text:
Stern-Volmer plot and calculation of rate constant for quenching
I did an experiment irradiating benzophenone in UV light and worked out the rate constant for the reaction. Then I repeated the experiment using various concentrations of a quencher and worked out their respective rate constants. Now I need to calculate the rate constant for quenching $k_\mathrm{q}$ using a Stern-Volmer plot. How would I do this?
The equation I have used before for the Stern-Volmer relationship is:
$$\frac{1}{I_\mathrm{f}} = \frac{1}{I_\mathrm{abs}}\left(1 + \frac{k_\mathrm{q}[\ce{Q}]}{k_\mathrm{lum}}\right)$$
where the x-axis is $[\ce{Q}]$, the concentration of the quencher, and the y-axis is $1/I_\mathrm{f}$, and to work out $k_\mathrm{q}$ you calculate the gradient/y-intercept = $k_\mathrm{q}/k_\mathrm{lum}$.
I think I would plot a graph of concentration of quencher $[\ce{Q}]$ on the $x$-axis against 1/rate constant on the y axis. But how would I work out $k_\mathrm{q}$?

Benzophenone is a notorious case where prompt fluorescence, phosphorescence and delayed fluorescence can mix. Have a look at Modern Molecular Photochemistry by N.J. Turro for some general information and Prompt and delayed fluorescence from benzophenone by
R.E. Brown and L.A. Singer for the original reference.
\[
\begin{align*}
\ce{BP &->[h\nu]\ ^{1}BP^{\ast}}\\
\ce{^{1}BP^{\ast} &-> BP + h\nu'\quad &(prompt\ fluorescence)}\\
\ce{^{1}BP^{\ast} &->[ISC]\ ^{3}BP^{\ast} &(intersystem\ crossing)}\\
\ce{^{3}BP^{\ast} &-> BP + h\nu''\quad &(phosphorscence)}\\
\ce{^{3}BP^{\ast} &->\ ^{1}BP^{\ast} &(repopulation\ of\ S_1)}\\
\ce{^{1}BP^{\ast} &-> BP + h\nu'\quad &(delayed\ fluorescence)}\\
\end{align*}
\]
Let's ignore all this (for a moment).
Fluorescence quenching, i.e the observed decrease of fluorescence intensity in the presence of a quencher, may occcur through two different pathways:
Static fluorescence quenching
In this case, formation of a ground state complex between the fluorophore $\ce{F}$ a quencher $\ce{Q}$ decreases the concentration of $\ce{F}$, hence the concentration of the excited state $\ce{^1F^{\ast}}$ and the fluorescence intensity. This has no impact on the fluorescence lifetimes! These are the same in the absence ($\tau_0$) and presence ($\tau$) of a quencher.
\[
\frac{\tau_0}{\tau} = 1
\]
Dynamic fluorescence quenching
Here, quenching occurs from the excited state $\ce{^1F^{\ast}}$. With other words, the presence of a quencher introduces an additional, non-radiative pathway for the deactivation of the excited singlet state. Consequently, the fluorescence lifetime $\tau$ is shortened!
Let's try to derive the Stern-Volmer equation and see if that leads to some insights:
(You probably know all of this already, but I'm just in the mood...)
In an experiment on dynamic fluorescence quenching, the following processes and their rate constants have to be considered:
\[
\begin{align*}
\ce{F + h\nu &->[k_{abs}]\ ^1F^{\ast}\quad &\mathrm{(excitation)} } \\
\ce{^1F^{\ast} &->[k_{ic}]\ F + heat\quad &(internal\ conversion)}\\
\ce{^1F^{\ast} &->[k_{f}]\ F + h\nu'\quad &(fluorescence)}\\
\ce{^1F^{\ast} + Q &->[k_{q}]\ F + Q'\quad &(quenching)}\\
\end{align*}
\]
The fluorescence quantum yield $\Phi$ is given as the quotient of emitted ($I_{em}$) and absorbed ($I_{abs}$) photons:
\[
\Phi = \frac{I_{em}}{I_{abs}}
\]
The number of emitted photons depends on the concentration of the fluorophore in its excited singlet state $\ce{^1F^{\ast}}$ and the rate constant $k_{f}$:
\[
I_{em} = k_{f}\cdot[^1F^{\ast}]
\]
The number of absorbed photons depends on the concentration of the fluorophore $\ce{F}$ and the rate constant $k_{abs}$:
\[
I_{abs} = k_{abs}\cdot[F]
\]
The rate of absorption processes is difficult to measure, they're bloody fast. Fortunately, we don't have to! Under steady-state conditions, population of the excited singlet state and its deactivation balances out.
Consequently, we can express $I_{abs}$ as:
\[
I_{abs} = k_{f}\cdot[^1F^{\ast}] + k_{ic}\cdot[^1F^{\ast}] +k_{q}\cdot[^1F^{\ast}]\cdot[Q]
\]
The expression for the fluorescence quantum yield thus is:
\[
\begin{align*}
\Phi &= \frac{k_{f}\cdot[^1F^{\ast}]}{k_{f}\cdot[^1F^{\ast}] + k_{ic}\cdot[^1F^{\ast}] +k_{q}\cdot[^1F^{\ast}]\cdot[Q]}\\
&= \color{\red}{\frac{k_{f}}{k_{f} + k_{ic} +k_{q}\cdot[Q]}}
\end{align*}
\]
The latter is pretty convenient since we don't have to think about the pesky $[^1F^{\ast}]$ anymore! YAY!
If we do perform a fluorescence measurement in the absence of the quencher,  $[\ce{Q}] = 0$, $\Phi_0$, the fluorescence quantum yield under this condition is:
\[
\Phi_0= \frac{k_{f}}{k_{f} + k_{ic}}
\]
The effect of a quencher $\ce{Q}$ on the fluorescence is expressed as:
\[
\begin{align*}
\color{red}{\frac{\Phi_0}{\Phi}} &= \frac{k_{f}}{k_{f} + k_{ic}} \cdot \frac{k_{f} + k_{ic} +k_{q}\cdot[Q]}{k_{f}}\\
&= \color{\red}{1+ \frac{k_{q}\cdot[Q]}{k_{f} + k_{ic}}}
\end{align*}
\]
Too bad that we don't know all these rate constants and using just a steady-state spectrometer, we won't be able to measure them! And for a lab chemist, there's too much stuff in the equation anyway! Bummer!
But we can argue about what influences the fluorescence lifetimes, which, btw, can be measured!
Lifetimes are pretty much the inverse of the (sum of the) rate constants.
Without the quencher, the fluorescence lifetime $\tau_0$ is:
\[
\tau_0 = \frac{1}{k_{f} + k_{ic}}
\]
In the presence of a quencher, the fluorescence lifetime $\tau$ is:
\[
\tau = \frac{1}{k_{q}\cdot[Q] + k_{f} + k_{ic}}
\]
If we plug that in again and rearrange a bit, we get:
\[
\begin{align*}
\frac{\Phi_0}{\Phi} &= 1 + \tau_0\cdot k_q\cdot[\ce{Q}]\\
&= 1+ {K_{SV}}\cdot[\ce{Q}]
\end{align*}
\]
We name $K_{SV}$ the Stern-Volmer constant!
Summary

Plotting $\frac{\Phi_0}{\Phi}$ or the quotient of the areas of the fluorescence spectra vs the concentration of the quencher should give a straight line, that cuts the $y$ axis at 1.
The slope of the curve is the Stern-Volmer constant $K_{SV}$.
The real rate constant $k_q$ for the reaction of the excited state $\ce{^1F^{\ast}}$ with the quencher can only be obtained from $K_{SV}$ if the fluorescence lifetime $\tau_0$ has been obtained from a time-resolved fluorescence measurement!

If you plot such a graph you should obtain a straight line with some slope $a$ and intercept $b$: $y=ax+b$. You can see that also Stern Volmer relation is a linear function (of concentration). Thus, $k_q$ can be calculated by equating value of slope you obtained ($a$) to $k_q/klum$. Such procedure is simple and important to understand as can be used to determine many quantities which occur in linear equations.

I don't know what klum is. Is there a way to work it out?

I find that the commonly used definition of quantum yield as $\Phi = I_{em}/I_{abs} $ is not very intuitive. It is easier to realise that the yield is the $\Phi =$ (rate of fluorescence)/(sum of rate for all other processes) which leads directly to the first equation above in red.


text:
How to calculate the solubility of AgCl in a MgSO₄ solution?
The solubility product of $\ce{AgCl}$ in water is given as $1.8\cdot10^{-10}$
Determine the solubility of $\ce{AgCl}$ in pure water and in a $0.25~\mathrm{M}$ $\ce{MgSO4}$ solution.
I solve to the $K_\mathrm{{sp}}$ and got $1.34\cdot10^{-10}$
for Part B I was told to use the activity and am not sure if I did it right,
my new $K_\mathrm{{sp}}$ turn out to be $1.123\cdot10^{11}$
could someone help me with part B?

On Chemistry StackExchange some personal effort on homework questions needs to be shown.

In pure water, as you calculated, for the reaction
$\ce{AgCl -> Ag+ + Cl-}$ with $K_\mathrm{sp} = 1.8\cdot10^{-10}$
 the concentration of $\ce{Ag+}$, i.e. $\ce{[Ag+]}$ will be $1.34\cdot10^{-5}~\mathrm{M}$.
In a solution of $0.25~\mathrm{M}$ $\ce{MgSO4}$ (which dissociates almost completely), there's an additional equilibrium here:
$\ce{Ag2SO4 -> 2Ag+ + SO4^2-}$ with $K_\mathrm{sp} = 1.4\cdot10^{-5}$
Looking at the $K_\mathrm{sp}$ of both the reactions, we can see that $\ce{AgCl}$, having the lesser solubility, decides the $\ce{[Ag+]}$. Here we can even check if the concentration of $\ce{[Ag+]} = 1.34\cdot10^{-5}~\mathrm{M}$ and $\ce{[SO4^{2-}]} = 0.25~\mathrm{M}$ exceeds the solubility limit of $\ce{Ag2SO4}$. It doesn't.
$(1.34\cdot10^{-5})^2 \cdot 0.25 = 4.5\cdot10^{-11} < (K_\mathrm{sp})_{\ce{Ag2SO4}}$
The final composition of the solution will be $\ce{[Ag+]} = 1.34\cdot10^{-5}~\mathrm{M}$, $\ce{[Cl- ]} = 1.34\cdot10^{-5}~\mathrm{M}$, $\ce{[Mg^2+]} = 0.25~\mathrm{M}$ and $\ce{[SO4^2- ]} = 0.25~\mathrm{M}$.
In conclusion, the $\ce{[Ag+]}$ is the same for both the cases.

