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Why aren't lower aromatic organic compounds taught?
Why aren't aromatic compounds with 3, 4 or 5 carbon atoms taught at high school level chemistry? We are instead directly taught about aromatic compounds with 6 carbons/benzene rings. Why?
My core question is: is because the lower members are tough to understand? Do they have any strange properties?

Hi tvamsisai, I'm afraid this question is rather subjective (and depends on local cursus, etc.). It seems to me that the simple reason is that benzene-based aromatics are the most common of these compounds…

F'x: maybe, but its surely not off topic.
ah, availability! that partly answers. are lower members tough to study/learn/understand/teach?

I was taught these at high school level. At first, benzene was taught, and then these compounds were used as examples for aromaticity. Including the crazy positively-charged triangle.

This is a very interesting question, as I took two years of chemistry in high school, and was taught no organic chemistry whatsoever.

single-charged quadruple triangle known as tricyclopropylcyclopropenium is stable enough to form salts with other aromatic ions to form a hydrocarbon salt from https://pubs.acs.org/doi/10.1021/jo00290a034

It is simply impossible to teach everything about anything to everyone at any level. The purpose of High school is preparation for a successful life and hopefully an awakening of interest in the many facets of our surroundings. It seems to have started to work with you.  There is a lot more to Chemistry than simple aromatic compounds.

Hueckel's rule (i.e. that aromatic compounds have $4n+2$ $\pi$ electrons) places limits on aromaticity in cycles with less than 6 carbons.
As there are only 3 cases to look at, let's enumerate them:
5 carbons: cyclopentadiene ($\ce{C5H6}$) is not aromatic, but the cyclopentadienide anion ($\ce{[C5H5]^{-}}$) is. This species is well known in organometallic chemistry, due to its propensity to act as an $\eta^5$ ligand and form sandwich complexes with transition metals. The most famous of these is no doubt ferrocene, where two cyclopentadienide ligands sandwich an $\ce{Fe^{2+}}$ ion. This ion is didactically problematic and in my experience not covered in high school or early undergraduate curricula because most of its interest lies in its sandwich bonding mode with metal ions, which cannot be easily reconciled with the simplified VB/VSEPR/hybridisation theory used to teach introductory organic chemistry. The same goes for the under-loved tropylium cation ($\ce{[C7H7]^+}$), which has 7 carbons.
4 carbons: cyclobutadiene is antiaromatic, and as such only exists at very low temperatures. If I recall correctly, it possesses two unusually long bonds and two unusually short bonds, quite at odds with the bond equivalency found in benzene. This molecule would have to lose 2 e- to become aromatic. It does so happen that the cyclobutadienyl ligand is known in $\eta^{4}$ metal complexes such as $\ce{(C4H4)Fe(CO)3}$ (1) - I was not aware of these complexes until an expert in the field pointed them out to me.
3 carbons: I was not sure about this, but apparently the cyclopropenyl cation is a thing and is aromatic. Cotton (2) also mentions this being known as a $\eta^3$ ligand. I suspect that it is not taught because it's too weird.
In conclusion:
The smaller aromatic rings are principally known as polyhaptic ligands to metals due to their non-neutrality. This places them outside of the comfort zone of the theories of molecular structure and bonding used in introductory organic chemistry, and as such they are usually left until later.
(1) Cotton, F.A., Wilkinson, G.; Advanced Inorganic Chemistry; Wiley Interscience, 5e, (1988), p. 1170
(2) Ibid, p. 1168

We were taught about the cyclopropenyl cation, and they even showed us some reactions where its aromaticity makes a difference (so yeah, it's a thing). But yeah, it's weird.

@Manishearth - Most chemistry educators will be suddenly consumed by a pillar of eldritch flame if they attempt to reveal any unusual organometallic bonding patterns to any student before second year, due to the dark pact they enter into.

An excellent answer, but perhaps for completeness 7-membered rings should be mentioned, even if not specifically asked in the question. Cycloheptatriene loses H- when treated with an appropriate oxidant, to form the tropylium cation C7H7+ which is aromatic because it has 6 pi electrons.

As a university academic I find that I have the problem of too little teaching time, I imagine that the same problem exists in school chemistry. One of the ways in which teaching can go wrong is by trying to stuff too much content into a course. By doing so it will reduce the ability of the students to reflect on the content. By adding the interesting aromatics with 3, 5 or 7 carbons in a ring. I worry that it will reduce the ability of a high school student to understand aromatic compounds by overloading them.


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Understanding how mass spectroscopy works
I’m trying to get a deeper understanding of how mass spectroscopy works. Most tutorials and textbooks I’ve encountered omit certain details about the process and I’m hoping someone out there who understands the process can fill in the gaps. I’ll start by explaining the way I currently understand the process and then list where I start to get confused.
1 First a particular molecule is bombarded with a beam of electrons (ionization step). This step works to free an electron from the molecule under analysis which generates a cation.
Note: It’s important to remember that the only reason why this molecule was ionized is to make use of a magnetic field to bend its direction. Only ions will be affected by this magnetic field which allows us to pinpoint the molecule in question.
2 Next, these ions are accelerated by an electric field toward a magnetic field that bends the moving ions by a certain amount.

Question 1: What’s the difference between an electric field and a magnetic field? And why does the magnetic field bend the ions while the electron field does not. Is this just a consequence of the shape of the apparatus? Could you use a magnetic field to accelerate a particle and a electric field to do the bending? I’m just really confused as to the difference between electric fields and magnetic fields.

3 The ions that bend will travel around a tube by a certain degree and hit a detector. The amount of bending will tell us about the mass of a particular molecule. Heavier molecules will bend less and lighter molecules will bend more.

Question 2: I’ve come across a tutorial that said that what this detector is actually measuring is the mass to charge ratio. How is this calibrated and how does this step allow us to measure the mass of a single molecule. Won’t there be multiple particles hitting the detector at the same time, will this not affect the mass reading of a particular molecule? I'm just confused about this final step and how we can get an accurate mass reading for a particular molecule.

Any help understanding this concept would be appreciated.

What is the difference between mass spectroscopy and mass spectrometery

Question 1: What’s the difference between an electric field and a magnetic field? And why does the magnetic field bend the ions while the electron field does not. Is this just a consequence of the shape of the apparatus? Could you use a magnetic field to accelerate a particle and a electric field to do the bending? I’m just really confused as to the difference between electric fields and magnetic fields.

Electric fields accelerate charged particles (linear acceleration). Magnetic fields deflect them (angular acceleration).
Charged particles respond to electric fields and magnetic fields differently. Electric fields accelerate charged particles linearly in the direction of the flux. For example, if an electric field exists between two plates, one with a buildup of positive charge and the other with a buildup of negative charge, then positively charged particles will accelerate from the positive part toward the negative plate. The acceleration of the particle ($\vec{a}$) depends on the strength of the field ($\vec{E}$), the mass of the particle ($m$), and the charge on the particle ($q$). Since the acceleration is a scalar multiple of the electric field, they have the same direction.
$$\vec{F}=q\vec{E}$$
$$\vec{F}=m\vec{a}$$
$$q\vec{E}=m\vec{a}$$
$$\vec{a}=\frac{q}{m}\vec{E}$$
Magnetic fields are orthogonal to electric fields. Magnetic fields alter the angular acceleration of charged particles (i.e., they deflect them). Thus, charged particles have a circular trajectory in a magnetic field. The radius of the circular path ($r$) is dependent on the strength of the magnetic field ($\vec{B}$), the charge of the particle ($q$), the mass of the particle ($m$), and the velocity of particle ($\vec{v}$). The radius of curvature is proportion to the mass to charge ration ($m/q$).
$$|\vec{F}|=m\frac{|\vec{v}|^2}{r}$$
$$|\vec{F}|=|q\vec{v}||\vec{B}|$$
$$m\frac{|\vec{v}|^2}{r}=|q\vec{v}||\vec{B}|$$
$$r=\frac{m}{q} \frac{|\vec{v}|}{|\vec{B}|}$$
The acceleration of the particle depends on the velocity, magnetic field, radius, and mass. Since acceleration in a magnetic field is a scalar multiple of the cross product of velocity and field strength, the direction of acceleration is always changing.
$$\vec{F}=q(\vec{v}\times\vec{B})$$
$$m\vec{a}=q(\vec{v}\times\vec{B})$$
$$\vec{a}=\frac{q}{m}(\vec{v}\times\vec{B})$$

Question 2: I’ve come across a tutorial that said that what this detector is actually measuring is the mass to charge ratio. How is this calibrated and how does this step allow us to measure the mass of a single molecule. Won’t there be multiple particles hitting the detector at the same time, will this not affect the mass reading of a particular molecule? I'm just confused about this final step and how we can get an accurate mass reading for a particular molecule.

Now that we know the relationship between radius of curvature and mass-to-charge ratio, this question is easy. The detector counts the number of particles that hit it. That is all that it can do. That is all that it needs to do. The hard work of separating the ions by m/q is done by the mass analyzer, which is the magnetic field in curved tube. The magnetic field scans (quickly) over a range of $\vec{B}$ to alter the curvature of the ions by their m/q. Since the tube has a fixed curvature, only the ions with the correct m/q will pass to the detector. All others will hit the walls of the tube since their radius of deflection would be too large or too small. Thus, since the instrument controls the strength of the magnetic field, and physics determines what m/q will have the correct deflection at that field strength, there is no need to calibrate the detector.
More complex mass analyzers exist, for example the quadrupole mass analyzer. Instead of the typical bipolar magnetic field, it generates a quadrupolar magnetic field. Ions are thrown all over the place to hit a spherical detector. The physics is more complex, but the trajectory once again depends on m/q, so the detector only has to count hits in certain spots.

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How does sulfuric acid catalyze the reaction of potassium chlorate and sucrose?
Part of the total reaction is mentioned here:
http://chemistry.about.com/od/demonstrationsexperiments/ht/instantfire.htm
$\ce{2KClO3(s)} + \text{heat} \rightarrow 2\ce{KCl(s)} + 3\ce{O2(g)}$
The rest is a mystery. (And where does the heat come from?)
I regret to say that I have delayed acquisition of organic chemistry principles that might have answered this question already but can't help but wonder why, right now.
I feel that once the role of sulfuric acid is cleared up, the reaction with sucrose not described above will explain itself.
(And no I am not going to run around burning things down with this knowledge.)

