text:
Why does adding an acid speed up electrolysis of water?
I've read in several places that when electrolyzing water into $\ce{O2}$ and $\ce{H2}$, that adding a salt or an acid will speed up the process. Why is this?

Primarily due to

Pure water conducts an electric current very poorly and, for this reason, is difficult to electrolyze.

(Source: Water Electrolysis)
and that salt and acids are able to act as an electrolyte catalyst as most salt and acids dissolve (hence dissociate) in water, specifically, according to the Wikipedia page,

If a water-soluble electrolyte is added, the conductivity of the water rises considerably. The electrolyte disassociates into cations and anions; the anions rush towards the anode and neutralize the buildup of positively charged $\ce{H+}$ there; similarly, the cations rush towards the cathode and neutralize the buildup of negatively charged $\ce{OH-}$ there. This allows the continued flow of electricity.

A diagram of the electrolysis of water with dissolved salt/acid ions is shown below:

Image source
Further information can be found in the article A Review of Water Electrolysis (Zoulias et al).

So the salt only helps exchange the water around the electrodes?

the dissolved salt or acid increases the conductivity of the water, allowing a more continuous flow of electricity compared to that of pure water.  More electric current = speedier electrolysis

@FriendofKim is there anything else you would like to know?

If I'm correct water is split up into oxygen and H+ on the anode releasing electrons in doing so. On the kathode electrons are supplied and splits water into H2 and OH-. H+ and OH- reacts and forms water. I don't see any flow of electrons, only two separate reactions.

@FriendofKim i'll add a diagram to help with an explanation

Thank you! I'm reading the source you linked first. Then I'll ask if there is something that was unclear!

I'd recommend using a strong base, e.g. NaOH. The reason is that if there's chloride in the solution (e.g NaCl as the easiest accessible salt or HCl as cheap strong acid), you'll produce chlorine gas instead of oxygen.
This is most suitable for industrial purposes, but not for home experiments ;-)

Yes, indeed. The best I have is $\ce{NaHSO4}$ (bath tub pH-lower). It says 95 % of the acid and 5 % inert ingredients.


text:
How explosive is hydrogen gas?
I want to electrolyze water to get oxygen and hydrogen. I want to mix them in a regular balloon and ignite it. How much hydrogen would you need in a 2:1 ratio with air for it to be dangerous? Is this generally a safe experiment?

This great video shows what happens.https://www.youtube.com/watch?v=qOTgeeTB_kA Not the explosion that Uncle Al describes.

First, let me say that I've enjoyed many times exploding soap bubbles of about one milliliter filled with hydrolysis gas. That is 1 cubic centimeter.
That will give you a sound that rings in your ears in a decent sized living room. You may wish to use ear protection for the experiment.
50 ml will have an effect in a lecture hall that not only wakes up everyone, but also may make people complain.
Now while the explosive limits of hydrogen in air range from about 18 -- 60 % the flammable limits are from 4 -- 75 %, in oxygen the limit of flammability goes all the way from 4% to 95% read: for practical purposes, hydrogen in oxygen is always at least a flammable mixture.
For comparison, gasoline in air is flammable roughly between 1.5 - 7%.
However, you start with a stochiometric mixture which is ideal for explosion (after all, that's why you do it, right?), and it is in oxygen, not in air. So even if you "only" have a flammable mixture reaching an ignition source (e.g. electric switch), chances are quite high that as the ignition proceeds through the mixture, it can reach a zone where the gas/air mixture is explosive. Also, because hydrogen is so much lighter than air, it tends to accumulate under the ceiling, so even if there is enough air in the room that an ideal mixture would not be explosive, chances are that there is an explosive layer of gas. (Same with gasoline or, practially more relevant: solvent from glue for parquet floor tiles)
The safety relevant points are:

the limits of explosion and flammability are, even in air, extremely wide.
limits of flammability are important even if there happens to be no explosive mixture: hydrogen is a small and light molecule that diffuses fast. If the air you breathe happens to have a hydrogen concentration within the flammability limits, any ignition will burn deep into your throat on its way to the lungs.

I'm actually planning on making very little of this just to get a little sound. I've seen people ignite large party balloons with hydrogen before. Are you saying it could have been a big disaster?

It could, but there is one thing that usually makes the balloon experiment less dangerous (and IMHO somewhat misleading as to what a good explosion is): the hydrogen diffuses really fast through the rubber of the balloon, while the oxgen stays in. As it takes a while to fill the balloon, the resulting mixture is far from ideal.

So you're saying that there is very little hydrogen left, and thus making the explosion less intense?

Yes. I've also had school chemistry lecture hydrolysis gas explosion. After the soap bubble experiment I can tell from the sound that the mixture was not ideal. The trick with the soap bubbles is that this is really fast. We made them 2 - 3 per second with an autogenous welding machine for dental technicians. It uses a canule as tip, and we just put that into soap water and immediately lighted the bubbles. When not bubbling, we kept the welding torch burning - that is the safest way.

Aha, okay. I'm only intending to make a very small amount, anyway. Thank you for pointing this out.

We hugely enjoyed that as kids (rule was: whenever we wanted to have a "boom" we should ask dad, so he could "participate" in the "boom" as well). Being a professional chemist now, I can really appreciate that rule...

Yes, it is indeed important to be careful. I'm just trying to figure out how this electrolysis will work. I'm electrolyzing water with $\ce{NaHSO4}$ as an electrolyte. However I don't figure out how the current is going to flow.

let us continue this discussion in chat

@cbeleites Actually, the thing that makes the balloon experiment safe is that people don't usually fill balloons with explosive mixtures because it is really dangerous, they use pure hydrogen. This doesn't explode but goes "phuut" due to the limiting rate at which the hydrogen can mix with air. It can still be pretty dangerous, but is nothing like as dangerous as the large, dangerous, loud bang that results from an explosive mixture.

@matt_black: good point, yes. It certaily applies to those who fill it from the gas bottle. From bottles the easiest way to get a good mixture of gas with oxygen into a balloon is via a torch. But hydrogen is AFAIK really a niche for welding/soldering (dental technology) so my guess is that even most workshops at universities don't have hydrogen torches.

Here is a neat experiment, but it unfortunately needs a welder to make the equipment, a metal sphere with a volume of about 100 mL. It should have a very small opening on top and a larger one on the bottom. It also needs feet to stand.

The metal container should be capable of some pressure, because the gas will explode inside and you don't want to hurt anyone. Additional you need bucket or bowl filled with water and hydrogen gas.
First, close the hole on the top of the container and place it in the bucket under water until it is completely full. Now you can fill the container with hydrogen gas until all water is gone from it. When you remove your container, make sure that the bottom hole stays on the bottom. As Hydrogen is much lighter than air, it will be trapped inside.
Place the container on a stable, nonburnable table/ bench. Have a lighter ready. Now open the top hole and ignite the hydrogen gas. Move away from the experiment. After a while you will hear a phenomenal boom (make sure you have your mouth open). Do not try this in a small room.
The trick behind this is in the ratios. As Hydrogen will burn on top, air will be sucked in at the bottom. As long as the mixture has not the right ratio, it will not explode but just burn at the top. Once enough oxygen has entered it will ignite the whole thing and do a phenomenal bang. The experiment is pretty safe.
The experiment is part of this lecture, which is in German unfortunately.

And this is with only 4 percent of oxygen. I wonder what would happen with a 1 to 2 ratio...

@Friendofkim According to this the 'concentration' of hydrogen in air should be about 59% when it explodes (you start at 100%). So Oxygen level should be about 8%. Nevertheless, if you have a 1:2 mix then cover your ears and keep your mouth open!

Ahh, I wrote the comment a bit too fast.. I was thinking about the burning percentage.

That experiment was performed at our school (yes, high school) multiple times using what looked like a simple old tin can (cylindric). Hydrogen was simply filled in directly from the bottle through the hose without any of the fancy water stuff. The teachers actually continued letting hydrogen flow through before turning it off and starting the explosion trigger.

Oxygen and acetylene mixed in a ballon will produce a very large explosive bang when lit and a loud noise and gratifying plume of putrid black smoke. Hydrogen and oxygen is going to be many times more explosive, not sure how many times more though.
I was on the back of a flat bed truck with a parade display and a weather ballon on a string about 10 feet above the tray (possibly less was twenty years ago) was accidentally ignited (not by me!) and a few of us were knocked to the floor/truck tray and the closest person had open flesh burns all the way down his arm which was extended in the air towards the weather ballon (may have even been holding a party sized ballon with acetylene as he was lighting them while holding them which is very dangerous and I in no way endorse (he was a lunatic this guy and I had no authority over him)).
So bottom line start very small with these kinds of experiments and be aware things can and will go wrong so always take as much precaution as you can.

The most important point presented above is, in my opinion, the explosive range of hydrogen mixed in air. It is uncommonly and dangerously large!
Real world point is, for example, I would NOT recommend recharging your fuel cell car in your garage! Accumulation of odorless and colorless H2 in a room followed by ignition from a spark from flipping on the light switch is a potential disaster for you, the car and the entire house!
Relatedly, if you enter a room and smell natural gas, do not touch that light switch! Either leave immediately or turn off the power to the house and then open windows and such. Why? Because opening the window impacts temperature which could trigger a thermostat to turn on the heating/cooling system (possible spark or flame).

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text:
Electrophilic Aromatic Substitution: Are protonated OH-groups ortho-/para- or meta-directing?
In Electrophilic Aromatic Substitution ($\text{S}_{\text{E}}\text{Ar}$) reactions groups that can donate an electron pair into the aromatic ring, e.g. $\ce{-OH}$ or $\ce{-NH2}$, are ortho-/para-directing.

In the case of $\ce{-NH2}$-groups one has to be cautious because if the $\text{pH}$ value of the solution is too low the $\ce{-NH2}$-group will be present mostly in its protonated form, $\ce{-NH3^+}$, which lacks a free electron pair and has a meta-directing effect because of its strong $-I$ effect. This brings up the question: What is the situation like with protonated $\ce{-OH}$ groups, $\ce{-OH2^{+}}$? In contrast to $\ce{-NH3^+}$ this group still has a free electron pair which could, in principle, interact with the aromatic ring ($\Rightarrow$ ortho-/para-directing) but it should also have a very powerful $-I$ effect ($\Rightarrow$ meta-directing).