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text:
Why does Fe(CO)₄ adopt a tetrahedral, as opposed to square planar, geometry?
Why does $\ce{Fe(CO)4}$ adopt a tetrahedral, as opposed to square planar, geometry? Here's my analysis:
Factors for square planar:

coordination: 4
$\ce{Fe(0)}$: d8
$\ce{CO}$: strong field ligand (gives large d-splitting)

Factors for tetrahedral:

oxidation state: 0 (gives weak d-splitting)
$\ce{CO}$: relatively bulky (?) ligand

Other thoughts:
The general guidelines for determining whether or not a row 1 transition metal complex is square planar or tetrahedral (given it  has 4-coordination and d8 electrons), are the strength and bulkiness of the ligand. Generally, strong field ligands like $\ce{CO}$ and $\ce{CN^{-}}$ gives square planar complexes. In fact, $\ce{[Ni(CN)4]^{2-}}$ is square planar. Thus, the only way I could explain why $\ce{Fe(CO)4}$ would be tetrahedral would be because of its low oxidation state. Is this correct?

The hypothetical $\ce{Fe(CO)4}$ complex is isoelectronic to the observed $\ce{[Ni(CN)4]^{2-}}$ complex and therefore would also be square planar.[1,2] Both are 16 electron complexes, i.e. one of the valence orbitals of the central atom remains unoccupied. Your assessment for $\ce{Fe^{\pm0}}$ as $\mathrm{d^8}$ is absolutely correct, also that carbonyl is a good ligand causing a strong field and hence a strong splitting. In a tetrahedral complex there would be triply degenerated HOMO, which would only be occupied by four electrons. This arrangement cannot be more stable than a planar complex. However, the iron tetracarbonyl complex as a singular entity has not been observed (to my current knowledge). The observed structures for the compound are various dimers and trimers.[3]
There is also a neutral iron carbonyl complex, but it binds five instead of only four ligands. Therefore $\ce{Fe(CO)5}$ is a trigonal bipyramidal 18 electron complex.[4]
The observed $\ce{[Fe(CO)4]^{2-}}$ complex[5] is isoelectronic to $\ce{Ni(CO)4}$ which is a tetrahedral 18 electron complex.[1,4] Without going into more detail, the iron in this complex is of course $\ce{Fe^{-II}}$ and therefore $\mathrm{d^{10}}$. A square planar arrangement would no longer be more stable and a complex with a higher symmetry and more degeneracy is preferable.

References

J. Demuynck, and A. Veillard, Theor. chim. acta,
1973, 28 (3), 241-265.
ChemSpider.com
Raymond K. Sheline, J. Am. Chem. Soc., 1951, 73 (4), 1615–1618
Dario Braga, Fabrizia Grepioni, and A. Guy Orpen, Organometallics, 1993, 12 (4), 1481–1483.
Henry B. Chin, and Robert Bau, J. Am. Chem. Soc., 1976, 98 (9), 2434–2439.

Note
Carbonyl is nowhere close to a bulky ligand. Apart from halides and hydrides it might as well be the smallest possible ligand.

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text:
Could I settile for galvanized steel over stainless for salt corrosion resistance?
Bear with me, I'll get to the chemistry soon.
I have a float tank that I want to attach a skimmer to.  Unfortunately, the enclosure is weakened by cutting a hole out of the ABS plastic.  The weight of the salt water (1.25SG) to somehow causes a bend in the plastic just enough that getting a good seal isn't possible without additional support.
The float tank consists primary of (pardon the units) 800lbs of $\ce{MgSO4}$ Heptahydrate raised to about 96 degrees Fahrenheit.  The pH is roughly 7.2.
My concern with supporting the ABS plastic is that the Epsom salt water may corrode the metal, if it somehow isn't protected by a liner.  Would I absolutely need to have stainless steel, or could I get by with galvanized (which is much cheaper).
In case it matters, I estimate the volume of water directly pressing on the skimmer opening to be about 9+ gallons (again at 1.25sg).
Thank you for answering non-chemist,
Ben

"pardon the units" No!

I'm a chemist, but this seems more like a question for an engineer.
Unfortunately I don't have some references to back up my answer.
If you use galvanized steel you have a protective layer on the outside of the steel. As soon as you handle it for attaching, welding... you destroy/scratch/bruise this top layer and the underlying steel gets in contact with the salt solution which is normally aggressive towards metal. This mostly is right for e.g. chloride salts in a mix with oxygen and I'm not quite sure how corrosive MgSO4 is.
Here is a paper about corrosive currents of steel in Saltwater/cement solutions, but that`s not my speciality: http://www.jmst.org/fileup/PDF/pass543.pdf
Stainless steel is more robust, because it's an alloy of steel and chromium. Therefor it's not just covered with a protective layer but saver as a whole. An I guess it's therefor recommended.
(As a side note, some years ago I had a field trip to a big chemistry company with lots and lots of pipelines. We had some time there with the technical stuff in charge of teh pipelines. They explained, that the most annoying pipelines were the ones with salt water, since it was more aggressive towards the pipes than acids, mineral oils, gases etc.)

I remember hearing that in some seawater piping applications, brass was the preferred material, not steel, but I couldn't find any good references for this after a quick Googling.

You may find it worth your time to acquaint yourself with the concept of streaming potentials.

That’s true about the welded joint for two galvanized objects.     The welded area needs to be galvanized after welding to provide the protective layer (anode).      Depending on the solution in contact with the galvanized and time and temp you have varying degrees of protection.    Galvanized pipes etc are ok in general weather but salt water for instance will consume the zinc or magnesium rather quickly and once that’s breached the chlorides will have a field day with the metal underneath.

I'm only a chemist by desire and curiosity but as a Practical Maintenance Engineer let me throw this out there.
Galvanized steel is a really bad idea.  Just about any commercial sheet metal is going to breakdown relatively quickly in your  environment.  Brass is the best of a bad list but, it too, will ultimately corrode.
Find someone with a plastic welder.  Harbor Freight carries one the would work for you....
Repair and Reinforce container with another piece of ABS.
You could use a number of different plastic pieces of a number of different plastic materials to build a frame around your tub.
Or
Owens Corning sells buckets, tubes, bottles, and boxes of (in Ultra Hi-Tech terms) pookey* stuff and substrates to repair ABS tanks for consumer use (think Koi ponds / waterscapes).  See what is available and check the SDS of products that look workable.
*Pookie, in this context; any repair product in any form that will stick well enough when cured to contain without seepage the chemicals that need to be contained, under the circumstances of the intended use without altering said chemicals.
Good luck!

And one could use stainless ( 316  SS , very common) fasteners to add reinforcement of polymer or Fiberglas composites around the cut opening.

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text:
How to rationalise the IUPAC name for tetrahedrane?
I'd like some assistance in understanding the IUPAC name for tetrahedrane. I understand the basics of polycyclic ring naming however this example has me confused.

Let's start with bicyclobutane:

The proper name for this compound is bicyclo[1.1.0]butane. The compound is numbered starting at one bridgehead and working around. The numbers in brackets represent the number of atoms between the bridgeheads.
Now, tetrahedrane, which has an extra bridge between carbon atoms 2 and 4:

Once again, we list all of the bridges in the brackets, but we have an extra bridge, and it is not between the main bridgeheads (1 and 3). We list this extra bridgehead last and then use superscript locants to indicate which ring atoms this bridge connects. This extra bridge is between carbon atoms 2 and 4, thus $0^{2,4}$: tricyclo[1.1.0.0$^{2,4}$]butane.
For more examples, including applying the rules to crazier ring systems, here is the IUPAC publication

link at the end is dead


text:
Effect of temperature on cooking
I'm poaching some pears.  The recipe calls for bringing the liquid to a boil and adding the pears, then simmering until done, about 25 minutes.  It then calls for removing the pears from the liquid and letting both come to room temperature, then putting the liquid over the pears again and marinating overnight.  Separating the pears and the liquid will let the pears cool more quickly and presumably prevent overcooking.  I have heard a rule of thumb that every 10C in temperature doubles the reaction rate.  As the pears simmered for 25 minutes, it seems that the cooldown (even if they stayed in the liquid) will not add much to the reactions of cooking. Is there any sense to extracting the pears from the liquid?

n fact, you answer yourself to the question. When you you take off the pears from the liquid, you stop the undergoing reaction. After that, you put again the pears into the liquid overnight to let the pears impregnate of the good taste of fresh pears and reabsorb  the minerals that passed to water.

@YomenAtassi: my thought is that even without removing them, the pot will cool fast enough that the additional cooking (compared to 25 minutes at 100C) is small.

I think  you can leave one pear in the hot liquid to cool down and you will see. I think the pear will become too soft and loose.

@YomenAtassi:  Unfortunately, I have only enough for dessert tomorrow, so I can't do the experiment and followed the recipe.  But I wonder....

(Note that I'm a rather bad cook, except in the lab!)

Is there any sense to extracting the pears from the liquid?

Yes! Yomen Atassi has already explained that you actually have to consider two (distinct) processes in the recipe:

Cooking the pears for a short period of will yield the right consistency. Additionally, some transformations happening during the cooking will possible break up some cells, convert some ingredients to even more tasty ones, and maybe even facilitate the later take up of other of other compounds from the liquid. If you cook for too long, the pears will be sludge.
Marinating means that the compounds dissolved in the liquid will permeate the pear and add to the taste. Imagine that the pear acts like a spongue and sucks all the good stuff in from the marinade. This needs time.

With a two-step procedure given in the recipe, you should get the best of both worlds and I hope that the dessert was really tasty.


text:
Why do carbon dioxide and sodium hydroxide not form sodium oxide and carbonic acid?
Why do carbon dioxide and sodium hydroxide not form sodium oxide and carbonic acid?
$\ce{CO2 + 2NaOH -> Na2O + H2CO3}$
as opposed to:
$\ce{CO2 + 2NaOH -> Na2CO3 + H2O}$
This is in the context of fractional distillation of liquified air, before which carbon dioxide must be removed using sodium hydroxide.

Take a look at Standard enthalpy of reaction and you can find the answer  by yourself.

Could you elaborate, so we can say what's wrong with reasoning behind your first reacion? You know that Na2O is very basic?

I've updated my question. I do know that sodium oxide is basic, and I was expecting a conjugate acid and a conjugate base to be formed.