A drop of sulfuric acid on the sugar will cause its dehydration. The reaction is highly exothermic and yields a highly porous carbon "foam", blown up by the steam simultaneously released. [You'll find lots of videos of this experiment, more or less reasonably conducted (with/without gloves and proper eye protection), mostly using larger amounts of sulfuric acids.]
The heat of the dehydration reaction will furnish the decomposition of the chlorate, releasing oxygen which, in turn, will start to oxidize the porous carbon mass to carbon dioxide. This reaction again is exothermic ($-394\,\mathrm{kJ\cdot mol^{-1}}$). Sugar might start to caramelize and release more water (as steam), more chlorate decomposes to yield oxygen, etc.

I think the sulfuric acid isn't really a catalyst here -- I think you use the heat from the reaction between sulfuric acid and sucrose (which will start happily at room temperature) to start the reaction between the potassium chlorate and the sucrose (which usually won't).
ETA: Once you've done that, it's pretty much a straight-up messy combustion reaction, only with the oxygen being provided from a solid source (thermal decomposition of chlorate to chloride and oxygen as the reaction you mentioned shows) rather than anything mechanistically clean. Compare, for example, this fairly typical demonstration just using hot chlorate and a sugar source: Oxidation of Sugar or Gummi Bear with Potassium Chlorate.

well that's a start I guess

The difference in the reduction potentials of chlorate for the first step is almost one volt between basic and acidic conditions. Wouldn't one think that might have something to do with it? The addition of acid possibly changes both the mechanism of the chlorate reduction and causes a much greater evolution of energy as the reaction proceeds.

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Where hydrogen belongs in the Periodic Table
Why do some people say that hydrogen should be above lithium in the periodic table and others argue it should be above fluorine?

This comes down to the fact that hydrogen has a $1s$ valence shell, capable of holding only two electrons.
In some sense, hydrogen is like the halogens, in that it can achieve the electronic configuration of a noble gas (namely, helium, which has a full $1s$ level) by gaining an electron. This reduction yields the hydride ion, $\ce{H-}$. It should be noted that hydride does not have the same stability as the halogen anions (e.g. hydrogen has an electron affinity of 0.75 eV vs. 3.40 eV for fluorine).
In another sense, hydrogen is like the alkali metals, as it only contains a single electron in its valence shell. Whilst the alkali metals can attain noble gas configurations by losing their valence electrons, it's a bit abstract to talk about hydrogen doing the same by losing its single electron to generate a proton ($\ce{H^{+}}$)(1). Nevertheless there is strong analogy between the electronic structure of hydrogen and the alkali metals which motivates hydrogen being often placed in this group.
Periodic table makers sometimes get away with duplicating hydrogen above the alkali metals and the halogens, or putting it in no group in particular.

The simplest explanation is that hydrogen has a valence shell that can only accept two electrons. Hydrogen is thus in the unique position of being just as likely to want to gain an electron as to lose the one it has; either way, it reaches a stable valence state. For this reason, it defines the "center point" of the Pauling electronegativity scale; elements with a lower number are "electropositive" (they prefer to get rid of their valence electrons), while elements with a higher number are "electronegative" (they want more valence electrons than they have).
Elsewhere in the periodic table, alkali metals which have only one or two valence electrons want to get rid of them, to get back to a complete valence shell underneath. They'd need to gain 6, 7 or even more electrons to stabilize their existing valence shells, and that's unlikely (they don't have enough electrons to share, and their nucleus size won't support an additional 7 electrons at that energy level). Halogens, on the other side of the table, have a nearly-complete valence shell, and just need one electron to get there, so they'll take it from anything that'll give it up (and in the case of some of the more vigorous halogens, like fluorine, they'll even take it from things that won't give them up, like other halogens and even noble gases).
Hydrogen, however, can do either one. It can lose its electron and become a naked proton, the H+ cation, or it can gain an electron and become a H- anion. Which of these it is more likely to do depends on what it's reacting with. With alkalis, it'll happily accept the extra proton and form a hydride, such as LiH. With halogens, it'll happily give up the proton and form a hydrohalic acid like HF, HCl or HBr. With other elements, it'll share, generally forming "polar covalent" bonds where the other atom in the bond attracts the electrons more closely. Hydrogen's very "zen" about the whole thing; give one up, take one in, share, it's all good.


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Aflatoxin versus Benzene, permissible levels
Alfatoxin and benzene are carconogenic, and this is the major concern for low, chronic doses. They are regulated to the few parts per billion level in food. However, according to http://toxnet.nlm.nih.gov/cpdb/pdfs/ChemicalTable.pdf,
aflatoxin is three to five orders of magnitude more carcinogenic, as evident from the lower doses needed to induce tumors in rodents.
Why are the permissible thresholds similar?

If this doesn't get any joy here, we can consider moving it to Biology.SE if you want.

I suspect that chemophobia may play a role: benzene is PETROLEUM and so must be poison, but aflatoxin is more "natural".

Well, the social response to chemicals isn't something that either Chemistry or Bio.SE is really in a position to answer, though.  I'm just looking at it from a perspective of toxicology possibly being more on topic on Bio.  I don't think anyone would have a problem if you want it to remain here.

I was going to write off your conjecture as cynical, but there's probably a kernel of truth to it. Most people simply have not heard of aflatoxin, despite the fact that it's virtually the mad cow disease of the peanut world. It's difficult to petition regulatory agencies for tighter controls over something you don't know exists. What gets me is that in the PDF you link, the TD50 for rats is less than 1% of that for mice. What's up with that?

It is usually assumed that there is no finite threshold below which carcinogens will not exert an effect. Therefore, there may be no safe exposure level below which there is no health risk.
Accordingly, a target value that is based solely on possible health risks (considering exposure over a lifetime, possibly even demanding an adequate margin of safety) would be zero.
Nevertheless, permissible limit values for carcinogens should not be set as low as possible since the exposure should not be reduced as much as possible at all costs. Instead, the potential exposure and limit values should be as low as reasonably achievable taking into account economic factors, other benefits and disadvantages, and the ability to detect and remove contaminants.
Hence, there may be similar limit values for various substances with different carcinogenic potentials. On the other hand, there may be different limit values for one substance depending on the exposure pathway (e.g. for different foodstuffs).


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Significant Figures Interpretation
Please correct me if I’m wrong, but I’ve read that the reason why we use significant figures is to avoid making the result of a calculation more accurate than the starting values prior to the calculation. For example, if we were seeking the ratio of two weights, lets call them A and B, using a cheap scale and an expensive scale we could get the following varied results:
Cheap Scale
$$A = 34 \space oz$$
$$B = 23 \space oz$$
Expensive Scale
$$A =  34.0000 \space oz$$
$$B = 23.0000 \space oz$$
Computing the ratio $\frac{A}{B}$ using a calculator we get the following value:
$$\frac{34}{23} = 1.47826$$
Now this is a correct ratio depending on the scale used (ie the cheap scale could not have produced such an accurate measurement given the starting values)
This leads to the following question:

Why are the zeros behind a decimal point ignored if they are just place holders?

For example, the value 0.000023 oz only has two significant figures, but isn’t that value something only a very fine tuned scale could measure (keeping in line with the previous weight example). So why are these zeros ignored?

This is part of the reason I really don't like the way we teach significant figures. What's really important isn't the number of digits you have, but rather the uncertainty in your measurements. I think significant figures are taught as an approximation to the proper uncertainty analysis, but it really seems to be not worth the trouble.
What's really going on is this: when you say $A = 34$ oz, you're actually saying that $A$ is closer to 34 oz than it is to either 35 oz or to 33 oz. In other words, you mean $A = (34 \pm 1)$ oz. Likewise, when you say $A = 34.0$ oz, you're really saying $A = (34.0 \pm 0.1)$ oz. And in the final example you give, when you say $A = 0.000023$ oz $= 2.3 \times 10^{-5}$ oz, you're actually saying $A = (2.3 \pm 0.1)\times 10^{-5}$ oz.
The more precise way to do things is to use the uncertainty you have for each measurement, and use the rules for propagation of uncertainties to get the uncertainty in your final answer. One advantage of using the uncertainties is that your uncertainty won't always be 1 unit of your least significant digit, but the rules for propagating significant digits essentially imply that. Using sig figs usually gets close to the answer you would get from properly using uncertainties, but there are some cases where the result is really weird.
To answer your main question directly, you ignore those placeholder zeros because they don't affect the fractional uncertainty of the quantity you're looking at. Keeping them would give you an answer very different from the proper uncertainty analysis.

but isn’t that value something only a very fine tuned scale could measure?

Not necessarily. The cost of an instrument is much more closely related to the relative precision of its output, rather than the absolute precision. If all you're trying to do is weigh things with masses between $1 \times 10^{-5}$ and $1 \times 10^{-4}$ oz, then you don't care at all if your scale can measure 1 oz. But if you have objects that are 1 oz and $1 \times 10^{-5}$ oz, and you care about the difference, then you'll need a really precise scale.

If A=34, and this implies it is closer to 34 than to 33, or 35, the the error implies is 0.5 and not 1. Or am I wrong here?

@Martin-マーチン I think we should interprete closer as most probable. So we are assuming this human or machine is near the mean, but it could be 33 or 35

In your example 0.000023 oz does indeed have two significant figures. It is indeed probably something only a very fine tuned scale could measure.
Now consider that you add that mass to another sample that you measured with the much less accurate scale, let's say the other scale measured that other weight as 23 oz. Both have two significant figures, but considering the weight read by the inaccurate scale, you have no legitimate idea whether it is really 23.28125 oz or 22.87284 oz, or some other number for that matter.
What you do know is that it probably falls within some range of values, we usually look at the last significant digit and if the uncertainty is not mentioned we can assume it is +/- 2%. That means the value 1.0 oz might be somewhere between 0.98 oz and 1.02 oz.
So now when we add the two masses together, we do not want decimals since 23 oz has none. The result is 23.000023 oz. Which is 23 oz knowing that we can only go down to the ones place. The reason is that you have no clue whether it is 23.38125 oz + .000023 oz, or 22.87284 oz + .000023 oz, or any other values for that matter. The fact is that these zero's are not ignored, they are giving us the magnitude of the number. But when we do something with them in conjunction with some other uncertain numbers, we cannot be sure of the result to high certainty. One rotten apple spoils the whole barrel.
Now in this scenario described above, even if we said 0.000023 oz has 6 significant figures. The other number 23 oz still has only two. Therefore we must go with 2.
When we divide or multiply we count the sig figs. 23 oz has 2 sig figs, so our result of multiplying should have 2 as well. So how many times greater is the inaccurate weight (23 oz) compared to the accurate weight (0.000023 oz). 23 oz / 0.000023 oz ~= 1000000 times greater. Notice that the number 1000000 has only 1 significant figure. We want two significant figures so we write. $1.0 * 10^6$


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The colour of fireworks
When the electrons that are excited by the energy of burning drop back into lower energy levels they emit light at the wavelength of colour.
When metals are burnt, metal oxides(some coloured) are formed.
When metals are burnt under high temperature, will they turn to plasma state? If so, should the metal ions be responsible for the colours and sparks?
Which one is the reason for the beautiful colours of fireworks?