I'd say, if the electron pair can be effeciently delocalised into the aromatic ring then its ortho-/para-directing power should outweigh the meta-directing power of the inductive effect. But I wonder if there even is any appreciable degree of delocalisation present because the mesomeric structures don't look very favourable:

Thinking about it from a frontier orbital perspective I would say: The orbital of the $\ce{-OH2^{+}}$ group's free electron pair will be much lower in energy than that of the $\ce{-OH}$ group's free electron pair. Thus, the energy difference between free electron pair's orbital and the aromatic ring's $\pi$-system might be too large to facilitate a good interaction. Therefore, I'd expect a protonated $\ce{-OH}$ group to be meta-directing.
But what is the actual situation when phenols or phenol ethers undergo ($\text{S}_{\text{E}}\text{Ar}$) reactions in strongly acidic solutions? Does one get the ortho/para or the meta product?
Edit:
I looked up the $\mathrm{pK}_{\mathrm{a}}$ value for protonated phenol: It is approximately $-7$. This makes me wonder if there are any reasonable (real world) situations where the solution is acidic enough to protonate phenol (or phenol ethers) completely (even $\ce{HI}$ with its $\mathrm{pK}_{\mathrm{a}}$ of $-10$ shouldn't be enough for that, right). I'm not sure if even sulfonation or nitration reach such low $\text{pH}$ values.
But @Uncle Al's answer made me aware that I forgot about Lewis acids and I'd say that Lewis acids like $\ce{AlCl3}$ that coordinate strongly to oxygen might lead to a state that will be very similar to a $\ce{-OH2^{+}}$ group (please, correct me if I'm wrong and the coordination of Lewis acids is not strong enough to be compared to the $\ce{O-H}$-bond). If this is the case, then what about Friedel-Crafts reactions which make heavy use of Lewis acids: Would those Lewis acids interfere with the ortho-/para-directing properties of phenols? Does anyone have experience with the directional properties of a $\ce{-OH}$ group in the presence of a strong Lewis acid, maybe even in the context of Friedel-Crafts reactions?

Interesting question. Regarding the possibility of quantitatively protonating phenol, it's certainly doable with, say, fluorosulphuric acid or carborane superacids. Whether anyone does that with regularity, I don't know. I don't think that comparing values of $pK_a$ works too well because they are rather poorly defined for extremely strong or extremely weak acids/bases. It's probably best to compare them with some extension such as the Hammett acidity function.

This article from the master of superacidic media George Olah will probably have some insight into phenol protonation and reactivity. And here's what seems to be an even more precise reference.

@NicolauSakerNeto Thanks for the hints. I agree that $\mathrm{pK}_{\mathrm{a}}$ values in this regime shouldn't be taken too seriously. But I think, they are at least evidence that under usual reaction conditions, phenols won't be protonated completely and in those situations I don't have to worry about a possible meta-directing property of a phenol.

@NicolauSakerNeto Thanks again. The second reference you provided seems to be exactly what I was looking for. It supports my view that protonated $\ce{-OH}$ groups are meta- rather ortho- / para-directing. But still information about the influence of Lewis acids on the directional properties of the $\ce{-OH}$ group would be interesting. Especially, since it would throw some light on how strong the coordination between oxygen and the Lewis acid is.

Excellent, I'm happy to help. Perhaps you could consider making your own answer with what you've learned!

Sorry for the unrelated comment: how did you draw such nice mesomeric structures ?

@mannaia I used ChemDraw for this. It is definitely the best program for drawing Lewis structures out there but unfortunately very expensive if you don't have a university that provides you with a license.

$\ce{O}$-protonated phenols indeed direct in meta, according to an article on that topic. As it occurs, phenols can be, depending on acid, protonated on oxygen, but also on carbon. They can even be diprotonated, when most powerful mixtures of $\ce{HF}$ and $\ce{SbF5}$ are used, but not twice on the same atom. $\ce{O}$-protonated forms can be brominated, hydroxylated and alkylated. $\ce{C}$-protonated forms don't posses such reactivity and are essentially Meisenheimer's complexes.

A four-coordinate $\ce{sp^2}$ oxygen with two positive charges is a very high energy state.  Aniline is a much weaker base than an aliphatic primary amine for leaking electron density into the ring.  Phenol in turn will be a much weaker base than a primary alcohol.  Deactivating the phenol will require a very Brønsted acidic medium indeed.
Lewis acids would be more effective.  Anhydrous $\ce{AlCl3}$ is a powerful oxophile. Adding a mole would turn off all the phenols.  One might have quite a time getting the phenol free of the aluminum during workup.  Note the large number of LAH workup recipes to avoid intractable aluminum oxohydroxide gels during workup.

Thanks for your answer. It made me aware of the very low pH necessary to protonate phenols completely. But unfortunately, you don't say anything concerning the ortho- / para- or meta-directing properties of protonated/Lewis-acid-coordinated Phenols. Could you add something concerning this topic to your answer. Please, have a look at the Edit in my question.

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text:
How to determine the reaction of a strong electrolyte?
I'm told the $\ce{KMnO4}$ is a strong electrolyte.
How can I figure out how the reaction looks like?

The reaction of what exactly?

$\ce{HMnO4}$ is s a strong acid (compare oxidation state with perchloric acid).  Its anion is then a weak base.  The salt is completely dissociated in water solution.


text:
Polymers with no molecular mass distribution?
What sort of ideal conditions are required - theoretically - to conduct polymerization reactions to make all the macromolecules end up with exactly the same molecular mass, just like the products in normal chemical reactions? How do living organisms manage to synthesize macromolecules, for instance proteins, different than humans do in the lab?

Living polymerizations can afford nearly zero polymer polydispersivity.  You can go anion (Kraton thermoplastic elastomers - and tell us about the one silicon atom in each polymer molecule), free radical (TEMPO or PROXYL capping of the free radical end, thermally activated), redox (copper halide), olefin metathesis polymerization (ROMP and ADMET), and other clevernesses.
Proteins are made from a DNA template via RNA, one AA residue at a time. There is no dispersivity.  Post-synthesis modification, e.g., glycosylation in erythropoietin, can be polydisperse.


text:
What is Bisphenol A used for?
I am trying to find out what BPA is used for in the industrial production of plastics. On sources such as Wikipedia I only found the information that it is used, but not why.
What are the advantages in using BPA in the production of plastics? Why is it needed?

Bisphenol-A is very inexpensive for its considerable formula weight - from benzene, propylene, air, acid; acetone (or other ketone or aldehyde) and acid.
1) Polycarbonate, as above.  If you don't like HCl, dimethyl carbonate will do.
2) Epoxy resin, by bis-etherifying with glycidyl chloride.  That is a huge market, including paints and coatings.
3) Bis-etherify with propargyl chloride.  That thermally B-stages (removes the shrinkage upon cure) then thermosets to an astounding thermally resistant super-crosslinked resin that adheres to any polar surface.
4) Bis-esterify with methacrylic acid to get into heavily cross-linked free radical or UV cures.
5) And if you really want something resistant, instead of acetone use fluorenone or a 2,7-disubstituted fluorenone.  Or be fiendishly clever and use 4,4'-dihydroxybenzophenone instead of acetone.  Now there is a tetrahedral tecton for synthesis!
Bisphenol-A is a claimed  endrocrine disruptor.  Eastman Kodak has immense production capacity for acetic anhydride (cellulose acetate photographic film).  It offshot into ketene dimer chemistry then mixed cis- and trans-2,2,4,4-tetramethyl-1,3-cyclobutanediol as a BPA replacement.  If you do not mind paying more and getting less, stuff as well as performance, the problem is solved.

Bisphenol A is the monomer used for synthesizing polycarbonates, sometimes via the following reaction:

(Source)
If you take a look at this, you can see that it has many different (read: non-polymer related) application possibilities, and is cheap to produce (making it the most preferred option when selecting the monomers).
The synthesis of BPA involves acetone and phenol, both very very cheap substances.


text:
What does negative pressure mean in the context of this manometer?

I have a basic question on a closed manometer:
I see that in this case the pressure of gas is lower than that of the liquid, which is why the liquid is higher on the left side. Now, what I don't understand is why do we say that the gas is at a negative pressure?  Wouldn't that mean there's no gas at all? Also, how do we know that it's negative? Usually, I only take the difference between the highest point and the lowest.

The manometer shown doesn't measure absolute pressure, just relative pressure. In this car the pressure of the gas in the closed tube versus the pressure of the gas in the bulb.
As you have drawn it, the pressure of the fas in the tube is higher. If you measure pressure in the bulb relative to the pressure in the tube the answer is negative. If you measured absolute pressure (which this manometer can't do because of the lack of any external reference in the closed system) both would be positive but one would be higher than the other.

If the manometer were open at both ends, the two fluid levels would be identical.  If the bulbed gas were at ambient pressure, ditto. That the fluid is higher at the bulb end says the pressure in the bulb was lower than ambient pressure, and lower still for the weight of the differential fluid head, height times density.
Why is the diameter of the tube, short of capillary tubing, irrelevant?  You can then tell us the "negative" pressure in the bulb.


text:
How do I determine the equilibrium concentration given Kc and the concentrations of component gases?
I feel my method for solving it is correct and I've checked my calculations yet I can't get the right answer.