Sodium oxide is not a base in either the Bronsted-Lowry or Arrhenius senses.  It is technically a base anhydride, meaning that it can be hydrated to yield a base:

$$\ce{Na2O + H2O -> 2NaOH}$$

Carbon dioxide can similarly be viewed as an acid anhydride, meaning it can be hydrated to form an acid.

$$\ce{CO2 + H2O <=> H2CO3}$$

The difference is that the hydration of sodium oxide is essentially irreversible.  One program I used estimated $K_{eq}=10^{39}$ for sodium oxide hydration.  In contrast the hydration of carbon dioxide is reversible, with $K_{eq}\approx1.5\times10^{-3}$.
The reaction that you are wondering about, $\ce{CO2 +2NaOH⟶Na2O +H2CO3}$, is essentially the dehydration of sodium hydroxide by $\ce{CO2}$ to form sodium oxide and carbonic acid, that is, it is reaction 2 minus reaction 1.  But the equilibrium constants I just mentioned show that sodium oxide has an "affinty" for water that is about 42 orders of magnitude greater than carbon dioxide's affinity for water.  So that reaction is very unfavorable.  The reverse reaction would be extremely favored: Sodium oxide would be a powerful reagent for $\ce{CO2}$ removal prior to air liquefaction, at least as powerful as sodium hydroxide.
If you are curious, here is the R code I used to estimate the equilibrium constant of sodium oxide hydration, using the package CHNOSZ:
> subcrt(c("Na2O","H2O","NaOH"),c(-1,-1,2),T=25)
info.character: found H2O(liq), also available in gas
subcrt: 3 species at 298.15 K and 1 bar (wet)
$reaction
     coeff         name formula state ispecies
2031    -1 sodium oxide    Na2O    cr     2031
1       -1        water     H2O   liq        1
1157     2         NaOH    NaOH    aq     1157

$out
      logK         G         H         S         V       Cp
1 39.01742 -53229.29 -57143.24 -13.24728 -36.04979 -40.8745

On a related note, on of the commenters asked me to look up the standard enthalpy of my proposed reaction. Any ideas why?

It would be very positive, a hint that the reaction is likely to be thermodynamically unfavorable.  (I say "hint" because favorability is determined by $\Delta G=\Delta H - T \Delta S$, so in theory it would be possible for an extremely large positive entropy change to counteract the very large enthalpy change, thus making $\Delta G$ negative and the reaction thermodynamically feasible, but since the reaction consumes a gas, it's entropy change is probably negative, which would only make even more disfavored.)

$\ce{O^2-}$ is a base according to Brønsted and Lowry’s definition as it takes up $\ce{H+}$ ions to form $\ce{OH-}$.  $-1$

Good point Jan.  I was going off of Wikipedia which says A base anhydride is neither an Arrhenius base, nor a Brønsted–Lowry base, since it does not accept protons and do not increase the hydroxide ion concentration of water. However, a base anhydride is a Lewis base, since it will share an electron pair with some Lewis acids, most notably acidic oxides.  It sounds like that is not completely correct.  If someone edits Wikipedia I'll edit my answer.


text:
How to calculate concentration of hydrogel nanoparticles
I have prepared hydrogel nanoparticles in aqueous solution. How to calculate concentration of hydrogel nanoparticles ?


text:
What happens if you make a gas of magnetic particles
I was reading todays SMBC which asked what happens if you make a gas of magnetic particles

So what does happen if you make a gas of magnetic particles?

You mean ferromagnetic? I'm afraid it won't be gas, more like dust cloud maybe.

What do You understand by "magnetic"?

Hmm, supposedly this question is unclear unless you provide the background you have of what "magnetic" is. Even so, it would be more appropriate at physics.SE.

If so please migrate, since the comic doesn't define magnetism I don't know what to say, in the same way asking what happens when you pour water on paper seems self contained and obvious until someone asks what material is the container made out of. Of course the answer may be that a magnetic gas isn't possible

Oxygen is a gas and it is paramagnetic, so is nitrogenoxide and nitrogendioxide. When talking about forromagnetic properties, that should strictly speaking not possible. Aerosols, like @Mithoron suggested might be the best you can do. However, I think it fits our site, but you might get more attention at physics.se. Let it play out for a while and flag for migration, if you want to try it there.

This answer with link to supercold lithium wasn't bad but was deleted ;(

It deepens what you mean by "magnetic". As another comment says air is already 20% paramagnetic because it contains dioxygen. But of you mean ferromagnetic, then it is probably impossible in a gas as gases don't have long range structural order. A dispersion of small magnetic particles would not be gas.

@Mithoron The news from MIT mirrored on phys.org and with special recommendation Gyu-Boong Jo, et. al., Science, 2009, 325 (5947), 1521-1524. Although, I am not 100% convinced we are talking about gasses any more.

Paramagnetism doesn't sound like what the comic is asking about because it's induced magnetism. The gas isn't really magnetic, so it would be ferromagnetism. In which case the answer to this question appears to be "it would no longer be magnetic"

It surprises me looking back that people saw a comic and yet still assumed I had something more to it. Suffice to say a magnetic gas could be gaseous iron, which in hindsight the answer would be "nothing would happen as the magnetic domains wouldn't be lined up in a gas". All the calls for clarification kind of ignore that it's literally a scene in a comic

Also the book the guy is reading on the toilet gives a hint at something, as well as an opportunity to explain why plasma is not a magnetic gas. So many learning opportunities squandered

If you heat up the magnet above Curie temperature (or more drastically just evaporates it), the magnetic spins will be randomly aligned and the magnetic field  is not produced anymore (at higher temperature).

Hmmm this looks like one of those questions where we could come to a different answer if we considered a gas of marbles. (We note the ideal gas law works for marbles in cylinders over a vibrator.)


text:
Is reduction potential equivalent to the ease of discharge?
Loosely speaking, is the reduction potential equivalent to the ease of discharge in the context of electrolysis?
i.e. is it easier to discharge an ion with a higher reduction potential as opposed to one with a lower one?

Easy is a big word.
Lower potential means it requires less energy, but that doesn't say anything about how quick the reaction will proceed.
A low potential reaction may still have slow kinetics and therefore not be "easy".
If you are comparing ions as to which will produce more product as a result of an electrochemical reaction, you need to find something with low potential and fast kinetics.

I'm not concerned with the speed of reaction, I just want to know which ions will be discharge and which won't.

That is exactly what I mean. Just because something is thermodynamically spontaneous, doesn't mean it will happen. It might be kinetically so slow that it appears as not happening.


text:
Determining [CO2] of a sample in ppm
I have 20mL gas samples and I need to determine the [CO2] in ppm. I have a Li-Cor 6200 with a closed loop, so essentially I will inject a volume of gas into the loop and the change in the output is how to determine the concentration of CO2.
To calibrate I'm using a standard of gas that is 5020ppm CO2, and injecting volumes of 50, 75, & 100 uL then recording the change in [CO2] in the closed loop.
After calibration I will inject 50uL of sample and record the change of CO2 in the closed loop.
But how do I set up the conversion? It is spring break and I am having trouble finding help around campus.
This is what I have done.
Taking the average change in CO2 output from 3 different volumes of standard injected, the output changes 0.114 per uL of standard gas injected (5020ppm CO2). So, (5020/.114)=44035.0877. Then, I just multiply the change in output per uL of sample injected by 44035.0877 to get the CO2 ppm of the sample.
My first sample I inject 50uL and the output changes by 20.2, so I divide to get the change per uL; (20.2/50=0.404). Then multiply, (0.404 * 44035.0877 = 17790.18 ppm CO2)
Is it really this simple, or is my thinking to simple?
Thanks. This is my first post, I hope I have followed the rules.


text:
the chemical mechanism for deviations from Raoult's law
Mixtures of ethanol and acetone have vapor pressures that are higher than predicted by Raoult's law, while mixtures of acetone and water have lower vapor pressures than predicted by Raoult's law.  Why is this?
That is, why is there a "positive deviation" from Raoult's law for acetone/ethanol solutions but a "negative deviation" for acetone/water?
Isn't an acetone-ethanol hydrogen bond stronger than an acetone-acetone hydrogen bond? If the reason for the positive deviation of the ethanol/acetone solution is that the H bonds there are weaker than ethanol-ethanol H bonds, then there wouldn't that imply that acetone/water solutions would also have a positive deviation, as water-water H bonds must be much stronger than water-acetone, right?
So is there some other explanation?

What do you mean by "positive" and "negative" deviations?  Does positive mean that the vapor pressure above the liquid is higher than Raoult's law would predict (or vice versa)?

Yes. That is what I meant to say.

Thanks much for clarifying.  One more question: is the deviation positive for both components?  I.e. is the vapor pressure of both acetone and ethanol higher than Raoult's law would predict, or just one component?

I couldn't find any information on that. They are just listed as examples under positive and negative deviation. It is also given that water and ethanol forms a solution that deviates positively like acetone and ethanol, if that helps in any way.

Here is a related (but not an exact duplicate!) of your question: http://chemistry.stackexchange.com/questions/6955/deviation-from-raoults-law-because-of-adhesive-and-cohesive-forces

Thank you for the link. Interesting analogy. But the problem is not that I don't understand the reason for the deviations in general, but it is that I can't make sense of this particular case. Is there something more to consider? Is it more complex than comparing the A-B interactions of different solutions with common components?

I think azeotroping is also important. It's not just the interactions in liquid mixture.

@Curt F.  No, we speak of both components (i.e. The solution as a whole) when we speak about the vapour pressure of a liquid. Sure, molecules from both components will vaporize, but we speak of the whole pressure above the liquid, if that makes sense.

@CurtF. the mixture of acetone and water causes decrease in volume but there will be no decrease in volume for ethanol-acetone mixture.

I edited this question to try to improve clarity.  @DhruvaPatil, let me know if the edits represent your question well.  If they don't, I'll gladly revert them.

First some general comments about Raoults law before discussing the particular solutions in question.
Raoult’s law suggests that the partial pressure of each substance above a solution is proportional to its mole fraction x, thus $p=p^ox$ where $p^o$ is the vapour pressure of the pure substance.
Experimentally there are deviations from Raoult’s law and they can be in both the ‘positive’ and ‘negative’ direction. A positive deviation means that p is greater than expected from Raoult’s law and thus a negative deviation has pressure that is less than expected. Alcohol/water mixtures show positive deviations and for example acetone/chloroform negative deviations.
The reason for these deviations is due to the different interaction between the molecules. The change in interaction on adding one liquid to another is due to the difference in interaction energies between them, thus if the energies  between the same type of molecules of type 1 and 2 is $E_{11}$ and $E_{22}$ (which can be positive or negative) we can write the difference in energy as
$$ \Delta E = c(E_{12} - E_{11}/2 - E_{22}/2)$$
where c is a constant to account or the fact that molecule of type 1 and 2 are different.  A positive value of $\Delta E$ indicates that the interaction between types 1 and 2 is less attractive than either molecule to itself and a negative value that types 1 and types 2 will prefer to interact with the other type of molecules than themselves.
All atoms have dispersion interaction (induced dipole-induced-dipole) due to their polarisability, and in addition may have permanent dipoles, (causing dipole-dipole  and induced-dipole energies) and hydrogen bond interactions. Clearly which contribute depends on the particular case being studied.
Starting with the chemical potential and the interaction energy it is possible to calculate the partial pressure as a function of mole fraction and energy $\Delta E$ and this is
$$ \frac{p}{p^o}=x\exp(+(1-x)^2\Delta E^o)$$
where $\Delta E^o = \Delta E/kT$ where k is Boltzmann’s constant and T temperature. This function is shown in the figure for different values of $\Delta E^o$. When this is positive there is a positive deviation from Raoult’s law, a $\Delta E^o = 1$ approximately describes the ethanol/ water data and $\Delta E^o = -0.5$ the acetone/chloroform data. (The maximum $\Delta E^o = 2$ since above this phase separation occurs)

figure: calculated pressure vs molec fraction x. Positive deviation $\Delta E^o > 0$ and negative deviation $\Delta E^o < 0$. Raoult's law $\Delta E^o = 0$
Now for the particular solutions mentioned, ethanol-acetone has a positive deviation and thus $\Delta E^o > 0$. It is not really possible without some (spectroscopic) experimental data to say what interaction causes this; most likely it is slight changes of each type of interaction, but probably mainly changes in dipole-dipole and hydrogen bonding. The data indicates that acetone-ethanol interactions are less favourable than either molecule with itself.
In the acetone - water data the interaction between these two types of molecules is more favourable in such a way as to overcome the sum of water-water and acetone-acetone interaction energies. This is somewhat unusual for water as it is a highly associated (i.e. hydrogen bonded) liquid and a second component tends to break up this association.
We must conclude, however, that the acetone is able to become solvated in such a way as to disrupt the water structure but still be of lower energy than the sum of water-water and acetone-acetone interactions. Again only spectroscopic evidence will indicate what the interactions are; without this we can only speculate but it could be that the main gain comes from removing acetone-acetone interactions and replacing them by water-acetone even though the water-water interaction is favourable, losing some of this interaction must be compensated for by gaining acetone-water ones.

