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Why are there no cheap diamond equivalents?
We recently learned in school that diamonds sparkle as the are very optically dense, meaning that it takes longer for light to pass through them, thus meaning that the light totally internally reflects when you look at a diamond, whereas in glass it would pass through at that angle.
Why have we not been able to create a solid material that is so optically dense? Is it because such materials are rare, expensive or hard to produce, or because no-one has looked into it?

Yes, it is. As far as I'm concerned, you can make Gold out of Lead; but it's not worth it.

As far as searching for "alternatives" or improvements for other things (e.g. bronze vs. steel vs. composites), improvement of one property is almost always is a tradeoff for something else. If you're making a structure of some sort, usually you just care about a couple properties (Young's modulus, tensile strength) and the others are irrelevant. Sometimes though, you care about lots of properties (color, castibility, patina, strength...) it's increasingly difficult.

And for crystalline materials, you're talking about fairly fundamental atomic properties that are nearly impossible to rationally control or manipulate.

There are four properties that make a diamond look like a diamond. I will compare them with the two best diamond simulants at the moment:

cubic zirconia $\ce{ZrO2}$ - a special ("cubic") form of zirconium oxide. Don't confuse it with zircon, a different mineral with the forumula $\ce{ZrSiO4}$.
moissanite $\ce{SiC}$ - the name for the naturally occurring form of silicon carbide, also used for the synthetic gemstones varieties of it.

So what are the differences and why none of them is an all-round replacement for a diamond?

Refractive index: determines the brilliance. This is what you were talking about when you said "optical density". This basically means the amount of light the diamond reflects to the observer. If you change the refractive index, you also have to slightly change the cut of the diamond to achieve optimal brilliance, and then your "diamond" will not look exactly like a diamond. The refractive index of diamond is 2.42. Moissanite is higher: 2.65 and cubic zirconia is lower: 2.17.
Dispersion: determines the fire. This is what you wanted to say when you mentioned that diamonds are sparkling. Fire is the property of diamonds to break the light into different wavelengths thus creating the characteristic play of colors that you get in diamonds. Here's the interesting thing: The dispersion of diamond is 0.044 and interestingly the dispersion of both cubic zirconia and moissanite is higher: 0.060 and 0.104, respectively. So theoretically, if you consider only fire (the play of colors), then they both look better than diamonds.
Birefringence: (the lack of) determines the clarity. When you look through a diamond at all angles, you can clearly see what's behind. This is also true for cubic zirconia. Both minerals are isotropic. On the other hand, moissanite is birefrigent - it will double the image you see through the mineral, causing it to look blurred and lose the visual clarity.
Hardness: while not directly related to optical properties, a mineral with higher hardness will retain it's sharp cut. If the cut is damaged, the gemstone will not look as good and lose its value. Diamond has a hardness of 10, moissanite 9.25 and cubic zirconia 8.25. Note that this scale is not linear: 10 is much harder than 9 compared to 9 and 8, for example.

As you can see, none of the alternatives fully satisfies the entire set of properties that make a diamond. Moissanite may be slightly better, but it's also more expensive. And remember - one of the most appealing things about diamonds is their rarity and natural occurrence. No synthetic simulant can replace that.

http://www.huffingtonpost.com/rohin-dhar/diamonds-are-bullshit_b_3708562.html diamonds are bullshit (to some)

@Jonny and it's not for others. The question was about diamond simulants, not whether diamonds are rare and whether they should be valuable. That said, I was never too fond of diamonds as gems: 1. They're (mostly) colorless and that's boring. 2. They're chemically too simple for my variety-in-elements-loving-brain. I'd rather have some tourmaline: Much more beautiful and also chemically interesting.

There are plenty of synthetic materials that would fall into the "cheap diamond equivalent" category. If we're only talking about the optical properties, cubic zirconia ($\ce{ZrO2}$) is probably the most familiar as it is often used in jewelry. It's index of refraction isn't quite as high and it's not as hard, but it's close enough that a fair amount of it is used to make jewelry. A more recent development is moissanite (either synthetic or natural $\ce{SiC}$), which is almost as hard as diamond and actually has a higher refractive index. Other synthetic materials with even higher indices like barium titanate and gallium phosphide exist, though generally their optical properties are used for things like lasers.
Of course, we have gotten quite good at making synthetic diamond itself. They are only used for a tiny fraction of diamond jewelry, though. It wouldn't surprise me to find a concerted effort to market natural diamonds as superior, given how lucrative the gem diamond market is, which might be preventing wider use for gems. Synthetic diamonds are widespread in industrial applications, however, where smaller diamonds are used to make tools and optical properties are largely irrelevant.

Use of cubic zirconia and synthetic diamond in jewellery is (I think) an economics/psychology question rather than a chemistry one. Most casual observers won't notice the difference by eye, but one of the purposes of jewellery is to be expensive. So people who want expensive jewellery won't buy a cheaper version no matter how similar it looks, since it won't hold the same value. The natural diamond is superior because it's difficult/expensive to obtain, not because of its optical properties!

You're probably right. I imagine if diamonds were cheap enough to stick them on everything like rhinestones, they would probably lose a lot of their appeal. However, from what I can tell, gem-quality synthetic diamonds are actually more expensive than mined diamonds and are mostly used for coloured gems as coloured natural diamonds are very rare.


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Comparing Basic strengths of different species
I was doing questions on organic chemistry. I came across tonnes of questions  where I was required to compare basic and acidic strengths.
Comparing acidic strengths is fine — I remove an $\ce{H+}$ and compare the stability of the anion because my teacher told me a strong acid has a stable anion and explained the reason to me from an equilibrium point of view.
When it comes to comparing basic strengths what do we do?
Can I add an $\ce{H+}$ to the species and compare the resulting stability by resonance, hyperconjugation and inductive effect and then can conclude the more basic one will have a stabler cation?
Really baffled.
Please do correct me if I am mistaken in my fundamentals.

The stability of the conjugate acid of a particular base is not the only thing that you should consider. You should consider the relative stability. Though for simple problems that will usually suffice. Often with bases protonation leads to loss of delocalization (e.g phenylamine).

This is a comment. Your approach assumes that an acid must have a hydrogen atom available to it. This is the Bronsted acid-base theory and neglects questions such as whether $\ce{BF3}$ is acidic or basic (in solvents other than water). Lewis acid-base theory is much better, imho, when it comes to organic chemistry. Acetic acid is a base in liquid $\ce{HCl}$, fwiw.

Comparing acidic strengths is fine - I remove an H+ and compare the stability of the anion coz my teacher told me strong acid has a stable anion and explained me the reason from equilibrium point of view.
  When it comes to comparing basic strengths what do we do ?

You follow the same general idea - the stronger the base, the weaker its conjugate acid. Consider a generic base such as $\ce{B-}$ reacting with water, the acid in this case.
$\ce{B- + H2O <=> HB + HO-}$
The conjugate acid of $\ce{B-}$ is $\ce{HB}$. And the stronger that $\ce{HB}$ is as an acid, the more likely it is to protonate hyroxide ion and favor the left side of the equilibrium, thereby making the base weak.

Can I add an H+ to the species and compare the resulting stability by resonance ,hyperconjugation and inductive effect and then can I conclude the more basic one will have a stabler cation?

Yep, that's it.

When do I know which hydrogen is to be removed(for basic strength) or where hydrogen is to be added (for acidic strength)?

To find where the proton is most likely to be added, determine where is a) most negative and b) sterically accessible. Also consider the stability of the conjugate acid.

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Acidity of oxonium ion
How was the pKa value (-1.7) for the oxonium ion determined? Why does it correspond to the minimum pH value accessible in water? What reaction does it represent?

There is a nice little article in the Journal of Chemical Eduction Negative pH Does Exist pointing out examples of negative pHs in water, for example pH = -3.6 in mine water in California.  So -1.7 is not the lowest possible pH in water. Keep in mind that pH is defined in terms of hydrogen ion activity rather than concentration.  This is emphasized in Paradoxes: Demonstrating That It Is Not True That pH ≡ -log[H+].  At high acid concentation, hydrogen ion activity greatly deviates from hydrogen ion concentration.  For example, in 16 molar HCl, hydrogen ion activity is 678!
And see Negative pH and Extremely Acidic Mine Waters from Iron Mountain, California (no paywall), which describes pH values even below -4.
In Hydronium ion activity in strongly acidic media. Remarkable agreement between independent estimates Am. Chem. Soc. vol. 95, pp 3055–3057, 70% sulfuric acid is found to have a pH of -9.8 (negative 9.8)!
However, ignoring activity and the true definition of pH, considering that pure water has a concentration of 55M, if all the water molecules were instead hydronium ions, -log(55) = -1.7.  Then in articles like Who Knows the Ka Values of Water and the Hydronium Ion? this value is attributed to the pKa of hydronium, with the supposed justification that:
$K_a = \frac{[\ce{H3O+}][\ce{H2O}]}{[\ce{H3O+}]} = [\ce{H2O}]$
The article New point of view on the meaning and on the values of
Ka(H3O+, H2O) and Kb(H2O, OH-) pairs in water is a serious consideration of the true meaning of the pKa of hydronium, and points out that the -1.7 value for the pKa of hydronium has no justification.
In conclusion, pH can be below -1.7 and the pK of hydronium isn't really -1.7.

How is -3.6 possible? Please elaborate.

@Marko, OK, I added to the answer, basically because hydrogen ion activity greatly deviates from hydrogen ion concentration in concentrated acids and pH is defined as -log of hydrogen ion acitivity, not concentration.

Sadly, can't access the: Paradoxes: Demonstrating That It Is Not True That pH ≡ -log[H+].

@Marko As a substitute see http://community.asdlib.org/imageandvideoexchangeforum/2013/07/31/activity-vs-concentration/ which illustrates one of the paradoxes: that [H+] can be cut in half, yet hydrogen ion activity increases is a given example

Another example, just check the wikipedia page for HCl http://en.wikipedia.org/wiki/Hydrochloric_acid and you can see that you can "easily" reach negative pH.

@Babounet the Wikipedia table isn't correct, the pHs are really even more negative than that at high concentration

Negative sure, but I am interested in less than -1.7.

@Marko OK, see http://digitalcommons.unl.edu/cgi/viewcontent.cgi?article=1488&context=usgsstaffpub should be no paywall, has pH -4 and below, I'll add it to answer

Why does the activity coefficient increase, when the concentration increases? Is it because of triple ions formation?

@Marko You have to study Debye-Huckel theory and further advances of that theory.  Activity coefficients initially decrease with increasing ionic strength, go through a minimum and then increase sharply.

How was the pKa value (-1.7) for the oxonium ion determined?
Why does it correspond to the minimum pH value accessible in water water?