Nitrogen and oxygen gases may react to form nitrogen monoxide. At 1500 °C, $K_c$ equals 1.0E-5.
$\ce{N2_{(g)} + O2_{(g)} <=> 2 NO_{(g)}}$
If $2.75 \times 10^{-2}$ mol $N_2$ and $2.75 \times 10^{-2}$ mol $O_2$ are sealed in a $0.886 L$ flask at $1500 °C$, what is the concentration of $\ce{NO_{(g)}}$ when equilibrium is established?
Notes: $K_c$ = equilibrium constant
a) 1.38E-7 M
b) 4.35E-5 M
c) 3.85E-5 M
d) 8.70E-5 M
e) 2.30E+4 M

I set up the problem with $(2x)^2$ as a numerator and $(.03104 - x)^2$ as the denominator
I took square root of both sides and did algebra and got $4.9*10^{-5}$
For the record the answer is $8.7 \times 10^{-5}$

Welcome to Chemistry.SE! We have the MathJax plugin installed here so you can format your chemistry and maths better. Could you be a little more specific as to which formulas you used - it would be best if you wrote down the whole calculation method you used.

I used k = [Product]/[Reactant] I'll write out all my calculations in a few minutes.

For a reaction in the form $\ce{ \mathit{a}\,A + \mathit{b}\,B <=> \mathit{c}\,C}$, the quilibrium constant $K_c$ is given by
$$K_c = \frac{[C]^c}{[A]^a[B]^b}$$
You might want to check yours.

I agree, that's how I attacked the problem I got (2x)^2 / (.03104 - x)^2 = K  ;; 2x/(.03104 - x) = sqrt(k) ;;; 2x = sqrt(k)(.03104 - x) ;;;; 9.8157e-5 - .00316x = 2x ;;;;; 9.8157e-5 = 2.00316x ;;; x = 4.9e-5

It's .03104 and not .0275 because its .0275 moles in a .886 L container so .0275/.886 = .03104 M


text:
What is the difference between the roles of a surfactant and a stabilizer in the process of nanocrystal growth?
The crystals that I am synthesizing use two different stabilizers CTAB and SDS, but they say one is used as a surfactant and since they are both stabilizers what is the need for both?

CTAB is a cationic surfactant, SDS is an anion surfactant.
These surfactants act to selectively stabilize ionic faces on the crystal as it grows, causing those faces to grow more rapidly. This leads to a change in geometry or crystal shape. CTAB causes one morphology being cationic, while SDS will cause another, its head ion having the opposite charge. Further, a mixture of the two will cause an intermediary morphology.
For instance, CTAB will cause an aqueous solution of thioacetamide and Lead(II) acetate to form PbS nanocrystals in a octahedron shape, while SDS will cause them to form a dendritic fern like shape, while a CTAB/SDS mixture (5:1) will cause a stellated octahedron or star shape to form.

Thank you for the help. I am actually fabricating PbS NC's for my senior project. Maybe you can explain to me or send my in the right direction how EDTA, SDS, CTAB, lead acetate and thiorea solution facilitate the formation of these NCs when only heated to 70 degrees C. Any information helps

Low-Temperature Synthesis of Star-Shaped PbS Nanocrystals in Aqueous Solutions of Mixed Cationic/Anionic Surfactants


text:
What is made at the copper cathode when electrolyzing water?
I just electrolyzed water with $\ce{NaHSO4 -> HSO4- <=> H+ + SO4^2-}$ with copper electrodes.
At the anode this reaction happened:

$\ce{Cu -> Cu^2+ + 2e-} \hspace{26 mm} E_0 = -0.34V$
$\ce{Cu^2+ + 6H2O -> [Cu(H2O)6]^2+}$

Can anyone confirm this last reaction? There was a blue-greenish substance falling to the bottom.
At the cathode this reaction happened:

$\ce{2H+ + 2e- -> H2} \hspace{27 mm} E_0 = 0V$

$\ce{2H2O + 2e- -> H2 + 2OH-} \hspace{5 mm} E_0 = -0.83V$
$\ce{OH- + H+ -> H2O}$

However, additionally, the cathode ($\ce{Cu}$) turned black. What reaction could have happened there? I electrolyzed at $12V$ using a car battery charger.

Look at your redox potentials for the half reactions.  12V is tremendous overkill.  At the cathode (reducing) you will make hydrogen to which copper is inert.  Could you reduce sulfate to sulfide or sulfur and have that react with copper?  CuS is black.
At the anode (oxidizing) you will oxidize water to oxygen - but one atom at a time.  It will be very reactive to the copper surface.  CuO is black.  You will also oxidize copper to Cu(II).  It coordinates four waters square planar.  The apical waters are distant and weakly bound given the Jahn–Teller effect,
http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Coordination_Chemistry/Coordination_Numbers/Jahn-Teller_Distortions
http://www.adichemistry.com/inorganic/cochem/jahnteller/jahn-teller-distortion.html
One possibility is to add a series resistor to lower the working voltage.

I'm thinking there will be so little $\ce{SO4^2-}$ around the cathode that there shouldn't be any $\ce{CuS}$.

Because the redox potential for $\ce{Cu -> Cu^2+}$ is lower than that of water, shouldn't it be the only reaction happening?

The copper half-reaction is 0.34 V.  You are putting in 12 V.   How many volts oxidizing is the fluorine half-reaction?  How many volts reducing is potassium metal?

Are you saying that the sulfate could be reduced to sulfide even though there is a lower potential needed to make hydrogen gas? At school we learned that this would never happen, even though I think that this may be an unlikely statement.

The standard reduction potential of potassium metal is -2.93 volts; you've got 12 volts.  "Never happen" is a relative statement.  If it smells of hydrogen sulfide, it's happening.  I wouldn't bet a grade on footnotes.   One day you will be at a lab bench.   Reality plays a dead hand - but its all trump.


text:
Alkene reactions with sodium tert-butoxide and CHBr₃
In a problem they gave me some reagents and they want me to figure out the end product.
The reagents given are:
2-butene reacting with sodium tert-butoxide and $\ce{CHBr3}$.
Now, what I was thinking was that the oxygen would pull the hydrogen from $\ce{CHBr3}$ making a positive charge on the carbon. (I'm not quite sure if the positive charge would be strong since the net dipole moment on $\ce{CHBr3}$ would be zero). However, from there, I immediately thought of a carbene reaction, yet, it has 3 bonds with $\ce{Br}$.
So, am I on the right track, and what are the next steps from here? Sorry if I didn't give you much to work with.

Strong bases react with chloroform $(\ce{CHCl3})$ and bromoform $(\ce{CHBr3})$ to form dihalocarbenes by alpha elimination.
The net reaction is as follows:
$$\ce{RO- + CHCl3 -> ROH + Cl- + CCl2}$$
The base removes the hydrogen atom from chloroform or bromoform to form an anion. Remember, when you deprotonate something, the electrons stay behind.
$$\ce{RO- + Cl3CH <<=> ROH + Cl3C:^{-}}$$
The trihalomethyl anion is unstable and fragments to lose a halide ion and generate the carbene.
$$\ce{Cl3C:^{-} <=>> Cl- + Cl2C:}$$
The mechanism using arrow-pushing formalism can be found at this website from the University of Liverpool.

That's a very nice use of JMol!

Dibromo carbene will form in the reaction for sure because tert-butoxide is a very strong non nucleophilic base and here dibromo carbene will form after the alpha elemination. For the next step I think there would be an insertion reaction of the carbene in the double bond of 2-butene to form a 1,1-dibromo-2,3-dimethylcyclopropane.

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text:
name of a pharmaceutical drug used in dentistry that has Ionic bonding? and how this bounding is made?
Hello I am looking for this information and I did not find anything, the closest thing I have found it is Fluorocarbon, but I think it is wrong, does anybody know?.
Help really appreciated I am still looking for this

Prescription strength fluoride (Acidul), though it probably is not considered to be a pharmaceutical.

The fluorocarbons that you mentioned do have polarized $\ce{C-F}$ bonds, but no ionic bonds. If you think ionic, think salts.
I'm not a pharmacologist, but what about the local anestethics used at the dentist? Lidocaine (1), mepivacaine (2), and articaine (3) are frequently used.

All these compounds have tertiary and secondary amino groups. (I've marked the relevant nitrogen atoms blue). When these compounds are treated with hydrogen chloride, the amines get protonated and the neutral compounds are converted to ionic hydrochlorides, $\ce{R3N ->[\textrm{HCl}] R3NH+ Cl-}$.


text:
Do chemists refer to water as "dihydrogen monoxide"?
Is the name "Dihydrogen monoxide" actually what chemists would use to refer to $\ce{H2O}$ (assuming there was no common name, "water")?
Of course, this is all over the internet. I'm a little skeptical though because the similar chemical $\ce{H2S}$ is called "hydrogen sulfide", not "dihydrogen monosulfide".

No, it's not. The "dihydrogen monoxide" name is used as part of a hoax. In the scientific community, there are chemical names for water, and which one is used in the literature generally depends on how it interacts with something else (hydroxic acid and hydrogen hydroxide were two I heard most often in acid-base reactions).
IUPAC, the standards committee that sets standard names for chemical structures, suggests "oxidane" as a starting point for the construction of other names for chemicals that are derived from water. However, they do not suggest "oxidane" itself be used to refer to plain water.
Most chemists would use "water", even when writing scientific papers.

Like most of the other professionals answering here, I've given a couple of talks on international conferences and published some articles in peer-reviewed journals.
I have never used the terms dihydrogen monoxide or oxidane and do not intend to do so in a serious, scientific context.
Typically, one talks and writes about water, aqueous solutions and even uses a (traditional) term like brine when referring to an aqueous, saturated solution of sodium chloride.

I agree with the other answers. No serious chemist uses any word other than "water" in whatever language the chemist uses.
However, the name does appear to be following the established rules for the systematic naming of binary main group covalent compounds.

Take for example $\ce{N2O5}$:

We list the elements in order of increasing electronegativity: nitrogen oxygen
We convert the second element's names in "-ide": nitrogen oxide.
We use prefixes to indicate the number of each element. Note the the "-a-" in "penta-" goes away to make pentoxide easier to pronounce: dinitrogen pentoxide.