How is interaction energy defined/calculated?

This is where these simple thermodynamic models break down. You could find the energy by fitting the data to one of the curves, and then compare with calculation of general van-der-Waals forces between molecules, i.e. any or all of ion-ion, ion-dipole, dipole-dipole, induced dipole -induced dipole and so on as appropriate.


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What gas is released during dissolving NaOH
What gas is released during dissolving solid $\ce{NaOH}$ in a high purity water? Water more than 10 M$\Omega$/cm and standard quality $\ce{NaOH}$.

If the water had some dissolved ammonium ions in it, those could evaporate when you raised the pH by adding NaOH ($\ce{NH4+ (aq) + OH- (aq) -> NH3 (g) + H2O}$). But that seems unlikely, and in any case, the smell would be very distinctive. I can't really think of any other obvious candidates for volatile cations one might find in water.

I don't think any gas is released.  You are probably experiencing dust or mist that contains NaOH itself, or tiny amounts of gas phase NaOH.
There is related discussion at a soapmaking website
http://www.tikvah.com/medical/lye.html
See also "Pneumothorax following inhalation of caustic soda fumes" Archives of Emergency Medicine, 1988, 5, 45-47

I will check it next time.

@JaroslavKotowski You might be able to detect NaOH in the air from a bunsen burner flame turning yellow above the open vessel.

That is simply air included in the evaporation/drying process.

The gas is not possible to breath, it maks me cough. And its temerature is same as ambient.

I agree the gas is likely air.  Breathing it is a horrible idea though, because it it likely to contain aerosolized sodium hydroxide, which could reach your lungs and do some damage.  Be safe!

Thx for the answer, try to avoid it if you prepare 100 l of 5% solution in an open tank.

I think it is due to the heat of solvation of NaOH, causing the solution to evaporate. If you dissolve NaOH with a large amount of solution, the "gas" wouldn't be released.

Thanks for the comments. Based on them, here what I observed: The thank is filled with 100 l of water then $\ce{NaOH}$ is added. If it is added slowly, no bubbles or gases were observed. We need to shorten the preparation time so we stirred it intensively. During stirring, air bubbles are introduced to the solution. After their release, a alkaline mist/water dispersion is released to air - that is what I was observed. Thanks again for your thoughts.

Ammonia is possible in the small amount.
$$\ce{NH4+ (aq) + H2O (l) <=> NH3 (aq) + H3O+ (aq)}$$

This answer is in no way different than the comment by Ilmari Karonen to the original question. Also,
Welcome to Chemistry.SE. Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$.


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Is there an experiment that visibly shows the effects of metastability?
I am attempting to create an experiment to show the effects of metastability to a public audience and get people excited about science. More specifically, I am wondering if there is any sort of liquid that can mimic the effects that water had in the book Cat's Cradle.
In the book, the idea was that a shard of a more stable polymorphic crystal of ice with a high melting point could "infect" water. This crystal would quickly spread, solidifying oceans and destroying most life on earth.
The basic question I'm asking is as follows. Is there any public display I could make to show a liquid quickly solidifying from a seed crystal? It is preferable (but not necessary) for the materials to be reasonably inexpensive and require no special equipment, so as to allow for the experiment to be done on a grander scale. Additionally, the experiment should have a low failure rate (unlike using supercooled water, which is difficult to control in large amounts).

Might be able to just do this with water if you can supercool it. I've heard that if you carefully time how long a water bottle is kept in the freezer, you can make it supercooled, then freeze it by shaking. If you can make that work, try adding an ice cube instead of shaking it.

Indeed... there are numerous YT videos showing the flash freezing of supercooled water.

I specifically stated in the question that I don't want to use supercooled water.

The standard demo for this is crystallization of sodium acetate trihydrate from its melt.  The trihydrate melts at 58 °C.  The molten liquid is easily supercooled, after which point dumping in a few crystals of unmelted stuff results in a rapid solidification of the mixture.
Several videos on YouTube demonstrate the reaction; Nurdrage's is my personal favorite.


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In which country is it allowed to practice non-clandestine amateur chemistry?
I travel a bit so am flexible where I live, and have taken a recent interest in chemistry to try to gain an understanding of physical reality. I find learning from textbooks to be good, but want to have the comfort of applying a lot of this knowledge in my own lab. To see and feel how it works first hand. From my understanding, the west has a huge paranoia with clandestine chemistry, which is of no interest to me. However, it seems a vast quantity of common reagents and catalysts in organic chemistry books are banned in western countries due to this.
Would someone know how it's possible to practice amateur chemistry from home, without fear of accidentally breaking the law by using some reagent I'm not supposed to? And if not, what would be the best way to gain hands-on experience like this, where I can have complete freedom to experiment with what I want to do? I find at university I'm limited to lab sessions, and if one day I gain employment with a lab I'd no doubt not be free to conduct my own experiments. I wish to emphasize, I'm financially independent from my chemical studies, so purchasing the equipment isn't the issue. It's the legality I'm concerned about.
Is there an option to do this, to pursue amateur knowledge in chemistry, or is it illegal? Is there a country in which it isn't illegal?

There aren't many reagents that are actually illegal to possess in the US. Scheduled compounds, which are mostly drugs, you would find trouble with, and manufacturing explosives might also get you into trouble. The big problem is really not violating local zoning laws.

Interesting, thanks Jay. I was just doing a bit of research and saw some horror stories of people being monitored and raided for simply purchasing acetone and things like that, even without manufacturing anything illegal. Perhaps I'm just being paranoid. Would it be best to notify the local authorities if you're planning to setup a lab, or would this not be required do you think?

I'm voting to close this question as off-topic because Chemistry.SE is not a place to ask a legal advice.

Your question is basically too broad to be answered in full detail. I suggest that if you want a more detailed answer about a certain part, you should open a question for that matter separately. That being said, I'll try to summarize my own experience as a hobby chemist.
First of all, are you actually studying chemistry? IMHO, if this is within your possibilities, it is the best way to gain experience and also do your your own research. You will have supervision and the best equipment available. You will get the space to do research in your own areas mostly during your graduate studies, but here (The Netherlands) you will also get some space as an undergraduate.
Consider the following points (in no particular order):

You say that finance is not a problem, but do not underestimate the price of all the chemicals and equipment you will need to do any bit of serious chemistry. Scales, vacuum pumps, glassware, solvents, a dedicated fridge, a specially constructed cabinet to store flammables, a fume hood with explosion proof suction, etc (and I'm not even mentioning NMR, mass spectrometers, IR, all of which are available to you as student). To run any decent lab from your backyard be prepared to invest several thousands of dollars (or euros). This can be done of course a lot cheaper, but then you will be limited in your possibilities or you will have to exchange safety for money.
Almost every chemical you will ever need when doing a normal synthesis (i.e. no explosive making or pharmaceuticals) is legally available in most countries (I heard however that some states in the US are quite notorious w.r.t. hobby chemistry, like Texas). You are even able to obtain chemicals that are clearly dangerous and should not be sold to amateurs, like osmium tetroxide, sodium cyanide (not everywhere), gaseous hydrogen chloride in large 110L pressurized tanks. There are a few exceptions:

Common precursors and preprecursors to illicit drugs in large quantities. Most notably phenylacetone and derivatives, piperonal, piperidine, methylamine, benzaldehyde, nitroethane, etc.
Chemicals that can be used to manufacture explosives in large quantities. Examples include: nitromethane and ammonium nitrate.
Chemicals that are severely dangerous to the environment or can be easily abused to poison people, although oddly enough some of the most hazardous are freely available. Examples are: elemental mercury, a lot of (organo) arsenic and other heavy metal compounds, a lot of radioactive compounds, some cyanides (may be available though from mining suppliers), etc.

Note: you are able to obtain most of the above mentioned chemicals (also in larger quantities) if you start a business and have it legitimately registered to the Chamber of Commerce. This business does not actually have to do anything, it can function as a facade. It still won't work for pharmaceutical precursors or very large quantities of chemicals however, but for a hobby chemist that is mostly no problem.
Note 2: At least here in The Netherlands some pharmaceutical precursors and preprecursors can be legally owned and synthesized, they are just not available commercially because of the high change for abuse. These chemicals however (e.g. phenylacetone) can be used in many other legal synthesis and hence it might be useful to synthesize them yourself.
Be prepared to spend a lot of time on this hobby, both studying theory and doing practical work. Independently doing (serious) chemistry is not easy and you will have to rethink and adjust your setup a lot of times. Also the possibility of failure or accidents is higher without supervision. It is not entirely without risk.

I would add that hydroiodic acid is also a controlled chemical.  Many others are on the list too, but HI is so simple and easily/accidentally prepared (just mix potassium iodide with an acid!) that it's worth knowing that its essentially illegal to do so in the US [unless you have clearances etc.].

@CurtF. I can buy HI here no problem. Albeit a bit expensive (€339,82 a liter). I guess Europe and the US are different what regulation concerns. Mostly I would not worry about legal problems unless you actually try to make explosives or illicit drugs. Last tip to add is of course that in case of doubt there is nothing against contacting the local police and inform about the regulation - if you really want to be sure.

In the USA there is a list of 41 special federally regulated chemicals.
Actually, there are two lists, list 1 having 30 chemicals and list 2 having 11 chemicals.
Many of these are common lab chemicals including:
Acetone
Ethyl ether
Potassium permanganate
Toluene
Hydrochloric acid
Sulfuric acid
Iodine
If you are planning to use these, I would recommend studying
Title 21 parts 1300-1321 to make sure you are in compliance
There is also the Extremely Hazardous Chemical list that has seperate reporting requirements.
There is an EPA List of Lists that tries to catalog the laws and regulations that apply to various chemicals.
There is a relavent article Underground Science Chemistry hobbyists face a labyrinth of local and state regulations in Chemical and Engineering News Vol. 86, pages 38-40.