The colours of fireworks are caused by the electrons emitting energy when dropping orbital levels. Metal oxides are solids, mainly dull colours (even the yellows and reds are not in the same class as firework colours), and could not cause bright light in the air. The temperature of fireworks is far below that needed to form a plasma.


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Why can sulfur act like a catalyst?
In the reaction of combustion of gunpowder,

Sulfur is used to speed up the reaction and acts like a catalyst in that it increases the rate of reaction. However, unlike a catalyst, it is used up in the reaction.

I understand that transition metals are often used as catalysts but why can sulfur act like a catalyst?

It's not a catalyst here. It speeds up the reaction, which is why it acts like a catalyst.
Here is the reaction without sulfur:
$\ce{6 KNO3 + C7H8O -> 3 K2CO3 + 4 CO2 + 2 H2O + 3 N2}$
And here it is with sulfur:
$\ce{4 KNO3 + C7H4O + 2 S —> 2 K2S + 4 CO2 + 3 CO + 2 H2O + 2 N2}$
(Source)
Sulfur decreases the ignition temperature, as well as probably contributing a bit to the rate. Here, it is effectively turning $\ce{CO2}$ and $\ce{K2CO3}$ to $\ce{CO}$ and $\ce{K2S}$. If you take into account Le Chatelier's principle, the conversion of the products of the basic reactants speeds it up. (The decrease in concentrations/activities of the products due to their conversion means that the equilibrium shifts in the forward direction to compensate for it.)
So it is speeding up the reaction and making it easier to carry out. Not a catalyst.
A note: There's no reason why any element/compound can't act as a catalyst. Transition metals are generally catalysts in a certain type of reaction, and they generally catalyze the reaction by adsorbing the gases. Catalysts can work in many other ways—generally, they form intermediates in the reaction and present an alternative path via the intermediate. ($\ce{V2O5}$, for example, is a catalyst in many reactions involving organic compounds since it forms cyclic intermediates with various groups).

Can you elaborate on how Le Chateleir's principle comes in? I didn't get that part :-)

@Brian: done. I'm not really sure if it has an effect here, though. It may just be a thermodynamic phenomenon.


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How can I control the oxygen in an erlenmeyer flask in a bioprocess?
I need to control oxygen in a erlenmeyer flask in a bioprocess.  I need to remove overall oxygen in a erlenmeyer flask, and after I need to add oxygen in controlled amounts to a culture in a shaken flask.  Someone knows how can I do it?
Thanks for advance for your ideas.

How are you shaking the flask? What kind of flask is it?

Hello, is an erlenmeyer flask.  http://en.wikipedia.org/wiki/Erlenmeyer_flask

Controlling oxygen in a flask can be achieved by bubbling a purge gas of some description through it. This will entrain and remove oxygen from the solution as well as creating an oxygen-depleted headspace.
If you have a regulated nitrogen supply, a rubber stopper with a hole in it, an open-ended pasteur pipette and a length of hose, you can bubble nitrogen into the solution to purge oxygen. This requires the vessel to be open somewhere else to equalise pressure, which is achievable with a multi-neck flask but not with a standard Erlenmeyer flask. You can get around this with a stopper with two holes. Continual positive pressure should keep the external atmosphere out.

Depending on how vigorously you want to shake your flask, this may or may not be reasonable. If you are using a stirrer bar you will obviously want to give the pipette sufficient clearance.
As far as controlling addition of oxygen, this would seem to be a bit trickier as you are probably not going to be providing continual positive pressure. You might be able to get around this, depending on the specifics of what you are doing, by pumping in oxygen in the same way and using a balloon or a gas syringe or something to deal with the overpressure.

For clarity, I'm going to separate this into two parts: purging the oxygen from the flask, and filling the flask with a particular mixture of gas(es). It's the same idea in both cases, but the first is much easier than the second.
Purging the flask
This is usually done with Nitrogen, as it is one of the cheapest inert gases. You'll take a line from your nitrogen tank, pass it into your flask, and open it. You will need to have a second opening in your flask so pressure doesn't build up, as Richard Terrett describes in his answer. As long as you have the nitrogen flowing, there will be enough positive pressure in the flask that oxygen won't get in from the room. The headspace in your flask should purge quickly, but your culture will take longer. Agitation, either by shaking the flask or by stirring the culture, will make that go much faster.
Establishing a given atmosphere
This is really the same process as above: instead of pure nitrogen, you pump in the mixture of nitrogen and oxygen you want. The system will eventually equilibrate to the mixture you're pumping in. The hard part is getting the mixture you want. When I did this in the past, we used digital mass flow controllers (DMFC's). We would flow oxygen through one DMFC and nitrogen through another. The output of the two DMFC's would mix together before going into the flask. Once you have it set up, it's just a matter of setting the two DMFC's to the mixture you want. The big difficulty with that is that DMFC's are expensive. I don't know how much ours were, but they were made by Bronkhorst, and it looks like that type of DMFC starts at around $2000 each. The nice thing about using DMFC's is that you can control the oxygen composition to within less than 1% by mass.
Depending on how precisely you need to control the mixture, you might be able to get away with a simple flow meter. Again, you would run each gas into its own flow meter, and use that to control the output of each gas. Then you set the two flow rates so that the composition and the total flow rate are what you want them to be. I don't know how precisely you would be able to control the composition, but it might be enough for your application.
Regardless of the atmosphere you're running into the culture, you can either run the purge line into the culture itself, or just into the headspace. I have done it both ways; running it into the culture works faster, but might not be feasible depending on your setup. In one case I worked on, the culture was only a few millimeters thick, and there just wasn't room to stick a needle from the purge line into it. In that case, just pumping gas into the headspace and agitating the culture was sufficient because of how shallow the culture was. In my more recent work, I have a much deeper volume of fluid to purge. In this case, running the purge line only into the headspace probably would be too slow for my purposes, but it's pretty easy to run a needle from the purge line into the fluid without impeding anything else.

Nice answer. Do you recommend bubbling nitrogen through the culture or simply pumping it into the headspace and agitating?

I have done it both ways. I'll add some information on that.

@RichardTerrett When working with chemical substances, reacting with oxygen, it is enough to have it pumping and agitating the mixture with magnetic stirrer. In special cases it may be useful to degase solvent first: put it under vacuum for a while and then move into your flask using syringe.


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Bonding and anti-bonding orbitals in the light of time-dependent Schrödinger equation?
In organic chemistry, people draw 2p orbitals like this:

and then they explain how the orbitals combine to non-bonding (π*) or bonding (π) molecular orbitals, like this:

depending on whether the orbitals overlap out-of-phase (blue + white) or in-phase (blue + blue, or white + white).
But in reality, the orbitals are time-dependent solutions of the time-dependent Schrödinger equation, so they are not really like the first image above, but more like this:

So, what does it mean, when the orbitals overlap in-phase or out-of-phase? As far as I understand, the phase difference of the orbitals of two random atoms could be anything from 0 to 2π, and the cases of perfect overlap (0 phase difference) or perfect anti-overlap (π, or half-cycle, phase difference) should be quite improbable.

When you construct such molecular orbitals as a linear combination of atomic orbitals, you are simply choosing the AO as a basis set to project the Schrödinger equation onto. You are not actually adding them up, neither with nor without their phase. Each MO will then feature a phase of its own, depending on its energy.

@F'x - you should consider making this an answer.

@RichardTerrett I should find time to expand it… or someone should, and make it an answer :)

As F'x has indicated, the sum of atomic orbitals $\phi_k$ (AO) that form the molecular orbital $\psi_j$ (MO)
$$ \psi_j = \sum_k c_{jk} \phi_k $$
is just a solution (an approximate solution) to a self-consistent variational procedure to solve the time-independent Schrödinger equation, obtained after choosing a particularly well suited basis to represent the solution. They approximately solve this equation:
$${\sf F}\psi_j = \epsilon_j \psi_j$$
where ${\sf F}$ is the Fock operator or if you wish, an effective one-electron energy operator and $\epsilon_j$ is an effective "one-electron" energy that gives a good approximation to the ionization potential from that orbital.
Then you have to construct your overall n-electron wave function $\Psi({\bf x}_1,\ldots,{\bf x}_n)$, which in the simplest case (the one that follows from the minimal molecular orbital theory) is a Slater determinant. You can notice that for this wave function the coefficients $c_{jk}$ are not unique. Following the general properties of determinants, by a similarity transformation of the basis you can get a new set of coefficients $c'_{jk}$ that will not solve the Fock equation, but after which you obtain the same overall electronic wave function.
In everything that I have said so far, there are no time-dependent phases: we only have one "real" (observable) energy, an eigenvalue of the time-independent Schrödinger equations for the electronic Hamiltonian
$$ {\sf H}_{elec} \Psi({\bf x}_1,\ldots,{\bf x}_n) = E \Psi({\bf x}_1,\ldots,{\bf x}_n)$$
for a fixed geometry. Now is where you can put the time-dependent Schrödinger equation into work. In principle you could create electronic superposition states combining $\Psi({\bf x}_1,\ldots,{\bf x}_n)$ of different energy. These are the ones that would evolve in a manner similar to what you have assumed, with time-dependent evolving phases.