We need this sort of system to give us unambiguous names for binary compounds, especially when, for example, there are multiple oxides of nitrogen: $\ce{N2O}$, $\ce{NO}$, $\ce{N2O3}$, $\ce{NO2}$, $\ce{N2O4}$, and $\ce{N2O5}$.
Note that we rarely use the "mono-" prefix: $\ce{NO2}$ is nitrogen dioxide.
And we never use prefixes with the binary hydrides. The acidic ones are all named like ionic compounds, as are the metal hydrides, and the non-acidic ones all have common names that are used so frequently it is silly to use more complex names:

$\ce{NaH}$ sodium hydride
$\ce{BH3}$ borane
$\ce{CH4}$ methane
$\ce{NH3}$ ammonia
$\ce{H2O}$ water
$\ce{HF}$ hydrogen fluoride
$\ce{PH3}$ phosphine
$\ce{H2S}$ hydrogen sulfide
$\ce{HN3}$ hydrogen azide

Water would thus be hydrogen oxide if anything.

Even though there are other possible oxides of hydrogen (or hydrides of oxygen), they have different names based on the anions:

$\ce{H2O}$ - water (hydrogen oxide)
$\ce{HO2}$ - hydrogen superoxide
$\ce{H2O2}$ - hydrogen peroxide
$\ce{H2O_n}$ - trioxidane (n=3), tetraoxidane (n=4), pentaoxidane (n=5)

Depending on the context, and how one wished to look at the compound and its function in that context, one might also (still in principle, hypothetically), write HOH and call it "Hydrogen hydroxide."

"Note that we rarely use the 'mono-' prefix: $\ce{NO2}$ is nitrogen dioxide."  That's not quite right.  The IUPAC naming rules for binary (two-element) covalent compounds specify that the mono- prefix is not used for the first element, but is for the second.  Hence  $\ce{CO}$ is carbon monoxide, rather than carbon oxide or monocarbon monoxide.


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Chemical Potential of Idea gas?
Let's say we have a mixture of different ideal gases. I found that the chemical potential of a single component of the mixture is
$$\mu_{i}=g_{i}^{0}+Ln(\frac{f_{i}}{P_{ref}})$$
where $f_{i}$ is the fugacity of the component. I also found somewhere else that the chemical potential of pure substances is the molar gibbs property. Would it be safe to then say..
$$\mu_{i}(T,P)=g_{i}(T,P)$$

Yes, that relation should be true.

Yes, by definition chemical potential is related to Gibbs potential in the following way:
$\mu_i = \left(\frac{\partial G}{\partial n_i}\right)_{p,T,n_{j\neq i}} = g_i$
Those two notation are synonyms. You will find very detailed and precise explanation of this in: Atkins, Physical Chemistry.


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Alcohol irregularities in Hydrogen NMR
We've started learning about NMR in my ochem class and my instructor told us that in Hydrogen NMR an alcohol can have a fairly wide range of possible values - usually from about 2 to 5.  He said that because of this the actual ppm value is not very diagnostic - but the fact that alcohols tend to have a thicker peak is.
He told us not to really worry about why/how there is such a range or why the reading tends to be thicker, but now I'm curious.
I speculate that there can be such a wide range of possible values because Oxygen is fairly electronegative, and depending on its environment it will exhibit different levels of shielding/deshielding.  I have no idea about why it tends to be a thicker peak.
So I guess I have two questions:

Is my speculation as to the range of possible values correct?
Why is the peak generally thicker with alcohols than with other functional groups?

Alcohols have labile protons which are exchangeable through acid-base equilibria with other similar protons, such as other alcohols, amines, amides, thiols and especially water. This reaction is fast enough, on the NMR timescale, that it can be observed as a reversible exchange process. The most common scenario is that there will be some amount of water in the sample, and so the alcohol and water 1H atoms exchange. The reaction is typically so fast that you don't see the water and alcohol as separate peaks, but as a population-weighted average. Therefore the actual position of the peak depends on a number of variables; temperature, solvent, starting chemical shifts of the exchanging species, populations of the exchanging species, rate of exchange, solvent and so on.
When you say 'thicker', I assume you mean a broader linewidth. Again, this is a consequence of the exchange process, and that you are seeing an average peak for the -OH and (say) H2O. Sometimes this broadening is so large that you can only observe it as a broad hump in the baseline of the spectrum. Again, the linewidth will be a function of the starting chemical shifts of the two components, the rates of exchange, and their relative populations.
It is actually entirely possible to see the alcohol peak with a narrow linewidth, similar to the CH2 and CH3 linewidths. To do this, you need to be able to eliminate the exchange process, either by excluding water from the sample( run ethanol in very dry dmso for instance), or cooling down the sample sufficiently to slow down the exchange (probably not feasible for most common solvents with ethanol and water though). In this case, you will observe your alcohol as a narrow triplet (approx 1.7Hz coupling by memory), with the CH2 appearing as a doublet of quartets. A great example of observing alcohol protons is to run sucrose dissolved in pyridine-d5. All OH protons, and their couplings are observable.
It is really important to understand that the integrals of these peaks are still valid. In the absence of any exchange (where your alcohol peaks are sharp), the integration of an  ethanol 1H spectrum should still give you 3:2:1. Where there is an exchange process occurring with (say) water, the integrals will be 3:2:X, where X=(sum of -OH plus water). This method of analysis can be used to verify alcohol content of spirits and wine, for instance.


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aldehydes are easily oxidised but ketones are not
Aldehydes are easily oxidized to carboxylic acid but ketones are difficult to oxidise. Most books say that the hydrogen directly bonded to the $\ce{C=O}$ in the aldehyde is what aids the oxidation process. However, can someone please tell me why this is so.

This question may already have an answer here: http://chemistry.stackexchange.com/questions/4416/oxidation-of-ketones/4418#4418

In the oxidation of aldehyde, the hydrogen attached to carbonyl compound is replaced by an $\ce{-OH}$ group resulting in the formation of carboxylic compound. In case of ketones, bulkier alkyl groups ($\ce{-R}$) should be replaced by an $\ce{-OH}$ group to form carboxylic compound.
We know that hydrogen abstraction is much easier than that of bulkier alkyl groups. Likewise, after the oxidation of the aldehyde, the carbon chain containing the $\ce{C=O}$ group remains same but due to loss of $\ce{-R}$ group during oxidation, the length of the carbon chain decreases in ketones.


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Table Salt vs Kosher Salt difference
Does kosher salt has a different structure than regular table salt or it is an issue of crystals size?
Why some recipes want us to use one instead of another?

There is no chemical difference between table salt and kosher salt - both are sodium chloride. Kosher salt just has bigger crystals. However, some table salt brands are iodized, meaning that iodide salts have been added to help us maintain a healthy dietary intake of the essential nutrient iodine. Kosher salt is never iodized.
Sea salt is a different matter. Since sea salt is obtained from sea water, it may have other minerals in it, which perhaps alters the taste.
Kosher salt is used in some recipes as a matter of taste - the bigger salt granules take longer do dissolve, and so there may still be little pockets of salty crunchiness. Kosher salt is also used when this property is necessary, for example in the curing of some meats. For more on the differences in culinary use between the various types of salt, you should consult Seasoned Advice.

Thanks for the answer. I tasted kosher salt and the regular salt of the same ammount: the kosher is less salty than regular - I could not swallow so bitter it was.

Was your table salt iodized? I find iodized salt to have a slightly different taste. Alternatively, the stroner taste comes from it taking longer to dissolve.

I am not sure about iodation of salt - there is no information on the package. One thing I can say for sure - the kosher salt is different, it even more clear and almost transparent and its taste is "mild" even I make a powder of it... I am looking for a reference

Maybe try a blind experiment to see if you don't imagine the difference.

I need someone else to experiment coz I feel that there is difference

after reading many explanation I've found that the answer is correct and precise.

All types of salts with names like Kosher salts, Himalayan pink salt, table salt, sea salt or iodized salt are mainly sodium chloride (NaCl). There might be trace element differences
hence a difference in taste.
The key information is missing from the accepted answer as well as the Wikipedia link. Kosher does not mean that the salt crystals are larger. Kosher is just a religious label which shows that the food item has been prepared in conformity with the Old Testament's laws. The label "K" or Kosher of food products indicates that this salt does not contain any porcine product, nor it has come in contact with them and other additives which do not conform to biblical laws. Kosher is almost similar to the label "Halal" (=permitted according to Muslim laws) on many food and cosmetic products.


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Reasons behind the different colors of a solution containing Iodine

Q) Iodine dissolves in a variety of solvents, forming solutions that are
purple, brown or any intermediate shade. This property is due to:

Hydrolysis of iodine
Formation of crystalline adducts
Charge transfer between $\ce{I-I-}$ and the solvent
Charge transfer between $\ce{I-I+}$ and the solvent

I am completely aware of which concept does option 3 and 4 allude to. I know about the formation of $\ce{I3-}$ when iodine is dissolved in solution of iodides. This fact, according to me, is what option 2 refers to.

But the answer given is 3

Can someone provide an explanation as to what concept is involved here and how can we predict the answer? I read about charge transfer complexes, but the Wikipedia page was too complicated for me to understand, and also the application of that to the question under consideration was not given.

Iodine in an inert solvent is purple.  Electron donation alters that.  Linear starch (amylose)-iodine is dark brown.  Branched starch (amylopectin)-iodine is red.  The Wikipedia page shows iodine in chlorinated solvent (very weak electron donation) then addition of strong electron donor triphenylphosphine.
If an electron donor is reactive, then the interacting iodine molecule looks like iodonium (elecron accepter).  Initial interation is (4), post-mixing is (3).
Put a tiny iodine crystal into a melting point capillary, evacuate under vacuum, flame closed.  You will see a very pale purple vapor at equilibrium.  Crack the seal and the vapor is suddenly deeper purple.  Why?  Take a sealed cap and an evacuated sealed cap, and insert the vapor volumes into the beam of a green laser pointer. interesting. Take care that the optical geometry does not reflect back into your eyes.

Can you please expand on why this Charge transfer takes place? And why does it affect the perceived colours?


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In which state does the bifluoride anion exist?