I never understood that list of extremely hazardous chemicals. It contains some very commonly encountered chemicals like ammonia, chloroform, cyclohexylamine, phenol, nitrobenzene... We had to work with nitrobenzene and phenol in our very first practical as an undergraduate (lol). I understand that they are nasty, but I would reserve the term extremely hazardous rather for things like osmium tetroxide, ozone, carbon monoxide, etc.

@Jori I think that legislation was in response to the Bhopal tragedy, so it is from an industrial scale point of view


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How much does the quality of glassware matter, really?
Discussing about the pros and cons of glassware from different manufacturers for chemistry experiments easily becomes a heated debate. Well, that's how it seems if you read the chemistry related blogs, discussions, questions, etc. over the internet. How much does it really matter what kind or type of glass one is using? How big the differences are across different manufacturers?
It would be nice if the answers also elaborated on the differences between laboratory and home chemistry.

Great question but it is rather ambiguous, since there are window glasses and glass storage's, glassware for food and appliances and building things as well. Each one of them is made differently to accomodate to different needs. The quality of the glassware, essentially, depends on the use of the glass. I can go on and list different types of applications for glass. Check here for more information.

The question says for chemistry experiments. I suspect that it is a rare chemistry experiment that deals with making windows.. Though, we can restrict the question to educational and research chemistry experiments.

The difference that I see between different manufacturers is the thickness of the glassware. From one company, it is about half as thick as it is from another (and about 10 times more fragile).

@per, do you mean the quality of the glass material itself, or the grading of the glassware (e.g., grade A versus grade B volumetric flasks or graduated cylinders)?

In addition to differences in the absorption spectrum, different types of glass are more or less able to deal with temperature or mechanical shocks. But that is less about manufacturer: you'll find different types of glassware from the same manufacturer for different purposes (e.g. Fiolax and Duran are both by Schott).
As @LDC already commented, some glassware is thin walled and more mechanically fragile. OTOH, thick walled glass is more likely to break with a temperature shock. So e.g. for test tubes you have to decide whether you want to have them

suitable for local heating => thin wall borosilicate glass
suitable for centrifuge => thick wall borosilicate glass
cheap => soda glass
UV transmittant => quartz
suitable for huge temp shocks => quartz

If this question is about whether your borosilicate glass ware is called Duran or  Pyrex (or Jenaer or Ilmabor or Rasotherm) I don't see that much of a difference.

I am not sure about other brands, but Pyrex made in America is actually soda-lime glass. Anywhere else it should be borosilicate. So, if you get Pyrex and want high quality, buy stuff manufactured outside of America.

@ChemBird You're talking about Pyrex kitchenware. The consumer products portion of Corning was spun off, and Pyrex brand kitchenware is now made by a number of different (non-Corning) companies (with the US one using soda-lime glass). Corning held on to the Pyrex trademark for labware, though, and still uses borosillicate.

You've likely heard of the common and useful reagent, diazomethane $\ce{CH2N2}$, which methylates carboxylates to their esters without acidic conditions, forms carbenoids and therefore cyclopropanes and other structures. You have almost certainly been advised about that reaction as such compounds give even careful experimentalists pause about the quality of glassware used and the cost-benefit of using it.
New glass of the highest density and engineering is requisite. Even finely scratched, apparently new glass, regardless of temperature tolerance, shock and so on will not only ruin your diazomethane reaction but also potentially rob you of your sight, hearing, several digits, and appendages and can end your life in an instant. A trace of water or even causing a dilute solution to freeze and form crystals can trigger detonation. Consider spilling it..!
Small imperfections in the wrong sort of glass surface may not be visible but can cause detonation [1]. In a closed vessel, violent explosive discharge in all directions have caused many to pay with their lives for the cavalier attitude I hear echoed in the QA about warnings of equipment quality and handling, especially glassware. There are safer variants but a "home" (your term) chemist won't have access to that.
How much does it matter? Does your life matter? Ever run an ozonolysis reaction? Isolating an intermediate molozonide prematurely can cause similar destruction to the careless whether using low or high thickness but inferior quartz or glass or even steel.
A sealed glass tube with a hard cap are unusually thick due to the temperature extremes they were designed to tolerate in dewars and heating vessels. But these extreme uses can have extreme outcomes.
K. Barry Sharpless, Nobel laureate was making his rounds around an MIT lab in 1970, sans eye cover. When a student of his pulled a sealed-tube reaction out of liquid nitrogen to display it, the rapid warming of the contents in ambient air did not leave any time for Prof. Sharpless to evade the barrage of miniscule glass shards that expanded outward from the rapidly warming solution at tremendous force, enough to shatter. He (Sharpless) did not lose his life (he would've otherwise missed his share of the recent Nobel prize for his work in catalytic asymmetric induction. The prize isn't awarded posthumously). Instead, he lost an eye.[2]
Use these chemicals with "home-ware" and cheap glass, in your trailer (I kid you, sort of), with no hood, no blast shield, coat, thick gloves and expect a short life for you and anyone unfortunate enough to be near you.
1) Handbook of Reagents for Organic Synthesis – Reagents, Auxiliaries and Catalysts for C-C Bond Formation, edited by Robert M. Coates and Scott E. Denmark, John Wiley & Sons, Inc., 1999, ISBN 0-471-97924-4
2) How dangerous is chemistry? Nature 441, 560-561(1 June 2006) doi:10.1038/441560a

Obviously it depends what you are doing.
I often use glassware that is kitchen quality and comes from kitchen suppliers because it is way cheaper. You can get perfectly good graduated Pyrex beakers with handles from kitchen suppliers at 1/10th the cost of what it would cost to buy it from Cole-Parmer. This is true of many other everyday items.
On the other hand, I have had bad experience with made-in-china soda glass chemware. Often the items are thinner than normal and tend to break easily. Once I bought a lot of test tubes that were sourced in China and they are very fragile and break easily. Better quality glassware has thicker walls so it is more resistent to breakage.
Another issue is annealing. With borosilicate and other tempered glasses, they need to be annealed to be shock resistant. If you buy from a sketchy supplier the items might not have been annealed properly, which will make them more likely to fracture in use. If you are dealing with hot acid or items like that, it is important to have rugged, shock resistant glass.
Unfortunately, it is hard to tell the differences in quality without a lot of experience. One manufacturer can make good items one year, but then start skimping and process lower quality items the next. Also, note that certain manufacturers are known for certain items and those items will often be much higher quality than their non-specialty stuff which is just there to "round out" their catalog. For example, (to use an analogy from hand tools) Stanley is known for their hammers, so you would not be advised to buy a drill from them. Same principle for glass blowers.

The simple glass is not recomended to use in laboratory IF you are using substances thay may emit Ultra Violet light, because the glass absorbs it and we know that that kind of light has a really high energy. Therefore quartz ( silicon dioxide ) is more suitable. And of course, radioactive elements will need special conditions.

Quartz also helps if you want to deliver UV light into the glassware (photochemistry).


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Fluorescein synthesis
What is the mechanism of the fluorescein synthesis performed by von Baeyer?

just google "Fluorescein synthesis, resorcinol"

This reaction is considered a modified Friedel-crafts acylation. The reaction is formally a series of electrophilic aromatic substitutions, with hydroxyl leaving groups picking up protons (1), however the Lewis acid catalysis obviously plays a role in the mechanism. I think it increases the electrophilicity of the phthalic anhydride in the first step, and facilitates the elimination of the hydroxyl groups. For the mechanism without the involvement of the zinc(II) chloride, see (1). Here is a mechanism emphasizing the role the Lewis acid might play in catalyzing the reaction:

This reaction works in a similar way to the Nencki reaction (2)
In order to get the form of the molecule shown as the product of your reaction, the $\mathrm{pH}$ needs to be increased in order to open the lactone ring.(3) This is fluorescein-A, the form of the molecule that actually fluoresces.

References:

Jag Mohan, In Organic Analytical Chemistry: Theory and Practice; Alpha Science International Ltd.: Pangbourne, England, 2003, p. 497 (ISBN: 1-84265-119-6).
M. Nencki, N. Sieber, "Ueber die Verbindungen der ein‐ und zweibasischen Fettsäuren mit Phenolen," Journal für Praktische Chemie 1881, 23(1), 147-156 (https://doi.org/10.1002/prac.18810230111).
Clifton J. Stephenson, Ken D. Shimizu, "A fluorescent diastereoselective molecular sensor for 1,2-aminoalcohols based on the rhodamine B lactone–zwitterion equilibrium," Org. Biomol. Chem 2010, 8(5), 1027-1032 (https://doi.org/10.1039/B918823A).


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Possible Synthesis of MVK from 2-Butanone?
My professor wanted me to set up a lab to synthesize Methyl Vinyl Ketone (AKA Butenone) from some MEK we have lying around (2-Butanone).
Now, I know there are multiple ways to do this, but the tricky part is finding out a safe way to go about this as MVK has a very low flashpoint.

I figured we could do an acid catalyzed alpha halogenation with Bromine and some Acetic Acid, and then follow up with an Elimation using a bulky base such as tert-butoxide.
The only problem is I have no idea how long, how much, or how cold/warm these materials need to be for the reaction to occur.  Ideally the entire reaction would take place in an ice water bath due to the danger, but at the same time I don't want to take five hours synthesizing one compound.
Let me know what you think, and feel free to recommend a different (but safe) route of synthesis, as our lab has plenty of chemicals to work with.  By the way, we only need to create about 10-20ml of the stuff, but more is better.

The safest way is just to buy it. ~$120 for 100 mL.

Btw, welcome to chemistry.SE! And yeah, I don't have much insight into the topic, but I think I agree with Jerepierre. Purchasing is indeed safer than synthesizing, and probably cheaper.

Ha! Yeah that's the thing, our current supplier cannot get us the stuff until May, and we kind of need it right now.  So ideally we could make it in lab this week.

It can never be made "safe", as methyl vinyl ketone is itself extremely toxic.

@user97462 - Why not look for another supplier? There are a number of ways to look for suppliers. For example, ChemExper finds 103 possible suppliers in multiple countries.

I understand that, but the entire reaction would take place in a fume hood and in an ice bath.  What I want to know is if acid catalyzed alpha halogenation works at such low temperatures.

BTW, I cannot change my school supplier, they are heartset on the one they already have.

The flash point of MVK is -7 °C, that of MEK is -9 °C. Note that the flash point of diethyl ether is -45 °C. Running reactions under nitrogen or argon, using an "Intensivkühler" (combination of Liebig and Dimroth condenser) is pretty safe. As far as the toxicity of MVK is concerned: It's the same, regardless whether you synthesize or buy it! But as pointed out by jerepierre and Ben Norris, any attempt to synthesize MVK form left-overs in the lab is more like an occupational therapy and not really reasonable.