It is important to note that the phases of the MOs are irrelevant. Assuming that the Hamiltonian can be solved with a Slater determinant $\Psi_0=|\phi_1(1)\dots\phi_N(N)|$ of MOs $\phi_i$ with eigenvalue $E:= \sum \epsilon_i$, where $\epsilon_i$ are the eigenvalues of the Fock operator, then your time-dependent solution will be
$$ \Psi(t) = e^{-\frac{Et}{\Im \hbar}} \Psi_0.$$
Now you see that no phase can be uniquely attributed to any one MO. Canonically the MOs are chosen to be real-valued.


text:
Selectivity Coefficients of Ion Exchange Resins - Selectivity Reversal with TDS / Ionic Strength
I'm currently studying ion exchange resins and generally getting a bit stuck on some points of information which I do not fully understand / are not very well explained within the literature.
I understand that resins having varying selectivity or affinity for a particular ion and selectivity generally increases with increased valance number, atomic weight, reduced hydrated radius, etc.
I have seen some information that selectivity reversal will occur when the solution has high TDS e.g., sulphate will have a far higher affinity than chloride for a strong base anion resin within a solution with low TDS; however, at high TDS the affinity of these ions for the resin will be reversed - chloride will have a greater affinity than sulphate.
Any idea whether this is correct and what the mechanism causing this change is?

You don't need to debase yourself, we're all friendly here.  Even if it were a basic question (which it really isn't), everyone has to start somewhere.


text:
What is the product when Lithium solid is dropped in water?
I am working on reaction predictions for the AP Chemistry and I need some help with this problem. On a practice quiz I tried to use the rule of "Metal + Water = Metal Oxide" so I answered with $\ce{2Li + O^{2-} -> Li2OH}$. I got zero points for this answer meaning both reactants and products were wrong. What is the correct answer and how would you get to the correct answer?

I think that's what you were trying to do with the "exponent" on the oxygen, if not feel free to fix it to reflect the problem.  It would also be good if you explained your rationale for choosing that answer, so someone can help correct your process.  Thanks!

I wrote how I got the answer, thanks for the exponent formatting!

Hi, this question will probably be closed on account of it being too localised and homeworky. A general pointer, however, is that your products and reactants need to be stoichiometrically equivalent ('balanced'). You have hydrogen in your products but not in your reactants. Your reactants do not actually include water. Moreover, the alkali metals are famous for producing a certain... explosive gas... when reacted with water.

@RichardTerrett: Nah, he has shown some work (he could show more though), and it's a simple conceptual error. It probably won't be closed :)

@RichardTerrett Ordinarily, and in its original form, I would have agreed, but since this was a chance for someone knowledgeable to correct his rationale I think it could be useful in the future.

I think the major error you made is applying what you call the "Metal + Water = Metal Oxide" rule.  This may apply to some metals, but not all.  Specifically, Lithium is a Group 1 metal, the so-called Alkali Metals.  This means that Group 1 metals tend not to form oxides with water, but rather form hydroxides with water.  That is, the metal ions bond with hydroxide ion, $\ce{OH-}$, which has a -1 charge.  If you have ever seen this reaction (lots of YouTube videos if you have not done the reaction yourself), you will see that lithium in water produces lots of gas.  And in some videos, you will see that this gas can be set on fire!  Given the elements involved, that's a good indication that the gas is hydrogen, a very flammable gas.
So the most likely equation for lithium in water forms an hydroxide of lithium along with hydrogen gas.  Is that enough information for you to propose a different equation than what you answered?

I think you here looking for this:
$$\ce{aM(s) + bH2O(l) -> M_a(OH)_b(s) + $\tfrac b2$\,H2(g)}$$
Then, if you remove the water from the medium you are left with the $\ce{LiOH}$. The thing is, the lithium breaks up the water and molecular hydrogen is released. Nevertheless, this ionic compound ($\ce{LiOH}$) is soluble in water, hence the aqueous forms.


text:
What is the difference between hand wash and body wash soap?
I do not like body wash always with strong favor. I like hand wash soap to wash my body. Are they difference at all?
Thanks for the comments, so that I add the following information
I am really sorry that I posted the question using my phone at my office parking lot right before I went to work this morning, so not much detail was added
To my limited knowledge, Body wash and hard wash soaps are detergent, they both are OK to use on skin. If that is the case, why body wash soap is body wash soap? What I mean is that from the chemistry point of view, what makes body wash soap difference from hand wash soap?

Ingredients should be on the side of the bottle.

Hi Marco, could you perform a bit of research and edit the question to reflect what you have found the differences to be, and then perhaps ask for which ingredients might be more beneficial for removing dirt or grease?

(My sister-in-law works at Unilever, and she makes these soaps)
These are all detergents.  The primary differences are harshness (hand soaps can be harsher, as hands are often more soiled), lathering, concentration, perfuming/scents, and marketing.
(Emphasis on the marketing.)

Basically, they are the same thing. But, by adding some ingredients and changing the pH, soaps can act like mild or more harsh detergents. There are a lot of different types of soaps in the market; like ones especially made to treat baby skin, for face skin, body, hands, etc.

In common composition they are the same but there may be a little difference due to the antibacterial materials added and the smell.

What is the difference between dish soap, handsoap, body wash and shampoo? And can they all be used interchangably? : NoStupidQuestions

Dish soap is focused on your dishes, as the name implies. It is meant for clearing out things like grease and oil off your plates by breaking down fats.
Laundry detergent is still soap but with things like water softeners, bleach and enzymes added in to further break down the same grime, as well as things like plant chlorophyll.
Shampoo is to remove the unwanted build-up of dirt or grime in the hair without stripping out so much sebum (natural body oils) as to make hair unmanageable.
Hand soap is to reduce barriers to solution, and increase solubility. Water is an inefficient skin cleanser by itself because fats and proteins, which are components of organic soil, are not readily dissolved in water. Hand soap contains chemicals that break those down.
Theoretically they could be interchangeable, but they all contain specialized chemical mixes to be the most efficient at their respective jobs that using them not for their intended purpose would be counter-productive. i.e. Using laundry detergent on your hair may remove the oils that shampoo would not and damage your hair as a result.

ELI5: What's the fundamental difference between hand soap, dish soap, body wash, bar soap and shampoo? : explainlikeimfive

They're each focused on a different purpose
Dish soap should cut through grease and oil, that's its primary job. It doesn't necessarily need to smell nice or be soft on the hands as long as it plows through stuck on grease
Hand soap should be decent at getting stuff off your hands, but also needs to be soft on the hands. It should be neutral or smell nice.
Shampoo should remove the grease from your hair, but not be so aggressive it dries it out or changes its color which may be acceptable with something like a dish soap
Different tools for different jobs. If you want it to be soft on hands then it can't be super aggressive, if you want it to be super aggressive it'll do bad things to soft things like skin and hair.

ELI5: what is the difference between all types of soap. i.e. shampoo, hand wash. Body wash, bar soap, dish soap, detergent etc... : explainlikeimfive

Most modern liquid "soaps" are not technically soaps at all, which is to say they're not produced from mixing vegetable or animal fats with a strongly alkaline solution such as lye or potash. Instead they're a blend of (usually) petroleum-derived surfactants such as sodium lauryl sulfate with other chemicals to produce a detergent that matches the desired use.
Shampoo is designed to be gentle on the keratin which forms hair, have strong foaming properties to be more easily worked through the fine strands, remove common hairstyling products, and - especially for those of us with more than a couple inches of hair - have specific effects on the texture of the hair. It has a fairly low concentration of surfactants so that it rinses out quickly and you're not in the shower forever trying to get it all out of your hair.
Hand and body wash is usually formulated with a mild surfactant to avoid skin irritation, plus various ingredients that can moisturize the skin, add scent, improve lather, etc. Lathering agents are generally surfactants as well, so there's a careful balance here between getting a nice lather and not drying out skin. Hand washes are usually less foamy since they don't need to cover much surface area and are used frequently throughout the day, while body washes tend towards more foam since they need to cover a lot more surface area and are used less frequently. They're both a bit more concentrated than shampoo, since it's easier to rinse soap off of skin than hair, and in the case of body wash, most consumers pour it onto a sponge/pouf/washcloth/etc. before applying it to the skin, which spreads it out thinner than applying it directly. Face washes are their own magical category and can include all sorts of fun chemistry like ceramides and multivesicular emulsions, alpha hydroxy acids, benzoyl peroxide or salicylic acid, etc. - and as an acne sufferer, it definitely makes a difference.
Bar soaps are one area where true soaps are still relatively common. Moisturizing ingredients can also be added, and the naturally occurring glycerine is also somewhat moisturizing. However, traditional soaps also have a lot of limitations. They have a fairly narrow range of environments in which they're effective, needing hot water with a low mineral content to function, and must be rinsed a second time with clean water to avoid deposits.
Laundry detergents are commonly formulated these days with specific surfactants designed to work well with cold and hard water, both of which decrease the effectiveness of traditional soaps. They're also highly concentrated, since they're going to be diluted by the large volume of wash-water, which is why just a few drops of liquid detergent on your hands will take much longer to rinse off than an equal amount of hand soap. This is also why ideally you should fill the washer with soap and water so they can mix, then add the clothes.
Dishwasher detergents aren't worried about gentleness, since they don't come in contact with skin or organic fibers, so they can use harsher detergents and often include abrasives, but do rely on hot water to be effective. For similar reasons to laundry detergents, they're highly concentrated, but they use surfactants that are more effective on metal and ceramic.
Dish detergents meant for hand-washing dishes have to balance removing grease, starches, sugars, etc. from food with not completely stripping the natural oils from skin, which is a bit tricky - the oils that keep your skin nice and pliable aren't any different from the oils of any other animal, chemically speaking. They're also pretty highly concentrated for that "grease-fighting" effect, and so that your washcloth/sponge/etc. doesn't need more soap on it after every dish you wash.

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


text:
Covalently bonding amide to glass
I am unfortunately not very good at chemistry, so I am having trouble figuring out where to start on this project.  I need to have some way of get some amide functional groups stuck on a substrate (most likely some glass beads or slides) so I can try to show some interactions of the amide with a molecule I am interested in.
Does anyone know of any literature that shows a good way to functionalize silica with amides?
Thanks for any help with this, I really appreciate it.
EDIT: I found some of this stuff ((3-Aminopropyl)triethoxysilane) in our lab.  Could I just put a glass slide in it and leave it over night?
Then maybe come back and wash acetyl chloride over it as in this reaction?