Note that an oxonium ion is simply an ion in which an oxygen carries a positive formal charge. Many oxonium ions exist. One of which is hydronium ion.
The hydronium ion corresponds to the strongest acid able to exist in water for significant periods of time because a stronger acid will be immediately deprotonated by some base in the system. In other words, hydronium ion can exist in equilibrium with water.
$\ce{H3O+ + H2O <=> H3O+ + H2O}$
The reaction of a stronger acid with water will however favor the products.
$\ce{HCl + H2O -> H3O+ + Cl-}$

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Understanding and usage of Arrhenius equation
A widely used rule-of-thumb for the temperature dependence of a reaction rate is that a ten degree rise in the temperature approximately doubles the rate. This is not generally true, especially when a strong covalent bond must be broken. For a reaction that does show this behavior, what would the activation energy be?

What activation energy would lead to a doubling in the reaction rate
  for a 10 degree increase in temperature, using the Arrhenius equation?

You’re right it is not generally true. Let’s set up an equation to solve the problem.
The Arrhenius equation is
$$k=A\operatorname{e}^{\frac{-E_\text{a}}{RT}}$$
In this case
$$2=\frac{A\operatorname{e}^{\frac{-E_\text{a}}{R\cdot(T + 10\ \mathrm{K})}}}{A\operatorname{e}^{\frac{-E_\text{a}}{RT}}}$$
or
$$\ln 2 = \frac{ - E_\text{a}}{R\left( T + 10\ \text{K} \right)} - \frac{ - E_\text{a}}{RT}$$
let’s pick a temperature around room temperature, 300 K is a nice even number, but you can re-solve the problem with any temperature you like.
$$0.69 = \frac{ - E_\text{a}}{R \cdot 310\ \text{K} } - \frac{ - E_\text{a}}{R \cdot 300\ \text{K} }$$
Using $R=8.31\ \mathrm{J/(K\ mol)}$ will produce an answer in J/mol.  Solving we find
$$E_\text{a} = 53\,568.49\ \mathrm{J/mol}$$
or,
$$53.6\ \mathrm{kJ/mol}~~(12.8\ \mathrm{kcal/mol})$$
So around room temperature a reaction that has an activation energy around 54 kJ/mol (or 13 kcal/mol) would show a doubling in rate for a 10 kelvin increase in temperature.

"So around room temperature a reaction that has an activation energy around 54 kJ/m (or 13 kcal/m) would show a doubling in rate for a 10o Kelvin increase in temperature." Does reaction order matter here?

@Dissenter No, reaction order doesn't matter.


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Order of nomenclature precedence among these functional groups
The order of precedence for functional groups I have so far is:

carboxylic acid
ester
amide
aldehyde
ketone
alcohol
amine
multiple bonds
ether
organic halide

I'm pretty confident that from 1 to 7, I have the order of precedence right. But from 8 onwards, I'm not too sure. Could someone help verify that the list I've got is right or help me correct it?

@L to the V are you talking about precedence in the context of IUPAC nomenclature?

You've mentioned 'these functional groups' in your statement. But, Halogen, nitro group are never functional groups.

From Master Organic Chemistry: It looks like your table is correct

Wait...how come alkene is above alkyne? I thought triple bonds were of higher precedence than double bonds.

I had trouble cross-referencing my textbook, but i came across other online sources that agree with my answer. Do you have any material that might suggest otherwise?

The so-called “Table of Functional Group Priorities For Nomenclature” shown on masterorganicchemistry.com tries to combine the seniority order of suffixes with prefixes and endings. This can be misleading and result in wrong names. In particular, there is no suffix ‘-ane’, ‘-ene’ or ‘-yne’ (these are endings), and multiple bonds (and all groups that follow in that table) are actually not considered in the seniority order. Please do not use such sources.

@Loong do you have a recommended resource? I'm open to altering this answer with a more legitimate source. Could be in the style of a community wiki.


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Why is a trialkylamine a better leaving group, in nucleophilic reactions, than a Cl group?
Why is a trialkylamine a better leaving group, in nucleophilic reactions, than a $\ce{Cl}$ group? In the same vein, isn't a trialkylammonium a much weaker acid than $\ce{HCl}$?
$\ce{R-NR3+ -> Nuc-R + NR3}$ (Stronger leaving group)
$\ce{R-Cl  -> Nuc-R + Cl-}$    (Weaker Leaving group)
Compared to:
$\ce{H-NR3+ -> Nuc-H + NR3}$    (Weaker acid, NR3 thus weaker leaving group)
$\ce{H-Cl -> Nuc-H  + Cl-}$     (Stronger acid, Cl- thus stronger leaving group)

The trialkylamine is likely a lot less nucleophilic than a chloride ion. This reduces the possibility of reverse reactions.

Wouldn't the same hold true with acids? If Cl is more nucleiphillic, by extension, it should also be more basic, no?

no. Iodide ion (I-) is way way more nucleophilic than fluoride ion (F-) in a polar protic solvent. Yet F- is clearly way way more basic than I-. Nucleophilicity is a kinetic property; basicity and acidity are both thermodynamic properties.

Leaving group ability refers to how well a leaving group can stay off whatever it came from.
Consider the following prototypical nucleophilic displacement reaction:
$\ce{R-LG + Nuc- <=> R-Nuc + LG-}$
There are two issues at play here:
1) Thermodynamics. How stable is the reactant with the leaving group, $\ce{R-LG}$, and the nucleophile, $\ce{Nuc-}$? How stable are the products, $\ce{R-Nuc}$ and $\ce{LG-}$? And when examining stability, we must look at not only the reactants and products but also their environments - i.e. the solvent(s).
2) Kinetics. What's the physical ability of the nucleophile to actually displace the leaving group? What's the physical ability of the leaving group to displace the nucleophile again?
Usually we're taught that leaving group ability has to do with acidity and basicity. The less basic the leaving group is, the better its ability to leave. This isn't a perfect rule of thumb, but it's acceptable.
That's the thermodynamic part of the question. If $\ce{LG-}$ weren't thermodynamically stable, then it would be inclined to attack $\ce{R-Nuc}$ and return to the status quo. (Or just grab a proton).
However, how capable is $\ce{LG-}$ of carrying out such an attack? Here we come to the kinetic part of the question. The bulkier the leaving group, the more likely it is to be sterically hindered and/or strongly solvated (trapped by solvent molecules). Trialkylamine  is a very genetic term, but when I think of amines with alkyl groups I think of quaternary ammonium compounds, which have alkyl groups that can contain upwards of 18 carbons (!). Such bulk affects the ability of the leaving group to line up properly with the substrate to perform nucleophilic attack. Nucleophilic attack often has specific steric demands; $\ce{SN2}$ requires backside attack; the approach of attacking nucleophiles with respect to flat carbonyl groups has an angle of 104.5 degrees, etc.

Here is a good summary of sterics and amine nucleophilicity.
https://books.google.com/books?id=tP4cqrYF7nwC&pg=PA231&dq=amine+nucleophilicity+steric&hl=en&sa=X&ei=nXOhVLiOD8iKgwSS5YPICg&ved=0CEUQ6AEwBA#v=onepage&q=amine%20nucleophilicity%20steric&f=false


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Two 2" solid steel bearings for stress relief hand-rolling. Safe? What's that smell? How to thoroughly clean?
I bought these two 2" solid steel ball bearings for the purpose of rolling them in my hands. Just for something to do while sitting at my computer, without those silly bells inside. (They're still a bit noisy, but it sounds much cooler than bells.)
Aside from being moderately scary (I cracked my ceramic sink while cleaning them, when I dropped one about six inches. Some shards exploded off of the underside...don't tell my wife, luckily it's not leaking), I'm wondering if there's anything I need to worry about health-wise.
I've read that they might have nasty lubricants on them, and that rubbing alcohol should clean it. I soaked them in alcohol for the night, then rubbed them down with an alcohol soaked washcloth, and then cleaned them with soap and water.
But they still smell, and it makes my hand smell as well. Not that strong, but significant and yucky, and I'm wondering if it's something to be concerned about, or if that's just the nature of steel.
How can I thoroughly clean them? Is the smell something to be concerned about, or should I just wash my hands when I'm done and forget about it? Is there a safe way to coat/paint them, without causing further health issues? Anything else I should know?
Thank you.

aliteralmind, welcome to Chemistry.SE, and great question. Here's something that's tangentially related ... http://chemistry.stackexchange.com/questions/7916/why-can-we-smell-copper

the deleted answer was actually quite good; I don't know why the owner deleted it. In any case the smell likely results from a redox reaction between iron ions and organic molecules on your hands.


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Why does the ionic product of water remain constant after addition of non-neutral solute?
In my textbook, it is given that the ionic product of water $K_\mathrm{w}$ remains constant even when a non-neutral solute such as an acid is added to it.
$$K_\mathrm{w} = \ce{[H3O+][OH-]}$$
When a strong acid is added to water, the concentration of $\ce{[H3O+]}$ increases. Yet the $K_\mathrm{w}$ must remain constant. But how? How does the concentration of $\ce{[OH-]}$ decrease?

$K_w$ is the equilibrium constant of this reaction:
$$\ce{2H2 O <=> H3O+ +OH-}.$$
So, it's constant at a fixed temperature. It equals $10^{-14}$ at $25~^\circ\mathrm{C}$. At this temperature $\ce{[H3O+]=[OH^{-}]}=10^{-7}~\mathrm{mol/L}$.
If you add acid to water, the above reaction equilibrium is  displaced in order to counteract the change in the concentration of ion hydronium, i.e. the reaction is displaced to the left. This means that  ions hydroxide will react with the excess of ions hydronium to give water. As a result: the concentration of ion hydroxide will be less than $10^{-7}~\mathrm{mol/L}$, and the concentration of ion hydronium will be more than  $ 10^{-7}~\mathrm{mol/L}$. But their product is constant. $\ce{[H3O+] \cdot[OH^{-}]}=10^{-14}$.

If excess kfhydronium reacted with hydroxide , won't conc. of both decrease. I.e. H+ conc. still remains $10^{-7}$ ?

This explains what happens but not why.