A literature method for the laboratory synthesis of MVK can be found in White & Haward, "The preparation and polymerisation of Methyl Vinyl Ketone" (1943).
The synthesis is via the acid catalyzed elimination of water from 4-hydroxy-2-butanone with conc. phosphoric acid. They synthesized the 4-hydroxy-2-butanone from acetone and paraformaldehyde polymer, but it is commercially available now. You can find extensive details of the experimental method in the text of the paper.

As pointed out in the comments, there is no way to make it safe, as methyl vinyl ketone itself is a very nasty compound.
I am also not an expert on organic synthesis, but I found some procedures, that might help you in adopting a synthetic route. The one you suggested might work, but be aware, that you are dealing with a methyl ketone, which can undergo the haloform reaction quite easily.
For example, in [1] they describe the bromination with bromine and absolute methanol at the methyl group for 3-methyl-2-butanone at a nice moderate temperature range 0-10 °C. The yield is 95%. But this is not what you want. Adding an acid as catalyst might work in your favour though, as after protonation the partial positive charge is better stabilised at the higher substituted carbon.
In [2] they use N-bromosuccinimide and aqueous hydrogen bromide to introduce the halogen in α position. They convert hexanoic acid with thionylchlorid to the acid chloride and subsequently to 2-bromohaxanoylchloride. The reaction takes place in the also quite nasty carbon tetrachloride at about 85 °C. This might also work for your system, it's a bit warmer than you expected though.
In [3] they use sulfuryl chloride in carbon tetrachloride to convert pantane-3-one to 2-chloropentane-3-one at about 45 °C with about 80% yield. How selective this will work with an asymmetric ketone is not clear.
Once you have obtained the 3-halogeno-butan-2-one, things will not get easier. I found a patent [4], where they convert 3-chlorobutan-2-one via pyrolysis with sodium acetate at 500-550 °C.
I could not find a reference for an elimination, but that should also work as long as you use a bulky base.
Wikipedia [5] has a different route for the synthesis, which might be a bit more economical and easier.
Well, good luck!

Org. Synth. 1976, 55, 24.
Org. Synth. 1976, 55, 27.
Org. Synth. 2002, 79, 251.
J. R. Long: Preparation of methyl vinyl ketone. US 2256149 A, 16. 09. 1941.
http://en.wikipedia.org/wiki/Methyl_vinyl_ketone#Production

Haloform reactions only happen in base according to Clayden et al. OP is using a acid catalyst.

@Jori I included this as a courtesy warning. What is your suggestion to improve the answer?

Wow thanks Martin, that is a seriously good answer, thanks for the help!


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In what ways (other than color) does black Aluminum Oxide differ from white?
I work in a (physics) lab in which an aluminum oxide coating is applied to the inner surface of a plasma confinement vessel (made of Cu) to provide an electrically insulating layer that can withstand high temperature environments. The product we use comes in a medium that allows it to be brushed or sprayed onto the metal surface (something like: http://www.zypcoatings.com/aluminum-oxide-coatings/  -- see the data sheets).
I'm wondering if a similar product might exist for black aluminum oxide. I've seen that black aluminum oxide powder exists, but cannot find information on a coating product that is black.
I'd like to understand what the difference is between the two: Does black aluminum oxide have the same physical properties as the white variant? Is it reasonable to expect that a similar suspension or solution of black aluminum oxide in water or a different solvent could be made? What makes one black and the other white? Any information on their differences would be helpful.
Thanks!

The aluminium oxide that appears black seems that way because of the small particle size; the very small powder size promotes lots of light scattering. I would think that suspending fine particles would be easier than larger ones, it just might not be cost effective to use the very fine powder to make the coating solution for commercial sale vs custom order.


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Is there a general / simple formula for Slater's rules?
I just learned how Slater's rules work on Wikipedia. These rules really are very simple. But the presentation of the rules seemed not very efficient. I would think there would be someway to set up and equation of some sort such that all you have to do is plug in the relevant quantum numbers and the element number and use some sort of formula to just compute Slater's rules more easily.
Is there such an equation / formula? If so, I'd like to see it. It would be more convenient than memorizing the rules in verbal form.

Sadly, no. It wouldn't be too hard to program a "get Slater shielding" script. But it wouldn't be terribly useful.
First off, let's back up briefly. Slater's rules allow you to estimate the effective nuclear charge $Z_{eff}$ from the real number of protons in the nucleus and the effective shielding of electrons in each orbital "shell" (e.g., to compare the effective nuclear charge and shielding for say $3d$ vs. $4s$ in transition metals).
Slater wanted to use these for quantum calculations, such that the radial wavefunction would be something like:
$$\psi_{ns}(r) = r^{n-1}e^{-\frac{(Z-\sigma)r}{n}}$$
So the rules he devised were fairly simple and produced fairly accurate predictions of things like the electron configurations of transition metals and ionization potentials.
Later, others performed better optimizations of $\sigma$ and $Z_{eff}$ using variational Hartree-Fock methods. For example, Clementi and Raimondi published "Atomic Screening Constants from SCF Functions." J Chem Phys (1963) 38, 2686–2689. (That article covers elements up to Kr.)

It's pretty obvious that no simple formula will cover these curves.
These days, basis set optimization is done by computer, and I doubt many people even glance at the exponents.
In my mind, Slater's rules serve two main purposes at this point:

They're relatively simple to teach and learn and allow quick qualitative predictions, particularly for electron configurations in the transition metals.
They emphasize that exponents for multi-electron atoms must be empirically derived.

I still teach them, particularly, for the first reason. To a lesser degree, they also reinforce the idea of effective nuclear charge and provide some semi-quantitive basis for that concept.

So basically it is a slightly verbose / tedious rule of thumb that while very useful isn't physically accurate enough to be used for things requiring high precision. Correct?

I wouldn't say it's tedious. You can generally get the shielding constant in a minute or two. Considering he published in 1930, it's exceedingly good. Check the optimized exponents by Clementi, which aren't that different. Not many things in science stand up to 85 years.

But no, for high precision, no one uses Slater rules because no one uses minimal basis sets like this anymore. With Moore's Law, I can probably do QM on a smart watch.

Fair enough. I used it today. Maybe it just felt tedious for the 3 minutes it took me. I have never worked with minimal basis sets. Do you know a good introductory textbook that discusses them? I googled it, but I'd be interested in learning about them and related topics at a similar complexity level. I'm more familiar with physics, but I read chem too.

I'd recommend inorganic chemistry textbooks. We use Miessler and Tarr's Inorganic Chemistry which covers this.


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Is there a list of known rate laws for reactions?
I've been scouring the web for a website/pdf that lists/tabulates reactions and their corresponding rate laws. Something like:
\begin{align}
\ce{A + B &-> C} & \text{rate}&=1.234[\ce{A}]^2[\ce{B}] &&\text{at } 25~^\circ\mathrm{C}\\
\ce{D + E &-> F} & \text{rate}&=5.678[\ce{D}][\ce{E}]^2 &&\text{at } 25~^\circ\mathrm{C}\\
\ce{G + H &-> I} & \text{rate}&=9.001[\ce{G}][\ce{H}] &&\text{at } 25~^\circ\mathrm{C}\\
&\cdots&&\cdots&&\cdots\end{align}
Is there such a list of known rate laws for reactions?

If this exists, I would be euphoric.

Like this? http://kinetics.nist.gov/kinetics/index.jsp Though it seems there is no data for reaction in the liquid phase, which is probably what most chemists are interested in.

@NicolauSakerNeto That's similar to what I'm thinking, but I'm looking more for a ready-made list of the rate laws, i.e. not requiring you to enter the reaction.

Integrated Rate Laws in Chemical Kinetics are very well known (http://www.csus.edu/indiv/m/mackj/chem142/kinetics.pdf).
Here is a good article with the list of the constants: Table of Recommended Rate Constants for Chemical Reactions Occurring in Combustion by Francis Westley

How does the first link answer the question? It has 22pages of a PPT and only a handful of examples are given, the rest is just textbook theory on kinetics.

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


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How exactly do alcohol molecules dissolve in seawater?
The book "Lehrbuch der anorganischen Chemie" by Arnold F. Hollemann on page 47[link here][1] states that dropping 0.8g Alcohol (M = 46g/mol) in a sea or ocean would result in every litre seawater containing 8 molecules alcohol.
This completely baffles me. How exactly does this happen?
[1]: https://books.google.de/bookshttps://books.google.de/books?id=eGkvSDAqY9gC&pg=PA47&lpg=PA47&ots=MFR61amS0x&focus=viewport&dq=ein mol alkohol in den weltmeeren&hl=de&output=html_text

Here is my calculation.
The ocean has a volume of 1.3 billion cubic kilometers or 1.3e+18 cubic meters or 1.3e+21 liters. 0.8/46=0.0173 moles of ethyl alcohol x 6.02e+23 molecules per mole = 0.1047e+23 molecules, which if dissolved evenly in 1.3e+21 liters would give 10.47/1.3 or approx. 8 molecules of ethanol per liter.


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What is the reasoning behind the increasing pKa for organic compounds?

Arrange the following according to increasing $\mathrm{p}K_a$.

Cyclohexanol
Phenol
Cyclohexane carboxylic acid

Please tell me the logic along with the solution.
I am confused between phenol and cyclohexanol because i guess phenol is resonance stabilized-the phenoxide ion

No its not a homework question.Chemistry is my favourite subject.I am just practising organic chemistry myself before the new session begins and i got stuck with this question.

Even it is not homework for you, the policy for these kind of questions remains and it is very likely that this gets closed. Please read through the links I posted earlier, more detail is given there. For starters, why don't you look up the $\mathrm{p}K_a$ values and include them in the post, so that people willing to help you won't have to do it. Here is a table that might help.

Cyclohexanol is an alcohol. An alcohol contains an $\ce{OH}$ group connected to a tetrahedral carbon.
Phenol is slightly different; it contains an $\ce{OH}$ group connected to a trigonal planar carbon that is part of an aromatic ring.
cyclohexanecarboxylic acid contains a very different functional group, carboxylic acid.
Cyclohexanol has a $\mathrm{p}K_a$ of about 18. It is less acidic than water. It can give up a proton, but the proton is much more likely to be bound to the oxygen than disscociated. Phenol has a $\mathrm{p}K_a$ of about 9 and is considered mildly acidic. cyclohexanecarboxylic acid has a $\mathrm{p}K_a$ of about 4.9; its acidity is similar to hydrofluoric acid.
Let's compare stability of the conjugate bases of cyclohexanol, phenol and cyclohexanecarboxylic acid:

In cyclohexanol, the anion formed by loss of a proton is localized on the oxygen. There is no resonance stabilization.
In phenol, the anion formed by loss of a proton is delocalized. Resonance structures show the negative charge can be shared between the oxygen atom and three of the carbons in the benzene ring.
In cyclohexanecarboxylic  acid, the anion is also delocalized. This time the negative charge is shared between two different oxygen atoms.