Statement 1. The $\ce{HF2-}$ ions exist in solid state and also in
  liquid state but not in aqueous state.
Statement 2. The magnitude of hydrogen bonding between $\ce{HF}$ is weaker than that between $\ce{HF}$ and $\ce{H2O}$

Both are correct and 2 explains 1
Both are correct but 2 does not explain 1
1 is correct and 2 false
2 is correct and 1 false

The answer given is 1.
I know the second statement to be true. $$\Delta H (\ce{H-F...O})=40{-}45\text{ kJ/mol},\Delta H(\ce{H-F...F})=20{-}23\text{ kJ/mol}$$
But I am completely unsure about statement 1 and the relation of the two statements. I think the existence of the afforementioned ion in aqueous solution must not be absent, as I have read somewhere that this ion acts as the conjugate base of the molecule $\ce{H2F2}$ which dissociates into hydrofluoric acid.

Statement one is only true for diluted aqueous solutions.

Wikipedia page for for bifluoride anion. The question's answer appears to be true, with explanation and references.
The fluorohydrogenate group $\ce{-HF-}$ in anions such as bifluoride can assimilate a proton by recombination:
$$\ce{HF2- + H+ -> H2F2 -> 2 HF}$$
Because of this capture of a proton ($\ce{H+}$), bifluoride has basic character. Its conjugate acid is the reactive intermediate, μ-fluoro-fluorodihydrogen ($\ce{H2F2}$), which subsequently dissociates to become hydrogen fluoride. In solution, most bifluoride ions are dissociated.
$\ce{HF2- + 3H2O}$ is in a favored equilibrium with $\ce{2H3OF + HO-}$
This makes for an interesting oxygen coordination sphere re the other lone pair of electrons.  One presumes it is hydronium fluoride in which the fluoride is bonded to a proton with a delocalized charge.
I've worked with aqueous ammonium hydrogen bifluoride as a silicate etchant. The solution has tremendous surface tension.  A drop of Fluorad surfactant was necessary to have things work.  $\ce{HF}$ will reach out and eat you.  $\ce{NH4HF2}$ is fairly benign.  No skin contact!  Don't work in Pyrex.


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Protactinium-234 half-life
This is going to seem like a rather daft question, but what is the definitive half-live for protactinium-234 (234Pa)?  I'm seeing lots of different answers in various academic papers etc.
For instance, in Bonotto & Andrews (2000) the half-life of protactinium-234 (234Pa) is given as 1.18 minutes, but in Edwards et al. (2003) they report the 234Pa half-life as 6.67 hours.  So which one is it?
Bonotto, D.M. & Andrews, J.N. The transfer of uranium isotopes 234U and 238U to the waters interacting with carbonates from Mendip Hills area (England). Applied Radiation and Isotopes. 52, 965-983. 2000.  DOI: http://dx.doi.org/10.1016/S0969-8043(99)00151-7
Edwards, R.L., Gallup, C.D. & C Uranium-series Dating of Marine and Lacustrine Carbonates.  Reviews in Mineralogy & Geochemistry. 52, 363-406 . 2003. URL: http://www.minsocam.org/msa/rim/Rim52.html

So which one is it?

Both! Proactinium-234 exhibits nuclear isomerism. The metastable form $\ce{^{234m}Pa}$ has a half-life time in the range of 1 min, whereas $\ce{^{234}Pa}$ shows a half-life time around 6 hrs.

In fact, if you check out the Wikipedia article on nuclear isomers, the $\ce{^{234}Pa}/ \ce{^{234m}Pa}$ pair is the oldest known example of nuclear isomerism.

Okay.  So the half-life length depends on the state of the nuclear spin.  With 234Pa having a 4+ state and 234mPa having a 0- state.  So out of the two, which half-life is more common?  Is it similar to branched decay, in that one pathway, in this case half-life rather than decay mode, is more common than the other?


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About the T-S diagram
Suppose we have a T-S diagram of some vapor thermodynamic cycle , say the rankine cycle , and we are interested for example in the composition of the liquid and the vapor in the middle of the condensation process ;  knowing only the entropy of both the liquid and vapor at that point (or the enthalpy) , how exactly do we get the composition at that point ?

Welcome to Chemistry.SE! This question sounds like a homework question; What are your attempts at the problem? Where exactly lies your problem in understanding these questions?

@tschoppi , definetly not a homework , i was googling the "lever rule" , but unfortunately i didn't find anything .


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How to properly denote imaginary magnitudes of vibrational frequencies
Vibrational frequency computations provide the magnitudes of vibrational frequencies for each vibrational mode of a molecule ($3N-6$ degrees of freedom for non-linear molecules and $3N-5$ for linear molecules).  Sometimes these magnitudes can be non-real (i.e. imaginary).  This is common when characterizing a transition state which contain one, and exactly one, imaginary mode of vibration.
Some popular software packages will print these magnitudes as a negative number.  This syntax has unfortunately propagated itself into common, everyday practice even if technically incorrect.
My question is, what is the best way to denote these imaginary magnitudes, especially in the context of a peer-reviewed journal?  Should we stick to using the minus sign convention (e.g. -15) with a note explicitly defining the meaning of the $-$ sign,  or should they simply be listed with an $i$  (e.g. 15$i$)?
(I've just been told to NEVER use the minus sign convention as this will incur the wrath of the reviewers.  Please feel free to verify.)

I would love to know why you would include the magnitude of an imaginary vibration. Since this "vibration" may not be observed, it has also no physical meaning. BTW: The minus results from harmonic approximations used in quantum chemical packages as it is defined as $i^2=-1$.

When you characterize small molecular clusters, people are interested in knowing the nature of the stationary point.  The magnitude of the harmonic vibrational frequencies can become important if the magnitudes are small (in this context, close to zero).  Small magnitudes can arise from numerical noise or perhaps an insufficient level of theory.  Say you used 3 different levels of theory (LOT) to characterize molecule X.  X is a minimum at LOT1 and LOT2 but is a TS at LOT3.  If the magnitude was only a few wave numbers at LOT3, one may expect it to be a minimum.  That is why we report them.

Okay. I did not understand that you meant a value for a wave number. I thought you were referring to the amplitude. However, the obtained value has no physical meaning whatsoever. Any mentioning this in the paper (not for recomputing purpose in the supporting information) should be avoided. If you are talking about a "real" TS, it should be one at all LOT and you should confirm it via IRC calculations.

Well, IRCs are only really necessary if you are trying to connect stationary points on a PES.  If you're just anchoring the PES of some small molecular cluster, the connection between the various stationary points may not be necessary (lies beyond the scope of the study).  Mentioning the Hessian index in a paper is indeed important if you are trying to illustrate the deficiency of a particular LOT (such as DFT) in characterizing stationary points.  We do high-level CCSD(T) type calculations in my lab.  These types of things are DEFINITELY worth mentioning.

Agreed, I did not know you were doing benchmark calculations. Anything recent worth sharing, I'd love to have a good read :)

I just would try to avoid the usage of the term "vibrational frequency" in those cases, and made a short mention to the force constants to be safe.

The suffix i to the positive wavenumber to denote an imaginary frequency is seemingly accepted by reviewers, see Photochemical transformations of 5-methyltetrazole. Matrix isolation FTIR and DFT studies as one example I could find in a very quick search.

Impact factor of 2.4? I can roll with that.

@LordStrykerI am not familiar with that phrase.

The journal you linked has an impact factor of 2.4 which means that the journal has 'enough clout' for me to assume that the convention that is accepted by the journal is going to be generally satisfactory elsewhere.

Thanks for the explanation! I just looked there because J. Photochem. Photobiol. A isn't behind a paywall for me.

I'm completely out of my depth here, but I would, since the minus sign convention seems to be frowned upon, look for other symbols:

As mentioned, with an $i$ denoting the imaginarity: $15i$
This is maybe not so beautiful if we're actually talking about the absolute value of the imaginary magnitude, because in mathematics the $i$ clearly denotes the imaginary part of an imaginary number.

If not yet taken, I would give my vote for the asterisk (*): $15^*$

Since we're talking about transition state magnitudes, maybe the double dagger ($\ddagger$): $15^\ddagger$

Of course this notation would have to be defined somewhere towards the beginning of your article, stating that this symbol denotes imaginary (transition state) magnitudes.
Using this method you could even have the option of distinguishing between normal state and transition state imaginary magnitudes: Asterisk for the former, double dagger for the latter.

I should have mentioned that this does indeed refer to the absolute value of the magnitudes.

Well, then I personally think you should find some other way of marking it as based on an imaginary magnitude. The $i$ is not technically correct ;-)
In any way, the reviewers will let you know whether they approve of your newly conceived notation or not...

I must know what is technically correct. Don't leave me on a cliffhanger!


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How can I safely carry out an electrolytic etch on stainless steel without producing relatively harmful by-products?
I propose to carry out an electrolytic etch on a stainless steel plate like so:

One of electrodes must be the stainless steel plate which is being etched and the surface layer must be removed.
I'm aware such an electrolytic etches can produce harmful by-products such as Chlorine (NaCl electrolyte), Chromium compounds and more.
If it is possible, What electrolytes and electrodes can I use to prevent the production of these substances?
Would it be possible to create safer Chromium compounds?
If I can't, what other non-electrolytic options do I have?

maybe titanium electrodes would solve for that ?

$Cl^-/Cl_2$/ electrode has +1.36 potential, that is significantly higher that $Fe/Fe^{2+}$, so you can use $NaCl$ safely, as iron (and chromium) will oxidize before chloride-ions.
I doubt the electrolyte will have significant amount of $Cr(VI)$ as $Cr(III)/Cr(VI)$ potential is higher than for iron.
If you really care, threat waste electrolyte with Zn granules and then excess of $Na_2CO_3$ solution, collect the solid products and handle them as metal-containing garbage.