In Layman’s views, you can also think of it this way: initially, you have water, a small amount of $\ce{H3O+}$ and a small amount of $\ce{OH-}$. If you add an external proton source, we can consider what will happen to each of these:
\begin{array}{lccc}\hline
\text{time point} & \text{acid} & \text{neutral} & \text{base} \\ \hline
\text{before a.} & \ce{H3O+} & \ce{H2O} & \ce{OH-} \\
\text{after a.} & \text{no change} & \ce{H3O+} & \ce{H2O} \\ \hline \end{array}
In words: Nothing will happen to $\ce{H3O+}$ — at least nothing notable; $\ce{H4O^2+}$ is too unstable in aquaeous solution to contribute meaningly. Water will be protonated to $\ce{H3O+}$ and hydroxide ions will also be protonated (statistically) by the incoming acid to give $\ce{H2O}$. Thus, you can consider the addition of acid to both increase oxonium ion concentration and reduce hydroxide ion concentration.
The equilibrium constant and equation is just a more formal and scientifically more correct way of putting this.

This was helpful! But I am still looking for an explanation of why the increase in $\ce{H3O+}$ and decrease in $\ce{OH−}$ happens in just such as way that the product of their concentrations remains constant, instead of… say, the sum of their concentrations remaining constant, which is what I intuitively thought should happen after looking at your table.

@lukejanicke That basically books down to physical chemistry and equilibrium theory whose formulae I know how to apply but not how to derive from scratch. Sorry …

I haven’t been able to find an answer yet. Someone asked this and was told the question is a duplicate of this, which I disagree with, but the answer might be in there somewhere.


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Equilibrium Constant of a Salt knowing the solubility
Calculate the equilibrium constant in $\ce\\M ^ 3$ of the heterogeneous system in aqueous solution:
$\ce{CaF2 \leftrightarrows Ca^{2+} + 2F-}$
Knowing that the solubility of the salt is $15.9 \ce{mg/L}$.

They are three and gave the same result.

Ok but i can't post a photo of my efforts. I tried three times with differents methods but they gave the same result that is wrong.

No problem. The point is, we want to keep the community clean from 'users' who only want to get the answer and in other words, want to make the community do their homework. The thing you should do is to show that you have understood the concepts behind solving the problem. e.g.: Say that first I should find the M for $\ce{CaF2}$ and then I should use the formula of measuring the equilibrium constant which is ....

@MARamezani Yes i did like this:
Solubility= radq^3 of Kps/4
Because Kps=2s^2 s (in this case because the salt gives 2 moles of F and 1 of Ca)
I found then the result of the Keq, and it is 2.910^-7
(this is one of the method, the other one was using molarity for the salt and...)
The problem is that three methods gave me this result and the book says that the correct answer is another.

The molar solubility $s$ equals:
$$s= \frac{15.9\times 10^{-3}}{78}= 0.2038\times10^{-3} \ce{mol/L}$$

According to the equilibrium: $\ce{CaF2 \leftrightarrows Ca^{2+} + 2F-}$, the  equilibrium constant is:

$$K_{sp}=4s^3$$

By substituting the value of $s$ in the above equation, we find:
$$K_{sp}=0.034\times10^{-9}$$


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Experiment to prove that water has air
In elementary school, there was an experiment  to show that water has dissolved air, which involved boiling some water and observing the drops at the surface of the container,  and this proved that water has air. Now, in senior high school, I was thinking about the same,  and I think this conclusion is wrong. Aren't the drops formed by the conversion of liquid water to steam due to evaporation? Or is it really the dissolved air that is escaping? What are your thoughts?

There are two ways that I interpret the question you're asking, and I'll try to address each in turn.

If your teacher boiled a container of water and held a piece of glass above it to collect drops of liquid water, he or she may have been trying to show that water doesn't disappear when it boils, that instead it turns to a gas that can move through the air and can be condensed back into a liquid.  Essentially, $\ce{H2O}$ is $\ce{H2O}$, regardless of the state of matter it occupies.
If instead the teacher wanted you to observe the bubbles of gas forming in a container of hot or boiling liquid, he or she might have been on the right track, but either failed to convey the correct message to you or had the wrong idea in the first place.
If I take water from the tap and heat it strongly, bubbles will form on the inside of the vessel I'm heating it in prior to it boiling.  These are bubbles of gas that are no longer soluble because the temperature of the liquid is too high (and its pressure has decreased as compared to the tap.)  The gas is dissolved air (mostly carbon dioxide, I suspect) with a large portion of water vapor.  In any case, they are bubbles of gas that were dissolved but are no longer.  You can see this happen any time you boil a pot of water on the stove.
Once the pot reaches boiling, any bubbles are almost exclusively gaseous water formed via vaporization.
An easier, if slower, way to see this is to run a cold tap until the water is as cold as it will get, then pour a glass.  As it sits (perhaps for an hour or two), bubbles will eventually form on the walls of the glass.  These eventually will dissolve again and the gas will escape through the top of the glass, so it's not something you can set up at night and come back to the next morning.  The bubbles are air that was able to dissolve in the high pressure/low temperature water in the pipes but not in the low pressure/high temperature water in the glass.

The setup was number 2. So from what i gather, at a lower temperature,  bubbles are mostly dissolved air, and at higher temperature,  its water vapour.

@SaurabhRaje  Yes, that's basically it.  Far below 100°C, bubbles that form are mostly dissolved air (with excess carbon dioxide, because it dissolves much more readily than nitrogen or oxygen.)  Closer to 100°C, they would include more and more water vapor because the vapor pressure of water increases as the temperature rises.  Above 100°C, the water boils and the bubbles are almost exclusively water vapor.

A German scientific TV show once showed a great experiment (alas, I cannot find any copy of it on Youtube). They had two pots of water. In each of them there was a large water-filled closed-top glass cylinder that reached out of the water so that its upper part was freely visible. They then boiled the water in both pots for some time; part of the gas was collected in these cylinders and made the water level in them drop.
Then, they let the water cool down again (this was recorded in fast motion). In one of the two cylinders (calling it cylinder 1) the water level rose back to the top, the other one kept a cushion of air at the top of the cylinder (cylinder 2).
The explanation: The water in cylinder 1 and its corresponding pot had been pre-boiled. Since gases are better soluble at low temperatures, boiling and re-cooling effectively removed all dissolved gases from the water of cylinder one before the experiment. Thus, the only gas to be observed at the top of cylinder 1 was water vapour.
Cylinder 2’s water had not been pre-boiled. There were still gases dissolved in the water which again escaped upon boiling. These gases were collected at the top of cylinder 2 along with water vapour. When the setup was cooled down, the water vapour recondensed (to a significant fraction, at least) but the previously dissolved gas remained on top.

Your description is somewhat vague ("boiling some water and observing the drops at the surface of the container"), but it does seem unlikely that such an experiment demonstrates that air dissolves into water.  If by 'drops' you mean gas bubbles, then yes: the gas bubbles in boiling water are substantially all water vapor, not air. You are correct: any drops of liquid forming on the inner surface of the container would be from re-condensed water vapor, itself evaporated from the boiling liquid, and would have nothing to do with air having been dissolved in the water.
What I would do to attempt to demonstrate that air dissolves into water would be to rig up a 2L clear plastic (NOT GLASS) soda bottle so that I could connect it to an air compressor. (I'd have to think a bit how to accomplish this.) Then:

Fill the bottle about half full with plain water
Cap the bottle & connect to the compressor
Wearing safety glasses and being very careful (the bottle could burst, or turn into a rocket if the cap comes loose), pressurize the bottle to 100psi (~700 kPa),
Let the bottle stand, pressurized, for a while -- thirty minutes, maybe? -- periodically swirling it gently.
CAREFULLY vent the pressure over 5-10 seconds, like when opening a bottle of soda that's been shaken.
If no bubbles appear, tap the side of the bottle several times.

The amount of air that dissolves in water increases with the pressure.  The hope would be that enough additional air dissolves under 100psi air, that when you release the pressure in step 5 and the amount of air the water can hold then drops back down to normal, the excess air will form a bunch of bubbles throughout the water.

By drops, i mean droplets, like ones of condensation, or for eg ones you may observe if you pour sparkling water in a glass container. Sorry for my language, english is not native

It's quite all right; I just want to be sure I'm answering what you're actually asking.

@SaurabhRaje Do you mean drops of gas or drops of liquid?  Drops of gas would be called bubbles, which would make your teacher's demonstration make a lot more sense.

@JasonPatterson, i mean drops of gas, or bubbles.

Saurabh, having read Jason's answer, I think I was misunderstanding your question all along...apologies!

@brian, no problem

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text:
Why is dU an exact differential and dq an inexact differential?
dU, dG, dH etc are all exact differentials and the variables themselves are known as state functions because they only depend on the state of the system. However, dq and dw for example, are inexact differentials. My questions is, what does this actually mean? I've been told that exact differentials are not dependent on the path (what does 'path' mean?) but inexact differentials are. I have been told this is related to line integrals but I am not sure how. Also is this relevant?

As opposed to an exact differential, an inexact differential cannot be expressed as the differential of a function, i.e. while there exist a function $U$ such that $U = \int \mathrm{d} U$, there is no such functions for $\text{đ} q$ and $\text{đ} w$.
And the same is, of course, true for any state function $a$ and any path function $b$ respectively: an infinitesimal change in a state function is represented by an exact differential $\mathrm{d} a$ and there is a function $a$ such that $a = \int \mathrm{d} a$, while an infinitesimal change in a path function $b$ is represented by an inexact differential $\text{đ} b$ and there is no function $b$ such that $b = \int \text{đ} b$.
Consequently, for a process in which a system goes from state $1$ to state $2$ a change in a state function $a$ can be evaluated simply as
$$\int_{1}^{2} \mathrm{d} a = a_{2} - a_{1} \, ,$$
while a change in a path function $b$ can not be evaluated in such a simple way,
$$\int_{1}^{2} \text{đ} b \neq b_{2} - b_{1} \, .$$
And for a state function $a$ in a thermodynamic cycle
$$\oint \mathrm{d} a = 0 \, ,$$
while for a path function $b$
$$\oint \text{đ} b \neq 0 \, .$$
The last mathematical relations are important, for instance, for the first law of thermodynamics, because while $\oint \text{đ} q \neq 0$ and $\oint \text{đ} w \neq 0$ it was experimentally found that $\oint (\text{đ} q + \text{đ} w) = 0$ for a closed system, which implies that there exist a state function $U$ such that $\mathrm{d} U = \text{đ} q + \text{đ} w$.

Not a complete answer, but the path is exactly what it sounds like. Say you are rolling a boulder up a hill. This increases its potential energy, which can be released by rolling the boulder down the hill. But the path you take to get to the top of the hill is irrelevant, only the height you raise the boulder matters for potential energy. So if you roll it up half way, let it fall back a quarter of the way, or any such combination of forward and back, none of this matters to the total change in potential energy (assuming perfect conditions, no friction, etc.).
With respect to Wildcat's great answer, this means for state functions the endpoints of your definite integral are all that matter: you could parameterize any path you want between the endpoints and the resulting (line) integral is the same.