Ethanol solution, soaked over night of APTEOS will work.  You won't get great coverage, but it will be likely good enough for what you want to do.
The subsequent chemistry is a bit more touchy.  The problem is acid byproduct which forms ammonium salt with the amine functions that is difficult to remove.  This leaves latent acid catalyst at your surface.
You are better served forming the amide from your APTEOS and coating with that product.


text:
Is it possible to compute the colour of *any* molecule?

Above are some spectral emission lines for hydrogen, helium and neon.
Using the Schroedinger equation, it's possible to derive the colours that hydrogen will emit when light is fired at it. A little more complicated (although beyond my rudimentary knowledge of quantum mechanics), the same can be done for helium, and so on.
Of course as we progress to molecules, even the relatively small $C_5H_{11}Br$, it's (I'm told) very hard to predict the colours emitted from first principles, and even then incorrect assumptions must be made (like the Born-Oppenheimer approximation), so upon progressing to larger molecules, we find that these simplifications diverge from experiment.
My question is this: with the advent of computers, are we able to (in practice, not theoretically in a few decades) accurately compute the colours given off by a molecule  by numerically solving the laws of quantum mechanics for complicated systems? Much like nonlinear differential equations can now be 'solved' with computers without any simplifying assumptions (like $\sin(x) \approx x$), is it in humanity's capacity today to approximate colours from first principles, so that computing time is the only restraint? Note the difference between an incorrect model of reality (Born-Oppenheimer), and one that converges to the correct one (even if slowly).

Why is the Born-Oppenheimer approximation incorrect?

It is possible to compute electronic spectra with decent accuracy. Time dependent density functional theory (TDDFT) permits fairly rapid calculation of UV/vis spectra at a quantum mechanical level of theory. Whilst DFT is not truly a first principles method, it can furnish electronic spectra accurate enough to be useful, depending on care taken in selecting appropriate functionals, accounting for solvatochromic effects, etc. Obviously, DFT uses the Born-Oppenheimer approximation, but the beauty of the approximation is that it's actually very close to reality.
See, for instance, this paper by Jacquemin et al., where they calculate $\lambda_{max}$ for a wide range of dyes using TDDFT and get a MAE of 22 nm using CAM-B3LYP (a long-range hybrid functional), reducing that to 10 nm with an empirical calibration curve. The paper only really cares about accurately reproducing the location of $\lambda_{max}$, however this paper by Quartarolo and Russo computes complete spectra with fairly reasonable accuracy. Not perfect, but the general shape of the spectra are reproduced adequately. Part of the divergence between the experimental and computed spectra probably comes down to higher order broadening effects not reflected in the simple Gaussian convolution of the computed data, but I'm not a spectroscopist.
Additional: as for computing the colour of any molecule, well, that does depend heavily on the size of your molecule and any mitigating factors such as symmetry elements. Most QM techniques scale fairly viciously with the number of electrons in your molecule, and whilst you can always reduce the size of your basis set and heap on less and less conservative approximations, accuracy will suffer.


text:
What does m± represent in the equation for mean activity of a salt solution?
I need to find the activity $a_\pm$ for a $\pu{0.0120 M}$ solution of $\ce{Na3PO4}$, and have calculated the activity coefficient, $\gamma_\pm = 0.389$.
I know that $$a_\pm = \frac{m_\pm}{m_0}\gamma_\pm,$$
but I must be plugging in the wrong numbers. What does the top $m_\pm$ mean? And isn't the bottom $m_0$ simply $1$?

The top $m$ is the mean ionic molality of the solution you have.  You must know the concentration since you need it to compute the activity coefficient.  The bottom $m_0$ is the standard molality which in general is $\pu{1 M}$.  It is there to make the activity dimensionless.
The required formula for $m_\pm$ is $$m_\pm^\nu = (m_+)^{\nu_+} (m_-)^{\nu_-}$$ where $m_-$ is the molality of the negative ions and $m_+$ is the molality of the positive ions. Also $\nu_-$ is the stoichiometric coefficient of the negative ions and $\nu_+$ is the same for the positive ions. Also $\nu = \nu_- + \nu_+$.
Thus in your specific case, where you have $\pu{0.0120 M}$ of $\ce{Na3PO4}$, we have
\begin{align}
\nu_+ &= 3 & \nu_- &= 1 & \nu &= 4 \\
m_+ &= \pu{0.0360 M} & m_- &= \pu{0.0120 M}
\end{align}
and hence
\begin{align}
m_\pm &= \left[(\pu{0.0360 M})^3(\pu{0.0120 M})^1\right]^{1/4} \\
&= \pu{0.0274 M}
\end{align}


text:
Is it possible to calculate chromacity coordinates using absorbance values?
I have absorbance values for blood plasma. Absorbance was measured in range of 380 – 600 nm every 5 nm ending up with absorbance values for 45 wave lengths.
I was browsing through literature, and  I came  across chromacity coordinates (XYZ) and this seems to be a very convenient way of summarizing data about color in just two data points.
I have been reading some literature about different colorimetric systems, but since my background is in genetics, it didn’t make much sense for me.
Therefore, I would like to ask you if it is possible to calculate these chromacity coordinates using my absorbance values and if so does anyone now some kind of software for this or exact formulas on how to do it?

What a great question! This said, I suspect you'd have better luck on phys.SE with this one as there are some people quite knowledgable in colorimetry who hang out there.

Actually, it is pretty obvious. The wiki article about CIE color space has precise definition how to calculate color. However, having absorbance values does not give spectrum of reflected light, only of light passed through. There is a difference.  Moreover, the layers of blood with different thickness will have different colors. So, short answer is: no, you cannot.

This is a partial answer to your question.  I have used chromaticity coordinates for various scientific endeavours (air quality mainly).  I found that it is a very powerful means of determining colour differences.
A bit of searching and there is somewhat of a precedent for using chromaticity for blood plasma.  An example is the article Features of Color Characteristics of Blood Plasma of Patients with Liver Cirrhosis in the Colorimetric System XYZ as Compared with Healthy People (Dolomatov et al. 2012) - even though it may be a different context to your study, but does show that it can and has been done before.  Specifically, their method was

by colorimetric method, based on absorption electromagnetic radiation in visible area.

There are several more examples of how it can be used.

Thanks and sorry for a late reply. I also found that article (those guy have some  very interesting papers on this topic ). However, in the meantime I changed my mind and went only for a specific range of wavelengths which correspond to peak values of plasma components which are of my interest.


text:
Is there a positive radical Aurous, Au (x+)?
Is there a positive radical Aurous, $\ce{{Au}^{x+}}$? If yes what is the value of x?
We have Auric as $\ce{{Au}^{3+}}$.

Aurous would appear to denote Gold(I), that is, $x=+1$, as detailed on this site, which gives a profile of Gold(I) chemistry. This atom is not a radical.
Google is your friend.

@ Richard: You mean AuCl (Aurous Chloride) does not dissociate in to ${Au}^{+}$ and ${Cl}^{-}$.

@RajeshKSingh - AuCl probably does dissociate to some degree. However, Richard means that it is not a radical in the sense that it does not have unpaired electrons. The electron configuration of Au(I) is $[\ce{Xe}]4f^{14} 5d^{10} $, which consists only of closed shells, i.e. no unpaired electrons.

You can definitely make gold radical cations in a mass spectrometer...


text:
What reactions will their reaction rates increase with time?
My textbook states that for most reactions, the reaction rate decreases with time. I wonder what reactions will have increasing reaction rates with time. I know that adding a catalyst will increase the reaction rate but the final reaction rate will finally be zero. So, what reactions will their reaction rates increase with time?

an explosion !!

This question is nonsensical because you misinterpret the "time". Reactions in general slow down because they run out of reactands, which happens "with time". Time per se will nor alter reaction rates.

@Georg I think it is fairly common to refer to extent of reaction as `time'. Moreover, it is possible (though rare) to have reactions that become faster when the reactants run out as shown in the answers below.

Reactions that are autocatalytic can have their rate increase as a function of conversion.
One example is the decomposition of cellulose acetate by hydrolysis, also known as vinegar syndrome. The reaction is acid catalyzed, and it produces acid. In theory, the reaction gets faster. In practice, diffusion and limiting reagents many prevent a rate acceleration. Acetylsalicylic acid decomposes similarly. In the following equation, $\ce{R}$, could be the cellulose backbone or the salicylyl group. The acetyl group $\ce{-COCH3}$ is represented by $\ce{Ac}$. Thus acetic acid is $\ce{HOAc}$.
$$\ce{R-OAc + H2O ->[\ce{H+ catalyst}] R-OH + HOAc}$$
Another example is the alpha halogenation of ketones or aldehydes, which is acid catalyzed and acid-producing.
$$\ce{CH3-CO-CH3 + Br2->[\ce{H+ catalyst}] CH3-CO-CH2Br + HBr}$$
A historically interesting example is the allotropic decomposition of tin, or "tin pest", in which tin metal transforms to a brittle form in cold weather. The book Napoleon's Buttons, which is about the influence of chemistry (wittingly or otherwise) on history, opens with the conjecture that tin pest is what really stopped Napoleon's invasion of Russia.
In addition, as this answer reminds us, the rate of the reverse reaction increases to a maximum during the establishment of dynamic equilibrium.

I will add. Bromation of acetone must be performed with care, allowing first portion to react over long time. Without it, the reaction has very good chances to go out of control. I want also to note, that some reactions produce heat, and heat is 'universal reaction accelerator.' For example, if you put some $FeCl_3$ into cold 30% $H_2O_2$ the reaction will be very slow. But once reaction  speed will exceed some limit, it will run out of control, accelerating because of it's own heat.

There are several, but the rate can only increase in time under certain limits, that is, assuming certain approximations. Obviously at "final time" (for instance, when the equilibrium is reached) the rate will not be increasing continuously.
In general, in order to have rates increasing with time, you need to have negative orders of reaction with respect to some reactant, that is, the velocity must depend with the inverse of the population (or less likely, other exponent). This is typical of chain reactions, the most typical case being explosive reactions.


text:
Why doesn't nitrogen monoxide dimerize?
Why doesn't Nitrogen monoxide dimerize even though there is an odd electron present whereas nitrogen dioxide does (because of the odd electron on nitrogen)?

