mean pairing energy is usually smaller than the splitting parameter, which results in low spin complexes regardless to the ligand type.
Therefore, a $\ce{Ru^3+}$ ion, which is $\mathrm{4d^5}$, will form low spin octahedral complex, that has spin allowed transitions with intense colors.
About charge transfer. It's possible, but I personally don't know.


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E-glass degradation in water
At the moment, a question puzzling me is why and how glass fibers degrade in water environment? As the degradation of the glass fibers occurs, its reduces the strength of them. As far as I know, the degradation occurs slower in seawater (containing ions) than in distilled water, which in my opinion indicates that process is related to dissociation of glass material.
How does this degradation occur and why is it different for seawater and distilled water? Do I understand correctly, that E-glass is slowly partly dissociating into ions in water environment? Or is it ion-exchange or some other mechanism?

Simply the glass degrades because it dissolves in the water.
First the how. There are two factors which impact this process which is the dissolution of sodium and the dissolution of silica. The dissolution of sodium occurs first as evidenced by a plot of pH vs time for a glass powder in water and proceed via electrophilic attack:
$$\ce{Si-O-Na_{(glass)} + H2O_{(l)} \longrightarrow Si-O-H_{(glass)} + OH-_{(aq)} + Na+_{(aq)} }$$
The second step is the dissolution of the silica which proceeds via nucleophilic attack:
$$\ce{(-O-)2Si(-O-H)2_{(glass)} + H2O_{(l)} <-> Si(OH)4_{(aq)} }$$
The solubility of the silica is, of course, low, but measurable and significant. This process at high temperature increases the solubility of the silica (hydrothermal) and is how natural quartz crystals were formed (hydrothermal vents).
Second you say that degradation occurs slower in seawater than distilled water and that, in fact, is correct. At first, one might suspect that the dissolved sodium reduces the dissolution due to a common ion effect or reduced chemical potential but is actually not the case. The operative word here is seawater.  Seawater already has a large amount of dissolved silica in it from the rocks and sand that lines coastlines and ocean beds such that it is saturated. Since the second reaction is reversible the increased concentration prevents the silica from being leached appreciably faster than it can be redeposited thus reducing the corrosion of the fiber

Thank you very much! Could You please provide some relevant references? Some sources, where I can read more on the topic?

Do a search on aqueous corrosion of glass.  These reactions are well established in the glass science community. Unfortunately my information comes from lecture notes rather than specific papers so I couldn't point you anywhere beyond that.