Resonance delocalization plays a clear role in stabilizing the conjugate base formed after loss of a proton. Despite the similar bond polarity in cyclohexanol and phenol, the proton is much more tightly bound in cyclohexanol. The greater stability of the phenolate anion compared to the cyclohexanoxide anion makes it easier to remove a proton from phenol than cyclohexanol.
The atoms onto which the charge is delocalized still plays a major role in determining stability. Delocalization onto an additional oxygen atom may be more stabilizing than delocalization onto several carbons.
For more details, please visit: http://employees.csbsju.edu/cschaller/principles%20chem/acidity/acid%20distal.htm


text:
How to rationalise the trend in the covalent radii of the transition metals?
Why do the covalent radii of transition series decrease at the start rapidly, then become almost constant and at the end of series begin to increase from left to right in periodic table?


text:
why colour of CuSO4 and CuCl3 is different?
These two are simple ionic compounds , not complexes [if I do not consider water of hydration] then how can I rationalise the difference in colour?

Cu3+ is still a different color than Cu2+, isn't it?

@Lighthart Honestly, I've never seen copper(III) in solution. Have you? It's not that copper(III) wouldn't exist, such as in ores, but as a naked, solvated cation?

Do you really mean $\ce{CuCl3}$ or is it just a typo and you thought in $\ce{CuCl2}$? On a second note, copper(II)sulfate itself undergoes a colour change when converting the pentahydrate to the anhydrous form.

LIGAND FIELD THEORY.  Colors in these ions are based on electron transitions between non-degenerate d-orbitals. The energy of these transitions (thus the color) is a function of the metal oxidation state, the type of ligand/counterion, and the molecular geometry. Hydration or not d-orbitals are still going to distort based on those factors.


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how to calculate the gas volume per purge when I change the cylinder?
I have some gas cylinders in a gas cabinet. Every time I change the cylinder, I use N2(100%v/v, 60L/min) to purge the residue gas in the pipe to a local scrubber. I want to know how many gases I should treat per purge?
follow is my customer give me the condition：
gas name                 flow rate         pressure(MPa)
F2(5%F2,
Blance Kr/Ne 95%)      max 4.4L/min         14.8
BCl3                    same↑                0.0627
WF6                     same↑                0.04
ClF3                    same↑                0.0588
N2                       60                   ─
Here is someone give me the solution：
100% gas volume (L/purge)= 0.07(L)*[pressure (pa)/98066.5(pa)]*1(cylinder/purge)*concentration%
How come?
In my thought, it use the formula P1V1=P2V2, where V2 is what we want to calculate. P1 is the gas pressure in the cylinder, P2= the pressure in the pipe.
But 0.7(L)? Why? And why P2 is not 1 atm = 101325 (pa) but use 98066.5(pa)=1kgf/cm^2
BTW, the pressure(MPa) value let me confused. Generally, how much pressure is the gas cylinder for each kind of gas above? The value is closed to the pressure of a new cylinder or to a used-up cylinder?


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when do we indicate the "+ or -" sign in uncertainty
how can we know when to indicate "+ or -" sign in uncertainty calculations because I noticed some problems has been solved by this sign and some has not.

Can you give us examples of what you mean by some problems have been solved by this sign and some have not? As far as I am aware, if you have uncertainty and you know what it is, you indicate it.

To me uncertainty is + AND - anyway...

when calculating the uncertainty of more that one problem .. I noticed that some final answers involve the sign of + or -
to illustrate, when calculating uncertainty in position there is NO "+ or -" sign .. while in calculating uncertainty in momentum, there is.

it's just because when indicating uncertainty, it's implicite. So if there is no signs, it means +/- anyway.

You either use both "+" and "-", $\pm$, or neither in which case it is implied that you mean $\pm$.
Usually uncertainty is expressed in a way that is symmetric about the mean, such as $123\pm4$, meaning to a certain degree of confidence the true value is between 119 and 127.
However, you could write $123^{+5}_{-3}$ to indicate the best estimate of the true value is 123 and to a certain degree of confidence the true value is between 120 and 128.
For example, if you were highly certain that an object is a cube, and measure its edge as 100 $\pm 20$, then it might be better to express the volume of the cube as
$1.0^{+0.7}_{-0.5} \times 10^6$


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Diluting down solutions method
I'm designing an experiment and it involves testing various concentrations of solutions, and I will be testing 5 in total. I proposed to make up a stock solution, and distribute it evenly in 5 different beakers and add different amounts of water to each solution but my teacher doesn't like this method and says I should simply dilute down but I don't know what he means. So if I had a 1.00 mol/dm$^{-3}$ solution and wanted concentrations of 0.800, 0.600, 0.400 and 0.200  mol/dm$^{-3}$, how do I dilute it down? To be honest, the concentrations don't really matter, it's just the method of making them - would it be easier to dilute them by half each time? Each solution is going to be used up so I can't simply reuse them...
Ultimately I will need 200ml of each solution in the experiment design, and the precision of the concentrations is not very important. How can I dilute one stock solution to 4 other solutions?

It depends on what degree of accuracy you need and what order of magnitude the final concentrations will be.
If you need high accuracy and concentrations are in the range you are describing, I would put a class A volumetric flask on a balance, add the appropriate weight of solute, add water to not quite the full volume, dissolve the solute, and add water to the volumetric line, seperately for each solution.
If you need concentrations much lower than what you are saying, when weighing becomes your largest source of error, then it is more accurate to make a concentrated solution, and the dilute the concentrated solution.  For example, you would use a class A TD (to deliver) volumetric pipette to measure the concentrated solution into a class A volumetric flask.

Thanks @DavePhD . My degree of accuracy does not really matter at all, as this is just an experiment design. It's more about having 5 solutions of different known concentrations. How can I use a 1 mol/dm^3  solution to make 4 other solutions then? Ultimately, I will need 200ml of each solution.

@Jim "known concentrations" and "accuracy does not really matter at all" are opposites.  You could take a graduated cylinder, fill to 160ml with concentrate, add water to 200mL, then repeat for the next solution using 120mL concentrate, etc.

+1, but first add some solvent, then the solute or stock solution (the concentrated solution), then solvent to the mark.

@DavePhD already gave a good description of how to prepare the solutions.
Here are possible good reasons why the teacher doesn't like your original proposed method:

distribute evenly

does not work in practice (unless you mean: prepare a bit more, and then pipette 5 equal amounts).  But if you pipette, there is no reason not to pipette the amounts of stock solution that directly lead to the neeeded volume for the dilutions.

beaker

is no volumetric glassware, and in addition not the preferred glassware for mixing (though mixing with magnet stirrer is OK). Volumetric flasks are much better: you can close them and then thorough mixing is possible. If you don't have volumentric flasks, Erlenmeyers are better than beakers.

would it be easier to dilute them by half each time

That would lead to an exponential series of dilutions, like 1 mol/l, 0.1 mol/l, 0.01 mol/l etc. instead of the equispaced concentration series you describe. Also, serial dilution means that concentration errors build up: each dilution step adds error.

 Ultimately I will need 200ml of each solution

With volumetric flasks, you can directly prepare 250 ml of each solution. Your proposed method would lead to varying volumes. It is unlikely that the proper volumes are available as volumetric flask.
Calculating the needed volumnes for the volumetric flask + pipetting stock solution or the graduated cylinder version is easy. A plausibility check of this can probably be done on the fly in the lab without calculator.
Calculating the needed additional volume for your method is somewhat more elaborate. This is of course a bad reason if it  is only about laziness. But it is a good reason if it is about doing plausibility checks while doing lab work.
Last possible reason: working with volumetric flasks thorough mixing is easier and the volume is anyways more precise than with a graduated cylinder. (Far more precise if you are still learning the practical skills of lab work: with the graduated cylinder you neet to get both volumes right at the first attempt.
With a pipette getting the volume right is much easier. So only the flask volume needs to be right at the first attempt - and the flask has a neck that helps with precise volumina.)


text:
How it is possible to make Aqua Regia in the lab?
How to make "Aqua Regia" in the lab conditions? Which lab equipment must be used?

Typically, 1 part of concentrated nitric acid is added to 3 parts of concentrated hydrochloric acid.

Personal protection (goggles, gloves, lab coat) is vital.
Prepare and use it under a well-vented fumehood only - hands inside the fumehood, head outside!

UPDATE
The comment of Crystal Lettuce is very important and points to a crucial aspect:
Aqua regia isn't a mixture to keep bottled on the shelf: it has to be prepared and used immediately!

I would like to add: The vessel you prepare it in, should not be closed for several hours after combining the acids. A large amount of gas is produced which led to accidents in the past.

Indeed, it should be prepared in small amounts and only what's needed for a particular procedure (or cleaning). Even after use, the remaining material is highly potent.

There are also multiple guides to handling aqua regia safely. Notably, use only glass containers, funnels, etc. and add the nitric acid slowly to the HCl in the hood. The acids will become very hot. Neutralize with weak base (e.g., sodium bicarbonate) slowly.

I want to add that aqua regia will eat most latex and nitrile gloves. You will want some think butyl gloves for this.


text:
Saltpetre in the Garage
My neighbour warned me that saltpetre (potassium (sodium) nitrate) is danger in the certain conditions. Is it possible that it can explode in my garage?

Darius, is this about salpeter crystals "growing" on the walls of your garage or about bags of fertilizer?

Potassium and sodium nirate can be used to make black powder (gun powder), but they are not explosives by themselves. It is safe to store bags of salpetre in the garage.
If there's salpetre forming on the brickworks, I'd rather be concerned to keep the bags closed and away from that wall: sodium nitrate is rather hygroscopic and salpetre crystals on and between the bricks indicate that the wall is wet.


text:
Effect of effective nuclear charge increase on the stabilization of 2s and 2p orbitals
Are 2s or 2p orbitals more stabilized when going from left to right in the period? There are good arguments for both possible answers:
1) 2p orbitals are more stabilized because they penetrate less in the space of the 1s orbital, so the increase of the effective nuclear charge affects them more. When plotting the 1st ionization energy with the increase of the atomic number, it can be seen that the ionization energy increases with the increase of the atomic number, but the slope is bigger for O-F-Ne (3.97eV) than for B-C-N (3.12eV). 3s and 3p orbitals aren't stabilized so much (because they both penetrate into the space of the inner orbitals), therefore it is much easier to get a high oxidation state of sulfur than of oxygen.
2) 2s orbitals are more stabilized because the electron spends more time closer to the nucleus, hence the orbital is more affected by the increase of the effective nuclear charge. Proof: the MO diagrams for B, C and N differ from the ones for O, F, and Ne, because s-p mixing in B, C and N happens as the 2s and 2p orbitals are similar in energy. The energy difference increases towards the right (because 2s orbital is more stabilized), so 2s and 2p can't mix efficiently in O,F and Ne. If the 2p orbital was more stabilized, then the difference between 2s and 2p would be smaller through the period.
I really need some clarification here.

is this different from your previous question http://chemistry.stackexchange.com/questions/16522/periodic-trend-in-difference-of-energy-between-the-s-and-p-orbitals?rq=1

The answers are in accordance with the 2nd answer I gave here, but what about the 1st answer? How to disregard the arguments?