Highstaker's answer is almost correct.  The antibonding orbitals that are occupied in NO are in fact pi symmetry, but when the dimer forms that is no longer relevant.  It is a sigma bond.
The enthalpy of the newly formed sigma bond in the dimer is weak because the net gain in bond is off set by the loss of a very odd set of single-electron resonance forms available for NO monomer.
Given $\Delta G= -17 \:\mathrm{kJ/mol}$ , and that dimerization is entropically disfavored, when the total free energy is considered there is no gain since entropic effects are on the order of $10-30 \:\mathrm{kJ/mol}$.  Thus any small gain in enthalphy is offset by the loss of entropy.
Always look at these things through Gibbs free energy and not just enthalpy.

Just to clarify, entropic effects refer to T (delta S) and not just delta S right?

Yes, the net entropic "effect" on the direction of spontaneity is indeed T*(delS), which also explains the stability of of the dimer at lower temperatures.

Nitrogen monoxide's dimer exists, but it is very unstable. NO consists of dimeres only at low temperatures (about −163 °C). Solid NO seems to consist mostly of dimers.
Seems that unpaired electrons on antibonding orbitals can only pi-overlap. In NO the electronic density is shifted towards oxygen atom, and the "petals" of pi-orbitals don't have enough electronic density to support stable bond (some call it a half-bond). So the dimer bond is weak (enthalpy of formation is about 17 kilojoules)

One alternative rationalisation of why the dimerisation of $\ce {NO}$ is not as favourable as that of the dimerisation of $\ce {NO2}$ is that on each of the nitrogens in the $\ce {N2O2}$ dimer, there is a lone pair while for the dimer $\ce {N2O4}$, there are no lone pairs on the nitrogens. The presence of lone pairs destabilise the nitrogen-nitrogen bond in the former, thus it decreases the enthalpic favourability of the bond formation between two $\ce {NO}$ molecules, whereas this destabilisation is not present in the latter dimer.


text:
Which kind of plastic material would be suitable for a clutch cover on a motorcycle transition?
I'm going to have an online machine shop build a part for me using a CAD drawing that I specify.
Typically, these parts are made of aluminium, but this manufacturer also mills parts out of these plastic materials:

Polypropylene (PP)
Polystyrene (PS)
High impact polystyrene (HIPS)
Acrylonitrile butadiene styrene (ABS)
Polyethylene terephthalate (PET)
Polyester (PES)
Polyamides (PA) (Nylons)
Poly(vinyl chloride) (PVC)
Polyurethanes (PU)
Polycarbonate (PC)
Polyethylene (PE)

Polypropylene, I know, is the plastic used in water bottles, so I know that is not what I want.
This fabricated part will be used for the clutch cover on a motorcycle transmission and will have a raised section that needs to be tapped to allow attaching a gear shifter to, like in the aluminium version below.

I'd like something rigid and strong. Plexiglass comes to mind, but there could be a better alternative that I am unaware of, as long as I am not getting into something exotic that would cost more than aluminium.
Could suggest a good plastic for me, which would need to be oil resistant and withstand typical engine transmission temperatures.

I have cut back the question to it's actual core of what has been asked. That you ended up using aluminium like the answer suggested is nice, but it is superfluous information for the actual question. As such the 'update' actually kind of invalidates the answer, which is strongly discouraged on our site.

Most plastics consist of linear molecules with little inter-molecular interactions, so most plastics melts too easy. The only ones you mentioned having a chance are polyuretanes and polyamides. They have (CO)(NH) fragments, that forms strong intermolecular hydrogen bonds. So, they are ones of the strongest polymeric materials
And I will recommend aluminum anyway as it is much harder to screw things with aluminum.

Glass filled nylon molds well, wears well and is most resistant to heat with regards to the plastics you mentioned


text:
Which mutated enzymes act as better drugs than their non-mutated form?
Is there an enzyme which is used as a drug (let it be called version $A$) and a mutated version of it ($M$), where the activity of $M$ was found to be higher than $A$?
Any references on such a case would be greatly acknowledged.

A possible option:
"Site-saturation mutagenesis of Tyr-105 reveals its importance in substrate stabilization and discrimination in TEM-1 β-lactamase"

Another option: "Most Efficient Cocaine Hydrolase Designed by Virtual Screening of Transition States"

A third option: "Why Does the G117H Mutation Considerably Improve the Activity of Human Butyrylcholinesterase against Sarin? Insights from Quantum Mechanical/Molecular Mechanical Free Energy Calculations"

Design of High-Activity Mutants of Human Butyrylcholinesterase against (−)-Cocaine: Structural and Energetic Factors Affecting the Catalytic Efficiency

There are mutants of clotting factor VIII that have longer half lives and better activity than natural Factor VIII. Factor VIII is one of the missing proteins commonly involved in hemophilia. Hemophiliacs usually use injections of Factor VIII to prevent bleeding. The mutant form should last longer in the blood, with better activity, resulting in fewer injections.
source: https://encrypted.google.com/patents/US7211559
Since this is a 2007 patent, I don't know if this product is actually being used to treat patients, drug approval takes a long time.


text:
Can you create a buffer with a strong acid?
So I was wading through several textbooks yesterday reviewing the concepts of buffers. I found them lot more challenging than I used to. One thing in particular disturbed me: The books and internet all said "Yeah, buffers work pretty great in as long as the  $\rm pH \approx pK_a\pm 1$." But I am not one to leave these things alone. I finally found the Quantitative explanation I was looking for over at buffer capacity on Wikipedia.
However, the graph inside the link says the buffering capacity becomes great when you solution becomes very acidic or very basic, as well. And, of course, looking at titration curves there seem to be a few "buffer" zones e.g. where the solution is highly acidic or highly basic, the $\rm pH$ doesn't change as much after a certain point:

So, obviously there is something wrong with my reasoning, since nobody calls those areas buffer zones. Is it because it is not useful to have such a highly acid/basic solution? Or is it just because technically, there is no buffering, just small $\rm pH$ changes? Or maybe something else entirely...

I totally agree with trb456. I'd like to add that what you're seeing in the figure you provided is actually a titration of water (that starts out at acidic pH). Remember that water is both an acid and a base. When you add NaOH as in your figure, that titration curve is what you see. That's why the inflection point occurs at pH = 7.0. You add enough NaOH to neutralize the nitric acid and get to pH 7, then as you add more NaOH you increase the concentration of hydroxide ion - increasing the pH. Also, remember that pH is calculated on a log scale, which helps explain the shape of these plots.

I think the issue with buffer systems is that they are resistant to changes from strong acids/bases in their normal state.  For example, biological systems often are buffered so that they have some resistance to changes in pH, which can wreck havoc on biochemistry.
Recall the basic buffer equation:
$\ce{HA <-> H+ + A-}$
where $\ce{HA}$ is a weak acid, and there is an excess of $\ce{A-}$, the conjugate base. Because $\ce{HA}$ is weak, it is a relative poor proton donor, but $\ce{A-}$ is a relatively good proton acceptor.  So if you now add a strong acid, the $\ce{H+}$ it donates reacts with the $\ce{A-}$ to produce $\ce{HA}$, shifting the equilibrium to the left, rather than to the right as one might expect.  Or perhaps, preventing as much shifting to the right as might be expected. This gets to your issue of buffering capacity.
This line of reasoning does not work with strong acids/bases because they donate/accept protons with gusto, and thus are not resistant to shifting the equilibrium.  You would need to be way at the extremes of pH for there to be a small change to pH, as your graph shows.  In contrast, buffers are resistant to pH change in their normal state.

What do you mean by "normal state"?

Buffering capacity is just the amount of acid (strong enough to protonate everything) you must add per mole to bring about a unit pH shift. For example, the buffering of pure sulfuric acid at its second pK is, theoretically, 0.57 Eq/mol-pH and falls off very rapidly at pH values removed by more that 1 pH - just like any other acid.  On the other hand a solution of water with sulfuric acid at mole fraction 0.1 has buffering of 0.08 but does not exhibit a peak at pH 1.99. This is because 0.1 mole of acid has buffering capacity of 0.057 while 0.9 mol of water has buffering capacity of 0.02. The buffering of the water is approaching the buffering capacity of the acid. At higher pH where we make buffers with weaker acids the buffering of the water itself is much smaller and so the acid dominates ad the peaks near the pKs become evident.
This is largely a problem with semantics. We are taught that a buffer is something that holds pH near a particular value by absorbing the 'stress' of added acid or base. When we speak of the high buffering capacity listeners tend to assume that that there is a peak in any case where in many cases there isn't.
At low pH added acid goes to protonate the solvent, not so much the anion of the solvent. The nice Gaussian shaped buffering vs pH curve goes away.


text:
Why can't we use fusion to make ourselves rich?
As we know on fusing two hydrogen atoms we can convert it to helium and by breaking them we get two $\ce{H}$ atoms. So why can't we convert mercury to gold by removing 1 proton per atom?

The quick answer is that (a) We can't do this in bulk yet, only to a few atoms, and (b) it's very, very expensive to do this. Hopefully someone else can expand on it.

@Manishearth - For nuclear transmutation in the most generic sense (be it fusion, fission, spallation, etc.) that's not really true - Tc-99m, Am-241 and Pu-239 all spring to mind as valuable materials which are synthesised economically in macroscopic quantities.

@RichardTerrett: True... but not gold, iirc.

While it is possible to synthesize gold from mercury-196 (exists as 0.15% of natural mercury) the resultant gold will be radioactive and will not be gold for long!
There is actually the spallation neutron source which creates gold from liquid mercury.
Wikipedia

Why can't we use fusion to make ourselves rich?

Because at the current stage of technological progress making gold from other elements by fusion in large quantities is so costly (energy-wise and therefore money-wise), that it is just extraordinarily unprofitable to do that.

The premise of this question is false.  We don't fuse two hydrogen atoms to make Helium, we fuse two deuterium (Hydrogen-2) to make helium, and it is definitely not going to split back to two deuterium atoms under any circumstances, it is too expensive energetically.  This process is called fission.
Highstaker's answer is partially correct, and Meditate's answer is almost correct.  The process Mediate describes is radioactive decay, not fusion, or even fission.  Even using fusion processes, 'atom smashing' tends to lead to radioactive products because of the high energy (Hightaker comments) required to overcome electrostatic repulsion.

sorry little bit confused guy in chem a pretty tough but interesting subject for me that's why I make these mistake well thank's for the correction it help me in futer not to these all silly things

No apology is required.  Asking these questions is the purpose of stack.


text:
Phenomenon where fumes can relight an extinguished flame
I lit a mosquito coil and used it to melt down small down candle wax which had fallen on the floor. The coil then produced some dense fumes. I directed the fumes to the flame of a candle. As soon the fumes reached the candle, my coil caught fire again and started burning. I mean the flame actually traveled the path of the fumes and re lit my mosquito coil. I just wanted to know what this phenomenon is called? Why does this happen?