Sand or shot blast it instead.  Rub it with crushed aluminum foil and saltwater.
Everything is harmful.  Voltage determines products.  Use the minimum voltage (slow way to travel).  Don't use a chloride electrolyte.  (TEST FIRST!) Use hydroxide, carbonate, (nitrate - Homeland Severity and OSHA will beat on you), sulfate, electrolyte (note solubility of metallic byproducts after reaction with electrolyte).  The larger the charge on the ion, the slower the kinetics.  Vent the chlorine plus inert air through lye solution to make bleach, or through a solution of reducing agent.
Cr(III) is OK, Cr(VI) is a carcinogen. Reduce spent electrolyte with sulfite, Vitamin C, zinc granules, FeCl2, shredded scrap iron, steel wool, etc - and have a way to test for Cr(VI) residual.
Before you vent or sewer anything, talk with your Safety Officer.  The EPA and OSHA are carnivorous.  it's always a good idea to be a good upstream neighbor.

Thank you for that important info. I think I have come to the conclusion that stainless steel requires a bit more experience than what I have at the moment. What about a non-electrolytic etch for Aluminium using an aqueous Copper sulfate + Sodium Chloride solution. http://www.nontoxicprint.com/etchzincsteelaluminum.htm

Green Death solution in general, up to 90 C.  You could add more copper and iron chlorides, more HCl, and some ntric acid for its oxidizing power.  For shallow etches, http://www.metallographic.com/Etchants/Etchants.htm, http://www.metallography.org/EtchCD.htm


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Dry ice and acetone
In my organic chemistry class there are several reactions that I’ve been told are generally conducted at $\pu{−78^\circ C}$, which is apparently also the temperature of dry ice in acetone.  This is all well and good, but I don’t really get what that actually means. Do you put the dry ice in the acetone and then toss in your reactants?  Do you add the dry ice and acetone and it makes the surrounding air colder?  Something else?
I’m just not sure how this is actually accomplished, and if you do add the reactants to the dry ice/acetone mix then how do you avoid reacting with either of those things?

You put broken up Dry Ice in in an evaporating dish in a bigger evaporating dish lined with glass wool, the whole atop a magnetic stirrer base, then carefully add acetone. Fizz.  Add and clamp your flask with stir bar under dynamic inert gas into the slush.  Go for it, solvent(s) and reactants.  Ace glassware has problems with star cracks.
Don't try this in a styrofoam bucket.  Note that acetone vapor is wildly flammable.  Cheesecloth atop NOT touching the slush for more insulation and to keep condensing moisture out.  Never plan to fill a flask more than half-full, because the universe hates you.

In some of the labs I interned in, liquid nitrogen was used to pre-cool the acetone (takes some trial-and-error) because dry ice had to be bought and liquid nitrogen was donated, i.e. essentially free. Some acetone would be lost in the process, though.

It's not usually the universe - just Murphy.

The mixture of dry ice and acetone is used outside your reaction vessel to keep the vessel cool just as a water bath is sometimes used to provide uniform heat to a reaction vessel. The only role is to provide a uniformly cool environment for the reaction vessel.
Acetone isn't the only choice here and isn't essential (anything still liquid at the temperature of dry ice will do). Acetone is cheap and (relatively) harmless which is why it is commonly used.

A liquid is needed to efficiently couple heat transfer.  Alcohols go viscous.  Nothing else improves upon acetone for price, odor, and toxicity.

Above two excellent answers have addressed your concern brilliantly. Thus, for yours and some other readers' benefit, I'll address why acetone/dry ice bath maintain at $\pu{-78 ^\circ C}$ temperature, which make it so popular in organic chemistry research laboratories worldwide.
Look at the phase diagram of dry ice (solid $\ce{CO2}$) below (from Wikipedia):

It clearly shows the $\ce{CO2}$ sublimation point at $\pu{1 atm}$ is $\pu{-78.5 ^\circ C}$. That means either sublimation or deposition (the reverse of sublimation) occurs at $\pu{-78.5 ^\circ C}$ ($\pu{194.65 K}$) at $\pu{1 atm}$ pressure. Thus, you may use any liquid with its melting point below $\pu{-78.5 ^\circ C}$ with dry ice, so the heat transfer occurs between dry ice and liquid such a way that the minimum equilibrium temperature of the mixture will be at $\pu{-78.5 ^\circ C}$ and maintain at that temperature as long as you supply enough dry ice. The solvent of choice could be acetone, methanol, ethanol, 2-propanol, hexane, and pentane, all of which have melting points below $\pu{-78.5 ^\circ C}$ (e.g., melting point of acetone is $\pu{-94 ^\circ C}$).
Additionally, if you need to maintain temperatures above $\pu{-78 ^\circ C}$, you can use a solvent with a freezing point exactly the required temperature, which is obviously above $\pu{-78 ^\circ C}$. For example, if you need to perform a chemical reaction at $\pu{-60 ^\circ C}$ instead of $\pu{-78 ^\circ C}$, you should choose diisopropyl ether, which freezes at $\pu{-60 ^\circ C}$ with dry ice. However, you must add dry ice slowly to avoid freezing the entire mixture, because when temperature reached $\pu{-60 ^\circ C}$, the solvent starts to freeze. Thus, keep adding dry ice as needed, just to maintain the solid-liquid slush.


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Can luciferin + ATP → luciferyl adenylate + PPi be reversed?
Reading about the light emitting reaction of luciferin + ATP, catalysed by luciferase I am confused about the following.
On this webpage the partial formula is written as:
luciferin + ATP → luciferyl adenylate + PPi
which seems to me like a one way reaction.
But on searching on Google "luciferase ATP equilibrium" I found websites like this one and this one which seem to indicate that the reaction is reversible, as there is an equilibrium.
Is it possible for this reaction to take place in reverse? And what about the link on books.google.nl showing a reversible reaction with light involved? What about that?
Thank you for clarifying.
EDIT 2014-02-25:
This article: "Kinetics of inhibition of firefly luciferase by dehydroluciferyl-coenzyme A, dehydroluciferin and L-luciferin" by Luis Pinto da Silva etc (Link) shows the following reactions:

Where Luc is firefly luciferase, LH2 is luciferin, L-CoA is dehydroluciferyl-coenzyme A, L is dehydroluciferin, ATP is Adenosine triphosphate, AMP is  adenosine-5`-monophosphate, PPi is inorganic pyrophosphate, O2 is oxygen, CO2 is carbondioxide.
Looking at the last equation, it seems it is possible to form ATP as a product, as it is on the right side of the arrow. Remarkable, I would say. still I do not see any luciferin (LH2) on the right side of any equation, so the answer to question of this topic remains no thus far.

In theory every chemical reaction is reversible.
The reaction that you mention does not emit any light yet, that comes later with the following reaction:
$$ \text{Luciferyl adenylate} + \ce{O2 ->} \text{Oxyluciferin} + \text{AMP} +h\nu $$
Since the reaction you mentioned occurs under loss of pyrophosphate ($\ce{PP_{i}}$) which is where the energy was stored beforehand, I guess it's somewhat hard to get it back.

It's almost like hoping that
$$ \ce{2CO2 + 3H2O -> C2H5OH + 3O2} $$
occurs under room temperature. Which it doesn't, because $\ce{CO2}$ is such a thermodynamical black hole. I think the same applies to your question.

Simple answer. I like it.

Reactions that are metabolically irreversible are usually so, because mechanisms are in place to keep the mass action ratio (Q) far away from the equilibrium constant (K). Recall that equilibrium occurs when Q = K.
The change in Gibb's free energy for standard conditions ($\Delta G°$) determines if the reaction is "inherently" thermodynamically favourable or not. But in reality the concentrations of reactants will be far away from their assumed 1 M values. Remember that the actual change in Gibb's free energy is related to $\Delta G°$ by the following expression:
$$\Delta_rG=\Delta_rG° + RT \ln Q$$
In this particular reaction, hydrolysis of ATP to release $\text{PP}_i$ makes the reaction exergonic / spontaneous ($\Delta G<0$). But in a hypothetical scenario with no substrates, and high concentrations of products (which would make $Q$ large, and thus $\Delta_r G$ positive), the reverse reaction would be spontaneous.
Now, if the reaction is practically irreversible, it's because things are being done to prevent such a situation from happening. For example, pyrophosphatase hydrolyses $\text{PP}_i$ to $2\text{P}_i$ at a high rate.
In many metabolic pathways, irreversible reactions are also catalysed by enzymes, that are inhibited by either their product, or a product downstream in the pathway, thereby inactivating the enzyme long before equilibrium is reached.


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How to make Sodium dichromate
I've been trying to make Sodium dichromate $\ce{Na2Cr2O7}$ by following these steps:

Dissolve an inox spoon in $\ce{H2SO4}$
Filter the solution
Dissolve the obtained sludge in $\ce{NaClO}$
Heat the solution until slurries appear
Filter the solution

The solution I obtained was yellowish (light yellow), I tried to add some acid to see the orange color of the (hypothetical) dichromate but nothing happened!
I added some ammonia before acidifying the solution, nothing as well.
1st: Is this method the right one to obtain chromate/dichromate?
2nd: If yes, why didn't the orange color appear?

While inox spoon's may indeed contain chromium, this reference states "at least 10.5 mass%" and "up to 26% is used for harsh environments".  So, it is still a small fraction of your spoon that constitutes chromium.  So it is more likely your first solution became yellow in color because of an other major constituent, iron. Yet I doubt a moderate oxidant like bleach is able to oxidize metallic chromium to yield a significant amount of chromates, $\ce{CrO4^{2-}}$ and dichromates $\ce{Cr2O7^{2-}}$, the later a known strong oxidizer.

bleach is able to oxidate chromium to chromate in strongly basic conditions, i.e. under excess of bleach in concentrated sodium hidroxide. However, the method itself is questionable, as inox contains too small amount of chromium.

Back in the days when cancer wasn't invented yet, people used to clean glassware with hot solutions of $\ce{CrO3}$ or $\ce{Na2Cr2O7}$ in sulfuric acid. If you work with (di)chromates, protect yourself and the environment. Please wear gloves and don't dump the solutions in the sink.

Does one of these steps isolate iron from chrome? If not is there any method to do it? Remark that I'm not chemist but have some basic notions of chemistry.
I wore gloves throughout the handling, how do I get rid of these materials then?