2s orbitals are stabilized more than 2p orbitals by the effective nuclear charge because of better penetration.
Everything you wrote above is correct except for,

2p orbitals are more stabilized because they penetrate less in the
  space of the 1s orbital

It is correct that they penetrate less, but they are not stabilized more.  Because they penetrate less, they are stabilized less by the effective nuclear charge.  Electron penetration correlates with electron stabilization.
Edit: Response to OP's first comment

explain then why 2nd period non-metals can't easily form higher
  oxidation states while 3p elements can?

First off, second period non-metals can form higher oxidation states, for example $\ce{NO3^{-}}$ where the nitrogen is in the +5 oxidation state.  But I understand what you're getting at, and the answer comes back to the difference in effective nuclear charge that 2p and 3p electrons feel.  3p electrons are further from the nucleus and screened by all of the 2s and 2p electrons.  3p electrons are therefore held more loosely than 2p electrons.  This leads to lower electronegativities,

image source
lower ionization potentials

image source
and higher oxidation states being more common for third period non-metals compared to second period non-metals.
Edit: Response to OP's second comment

3p electrons penetrate well into the space near the nucleus (link)
  Therefore it
  should be easier for 2p elements to form higher oxidation states than
  3p elements, but it's not, how?

Your link (Rose-Hulman) to the 2p-3p radial electron distribution is good, it shows that the 3p electrons are "on average" further from the nucleus than 2p electrons - which is the key point.  Here is a 3-dimensional representation of the same thing.  Look how much further away the 3p electrons are from the nucleus than the 2p electrons.  Given that the 3p electrons are 1) better screened (more interior electrons) and 2) much further away from the nucleus than the 2p electrons, it will much easier to remove more 3p electrons than 2p electrons and achieve higher oxidation states.  This is consistent with the electronegativity and ionization potential arguments presented in my first edit.

Edit: Response to OP's third comment

the effective nuclear charge has bigger increase for 2p than 2s
  (except N-O) according to this chart. How so?

The key here is that the s orbital is spatially symmetric, it has a spherical pattern.  An electron in an s orbital can screen outer electrons from the nucleus equally in all directions.  On the other hand, a p orbital is directional, it is not spherically symmetric.  An electron in the 2px orbital will not screen outer electrons from the nucleus equally in all directions.
As we proceed from boron to carbon, that first 2s electron is effective in screening the second 2s electron because of its spherical orbit (2s effective nuclear charge increase=0.641).  However, the first 2p electron is not as effective in screening the nucleus because of its non-spherical shape.  So when we add the second 2p electron, it "feels" the nucleus more strongly than the second 2s electron did so the change in effective nuclear charge is greater for the second p electron than it was for the second 2s electron (2p effective nuclear charge increase=0.715).
You noticed the "anomaly" for the 2p electrons as we go from nitrogen to oxygen.  This is because with 3 2p electrons, all of the nitrogen 2p orbitals (px, py, pz) have an electron and produce a spherical electron distribution pattern.  The nucleus is better-screened by the 3 2p electrons in nitrogen, so when we add one more p electron to yield oxygen, the effective nuclear charge increase is noticeably smaller (N -> O 2p effective nuclear charge increase=0.619).

Okay, but how to explain then why 2nd period non-metals can't easily form higher oxidation states while 3p elements can?

I've edited my answer to address this point.

Thank you very much Ron. It's only I still have some doubts. 3p electrons penetrate well into the space near the nucleus https://www.rose-hulman.edu/~tilstra/Engineering%20Chemistry%20I/RadialDisributionFunctions.jpg. Therefore it should be easier for 2p elements to form higher oxidation states than 3p elements, but it's not, how?

See my second edit and let me know if you still have questions.

Sadly I still have. Back to the second period, the effective nuclear charge has bigger increase for 2p than 2s (except N-O) according to this chart  http://img.docstoccdn.com/thumb/orig/118492050.png. How so?

Very good question!  See my edit.

And an even better response, but I still can't form a full picture, because I need to clarify only this: If the 2p orbitals experience more the increase in the effective nuclear charge than the 2s orbitals, how does the energy difference between the 2s and 2p orbitals increase in a period? Here is the table that proves this http://chemistry.stackexchange.com/questions/16522/periodic-trend-in-difference-of-energy-between-the-s-and-p-orbitals. The energy difference patterns doesn't follow the increase of the effective nuclear charge you explained. How?

The effective nuclear charge (ENC) and the ionization potential (IP)measure different things.  The ENC is a theoretical estimate of the nuclear charge experienced by a specific electron.  The IP is an experimental measurement of how hard it is to remove an electron.  How hard it is to remove an electron will depend on 1) the ENC, 2) the amount of electron-electron repulsion and 3) any special stability gained by being in a filled shell.  So while I would expect there to be some correlation between ENC and IP, I wouldn't expect complete correlation.

Doesn't the effective nuclear charge account for the electron-electron repulsion? So which of these two (ENC or IP) correlates best to the energy of an orbital? I think the ionization potential.

No, the ENC doesn't factor in electron-electron repulsion, it's not an "energy" term per se, it is just the number of protons a given electron "feels".  IP is a measure of orbital energy.

I think I understood, 2s are more stabilized through the period (because they penetrate more) and they feel bigger ENC while 2p feel better the increase of the ENC (because they penetrate less) but are less stabilized by this increase. Stabilization energy and ENC increase aren't well correlated.

"2p feel more the increase of the ENC (because they penetrate less)" and because of their shape they do not screen as well as s electrons.  I agree with everything else you wrote.


text:
Do epoxide rings react with bases?
We've learned in class that epoxide ring openings can be catalyized with acids, and I seem to recall either the professor or a classmate mentioning that they also react with bases, but some quick internet searching didn't turn up anything useful.  Do epoxide rings react with bases, and if so, what is the mechanism?
(Dis)claimer: this is just for my own curiosity/learning, and not being asked as a homework/test question (yet).

Do epoxied rings react with bases

Yes, they can be opened, e.g. with alkoxides.

what is the mechanism?

In principle, this is a $S_N2$ reaction, with the typical approach of the nucleophile from the back side. The difference to a non-epoxide case is that removal of the leaving group (bond breaking, ring opening) is facilitated by the ring strain.
But the general rules for the reactivity of alkyl-substituted centres in $S_N2$ nevertheless apply: the higher the substitution, the lower the rate. This is attributed to the steric hindrance of the respective centre.
Note that the regioselectivity is different under acidic conditions!

Thanks for the answer! Could you explain why the base attacks the least substituted carbon in basic conditions?  Or should I open another question?

@apocryphalauthor Not necessary, the matter of regioselectivity is perfectly within the scope of your question. I've tried to give an explanation in my updated answer.


text:
What is an orbital boundary surface?
Could anybody explain the following statement:
"The boundary surface is a common way to represent atomic orbitals, incorporating the volume in which there is about a 90 percent probability of finding the electron at any given time.
I don't understand what does the author means starting from "incorporating the volume.."

Other people might call them "isosurfaces." In reality, orbitals are more like "electron clouds."

It means that the boundary surface is a closed surface or set of closed surfaces, such that there is a 90% chance that an electron is interior of the closed surface and a 10% chance that the electron is exterior of the closed surface.
A better way of specifying how an orbital is represented is "equal probability boundary surface" which further specifies that each point on the surface must have any equal electron probablity density, and the surface must be a closed surface or set of closed surfaces with a 90% (or other percent) chance of containing the electron at a given time.
Without the "equal probability" statement there are infinite different boundary surfaces for each orbital (infinite different ways to enclose 90% of the probability density).


text:
Why water vapors exerts pressure rather than diffusing in air before reaching the boiling point of water?
I am discussing only two types of vaporization in my question that are, evaporation and boiling. We know that evaporation is the phenomena which describes vaporization of liquid under a certain temperature. And boiling point is the temperature at which vapors of liquid exceeds their pressure by $760\ \mathrm{Torr}$ and starts escaping. The liquid doesn’t absorb more heat and utilize all of the heat to change its state and is said to boil.
There are more details about these two kinds but there is no need to discuss them right now so we should go back to the question that asks that why vapors exerts pressure rather than diffusing in atmosphere.

What do you mean by "diffusing in air". If you consider evaporation diffusing, it does occur. However, boiling only occurs when the vapor pressure equals that of the atmosphere. However, water molecules are constantly lost even before the boiling point. I'm sorry if I misunderstood you but it would be nice if you could clarify.

NO you get me exactly right! you have made  my thought bit stronger. I must be doing a mistake!

I be mistaken, but are you referring to vapor pressure?
Vapor pressure  or equilibrium vapour pressure is defined as the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases (solid or liquid) at a given temperature in a closed system. The equilibrium vapor pressure is an indication of a liquid's evaporation rate. It relates to the tendency of particles to escape from the liquid (or a solid). A substance with a high vapor pressure at normal temperatures is often referred to as volatile. (http://en.wikipedia.org/wiki/Vapor_pressure)
What is a closed system?
A system in which no mass is transferred.
Thus, vapor pressure would not be exerted if the container is open and the liquid is constantly diffusing. It would only occur in a closed system.
Therefore, liquids DO diffuse (evaporate) in air, rather than exert vapor pressure.
You can leave a bottle of water, one with a lid and one without. The one without the lid will soon be without water, but the one with a lid would remain with water.
*Vapor pressure does relate to evaporation rate, which occurs in the open systems we normally observe in life

Vapors must exert a pressure on the blanket of atmosphere to get escaped. However during this rising-temperature process, some of the vapors get diffused in air. When there is a lid, vapors will exert pressure on the container and when there is no lid, vapors must exert pressure against the heavy blanket of atmosphere. When these pressures are equaled, all the vapors are free to escape now and liquid is said to boil. Irrelevant to the diffusion, some vapors must get collided with atmosphere particles and stay over the liquid and below the atmosphere and they are responsible for vapor pressure.

Your comment above is actually addressing my question. And, I got my answer from it.  If I am right at my explanation above then I am right to accept your answer.

Pictures are incomplete: they don't show molecules rebounding from others and pressing against all surfaces, both the container and the liquid.
The ideal gas laws can be derived from thermodynamics; see Kinetic theory. There's a nice animated GIF there that could help you to understand this.

Well... My pictures are not expressing my question.

As I understand it, the distinction between evaporation and boiling is that evaporation only happens at the surface of the liquid whereas boiling happens in the bulk of the liquid (or at least where it contacts the heat source). This is what forces the temperature to remain constant until the liquid is completely gone - each and every point within the bulk of the liquid where the temperature is high enough to produce a vapor pressure equal to or greater than the hydrostatic pressure in the liquid, will immediately form a bubble, rise to the top, and leave. Any time a bubble is formed in a region where the temperature is below the boiling point, the vapor inside the bubble has insufficient pressure to oppose the hydrostatic pressure (which for a small container is equal to the atmospheric pressure) and the bubble collapses.