Be careful, there are lots of confusion and misleading claims in introductory textbooks, like that "thermodynamics only applies to macroscopic objects", ignoring the whole field of nanothermodynamics or the thermodynamics of small objects; or that "thermodynamics only applies to equilibrium", ignoring that two Nobel Prizes for Chemistry were awarded to advances in the thermodynamics of non-equilibrium.
It is not true that an infinitesimal change in a path function "is represented by an inexact differential". Heat, as any other path function, can be represented by an exact differential. Precisely, one of those Nobel laureates has a book where heat is treated as an exact differential. The book is "Modern Thermodynamics: From Heat Engines to Dissipative Structures" by Kondepudi and Prigogine and the pair of authors explain this topic very well, so I will copy and paste the relevant part:

For a closed system, the energy exchanged by a system with the exterior in a time dt may be divided into two parts: $dQ$, the amount of heat, and $dW$: the amount of mechanical energy. Unlike the total internal energy $dU$, the quantities $dQ$ and $dW$ are not independent of the manner of transformation; we cannot specify $dQ$ or $dW$ simply by knowing the initial and final states. Hence it is not possible to define a function $Q$ that depends only on the initial and final states, i.e. heat is not a state function. Although every system can be said to possess a certain amount of energy $U$, the same cannot be said of heat $Q$ or work $W$. But there is no difficulty in specifying the amount of heat exchanged in a particular transformation. If the rate process that results in the exchange of heat is specified, then $dQ$ is the heat exchanged in a time interval $dt$.
Most introductory texts on thermodynamics do not include irreversible processes but describe all transformations as idealized, infinitely slow, reversible processes. In this case, $dQ$ cannot be defined in terms of a time interval $dt$ because the transformation does not occur in finite time, and one has to use the initial and final states to specify $dQ$. This poses a problem because $Q$ is not a state function, so $dQ$ cannot be uniquely specified by the initial and final states. To overcome this difficulty, an "imperfect differential" $\text{đ} Q$ is defined to represent the heat exchanged in a transformation, a quantity that depends on the initial and final states and the manner of transformation. In our approach we will avoid the use of imperfect differentials. The heat flow is described by processes that occur in a finite time and, with the assumption that the rate of heat flow is known, the heat exchanged $dQ$ in a time $dt$ is well defined. The same is true for the work $dW$. Idealized, infinitely slow reversible processes still remain useful for some conceptual reasons and we will use them occasionally, but we will not restrict our presentation to reversible processes as many texts do.
The total change in energy $dU$ of a closed system in a time $dt$ is
$$dU = dQ + dW \>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>(2.2.3)$$
The quantities $dQ$ and $dW$ can be specified in terms of the rate laws for heat transfer and the forces that do the work. For example, the heat supplied in a time $dt$ by a heating coil of resistance $R$ carrying a current $I$ is given by $dQ = (I^2 R)dt = VI dt$, in which $V$ is the voltage drop across the coil.

In more advanced formulations one always work with rates
$$\frac{dU}{dt} = \frac{dQ}{dt} + \frac{dW}{dt}$$
with heat and work rates given by
$$\frac{dQ}{dt} = - \int_{B(t)} \mathbf{q} \mathbf{n} dB$$
$$\frac{dW}{dt} = - \int_{B(t)} \mathbf{T} \mathbf{v} dB + \int_{V(t)} \rho \mathbf{F} \mathbf{v} dV$$
For a system enclosed in a volume $V$ with a boundary $B$, with the work performed by the body forces per unit mass $F$ and the contact forces $T$, $\mathbf{v}$ being the
velocity field, and $\mathbf{q}$ the heat flux vector on the normal $\mathbf{n}$ to the boundary, and $\rho$ the mass density.

To clarify, are you saying the the typical formulation of heat exchange is an inexact differential, but treating these changes as occurring over finite time we can produce a formulation that involves exact (though time dependent) differentials? Does this agree with the conclusions you reached here: http://www.juanrga.com/2016/12/what-is-heat.html

@Tyberius A change in any physical quantity only can happen is that quantity depends on time. The 19th century formulation of thermodynamics pretends quantities can change without using time as variable, which doesn't make sense. This pretension is the reason why concepts as inexact differentials have to be introduced in the classical formalism. Modern formalisms developed since 20th century use time and exact differentials for both heat and work.


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What is the mechanism for the creation of P(CH2OH)4+ cation?
I know that the initial reactants are phosphine and menthanal in an acid ($\ce{H+ A-}$) and that the first step in the reaction is the HOMO of phosphine (the lone pairs) and the LUMO of methanal (the π* orbital).
Then, the electrons that just moved to the oxygen atom react with $\ce{H3O+}$, so the oxygen is protonated. I'm not sure what follows, though.

The acid serves as a catalyst to protonate the formaldehyde oxygen.  This makes the carbonyl carbon more electrophilic, a more reactive target for the weakly nucleophilic lone pair of electrons on phosphorous.  Here is a diagram of the basic reaction mechanism.

The proton transfer is thought to be intramolecular according to this link.


text:
The use of ice water in an experiment
An experiment from Vogel that I want to conduct requires "a mixture of ice and salt of about -10 °C". Reading trough the text it appears that I need at least 1-2 liters of this. Obviously the salt is added to lower the freezing point of the water. I guess the interaction of water with the dissolved salt trough hydrogen bonding / lone pair donation is more favorable (less energetic) than ordinary water-water interaction.

With "salt", does Vogel mean just $\ce{NaCl}$?
Since I'm a hobby chemist I do not have the same fancy machinery as they have in the lab (those cool things that generate rough ice shards and in which everybody stuffs food against all safety regulations lol). All I can use is a small refrigerator with a tiny freezer. It's easy enough to make some ice cubes, but how do I go about making these ice shards that are used in the lab? I don't think grinding the cubes is very practical as it is a tedious job that takes a lot of time in which the shards have probably already melted (also from hitting them with a pestle continuously)...

1.With "salt", does Vogel mean just $\ce{NaCl}$?

Yep, it's cheap and available.

How do I go about making these ice shards that are used in the lab?

Wrap the ice cubes in a towel and hit the bundle with a hammer until you get whatever size you want.  That's how we did it in the lab.

I was imagining myself using a pestle and mortar to crush all the cubes, a towel is much more practical indeed. Thanks for the very helpful tip (it is the stuff that is so simple yet is in no book!) Was my comment about why salt is used correct?

Your welcome.  As to why the addition of a salt to water leads to a freezing point depression, it has more to do with concentrations and chemical potential.  See here,  for example

It's not addition of salt to water, but the addition of salt to ice that lowers the temperature of the mixture.


text:
Mechanism of dehydrogenation with nosylate
What is the mechanism of the attached reaction? How could nosylate do the dehydrogenate the species?

The reaction is from this synthesis (46th IChO preparatory problem 26):

I think this is quite a complex mechanism, but I would assume that the nitro group is being reduced. Could you add a reference as a starting point?

Edited my original post.

Thanks, that will help put it into context, but I am still thinking that a suitable mechanism is hard to propose.  I tried tracking down the original paper, which they used as source, but was unsuccessful due to language barrier.

So I guess that it would be rather written as a standard red-ox reaction.

@Marko The method of using 3-nitrobenzenesulfonate for that kind of dehydrogenation was invented by Gerhard Bachmann http://www.google.com/patents/US3487081 I don't know the mechanism, but if you can get access try looking at the following article and references cited http://onlinelibrary.wiley.com/doi/10.1002/jhet.5570320114/abstract I don't have access.

I don't have access either.

I checked Cignarella's J. Heterocyclic Chem. paper (1995, 32, 79). There is no mention of the mechanism. They cite a conference presentation by Bachmann as the first report of the method (no help there) and a couple of other med chem papers from the Cignarella group. I checked one of these (J. Med. Chem. 1989, 32, 2277), and there is no mention of mechanism.

Was that a question on the Olympiad or are you just curious?

I am just curious.

@jerepierre They usually don't bother with the reaction mechanism, as long as it is working. I would not even know where to start investigating this. Dehydrogenation often starts with a proton abstraction, but is a sulfonyl group or nitro group really strong enough to do that?

@Martin Without knowing the fate of the sulfonate reagent, I don't feel confident in any mechanism. If there were some transition metal to mediate the dehydrogenation, the nitro group could be the terminal reductant. The reaction is under basic conditions (hydroxide) which makes me think it's going through an enolate. I wonder if somehow the O-sulfonate is formed, and then an elimination occurs to form a carbonyl with formation of sulfinate. That's my best guess, but there are no similar reactions in SciFinder.

This is the reaction mechanism that I propose for this reaction. Though it does not show the complete reduction of the 3-nitrobenzene sulfonate, it could be reduced further through the nitroso-hydroxylamine pathway. This means that in total, a single nitrosyl anion could oxidize three molecules of the starting material.
Extraction of a hydrogen by the nitro group. The hydrogen is extremely acidic compared to an ordinary aliphatic $\ce{C-H}$ bond due to the massive conjugation in the resulting ion:

Amide-imidic acid tautomerization:

Resonance structures of the molecule, showing its extensive conjugation, and hence stability:

Radical reaction pathway and formation of the product. Every atom in product is $\mathrm{sp^2}$ hybridized and is conjugated over the entire system. It is much more stable than the starting material, and is the driving force of the entire reaction:


text:
Does the density of a liquid and gas phase of a substance(nitrogen or water) converge when reaching critical point
I know that the there is no difference in density in a liquid and a gas phase of a substance after the critical point. But I do want to know if this difference in density is more continuous or if it is more like a big step when reaching the critical point? I.e if we are like 0.01 Kelvin beneath the critical point of water (with the critical pressure). Is the difference in density still big at this point or is it near zero?

Does the density of a liquid and gas phase of a substance converge when reaching critical point?

Yes. This is exactly what happens. Let's take a look at a qualitative $P-T$ (pressure–temperature) phase diagram of an arbitrary substance:

source: Cansell, F. et al. Supercritical fluid processing: a new route for materials synthesis. J. Mater. Chem. 9, 67–75 (1999). 10.1039/A804964E
As you can see, moving directly from gas to liquid involves crossing the phase boundary. However, if you're doing it in the supercritical, the change in density is continuous. Maybe you can get a better, more quantitative view with this example of $\ce{CO2}$:

source: http://what-when-how.com/nanoscience-and-nanotechnology/microweighing-in-supercritical-carbon-dioxide-part-1-nanotechnology/
This one is also a a $P-T$ phase diagram, with equal density $(\rho)$ lines. You can see that the lines radiating from the gas–liquid boundary lines are not equal on each side. For example, 0.1 and 0.9. However, as you get close to the critical point, $\rho$ on each side of the line converges to a single value: 0.2 and 0.7, 0.3 and 0.6 and so on.
YouTube has some terrific examples of this happening. For example: http://youtu.be/GEr3NxsPTOA
What you see is a function of density - you can see the phase boundary between liquid and gas because they have differing densities. Notice in the videos how the phase slowly and gradually disappears as the density of both phases converges, effectively turning the $\ce{CO2}$ into a supercritical fluid.


text:
What is a covalent semiconductor?
Aren't all semiconductors' atoms bond in a covalent bond? How is a "covalent semiconductor" different from other semiconductors?