Note that the chromate ion $\ce{CrO4^{2-}}$ has chromium in the +6 oxidation state like dichromate $\ce{Cr2O7^{2-}}$. Chromate is yellow, dichromate is orange.
From the Wikipedia article on sodium chromate: sodium chromate is formed by reaction of dichromate with hydroxide:
$$\ce{Na2Cr2O7 + 2NaOH -> 2Na2CrO4 + H2O}$$
You may be making sodium dichromate, but the strongly basic conditions are converting it to sodium chromate.

When I add acid in excess, it normally (re-)convert to $\ce{Cr2O72-}$?


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Is melting of NaCl a physical and chemical process?
If I melt $\ce{NaCl}$, I get a molten salt. This means a liquid formed only by ions ($\ce{Cl^-}$ and  $\ce{ Na^+}$). That could mean that the crystal $\ce{NaCl}$ experimented a chemical change:
$$\ce{ NaCl->Cl- + Na+}$$ Like a chemical reaction... The ions have now more space to move and interact with much more ions than in the solid phase. I am trying to get an explanaition why the melting temeprature and the melting heat of molten salts show some erratic behaviour, I mean they do not follow a simple pattern such as that found for n-alkanes or n-alkanols.

In fact the lattice structure of $\ce{NaCl}$ is composed by $\ce{Na^+}$ and $\ce{Cl^-}$ even when is in the solid state we don't write it with the charge only for convention.
Ionic compounds are different from organic compound, they have complex lattice structure and are in fact compose by two species. Breaking a ionic bond required more energy than an organic compound where you deal only with intramolecular forces.

Your last sentence is a little bit dangerous if taken at face value. In an ionic lattice, the atoms have an "ionic bond" with every other atom in the lattice, no matter how far. It turns out however that most of the electrostatic attraction comes from the counterions in the immediate surroundings That attraction is very strong, however. In a gas of an ionic compound, you will actually often find ion pairs and a few larger agglomerates even at very high temperatures. Only at massive temperatures (>5000 K) should you expect the last ionic bond between ion pairs to give way, producing free ions.

@NicolauSakerNeto which sentence? Breaking a ionic bond required more energy than an organic compound where you deal only with intramolecular forces.? Do you mean when you melt salts you don't break ionic bond? Thanks for the comment

In a liquid ionic compound, you "break the bond" with the furthest ions, in the sense that the liquid has no long-range order as it does in the solid. But each ion is still quite tightly "solvated" by ionic bonds with nearby counterions. The attraction between two oppositely charged ions is so strong that it survives even when you turn the salt into a gas.

In sodium chloride, the ionic bond between a single $Na^+$ and a single $Cl^-$ is approximately half the lattice energy of the entire solid. That means an ionic bond energy of about $400 kJ/mol$, comparable to the strongest covalent single bonds. This arguably means the ionic bond still survives until about when you expect covalently bound molecules to thermally decompose into separate atoms, which requires huge temperatures.

@NicolauSakerNeto Thanks for the comment, I should explain better what I mean for two species what I would  underline is the fact that the interaction is not between identical compounds (such an organic solid) but between a cation and an anion (of course a series of each other).


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Intermolecular forces in secondary amines compared to primary amines

Secondary amines form hydrogen bonds, but having nitrogen in the middle of the chain rather than at the end makes the permanent dipole on the molecule slightly less. The lower boiling point is due to the lower dipole-dipole attractions in the dimethylamine compared with ethylamine.

Could someone please tell me why nitrogen, being in the middle of the molecule (as in secondary amines) causes the permanent dipole to be slightly less.

A primary amine has two $\ce{N-H}$ bonds and one $\ce{N-C}$ bond. A secondary amine has two $\ce{N-C}$ bonds and one $\ce{N-H}$ bond. Since $\ce{N-H}$ is more polarised than $\ce{N-C}$ we can expect ethyl amine to be more polar than dimethyl amine.

I would have gone for a less connected hydrogen bonding network as the primary engine.  Diethyl amine boils at 55.6 °C, 1.11 debyes.  n-Butylamine boils at 77.1 °C, 1.37 debyes (Organic Solvents, 4th Ed., Riddick, et al., 1986).
http://www.chemguide.co.uk/organicprops/amines/background.html
   Both appear to be important.

Same as a branched alkanes and alkenes, there is less compactness (surface area exposed to the other compound); between the alkanes/alkenes, hence less van der Waals attraction between them; as compared to straight alkanes/alkenes.  So, intermolecular forces of attraction is not too effective. This also applies here, having bulky groups (methyl, ethyl, propyl, …) attached to the nitrogen atom, make it more branched and less compactness (less surface area) , means less intermolecular forces, hence lower boiling point.
Guess that's the most obvious reason, because actually, the  electron donating group such as methyl, ethyl and propyl, would have given more electrons to the nitrogen making it more negative, hence expecting to have a more electronegative nitrogen; which might result in stronger dipole-dipole bond, but this isn't the case.  So, the only explanation that I could thought of is the one I explained above.

I'm not sure I quite follow this answer. In fact, I'm pretty sure that it is going entirely in the wrong direction.


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Molecule with 13 total electrons
Is there a stable molecule with total number of electrons equal to 13? Molecule can be either organic or inorganic

Aluminium metal? atomic number 13...

Apart from $\ce{Al}$ I am not sure whether they exist. You're pretty much always going to need period 2 atoms for this number of electrons and all the molecules I can dream up have an even number of electrons. Maybe someone more experienced in organic synthesis/plenty of knowledge on molecular bonds can explain why (if) this is the case?!

Nitric oxide NO is stable radical, but sadly, with 15 electrons.

There are quite some molecules that are persistent (= live long enough in solution to allow for a spectroscopical characterization, but cannot be isolated) or even stable (= can be isolated, bottled and/or used as a starting material) that have an odd number n of electrons, however $n \ne 13$.

Among these radicals are the Gomberg radical (1), the Koelsch radical (2), and the TEMPO radical (3).

A cyanide radical has 13 electrons. This stable radical was one of the first detected molecules in the interstellar medium, 1941 by A. McKellar, doi:10.1086/125159
Stable: persists in vacuum; no imaginary IR frequencies (negative eigenvalues). Reactive: diddles with something else, including another one of itself. Given the name, do your chair parade. The unpaired spin mostly resides on the nitrogen. Chemistry is passion, not compassion


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What is the melting point of diamond?
Textbooks and the online reference differ about this and there are more than two answers.

Some additional information can be found here and the links within.

melting point of diamond is 3823K

I wish this question would include a question …

Liquid diamond certainly sounds cool. It even glows!

Diamond (carbon) does not melt at 1 atm. It sublimes to vapor.
Using carbon's theoretical phase diagram below (from Wikimedia), "liquid diamond" could be achieved at about 10 GPa (99 thousand atmospheres) and 5000 K (4700 °C).

Edit:
In fact, heating up diamonds at 1 atm turns them into graphite first. Only further heating would cause sublimation, as expected by observing the phase diagram. Here's a video:
http://www.youtube.com/watch?v=7L7BV3IBfFA

It is 3783 K in reference books

Note that if heating it in the atmosphere, it will burn to carbon dioxide much before it sublimates to vapour carbon. HEATING DIAMONDS IS A BAD IDEA!

While I understand that the phase diagram you're using is mainly focused on the high temperature and pressure region, I can't help but feel that the boundaries of the metastable region could use some fine-tuning. Taken literally, it seems to be claiming that diamond isn't (meta)stable at room temperature and pressure(!). (Apparently, the actual thermal stability limit of diamond at low pressure is about 2000 K.)

Definition (wikipedia): Sublimation - is the phase transition of a substance directly from the solid to the gas phase without passing through the intermediate liquid phase.

Minor point:  The quoted temperature conversion is mathematically correct, but in all probability no more than two digits are significant.  Thus $5000 K=4700°C$.


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Evaporation velocity of ethanol: adding ethanol vapour to a gas mixture
I need to add a known amount of ethanol vapour to a gas mixture (He/Ar). To do that I let the mixture bubble in liquid ethanol kept at a known temperature (14 C).
The system is made of a large steel cylinder (d=20 cm; l=138cm) lying in horizontal position and half filled with liquid ethanol. the gas is bubbled trough the liquid on one side of the cylinder and extracted on the other side.
The gas-flow trough the system is usually pretty low (~100ml/min) but I would like to increase it to ~10l/min and I need to know if this leaves enough time for the ethanol vapour to approach its equilibrium vapour pressure.

Please note that I have never done this.
But are you sure that you want to rely on approximation/assumptions?
I realize that it needs additional changes to the experimental setup, but is measuring the ethanol concentration out of question? Gas sensors for ethanol are commercially available and a setup, in which they steer mass flow controllers is conceivable.
As a side note, you might be interested to see that ethanol sensors can be build in the lab:
A Novel Flexible Room Temperature Ethanol Gas Sensor Based on SnO2 Doped Poly-Diallyldimethylammonium Chloride.

Thanks for the reply! I would like to have at least an estimate, an upper limit on the maximum flux would be enough. I am also planning to measure the amount of evaporated ethanol after continuously flushing at high flux for a long time.
For purity reasons, I would prefer not to use and ethanol sensor, but that it certainly a good idea. On the other hand if a rough calculation shows the bubbler is big enough I can save some money! :)

The article you linked is interesting, but building such a sensor is beyond the reach of my lab, unfortunately.


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What are the maximum number of electrons in each shell?
In my textbook, it says that the maximum number of electrons that can fit in any given shell is given by 2n². This would mean 2 electrons could fit in the first shell, 8 could fit in the second shell, 18 in the third shell, and 32 in the fourth shell.
However, I was previously taught that the maximum number of electrons in the first orbital is 2, 8 in the second orbital, 8 in the third shell, 18 in the fourth orbital, 18 in the fifth orbital, 32 in the sixth orbital.  I am fairly sure that orbitals and shells are the same thing.
Which of these two methods is correct and should be used to find the number of electrons in an orbital?
I am in high school so please try to simplify your answer and use fairly basic terms.

You might find this answer of mine useful.