Welcome to Chemsitry.SE! This is an interesting question, and it has already attracted an answer. I edited your question to make it a little more readable and to remove some of your self-deprecating comments. You don't need to feel bad about something you don't know. I don't know what this phenomenon is either, nor how it works.

@BenNorris I tried to work with the title a bit, please reedit if you think I didn't capture the spirit of it.

It looks like you accidentally created a combustion mixture.  I checked the Material Safety Data Sheet (MSDS) for several mosquito coils, and in addition to the insecticide, one of them also contained the oxidizer potassium nitrate $\ce{(KNO3)}$.  The burning candle likely had some vaporized paraffin (fuel) in the flame, so when you directed the $\ce{KNO3}$ vapor into the flame, you had heat + fuel + oxidizer = combustion!  You literally set that vapor on fire.  I'd be careful with that: although the MSDS indicated a small amount of $\ce{KNO3}$ in the coil, you could still be injured if you were not prepared.

To add to this, the reverse can also happen - it is quite easy to relight a very recently extinguished candle through its vapour.

@trb456 thanks..but does the distance really matter in this? i mean suppose i stand somewhere far and direct those fumes will it still catch fire? is there a word for the vapor catching fire(is it some phenomenon?)

I don't think it matters specifically that the vapor caught fire.  What matters are the conditions of the system: you provided an oxidizer in the vapor along with heat and fuel from the candle.  That trio, no matter how it comes together, means combustion, which is why you need to be careful whenever you know you have two of the three in close  proximity.


text:
How do chemists find a procedure to produce a given substance?
How do they find out exactly what reactions they can use to produce some substance they want to produce? Is there some reaction database that helps them find a way to make the target substance out of available chemicals?
For example, let's say they want to produce ethanol. There are multiple ways to do this: they could use the fact that yeast can transform sugars into ethanol and carbon dioxyde and use distillation to produce concentrated ethanol, or use ethylene hydration to get the same thing.
The question is: how could they find those ways? How do they know what will be produced by a given reaction, or inversely, what reaction will yield a given substance? Some reactions are well known and were just observed and studied, but what about the obscure ones?
Thanks.

Chemists use many sources of information to do this.
Part of being trained as a chemists is learning, in general terms, what kinds of reactions are known and what their best uses are. That training can go 5-10 years beyond a bachelors degree and is very intensive.
Beyond what you can keep in your head, one of the most important information sources is the "primary literature", consisting of:

Patents
Journal publications

As an example, see this patent about ethanol manufacture.
Alternatively, see this very recent journal publication dealing a new catalytic reaction.
Having all of this information out there is a start, but it's a vast collection spread out over many publishers and 100 years or more. To find what you need, it's essential to have a good indexing system. This allows you to search large databases of reactions and molecules for references that exactly match, or which are close to what you need.
This last part - being able to reason by analogy - is critical to being able to get the most out of the primary literature. Maybe I don't find an exact match for the exact reaction I want to do. But I find something pretty close. Here's where my training and reasoning ability come in to either decide to adapt what I've found or keep looking for something better.
An example of one of these search systems is Chemical Abstracts Service, but there are others.
As a professional chemist, you're expected to keep up with all of the main journals in your field so you stay up to date. You do this by reading articles, talking to people, giving talks, and going to conferences.
There's another category of information consisting of 'reviews'. These are articles that summarize a large area of chemistry - for example, recent large-scale methods for the production of ethanol.
The discovery/invention of new reactions is a very active area of current chemical research.


text:
Is 2,3-dimethylprop-1-ene a possible organic structure?
Is 2,3-dimethylprop-1-ene a possible organic structure?
I am not sure how to approach this, what would the correct name of this molecule be?

2-methylbut-1-ene.
For future reference: an alkyl substituent at the 3 position of a propyl group probably indicates you did not select the longest carbon chain to name the parent molecule.
So your answer was not 'formally' correct.  That being said there was only possible structure to be drawn from the name you used, so correct or otherwise, it is effective communication.


text:
Are there any chemicals with this structural skeleton?
Is there a chemical that has the bond structure like below?

There are 5 atoms linking together.
Two of them have 4 bonds. (They are different atoms.)
Another two of them have 3 bonds. (They can be either same or different atoms.)
The last atom has only 2 bonds.
Suggested SMILES for unknown $\ce{X}$ elements is [X]2[X]14[X]3[X]1[X]234
Are there such kinds of chemicals?

To clarify - does the structure have to be composed solely of carbon atoms? With no hydrogens? Or is it a saturated alkane?

The OP indicates that they can be different atoms. A compound with formula $\ce{C2B2O}$ has the correct valences to accomplish that structure, but I am unsure if it exists.

@BenNorris If it does, that is going to be one unhappy molecule, what with how far from tetrahedral that central carbon is. I suspect it would not exist for very long...

Extremely unlikely. However, if you allow ligands, a metallic cluster with this graph as part of the structure may be found.

Similar motifs include [1,1,1]propellane and white phosphorus. Neither of which is noted for its stability.

Iron(III) Triethylenetetramine complex would look like this, except that there are two carbons between the nitrogen atoms that would bind to the iron center.

Expanding on @permeakra: You may also find a gold micro-cluster somewhere. I will also check my notes on exotic boranes later.

CCDC doesn't contain any molecules of such geometry. The closest topology observed is among pnictogens such as $\ce{P, As, Sb}$, which form tetrahedron units $[\ce{P4}]$, $[\ce{As4}]$, $[\ce{Sb4}]$ which are readily coordinating with Au, Rh, Ag and Cu by edge-sharing:

The first publication (1) tetraphosphido-ligand presents the following structure of $\ce{Rh}$-complex:

The most relevant structure best fitting the required connectivity matrix would probably be bis(tricyclo[$1.1.0.0^{2,4}$]tetraphosphane)-gold (2):

Thanks to @TAR86 comment mentioning exotic boranes, I also found a research (3) on ab initio calculations for $\ce{B4}$ cores obtained via the following proposed reaction:

They also obtained crystal structure for bis(tetrahydropyran)-(tetra(t-butyl)tetraborane)-lithium where $\ce{B4H}$ core is present:

(1) Lindsell, W. E.; McCullough, K. J.; Welch, A. J. J. Am. Chem. Soc. 1983, 105 (13), 4487–4489. DOI: 10.1021/ja00351a067
(2) Forfar, L. C.; Clark, T. J.; Green, M.; Mansell, S. M.; Russell, C. A.; Sanguramath, R. A.; Slattery, J. M. Chem. Commun. 2012, 48 (14), 1970–1972. DOI: 10.1039/C2CC15291F
(3) Neu, A.; Mennekes, T.; Englert, U.; Paetzold, P.; Hofmann, M.; von Ragué Schleyer, P. Angew. Chem. Int. Ed. Engl. 1997, 36 (19), 2117–2119. DOI: 10.1002/anie.199721171


text:
Electrophilic Substitution
I need some help in a electrophilic substitution reaction.

What will be the major product formed in this reaction and why?
I know that
(1) Activator wins over deactivator.
(2) A powerful wins over a weak activator.
(3) When a substituent possesses a lone pair, it it usually extends the product determining influence.
(4) Steric factor always plays a key role.
$\ce{NO2}$ is a powerful deactivator, so I think the middle ring will not undergo substitution. I can't decide which of the other two rings will undergo substitution and at what position.

Please see our homework policy. Could you show your own thought process/your own efforts in solving this reaction?

Ok, I will add my work.

I can answer this easily enough but it does you little good.  Draw the resonance structures which show the delocalization of + charge from the nitro.  Go ahead and make a carbanion on one of the rings to accomplish this.  When you draw these structures, it will become obvious to you which ring will be attacked.

@Lighthart can you elaborate on that procedure and maybe turn it into an answer.

I no longer am a practicing chemist, so I have no access to a structure drawing program; I'll do my best with words:
Accept the initial premise that the NO2 is strongly deactivating so only the 'end' rings will be attacked.  As we know o,p- substituion have functionalyl equivalent resonance forms, so we only have to consider 4 possible products: p-, m-substitutuion on the 3-ring and similar on the 4-ring.
Drawing all four of these structures and the reasonanc forms indicates that the p-,4-ring susbtitution and the m-,3-ring susbtitution are poor choices because of high energy intermediates.
The final question to resolve is whether or not the m-,4-ring is better than the p-,3-ring.  Based on the intermediates, this is a even, but the nitro group is in conjugation with the 4-ring, whle only in cross-conjugation with the 3-ring  You can see this by drawing a rsonance form with a carbocation at the para position, and rearranging the bonds via resonance all the way to the nitro group.  This can't be done with the with the 3-ring, meaning the nitro group deactivates the 4-ring more than the 3-ring.
Thus the product is likely p-substitution on the 3-ring.
Advanced:
The resonance for arguments require all three structures be coplanar.  This is not the case; biphenyls have some torsion and twisting because o-hydrogens bump.  However, there is still enough 'flatness' and orbital overlap to make some use of the resonance arguments presented above, albeit the magnitude of better and worse are smaller than they would be for more classical EAS problems.


text:
What barriers prevent plastic from becoming a fiber optic, data transmission medium?
This is a question that has bothered me for a while.  In the past many years, no company has been able to develop a material that can transmit light nearly as well as fiber optic glass, or at least not as cheaply.
After doing a bit of research, I found that although there is fiber optic plastic, the quality is considerably poor when compared to the standard glass medium.  I would assume that developing a highly-transparent, high-quality plastic that is cheap to manufacture would be considered the holy grail of data communication (not to mention, quite profitable.)
Why is it that in the many, many years that fiber optic data transmission has been around, no one has been able to develop such a form of plastic?  In fact, I've never even heard of such research making headlines on the more common tech news sites.

I think it's fine right here, so don't worry about that aspect.

In fact, I've never even heard of such research making headlines on
  the more common tech news sites.