No, most semiconductors are ionic.
I'm not sure I've seen the term "covalent semiconductor" before. Wikipedia has an article on covalent superconductors.
I think the main point would be that many if not most semiconductors are inorganic solid-state compounds. Consequently, most of these (e.g. III-V semiconductors are ionic solids, not really covalent ones. Consider GaAs or InP, which are typically classified as ionic.
That is the solid state electronic structure can be considered mainly composed of cations and anions (unlike organic chemistry where the electronic structure is considered mainly of neutral atoms).
(Incidentally, defining whether some of these compounds have purely ionic or purely covalent interactions is somewhat futile, since most bonding interactions are both somewhat ionic and covalent.)

Most semiconductors are inorganic solids but some polymers can be semiconducting and that is probably what the term "covalent semiconductor" is referring to.
The majority of semiconductor material are inorganic solids (as Geoff Hutchinson describes in his answer). The electronics industry is based on the use of silicon-based materials (often doped with small, controlled amounts of other elements to achieve the precise semiconducting properties required). Mixed compounds like GaAs or InP are also used. Arguing about the nature of the bonding in these compounds is is, as others have said, a little futile.
But there are a range of compounds that can be semiconductors that are more often thought of as "covalent". The most important of these are Organic Light Emitting Polymers (which, in devices, are called Organic Light Emitting Diodes or OLEDs). If the term "covalent semiconductor" means anything, this is the class of compounds it is referring to.
To simplify a little, the semiconducting property is created by delocalisation of electrons along the pi-system in the molecular chain as opposed to by the overall crystal structure of an inorganic solid.
There is a good Wikipedia article on the topic under the title "Organic Semiconductor". These are now becoming significant in display technology.

Pure silicon (Si) and pure germanium (Ge) are examples of covalent semiconductors. To explain, first take a look at the difference between ionic and covalent bonding.
In ionic bondig, the electron is given by the least electronegative atom and taken by the most electronegative atom. If the difference in this electronegativity of both atoms is rather large (so atom A wants to give badly, atom B wants to take badly), the bond is regarded as strictly ionic.
On the other hand, if there exists a bond between two atoms which have a similar affinity for an electron, the bond is regarded as covalent. This must be the case in pure silicon, as there are no other kinds of atoms present, so there are no different affinities. Silicon has 4 valence electrons and forms a crystal structure, sharing it's 4 valence electrons with 4 neighbors, thus all filling it's orbitals to the maximum of 8 electrons.
Silicon is an intrinsic semiconductor, meaning that without any modifications, some electrons may reach the conduction band. Adding dopants to a crystal structure such as this, can change these semiconducting capabilities. The difference with oxide semiconductors is that in covalent semiconductors, the effectiveness of dopants at a given temperature is solely determined by their ionization energies. In oxide semiconductors, it also depends on the extent of ionization of the structure. After all, the presence of the dopant in an ionic compound (oxide semiconductor) can be partly charge-compensated by other ionic defects. This is explained in chapter 2 of 'Physical Ceramics'.

A covalent semiconductor is one whose crystal structure contains covalent bonds. A covalent bond is one in which electrons are equally shared between the bonded atoms. Only silicon and germanium are purely covalent because all bonds are identical and thus, by symmetry, electrons are shared exactly equally. Compound semiconductors like gallium arsenide, GaAs, are partially covalent, there is some imbalance to how the electrons are shared in bonds. They spend more time with one of the constituent atoms than with the other one.
On the other end of the scale are ionic bonds and ionic solids where one atom in the compound rips the electron from the other atom. The electron then essentially becomes part of the negatively charged ion. NaCl is such a solid. I can’t think of any ionic semiconductors.
In Introduction to Solid State Physics, Kittel has a table of the “Fractional ionic character” of bonds in crystals. Silicon is 0 ionic (meaning fully covalent). GaAs is 0.31 ionic and NaCl is 0.94 ionic. All of the III-V semiconductors are in the range of 0 to 0.4. Some semiconductors like CdS go as high as 0.7.


text:
Why inert pair effect is seen in P- block elements only
why is inert pair effect specific to certain groups? Why heavier transition elements DO NOT show inert pair effect?

The inert pair effect is not limited to p-block elements. It does affect transition metals. However, the energy difference between d- and s-electrons is less than between p- and s-electrons, so the effect is less pronounced.


text:
How to decide acidic or basic nature of oxides
How can we theoretically decide the acidic or basic nature of an oxide? What are the reasons/ factors which make an oxide acidic, basic, amphoteric or netural?

You might find the answers given to an older question helpful.

In general, the electropositive character of the oxide's central atom will determine whether the oxide will be acidic or basic. The more electropositive the central atom, the more basic the oxide. The more electronegative the central atom, the more acidic the oxide. Electropositive character increases from right to left across the periodic table and increases down the column.
The trend of acid-base behaviour is from strongly basic oxides on the left-hand side to strongly acidic ones on the right, via an amphoteric oxide (aluminium oxide) in the middle. An amphoteric oxide is one that shows both acidic and basic properties.
This trend applies only to the oxides of the individual elements in the highest oxidation states for those elements. The pattern is less clear for other oxides.
We define non-metal oxide acidity in terms of the acidic solutions formed in reactions with water. For example, sulfur trioxide reacts with water to forms sulfuric acid.
In sum, acidic oxides are oxides of non-metals, and basic oxides are oxides of metals.
There are three non-metal oxides from the upper right portion of the periodic table, $\ce{CO}$, $\ce{NO}$, and $\ce{N2O}$, which have such low oxidation numbers for the central atom that they give neutral aqueous solutions.
Since the acidity of a cation rises rapidly with charge, d-block elements that exhibit a wide variety of oxidation numbers may have one or more oxides that exhibit only basic properties and one or more oxides that exhibit only acidic properties. The higher the oxidation number, the more acidic the corresponding oxide. Chromium is an example of such an element. $\ce{CrO}$ is basic, $\ce{Cr2O3}$ is amphoteric and  $\ce{CrO3}$ is acidic.

Consider an electropositive atom’s oxide and an electronegative one’s. And also consider them hydrolysed (i.e. $\ce{E=O + H2O -> E-(OH)2}$ or $\ce{E-O-E + H2O -> E-OH + HO-E}$).
Oxygen is very electronegative, so it should always be $\delta -$. But the electronegative atom will allow less electron density to be drawn away by the oxygen, so there will be less of a negative charge on the oxygen. This means that displacing the proton to give the oxygen more of a negative charge becomes more favourable. Thus, the oxide is acidic.
Now for the electropositive atom, oxygen now almost gets that atom’s entire electron denisty. This means that oxygen is just a tad too negative to feel well, so it will draw protons out of the surrounding solution to protonate itself. Thus, the oxide is basic.

It depends on what you mean by be acid or basic. Look at Lux-Flood theory ;)

@Shadock Brønsted-Lowry acid/base definition.

Never thought of acids and bases in this light ,really liked this answer.

There is a theory developed by Hermann Lux and Håkon Flood, named Lux-Flood theory to explain the basic or acid character of an oxide.
The rules are very simple.

An acid of Lux-Flood is an acceptor of $\ce{O^{2-}}$
A base of Lux-Flood is a donor of $\ce{O^{2-}}$

Examples
$\ce{CaO}$ is a base of L-F because $\ce{CaO}=\ce{Ca^{2+}}+\ce{O^{2-}}$
$\ce{SiO2}$ is an acid of L-F because $\ce{Si}$ has unoccupied d orbital then it can have a valence upper than two and then accepts $\ce{O^{2-}}$ ions.
Then they can react to give you $\ce{CaSiO3}$
EDIT
If you use Fajans' Rules you will find that $\ce{CaO}$ is more ionic than $\ce{SiO2}$ that's why this reasoning is correct.

To measure the strength of them it is common to use the scale of $\ce{pO^{2-}}=-\log(\ce{O^{2-}})$ like we do for the $\ce{pH}$.
It may exists amphoteric Lux-Flood compound but I have not examples in head right-now.

Your examples don’t really make sense. $\ce{SiO2}$ is also $\ce{Si^4+ + O^2-}$. Also, while silicon does have d-orbitals somewhere above in the sky, so does calcium. In neither do they take part in bonding in any significant way. I think, I understand where you want to go, but it is badly phrased.

@Jan I have a better explaination I will edit my post.

Very interesting point of view. I would like you to add examples with amphoteric oxides. Also, explanations that include the involvement of 3d orbitals in the chemistry of the 3rd row elements are highly discouraged.

@Marko for amphoteric coumpound there is the answer of Yomen I just saw. :)

I know, but I want wanted you to expand your answer, and threat the amphoteric oxides with the Lux-Flood theory.

@Marko I'll do if I find a bit of time :)

@Shadock No, the HSAB theory does not label acids stronger/weaker than others. It only talks about hardness and softness, which are not related to other acidity factors. E.g. ammonia is a hard base but a weak base. Amide is also a hard base but a much stronger base.


text:
Is it possible for Hydrogen to lose its electron?
I know the basics of Chemistry and one thing I've always wondered if it is possible for Hydrogen to give up it's one electron? I know Hydrogen is eager to share its election through covalent bonds, but is there a scenario where Hydrogen will give up its only electron?

Sure, then it is just a proton. Protons are very reactive though and will probably not form in solution, rather they get transferred from one molecule to an other which greatly stabilizes.

What happens to the size of hydrogen cation?It increases or decreases?

Hydrogen can lose an electron meaning it can be in the +1 oxidation state. However, just like any other cation or anion it never occurs free in condensed matter, it always is in contact with solvent and/or anions. Moreover, because of extremely small size of proton, it is an extremely powerful Lewis acid. Consequently, in common conditions proton would react with first electron pair it comes in contact with, up to and including inert gas electron pairs and covalent bond pairs.
On the other hand, hydrogen ions are quite easy to generate in electric discharge and/or under extreme heating. In fact, producing and confinement of super-hot plasma, consisting of hydrogen ions and electrons, is an area of active research for several decades and, well, producing and confinement of relatively cold plasma is not a problem. Confinement of several billions Kelvin hot plasma, however, is still a problem.

Or just get a glimpse of tomorrow morning's sunrise.  As in the plasma known as the Sun.

The night sky gives you plenty of opportunities as well ...


text:
Is carbon dioxide organic or inorganic?
Today in chemistry class we were discussing Organic Chemistry. We discussed what organic compounds basically are and then I asked the teacher whether $\ce{CO_2}$ is organic or not. She told that it is as it contains carbon and oxygen with a covalent bond. I told her it can't be as it is not found in animals (naturally). I am very confused about it.
I need some good reasons to agree with either explanation. (I have searched the internet already but found no great insights as of now).

It is commonly considered inorganic, but some of its properties have parallels in organic chemistry, so in some cases it can be considered as part of some series of organic compounds as a 'terminal' case. Same for methane and most $\ce{C1}$ and some $\ce{C2}$ molecules.

Originally, it was believed that living (whatever that is) things and the matter they were composed of would be distinguished from non-living things by the presence or absence of a life force/vital energy. Thus,  $\ce{CO2}$ exhaled from a cow would have to be different from  $\ce{CO2}$ gained by mixing mineral calcite with sulfuric acid gained by roasting sulfide ores. This turned out wrong, $\ce{CO2}$ from either source cannot be distinguished, chemically. As a tribute to tradition, components only containing  $\ce{CO2}$, especially without $\ce{H-C}$ bonds, are still called an-organic.

Does carbon dioxide know or care?

It is entirely arbitrary whether you call it an organic compound or not, though most would not.
The distinction you make that organic compounds should be found in living things is not a useful criterion. Moreover you are wrong that carbon dioxide isn't: it is both made and used by living things. Animals make it when they metabolise sugars to release energy; plants consume it when they build more complex organic molecules through photosynthesis. In fact most organic molecules are, ultimately, derived from $\ce{CO2}$.
Even more importantly most molecules considered organic are neither made by nor are found in living things. Chemists make new carbon compounds all the time (tens of millions in the history of chemistry) and most have never been made by animals or plants.
The organic/inorganic terminology is mostly very simple: covalent compounds containing carbon are organic. The only fuzzy area is around very simple molecules like $\ce{CO2}$ where the distinction doesn't matter much. So we would not normally think of diamond or silicon carbide as organic. But we might (though many would not) call calcium carbide organic because it contains a $\ce{C2}$ unit with a carbon-carbon triple bond.
However since the terminology is mostly very obvious and also somewhat arbitrary, it isn't worth much argument to sort out those very simple but awkward edge cases.

There is no exact definition of "organic" compound, although you can say that organic compounds must contain carbon.
There is no requirement that organic compounds can be found in living things, although the name organic comes from the fact that the first compounds in this class that were discovered did come from living things. That said,  $\ce{CO2}$ is in living things - all metabolic processes either consume or produce  $\ce{CO2}$ (with the exception of some thermophiles).
In general, ionic and "simple" molecular compounds that contain carbon are not called "organic," partly for historical reasons, but also partly because the category "organic" is more than just a name - it describes an entire set of bond types, reaction patterns, naming rules, and structure types. This is why there is no exact, simple definition - it is hard to draw a boundary around all of that and say "these other things don't belong."
Taking your example, $\ce{CO2}$ is not normally called "organic," even though it contains carbon, it is produced and consumed in biochemical reactions, it has similar properties to other organic molecules, and organic bonding principles can describe it. The reason for this is probably because historically,  $\ce{CO2}$ was discovered and named before organic chemistry existed as a field.
Some people say that in order to make it easy, any carbon-containing covalent compound is "organic" (your teacher is probably in this group). Some people say "any compound containing carbon and hydrogen" should be the rule. Both of these rules have exceptions, though, so they can't be used as exact definitions.

...and one could certainly argue even about the "must contain carbon" rule. For example, it does seem a bit arbitrary to deem hydrazine to be inorganic, but monomethylhydrazine to be organic. But of course, if you want to draw a sharp line, then you have to draw it somewhere.

Carbon dioxide is indeed strictly speaking organic, as it is a compound containing carbon. Your claim that it doesn't naturally occur in living organisms is not correct. For instance, it is the product of breathing.
That said, some chemists do not consider it to be organic as it has no hydrogen in it- a critical component of just about every organic compound.
You have to remember that the term "organic" is just a vague definition given by humans to some types of chemicals. Nature does not always play by our rules, especially when you're talking about borderline cases such as carbon dioxide.

"Carbon dioxide is [...] organic, as it is a compound containing carbon." By that argument, calcium carbonate is organic.

@DavidRicherby that's why I mentioned that the definition of "organic" is vague. Carbon dioxide is a borderline case.

@DavidRicherby. Moreover, if we consider organic compounds to (1) contain carbon, and (2) be of biogenic origin, a significant portion of the calcium carbonate in the world is "organic", as it is made by marine organisms. It all sums up to definitions made by us and the consensus in the scientific community.

Yes, the definition is vague and yes, carbon dioxide is borderline. But carbonates are not borderline. "Compounds containing carbon" is, therefore, doubly wrong as a definition: it fails to be vague and it includes compounds such as carbonates that are widely (universally?) agreed to be inorganic.

indeed, point taken.

$\ce{CO2}$ is commonly considered inorganic. You will rarely find a textbook on organic chemistry that discusses $\ce{CO2}$ and its properties. On the other hand every textbook on basic inorganic chemistry does.
Interestingly, urea $\ce{(NH2)2CO}$ historically was considered organic and its synthesis from inorganic reactants (ammonium cyanate, F. Wöhler, 1828) is widely regarded as the beginning of the end of vitalism.


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Are there two compound resins that tolerate over 1000°C for glass/carbon fiber?
I'm a hobby enthusiast that loves to play with electronics so my major is not in chemistry, there for I'm hoping that this forum is the right place to find aid in this question.
I've searched day in and day out, and when someone refers to "epoxy" with "extreme high temperature tolerance" they're generally speaking about 200-300°C, which to me isn't that much.
I can understand that depending on the context that is a lot, but there has to be something that is actually "extreme" in the sense that it can tolerate up to 1000 °C since formula 1 cars typically hit around 1000°C at the end of a break line and typically the thickness and heat dispersion of the setup just can't take more because of the limits within FIA.
So my question is, what is the technical term for high temperature epoxy's and what other options is there if not?

Siloxanes are available as bicomponent resins and there are siloxanes withstanding 300 Celsium. I'm very sceptical about polymeres stable at 500+ celsium for prolonged time. While some polyacrilonitrile-derived materials (pyron) can withstand extreme heating for some time, they still degrade over time. In case you want carbon-derived material, most likely you'll have to carbonize it moving to reinforced carbon. Various ceramics, glass and castable materials can be a valid alternative if stability above 300 C is required.

Thanks martin! @permeakra thanks for reminding me that my "it's all the same" isn't applicable hehe, sorry for this confusion. My intentions are to use either carbon fiber if it can withstand the heat or fiber glass strands and glue those strands in a epoxy or some sort that can withstand 500 or even 1000+°C.

@Torxed Carbon fiber certainly can withstand 1000 C as long as it is not exposed to air, otherwise it may start to oxidize... Slowly. Common glass fiber will not withstand such temperatures, it would fail somewhere between 500...700 I think. I don't think you'll find commercially available binary resing able to withstand over 300 C, but maybe you can use ceramics? Though it will require a proper furnace to use.

Can someone recommend a good 300 degree resin? -- I'm having trouble finding anything that is rated above 100 degrees Celsius...

According to Wikipedia Formula 1 disk brakes are made of carbon-reinforced carbon.

To succinctly answer your question, yes. However, they are not epoxies, and typically not available to a hobby enthusiast.

There is no such thing as a high temperature epoxy. The maximum service temperature of epoxies is around 200 C on a good day.

Higher temperature resin systems for 250 a 300 C are typically BMI-resins. These are rigid, thermoset plastics used in aerospace for composite airframe structures. BMI resins can be two-part, but are typically high-temperature thermal cures.

The next level of heat resistant resins are polyimides. These can range from 300 C to about 450 C operational temperatures. However, it depends highly on the application. Some thermoplastic polyimides will not thermally degrade until 600 C, but have a much lower softening temperature (Tg). Thermosets will maintain rigidity, but will begin to thermally decompose at a lower temperature. However, depending on the application, polyimides can carbonize to form carbon-carbon structures: https://www.techbriefs.com/component/content/article/tb/pub/techbriefs/materials/27033. Polyimides are not two-part systems and require high temperature (250-380C) cures.

Silicones. These are things like the fireplace sealant. They are not really high temperature, but will crystalize into ceramic-type structures at very high heat. They can be one or two-part. Things like this are typically used for glues, think space shuttle tile, and not composites, or as a precursor for carbon-ceramic composites.

There are some next generation hybrids, like polysiloxaneimides that are two part. They are mixtures of polyimides and ceramic silicones. Good for temperatures from about 300 C up to about 1200 C, but poor performance in the mid-range.

Carbon-carbon and carbon-ceramic. Made from polyimdes or silicones, these resins are fired at high temperatures (1000C+) to form ceramic-like materials capable with withstanding temperatures from 900 C to above 2500C. Two parts in the sense it is really a two-part manufacturing process.

When it comes down to it, there are a lot of options, including a few I didn’t cover like PEEKs and PBIs and Cyanate Esters. Most of these are not available to normal humans though, and there is a reason they tend to be reserved for aerospace and space applications. At the lower end of the BMI resins can cost hundreds a gallon, high-temperature polyimides can stretch into thousands per pound, and then all of these typically require special processing equipment, handling and safety measures. High temperature composites are not typically used on F1 cars as they are too expensive and not allowed by regulations - excluding exhaust systems and carbon-ceramic brakes. You also run into the problem that these high-temperature materials are not ‘consumer’ goods. So, unless you have access to a major university or corporation, there is no way to actually buy them.

Was torn between what to mark as "solved". But this answer contained a good comparison between the different options. Thank you for taking your time to answer this and welcome to this forum!