Tasks like that are called deveopment, not research, normally.
And no industrial company will talk much about such work
if its not sucessful.
In fact there are such organic fibers for minor purposes (eg the Laserjets
II and III used such a fiber to send a synchronising signal from the
laser scanning unit down to the control board, for about 10 inches.
30 years ago one could buy such fibres for experiments at radio shack)
If You read some literature on optical fibres, You will learn that
low scattering is the grail. And plastics scatter a lot, for
very basic reasons of polymer structure. Try to look throug a plate of
acrylic glass "edge on", and You will understand.

Good answer, but not perfect. I suspect there is room for more detail on the specific causes and some of the other effects (refractive index matters as does the specific absorption in the infra red range used for most data transmission).


text:
Sulfuric acid recovery
I know sulfuric acid is an important catalyzer and is used in various processes. My question is, how do I recover the remaining sulfuric acid? It will be impure, and I don't know how to do the "standard" procedure (is there one?)

In laboratory, nobody cares. It's too cheap to care about.
In industry, other catalysts are usually used (heterogeneous ones).

Of course in industry sulfuric acid is used! In all cases where some washing with water is needed anyway, there is no reason not to use sulfuric  or hydrochloric acid.

@Georg it is another reagent to transport, store, use and dispose. It costs.


text:
Influence of solubility on the complexation constants at different pH values
Given some small chemical compounds whose solubility varies with pH, and in the case of compounds that can form complexes, I have seen that the complexation constant varies with pH.
I think there might be two main reasons for this:

Chemical structure of the host system is affected by pH (for instance, protonation) and therefore it affects complexation constant;
It might be related to the small chemical compound, since its solubility might vary according to pH.

So I guess that in this case there are many chemical/physical equilibria involved, in this case mainly phase solubility and protonation for both species. I wonder what is the best approach for describing this, perhaps one can use different equilibrium constants for the mentioned equilibria, and from there derive a global constant that depends on pH. How do these two factors (protonation and phase/solvation) influence the value of the complexation constant?
As a particular case, I am mostly interested in complexation of flavonoids with cyclodextrins.

This is a much more focused question about solubility, pH, and complexation constants. I have edited the last line to make it clearly a question, since it reads like an invitation to discussion. An example of the types of molecules you are looking at would make this a great question. I know from your previous questions that you are interested in flavonoids. Is that true here also?

It is possible to calculate out all of the complexation and speciation reactions using the concepts of Aquatic Chemistry. I’m a PhD student in environmental chemistry and I TA a course in Aquatic Chemistry. Usually, when pH matters it’s because the protonation state controls the chemistry and can be described using equilibrium constants. I bet that your solubility issue is related to the deprotonated form being an anion and thus more soluble in water than the protonated, neutral form (very common for organic compounds). The protonation state can be calculated from the $\mathrm{p}K_\text{a}$:
$$
\ce{L- + H+ <=> HL}\\
K_\text{a}=10^{-\mathrm{p}K_\text{a}}=\dfrac{[\ce{L-}][\ce{H+}]}{[\ce{HL}]}
$$
Where $K_\text{a}$ is the equilibrium constant, and L is your compound. I would guess that your flavanoids and cyclodextrins might have multiple $\mathrm{p}K_\text{a}$s, you can write an expression like this for each one.
Complexation reactions can be written in a similar way. Let’s say that $\ce{Y}$ is the molecule your $\ce{L}$ molecule complexes with, and it is neutral when protonated ($\ce{HY}$).
$$
\ce{L- + HY <=> HLY-}\\
K=\dfrac{[\ce{L-}][\ce{HY}]}{[\ce{HLY-}]}
$$
Where $K$ is the equilibrium constant for this complexation reaction. Keep in mind that which species (protonation state) is in the expression matters, the value for the equilibrium constant is only valid for those species. Above, I’ve written that the deprotonated form of L ($\ce{L-}$) complexes with the protonated form of Y ($\ce{HY}$). This is where pH comes in. pH controls how much of L is deprotonated and how much of Y is deprotonated and able to participate in complexation, and $K$ controls how much of the available $\ce{L-}$ and $\ce{HY}$ form the complex. Complexes themselves can also have $\mathrm{p}K_\text{a}$s the one above ($\ce{HLY-}$) could probably be protonated to a neutral form $\ce{H2LY}$.
You might be able to guess that the mathematics gets a little complex at this point. It is possible to do these calculations by hand, but there are computer programs that can handle them. The most easily accessible one is Visual MinTEQ. The other hard part is having the data for the equilibrium constants, there may be data for your system, but you’ll have to look.
Just to summarize. Complexation constants don’t change with pH (they are, in fact, constant) it’s just that the speciation matters, and speciation is controlled by pH. The ‘global constant’ you refer to is pH, not something else.

excellent answer! thanks a lot. Could you also recommenda good book about aquatic chemistry?

W. Stumm and J. J. Morgan, Aquatic Chemistry: Chemical Equilibria and Rates in Natural Waters  Is the "classic" text for Aquatic Chemistry. J. F. Pankow, Aquatic Chemistry Concepts is also good.


text:
Integrated rate laws
My teacher gave us the equations for zero-order, first-order and second-order reactions as follows:
$$ [A]_t = -akt + [A]_0 $$
$$ \ln [A]_t=-akt+\ln [A]_0$$
$$\frac1{[A]_t}=akt+\frac1{[A]_0}$$
Where a is the coefficient of $A$ in the chemical reaction, but everywhere I've seen, the $a$ is not there. I have the Central Science 12th edition and there's no $a$ either, is my teacher right? Because that value does affect the answer.

From the reaction
$$\ce{2A -> B}$$
we have $a = 2$, which leads to the differential equation:
$$\frac{d[A]}{dt} = - k [A]^2$$
which then leads to the integrated rate law you've cited.  The $2$ is nowhere as a coefficient, it is related to the order of the equation, and determines which integral/form should be used.  Similar conventions also apply for 1st and zeroth order kinetics.  I disagree with your instructor.
It should be stressed that these rate laws apply to elementary reactions, not complex reactions.  Experimental measurement may differ from that predicted using an elementary approach, which then leads to the conclusion there are other steps involved.  In the case of OP's comment regarding oxidation of carbon monoxide with NO2, elementary reactions would predict a rate proportional to $$k [\ce{NO2}][\ce{CO}]$$ if the reaction is the result of a simple bimolecular collision. The fact that a different rate is observed indicates this process is more elaborate.

And I must disagree with you because the order of the equation has nothing to do with the coefficient of the reagent. Particular orders must be found according to experimental data... or so I've been taught.

You might want to reexamine that thought in the context of physical chemistry, and not just parrot 'what you've been taught'.  For any elementary reaction, the coefficient alone determines the order of reaction.  The probability of any two [A] molecules colliding to reaction MUST be proportional to $$[A]^2$$ (just think about it).  For any complex mechanism involving several steps, the apparent rate law may not have such a straightforward relation, but that is beyond the context of OP's question and therefor irrelevant.

I quote "For any reaction, the rate law must be determined experimentally. In most rate laws, reaction
orders are 0, 1, or 2. However, we also occasionally encounter rate laws in which the reaction
order is fractional  or even negative." Also "Although the exponents in a rate law are sometimes the same as the coefficients in the
balanced equation, this is not necessarily the case." And it's not beyond the context of my question.

Again with the quoting, and not with the thinking.  Elementary reactions follow, without exception the example I've shown.  Complex reactions do not.  An example of an elementary would be the coupling of radicals by combination.  An example of a complex rxn would be the acid catalyzed synthesis of ether from ethanol.  For radical combination, there is one step, and the math works flawlessly.  For acid-catalyzed etherification, there are three steps, and all must be considered, and the overall process may no longer have the same rate as a simple elementary reaction.

As an aside, but nonetheless relevant, simply quoting an unnamed source does not really strengthen an argument.  Please review Carey and Sundberg, Advanced Organic Chemistry, 3rd Ed. pp 183-196 if you need a full treatment.

I forgot to name the source, it's from the same book I mentioned in the question. Without exception? A really common reaction as $$NO_2 + CO → NO + CO_2 $$ is second-order, in which $$V=k[NO_2]^2$$

Yes, without exception.  What you have shown is not an elementary reaction, and therefore does not provide any evidence to dispute my claim.

let us continue this discussion in chat

It really doesn't "really" matter if you have $k$ or $2k$ as long as all parties involved know which you are using. The $k$ can "absorb" a constant term (which then changes the value of $k$...). Wikipedia notes some dis/advantages of this here: http://en.wikipedia.org/wiki/Rate_equation#Second-order_reactions

@Lighthart You may want to incorporate the first comment that you made into your answer, as I think it addresses the question directly.  Thanks for monitoring yourselves and taking this one to chat!


text:
Metal extraction by pyrolysis
There is a process named solar furnace pyrolysis which is used to reduce the oxides of various elements by concentrating solar energy. Oxygen is liberated but an oxide slag is left.
$$\ce{SiO2 ->[pyrolysis] SiO + 1/2O2}$$
How can the element can be extracted from the oxide produced?
I am making a project for NASA on space settlements in solar furnace pyrolysis. Can solar furnace pyrolysis be used for obtaining both metals as well as oxygen from the metal containing oxides left after pyrolysis? Can it be done by electrolysis? If possible, please provide with a link with your answer.

The answer appears to be forthcoming in the wikipedia article on silicon monoxide and references therein — to quote:

$\ce{(SiO)_{n}}$ irreversibly disproportionates into $\ce{SiO2}$ and $\ce{Si}$ in a few hours between 400 and 800 °C and very rapidly between 1000 and 1440 °C, although the reaction does not go to completion.

The article implies that $\ce{(SiO)_{n}}$ generates a passivating layer of $\ce{SiO2}$ in contact with dioxygen, so presumably the slag would be encapsulated in a thin layer of that.
Additionally, silicon is not generally considered to be a metal, but rather a semimetal or metalloid.

what about other metals that may be used in this process Mr. richard? can metal semi oxides be electrolysed the same way as metal oxides?

Housecroft and Sharpe, p. 339 indicates that Si is commonly prepared from SiO2 by smelting with a reducing agent (either carbon or calcium carbide). This is in common with the smelting of metal oxides. Purification of the silicon for use in semiconductor devices would require zone remelting, Czochralski pulling or the like.