In an oxygen-containing environment, no carbon-based polymer (which includes all epoxies) that I know of can tolerate 1000 C, at least not without significant degradation in mechanical strength.
There are two reasons for this:
One is that the carbon-carbon bonds will start to break down in large numbers at higher temperatures, and since the carbon-carbon to carbon-oxygen reaction is exothermic, if oxygen is around there will be no going back upon cooling.
The second is that even if the polymer doesn't chemically degrade, it will undergo a physical phase transition - it will melt, or at least soften. This can be compensated for by reinforcing with fiber or particles to make a composite. However, if it softens enough, a fiber composite will "delaminate" - the physical bonds holding the resin onto the reinforcment will come apart, and the composite will no longer be a composite.
Here is an interesting study that talks about both mechanisms of degradation, and lists several systems with their correpsonding melt and thermal decomposition temperatures. Polytetrafluoroethylene has the highest thermal decomposition temperature in this study, at 775 K (502 C).
Non-carbon polymers might be an option, although the tradeoff is that they are weaker and composite options are a lot more limited. Silicon-based polymers generally withstand higher temperatures than carbon analogues, but I wasn't able to find any that can approach 1000 C. Fluoro-polymers are another possibility - as you can see from the reference, they tend to have much higher stability in terms of thermal decomposition. There might be a fluoro-epoxy resin out there. One thing to keep in mind, though, is that the products of fluoro-polymer decomposition are particularly toxic and reactive, and so the applications where they could safely be used at high temperatures are probably very, very limited.
Edit in response to questions

One question regarding the "oxygen around", would this also include if layers of epoxy were applied as a extra layer ontop of the carbon or would it need to be a complete vacuum in order to prohibit oxidization?

If you had a layered carbon/polymer composite, the chemical reaction with the oxygen wouldn't start at the polymer-carbon interface, it would start at the surface exposed to oxygen. However, since the reaction is exothermic, once it starts burning, it would go quickly. The carbon would burn too at these temperatures. An inert atmosphere would prevent that reaction, but it would still thermally decompose - the difference would be that hopefully the decomposition wouldn't be as exothermic, and therefore would go more slowly. Now for epoxies, they have oxygen in the polymer structure. At high temperatures, that oxygen will react and the polymer will break apart whether there is an inert atmosphere or not.

Thank you for sharing! There's a lot to go on here. One question regarding the "oxygen around", would this also include if layers of epoxy were applied as a extra layer ontop of the carbon or would it need to be a complete vacuum in order to prohibit oxidization?

The answer was pretty long so I just wrote it as an edit.

This answer my questions enough to be considered a solution to my curiosity. I will simply have to look for other ways to do my applications as there will be oxygen in the working Environment and the material will degrade to quickly. Thank you for your in-depth answer and the study you pasted. Cheers!

I'm glad I could help - if you start another question that focuses more on your specific application, you might be able to get even better answers. There might be alternative materials/ways of doing things that you haven't considered yet...

Polyether ether ketone (PEEK) should be good to go up to 250 °C.
Polybenzimidazoles (PBIs) are stable up to temperatures of 500 °C, probably more. Protective suits for fire fighters are made from this material.
In order to withstand higher temperatures (nose cone of a Space Shuttle), carbon-reinforced carbon can be used.

Up to 250 °C is way to low for my application. 500 is also to low but not we're getting somewhere :) reinforced carbon is a valid choice so +1 for that. My concern however is the epoxy holding that reinforced carbon in place. As it's apparently the weak joint in any application because the carbon obviously can take a space trip without any harm :)

@Torxed I'm afraid I might not have read your requirements properly enough: there is no epoxy resin left in carbon-reinforced carbon. It's more like carbon fibres embedded in layers of graphite.

You can use fireplace sealant which can withstand 1500 °C to substitute the resin. This method works with carbon fiber but I don't know about fiberglass.
Credits: amazingdiyprojects, Integza (he build a pulsejet out of this material)


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Is this the correct equation for the mean free path?
$\lambda=\frac{KT}{\sigma p}$ is the equation given in my textbook but everywhere else I look it is different. Is it wrong? The derivation is as follows:
$$\lambda = \frac{v_{rel}}{z}$$ where $\lambda$ is the mean free path and $z$ is the collision frequency.
Since $z=\frac{\sigma v_{rel}}{KT}$ where $\sigma$ is the area where it is possible for two particles to collide.
Then substituting for $z$ gives the equation $\lambda=\frac{KT}{\sigma p}$. Have I gone wrong somewhere. I often see a factor of root two in the denominator but I don't see where that might come from.

Your derivation is correct. The $\sqrt{2}$ comes from $v_{rel}=\sqrt{2}v_{mean}$, where $v_{rel}$ is the mean relative speed (relative to the other molecules in the gas) and $v_{mean}$ is just the mean speed of the molecules in the gas.
In conclusion: note what the subscript says about the variable!


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Why do acids usually completely react with bases?
When a neutralisation reaction happens, for example, $ \pu{100 mol l^-1}$ of $\ce{HCl}$ with $\pu{100 mol l^-1}$ $\ce{NH_3}$, why does all of the base and acid get converted to salt? Why isn't there an equilibrium established between salt and acid-base?

Acids and bases don't necessarily completely react with each other. Consider your hypothetical reaction between hydrochloric acid and ammonia.
\begin{align}
\ce{HCl + H3N &-> H4N+ + Cl-}&
 \text{Extent:}& \approx100\%
\end{align}
The products are ammonium ion and chloride ion - the conjugate acid and conjugate base, respectively. So if the forward (above) reaction is complete, how can the reverse reaction between the conjugate acid and conjugate base be anything but insignificant? I.e.
\begin{align}
\ce{H4N+ + Cl- &<<=> HCl +H3N}&
 \text{Extent:}& \approx0\%
\end{align}
An equilibrium is established whenever the extent of reaction is not complete. Another good example would be an acetic acid solution. Acetic acid and water exist in equilibrium with acetate anion and hydronium ion. So no not all acid-base reactions go to an extent of 100%. Things to consider: stoichiometry and acid and base strengths.

Thanks for the answer. The reason I took ammonia was that it was a weak base, and for weak bases, dissociation is not complete. My question then was how ammonia gets incompletely dissociated when alone in water, but then completely reacts with HCl. So, what you've essentialy pointed out is that there is an equilibrium for strong acid-weak base reactions. Right ?

Ammonia being a base doesn't disassociate as do acids. Ammonia does react incompletely with water because water is a weak acid and incapable of protonating significant concentrations of ammonia. HCl however is a good few billion times stronger as an acid than water.

And no generally there is no equilibrium involving the reaction of a strong acid with a weak base because the base with be completely protonated.

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


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Distortion of graphene structure
What is the effect of rolling a graphene sheet into a carbon nanotube on the lattice constant and the geometry of the graphene?

Have you taken a look at the Wikipedia for the carbon nanotube yet?


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What are the physical properties affected in ZnO when there is a sulfur impurity?
I have a question. I work at a brass metal foundry where one of our waste products is $\ce{ZnO}$, which we sell. Because we have been making new alloys which contain sulfur, our $\ce{ZnO}$ has a very small sulfur impurity level and the people that we sell our $\ce{ZnO}$ to have been complaining that our $\ce{ZnO}$ makes their product yellow. Do you think this is due to the sulfur? What are the effects of very little sulfur on tons of $\ce{ZnO}$ dust? thanks!

Of course, if $\ce{ZnO}$ has a very small $\ce{S}$ impurity level, this could be the reason of the yellow colour of products.
As for the effects of very little sulfur on tons of $\ce{ZnO}$ dust, it all depends on the end use of $\ce{ZnO}$:The applications of zinc oxide powder are numerous. Most applications exploit the reactivity of the oxide as a precursor to other zinc compounds. For material science applications, zinc oxide has high refractive index, high thermal conductivity, binding, antibacterial and UV-protection properties. Consequently, it is added into materials and products including plastics, ceramics, glass, cement, rubber, lubricants, paints, ointments, adhesive, sealants, concrete manufacturing, pigments, foods, batteries, ferrites, fire retardants, etc...
Clearly, if $\ce{ZnO}$ is used in rubber industry, sulfur impurities have no harm effect.
If it is used in electrical applications (as a semiconductor), sulfur will affect the electrical conductivity.
If it is used in coating and paints applications, sulfur will affect the the color ....


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Can a single monomer be a macromolecule? If so, when?
In our science class, we're learning about the four fundamental macromolecules in every living organism: carbohydrates, proteins, lipids, and nucleic acids. Our class definition of a macromolecule essential is:

A macromolecule is a large molecule that is essential to all life, and present in all living cells. It is a polymer- a chain of monomers.

Our definition of a monomer is like so:

A monomer is the simplest building block of a macromolecule with the properties of that macromolecule. They can be strung together to produce a macromolecule (usually by dehydration synthesis).

I would have no problem with these definitions if not for my teacher mentioning once that some monomers can also be macromolecules by themselves. Because some monomers of certain macromolecules- such as the monosaccharide glucose vs. the disaccharide sucrose or the polysaccharide amylose - can act on their own as an essential and functional carbohydrate, they are macromolecules by themselves.
Is this true? For example, could glucose be a macromolecule by itself?
Thanks.

Welcome to chemistry.SE! You can always visit the help center or take a tour whenever you want, though it seems you didn't need this comment.

How can a lipid be considered a macromolecule? A lipid doesn't have repeating units like polymers (usually the fatty acids are different as well).

@LDC3 that's why I asked if some, not all, monomers could be macromolecules. I understand that a fatty acid can't be a full lipid.

When 2 or 3 fatty acids bond with glycerol, you have a lipid. Usually, in diglycerides, one fatty acid is saturated, the other is not saturated, and the third position has a phosphate bonded to choline (or serine, etc). Since lipids don't make larger molecules, then they would be monomers. About the only reason they would be considered macromolecules is if the definition has a low starting point (like 1000 daltons or 60 atoms).

I can't think of an example where a biological monomer would be a macromolecule.
Definitions of macromolecule vary, usually by molecular weight or number of monomers (repeat units).
Personally, I'd go with ~1000 Dalton for a minimum, but the original definition of 1000 atoms is a good start too.
In any case, no biological monomer, including glucose will function the same as a macromolecule.
Consider starch - a macromolecule of sugars. It doesn't dissolve as quickly as simple sugar and has different physical properties.
Update
To clarify my comments.. Macromolecules or polymers are made up of monomers the way words are made up of letters. So no, a glucose molecule isn't really the same as a macromolecule, just like "R" is not a word.
Yes, macromolecules can be used to make larger assemblies like microtubules, filaments, etc., much the same way that words can form sentences and paragraphs.
In the polymer literature there's even the concept of a "macromonomer" referring to a monomer that is already large in size.
In my opinion though, the basic constituents (monomers) are still amino acids, nucleic acids, sugars, etc. These are not macromolecules.

Well if you look at @user137 answer, then are there some exceptions to your answer?

It's a bit different. In this case, I'd consider these supramolecular assemblies. Yes, the protein is being repeated into a larger structure. But at least from my standpoint, it's like saying a macromolecule is like a word - made up of individual letters (monomers). Of course you can put words into sentences (supramolecular tubules, etc.) but the basic block is still the letters (DNA, sugars, amino acids).

Certainly as I read your question, it seemed as if you were asking "can a monomer act like a macromolecule" (i.e., can a letter act like a word). No, not really. Can we use words (macromolecules) to put together bigger things? Yes.

If you're a little looser with your definition of monomer, then Tubulin fits right in. Alpha and Beta Tubulin proteins bind to each other to form Microtubules which are an important component of cell's cytoskeleton and play roles in moving cellular components around the cell, especially during cellular division. And tubulin's not the only protein to act as a monomer, there is also Actin, which produces Microfilaments. I'm sure there are others, but I don't have the time to dig them up right now.

Macromolecule can be a monomer, but it's a little tricky because macromolecules are usually polymers by themeselves. I think that only common "monomeric" biomolecules which create oligomers, or even polymers are proteins - for example hemoglobin is present in erythrocytes as tetramer.


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Electroplating to create a scratch-free surface?
Can electroplating improve the texture of a steel ball bearing so that it will have a smooth surface free of scratches even a few microns wide? I'm trying to find a way to remove scratches (about 10 microns wide) from carbon-steel electrodes and have recently started looking into electroplating. I have already ruled out electropolishing and using rotary/vibratory tumblers.

In general, this doesn't really work. To level a surface by electrodeposition, material would have to deposit in recesses in the surface faster than it does onto surface features that stick out. Normally, the opposite occurs because things that stick out are more accessible for diffusion. Electropolishing works because it's doing the opposite process in solutions viscous enough to enhance the differences in diffusion.
Not to say that it's impossible to level surfaces by electrodeposition. It's been known for a long time that certain additives to plating baths can change the deposition selectivity to this end. However, the mechanisms behind these effects are not well known and it seems to be pretty difficult to achieve consistently. Also, you won't be able to electrodeposit carbon steel, so you would have some other metal on the surface, which may or may not adhere well. I wasn't able to find any specific method for carbon steel—electropolishing seems to be much more common.


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How effective is a level A hazmat suit for immersion in deadly substances?
I know that level A type hazmat suits can protect against minor splashes, but could somebody in one survive without harm after complete immersion for a few minutes in deadly chemicals such as sodium cyanide, high concentration (100%) hydrogen peroxide, and 80% hydrochloric/nitric acid?

Note that I'm used to standard lab safety measures including gas masks and masks mith independent air supply but I never used a Hazmat level A suit in research. You might want to take the following with a grain of salt!

Sodium cyanide

($\ce{NaCN}$) is a solid with a melting point around 560 °C. While it is a solid, jumping into a pool of it shouldn't do any harm. Level A suits are like "your personal bubble" with independant air support. They are however not that heat resistant: Do not jump into a pool of molten sodium cyanide!

High concentration (100%) Hydrogen Peroxide

3% solutions of hydrogen peroxide ($\ce{H2O2}$) in water are used for wound disinfection, solutions up to 18% are used to bleach hair. (In German, the term wasserstoffblond was used for bleach blonde).
Much higher concentrations or waterfree (100%) ($\ce{H2O2}$) is another league.
Hydrogen peroxide can decompose into water and oxygen:
$$\ce{2 H2O2 ->[\textrm{cat}] H2O + O2}$$
The reaction releases some heat, may happen spontaneously and is catalyzed by some metals. Add some burnable material that can be vapourized and you have a rocket propellant. Think Messserschmitt Me 163.
If the hazmat suit is tight, and the material withstands "bleaching" by the hydrogen peroxide there's still the question whether hooks, valves, visor frames or any other metal parts, that might be part of the whole gear will catalyze the decomposition.

80% hydrochloric acid

That does not exist in water. Water won't take up that much hydrogen chloride gas under normal conditions. Concentrated hydrochlorid acid (~ 38%) should be fine.

80% nitric acid

That's close to fuming nitric acid and again a pretty strong oxidant. I'd say that you're safe, but don't take my word for it ;)

Just a few follow up questions, since you seem to be rather knowledgable. Would a level B suit withstand these conditions? Which chemicals would the suit not survive?

@football-fsj Thanks for accepting my answer. Honestly, I wouldn't know the limits of level B suits. I'm sorry! Just a minor hint: Follow up questions in comments are frowned upon here. Please submit them as normal questions. And most important: Welcome to Chemistry.SE!

No, because Level A suits are not perfectly air-tight. They rely on positive air pressure to keep external fluids from getting in.
The pressure comes from exhaled air, which comes from a backpack tank or hose. The suits have one-way valves on the back that are protected by flaps. The zipper is also covered with a flap, but I don’t imagine it to be air-tight, either. Deep enough immersion would put enough pressure to stop the valves and make the zipper leak.


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Molecular structure of all periodic table element molecules exceptional cases etc
I am searching about some basics in chemistry. I was looking for the molecular structure of all periodic table element molecules.
eg: Hydrogen molecule: as $\ce{H2}$; structure:
I was able to find out several others too. But when I reach to certain molecules like $\ce{He}$ I was not able to find out the structure.
Does anyone know from where I can get the molecular structure of all periodic table elements. And if not, could you please explain the case of Helium and other noble gases. I am very poor in chemistry. Anyone showing me a reference on how to find the answer is welcome. Thanks in advance, and Happy New Year to all... :)

Welcome to chemistry.SE! Had you any questions about the workings of the site, visit the help center, or you can just take a little tour about how chem.SE works. I added the "reference request" tag so that your question doesn't get many downvotes. This link shows how good questions are asked. I recommend you take a look at what's written there.

All elements do not form "molecules", with itself, necessarily 2.Helium and other "noble gases" are unreactive (cannot form molecules under normal conditions)

The noble gases (group $18$) are rather inert, so their molecule is just a single atom.
The following elements usually consist of molecules made up of two atoms: $\ce{H2}$, $\ce{N2}$, $\ce{O2}$, $\ce{F2}$, $\ce{Cl2}$, $\ce{Br2}$, $\ce{I2}$.
Astatine ($\ce{At}$) should also belong to that list but it is not confirmed because of its short half-life.
Sulfur exists as ($\ce{S8}$); phosphorus exists as ($\ce{P4}$).
The structure of boron ($\ce{B}$), arsenic ($\ce{As}$), selenium ($\ce{Se}$), antimony ($\ce{Sb}$), and tellurium ($\ce{Te}$) are rather complicated.
Carbon ($\ce{C}$) has too many forms, but its most common forms are graphite and diamond, which is demonstrated below.
Silicon ($\ce{Si}$) and germanium ($\ce{Ge}$) both assume the structure of diamond.
Metals all adopt a metallic structure. Read more here.

Further
You might go to here to learn about the elements I skipped.

Gallery
Sulfur:

Phosphorus:

Carbon (as graphite):

The common structure of diamond, silicon, and germanium (left diamond, right graphite):

Credits:
All images are from Wikipedia.

Please do not use links to images on the web. Instead upload them to the sites contracted hoster. Before you do that, make sure you are allowed to use the image, i.e. can you redistribute it under creative commons by-sa 3.0 and always give proper credit to the original source.  For example: Can I upload it to my site?
Electronic redistribution is strictly prohibited.  Please fix your post accordingly, otherwise I am forced to delete it.

@Martin-マーチン Sorry, please give me a moment.

Also note that this is a [tag:reference-request]. Please refer to a (or more) publications to read up on.


text:
Oxidation of methane
In the reaction $\ce{CH_4 + 2O_2->CO_2 + 2H_2O}$ the $\ce{C}$ atom is oxidized. I can clearly see the addition of $\ce{O}$ atoms, however I don't see a change in Oxidation state. In the products, $\ce{C}$ oxidation state (O.S) was +4, and in the reactants, $\ce{C}$ O.S is +4. I believed an oxidation meant an increase in oxidation state.
What am I missing?

Actually in $\ce{CH4}$ with slightly more electronegtaive carbon it is in $-4$ state. With $\ce{CO2}$ the oxygen is more electronegative oxygen carbon is $+4$. Therefore oxidation is also defined as addition/replacement of a more electronegative element.

Electronegativities $$\begin{array}{}\bf H&2.2\\\bf C&2.55\\\bf O&3.44\end{array}$$


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Nuclear power: what material is the control rod made from?
What type of material is used in the control rod of a nuclear reactor? (Nuclear power plant)

Control Rods: They have the ability to capture the slow neutrons and can control the chain reaction at any stage. Boron and cadmium are the best absorbers of neutrons.
To start fission in $\ce{^{238}U}$ we need very fast neutrons having kinetic energies more than about 1 MeV. In a nuclear reactor, the probability that a neutron will cause $\ce{^{235}U}$ to fission is more for slow-moving neutrons. Thus there is less probability that the fast neutrons produced in fission will cause other fission to occur and a chain reaction will start. To slow down these neutrons (to initiate more fission events) in a nuclear reactor moderator is used which may be ordinary water or heavy water ($\ce{D2O}$).

There are various designs and materials for control rods, which are more technically correctly called control assemblies. The choice depends on the type and purpose of the reactor, the size of the core, the availability of other means for power control (e.g. boric acid in the reactor coolant, burnable poison rods, $\ce{UO2-Gd2O3}$ fuels, recirculation pumps, or thermal-hydraulic feedback), patent rights, and tradition.
The control assemblies of typical boiling water reactors (BWRs) are inserted from the bottom because, in a BWR, the neutron flux is larger in the lower part of the core and the space above the core is occupied by the water separator and the steam dryer. Therefore, the design should consider saving weight. BWRs usually use cross-shaped control assemblies. The four blades of one cruciform control assembly are inserted between four fuel assemblies. The blades are made of stainless steel and contain tubes that are filled with boron carbide ($\ce{B4C}$) as neutron-absorbing material. In addition, hafnium ($\ce{Hf}$) absorber may be placed in selected locations of selected control assemblies.
The control assemblies of typical pressurized water reactors (PWRs) are inserted from above. Each control assembly has multiple rod fingers, which enter openings at the upper end of a fuel assembly. These control assemblies are very heavy because, when the reactor is tripped, they have to quickly drop into the core against the pressure of the reactor coolant. The neutron absorbing material mainly is an $\ce{Ag-In-Cd}$ alloy (usually 80 % $\ce{Ag}$, 15 % $\ce{In}$, and 5 % $\ce{Cd}$) encapsulated in an austenitic steel cladding. These three neutron absorbers together effectively block the entire energy spectrum of fast, epithermal, and thermal neutrons. In other PWR core designs, the control rods are divided into two groups: so-called black assemblies made up of rods consisting of an $\ce{Ag-In-Cd}$ alloy in the lower section and of $\ce{B4C}$ in the upper part, and so-called grey assemblies containing absorber rods made of $\ce{Ag-In-Cd}$ alloy and steel.