Shells and orbitals are not the same. In terms of quantum numbers, electrons in different shells will have different values of principal quantum number n.
To answer your question...
In the first shell (n=1), we have:

The 1s orbital

In the second shell (n=2), we have:

The 2s orbital
The 2p orbitals

In the third shell (n=3), we have:

The 3s orbital
The 3p orbitals
The 3d orbitals

In the fourth shell (n=4), we have:

The 4s orbital
The 4p orbitals
The 4d orbitals
The 4f orbitals

So another kind of orbitals (s, p, d, f) becomes available as we go to a shell with higher n. The number in front of the letter signifies which shell the orbital(s) are in. So the 7s orbital will be in the 7th shell.
Now for the different kinds of orbitals
Each kind of orbital has a different "shape", as you can see on the picture below. You can also see that:

The s-kind has only one orbital
The p-kind has three orbitals
The d-kind has five orbitals
The f-kind has seven orbitals

Each orbital can hold two electrons. One spin-up and one spin-down. This means that the 1s, 2s, 3s, 4s, etc., can each hold two electrons because they each have only one orbital.
The 2p, 3p, 4p, etc., can each hold six electrons because they each have three orbitals, that can hold two electrons each (3*2=6).
The 3d, 4d etc., can each hold ten electrons, because they each have five orbitals, and each orbital can hold two electrons (5*2=10).
Thus, to find the number of electrons possible per shell
First, we look at the n=1 shell (the first shell). It has:

The 1s orbital

An s-orbital holds 2 electrons. Thus n=1 shell can hold two electrons.
The n=2 (second) shell has:

The 2s orbital
The 2p orbitals

s-orbitals can hold 2 electrons, the p-orbitals can hold 6 electrons. Thus, the second shell can have 8 electrons.
The n=3 (third) shell has:

The 3s orbital
The 3p orbitals
The 3d orbitals

s-orbitals can hold 2 electrons, p-orbitals can hold 6, and d-orbitals can hold 10, for a total of 18 electrons.
Therefore, the formula $2n^2$ holds! What is the difference between your two methods?
There's an important distinction between "the number of electrons possible in a shell" and "the number of valence electrons possible for a period of elements".
There's space for $18 \text{e}^-$ in the 3rd shell: $3s + 3p + 3d = 2 + 6 + 10 = 18$, however, elements in the 3rd period only have up to 8 valence electrons. This is because the $3d$-orbitals aren't filled until we get to elements from the 4th period - ie. elements from the 3rd period don't fill the 3rd shell.
The orbitals are filled so that the ones of lowest energy are filled first. The energy is roughly like this:
$$1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s$$
An easy way to visualize this is like this:

The pattern of maximum possible electrons = $2n^2$ is correct.
Also, note that Brian's answer is good and takes a different approach.
Have you learned about quantum numbers yet?

If not...

Each shell (or energy level) has some number of subshells, which describe the types of atomic orbitals available to electrons in that subshell. For example, the $s$ subshell of any energy level consists of spherical orbitals. The $p$ subshell has dumbbell-shaped orbitals. The orbital shapes start to get weird after that. Each subshell contains a specified number of orbitals, and each orbital can hold two electrons. The types of subshells available to a shell and the number of orbitals in each subshell are mathematically defined by quantum numbers. Quantum numbers are parameters in the wave equation that describes each electron. The Pauli Exclusion Principle states that no two electrons in the same atom can have the exact same set of quantum numbers. A more thorough explanation using quantum numbers can be found below. However, the outcome is the following:
The subshells are as follows:

The $s$ subshell has one orbital for a total of 2 electrons
The $p$ subshell has three orbitals for a total of 6 electrons
The $d$ subshell has five orbitals for a total of 10 electrons
The $f$ subshell has seven orbitals for a total of 14 electrons
The $g$ subshell has nine orbitals for a total of 18 electrons
The $h$ subshell has eleven orbitals for a total of 22 electrons

etc.
Each energy level (shell) has more subshells available to it:

The first shell only has the $s$ subshell $\implies$ 2 electrons
The second shell has the $s$ and $p$ subshells $\implies$ 2 + 6 = 8 electrons
The third shell has the $s$, $p$, and $d$ subshells $\implies$ 2 + 6 + 10 = 18 electrons
The fourth shell has the $s$, $p$, $d$, and $f$ subshells $\implies$ 2 + 6 + 10 + 14 = 32 electrons
The fifth shell has the $s$, $p$, $d$, $f$, and $g$ subshells $\implies$ 2 + 6 + 10 + 14 + 18 = 50 electrons
The sixth shell has the $s$, $p$, $d$, $f$, $g$, and $h$ subshells $\implies$ 2 + 6 + 10 + 14 + 18 + 22 = 72 electrons

The pattern is thus: $2, 8, 18, 32, 50, 72, ...$ or $2n^2$
In practice, no known atoms have electrons in the $g$ or $h$ subshells, but the quantum mechanical model predicts their existence.

Using quantum numbers to explain why the shells have the subshells they do and why the subshells have the number of orbitals they do.

Electrons in atoms are defined by 4 quantum numbers. The Pauli Exclusion Principle means that no two electrons can share the same quantum numbers.
The quantum numbers:

$n$, the principle quantum number defines the shell. The values of $n$ are integers: $n=1,2,3,...$
$\ell$, the orbital angular momentum quantum number defines the subshell. This quantum number defines the shape of the orbitals (probability densities) that the electrons reside in. The values of $\ell$ are integers dependent on the value of $n$: $\ell = 0,1,2,...,n-1$
$m_{\ell}$, the magnetic quantum number defines the orientation of the orbital in space. This quantum number also determines the number of orbitals per subshell. The values of $m_\ell$ are integers and depend on the value of $\ell$: $m_\ell = -\ell,...,-1,0,1,...,+\ell$
$m_s$, the spin angular momentum quantum number defines the spin state of each electron. Since there are only two allowed values of spin, thus there can only be two electrons per orbital. The values of $m_s$ are $m_s=\pm \frac{1}{2}$

For the first shell, $n=1$, so only one value of $\ell$ is allowed: $\ell=0$, which is the $s$ subshell. For $\ell=0$ only $m_\ell=0$ is allowed. Thus the $s$ subshell has only 1 orbital. The first shell has 1 subshell, which has 1 orbital with 2 electrons total.
For the second shell, $n=2$, so the allowed values of $\ell$ are: $\ell=0$, which is the $s$ subshell, and $\ell=1$, which is the $p$ subshell. For $\ell=1$, $m_\ell$ has three possible values: $m_\ell=-1,0,+1$. Thus the $p$ subshell has three orbitals. The second shell has 2 subshells: the $s$ subshell, which has 1 orbital with 2 electrons, and the $p$ subshell, which has 3 orbitals with 6 electrons, for a  total of 4 orbitals and 8 electrons.
For the third shell, $n=3$, so the allowed values of $\ell$ are: $\ell=0$, which is the $s$ subshell,  $\ell=1$, which is the $p$ subshell, and $\ell=2$, which is the $d$ subshell. For $\ell=2$, $m_\ell$ has five possible values: $m_\ell=-2,-1,0,+1,+2$. Thus the $d$ subshell has five orbitals. The third shell has 3 subshells: the $s$ subshell, which has 1 orbital with 2 electrons, the $p$ subshell, which has 3 orbitals with 6 electrons, and the $d$ subshell, which has 5 orbitals with 10 electrons, for a  total of 9 orbitals and 18 electrons.
For the fourth shell, $n=4$, so the allowed values of $\ell$ are: $\ell=0$, which is the $s$ subshell,  $\ell=1$, which is the $p$ subshell, $\ell=2$, which is the $d$ subshell, and $\ell=3$, which is the $f$ subshell. For $\ell=3$, $m_\ell$ has seven possible values: $m_\ell=-3,-2,-1,0,+1,+2,-3$. Thus the $f$ subshell has seven orbitals. The fourth shell has 4 subshells: the $s$ subshell, which has 1 orbital with 2 electrons, the $p$ subshell, which has 3 orbitals with 6 electrons,  the $d$ subshell, which has 5 orbitals with 10 electrons, and the $f$ subshell, which has 7 orbitals with 14 electrons, for a  total of 16 orbitals and 32 electrons.

The first shell can carry up to two electrons, the second shell can carry up to eight electrons.
The third shell can carry up 18 electrons, but it is more stable by carrying only eight electrons.  There is a formula for obtaining the maximum number of electrons for each shell which is given by $2n^2~\ldots$ where n is the position of a certain shell.


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Is Linus Pauling's General Chemistry still readable and up-to-date today?
I've found a Dover edition copy of Pauling's General Chemistry in the bookstore near me. I'd like to know if it's still readable today, I mean, is it still up-to-date? I have no background in chemistry, but the book was cheap and it seems to be a nice read.

I currently have the Dover copy of Pauling's book on my shelf.  It indeed is a nice read but it is in no way 'up-to-date' as it was written many years ago.  That being said, a lot of conceptual information presented in the book still apply just as much today as they did 6 decades ago (I would stray away from using the book as a source for benchmark measurements and such).  It is a book that should find its way onto any chemist's bookshelf, and it makes for a fascinating read into many areas of general chemistry.


text:
Are all organic molecules with odd-numbered electron count radicals?
This question was triggered by another question on this site, about molecules with a total number of 13 electrons.
I couldn't think of any (except the boring answer of $\ce{Al}$), and the answers to the question are all about radicals, instead of 'normal' molecules. So that got me thinking: are all organic molecules with an odd total electron count radicals or radical ions? If so, is there any fundamental reason for this?

Some of them are even "worse": they are radical ions ;) On the other hand, some species with an even number of electrons aren't "normal" either: Think in carbenes or nitrenes ;)

Good point! I just edited it into the question. I'm really curious whether there is any fundamental reason for the lack (absence?!) of 'normal' molecules with an odd total number of electrons

If you pair up all single electrons, you will always get an even number (any integer * 2 is even). If you don't, you will have an unpaired electron, and that's what a radical is.

@Brian mmm, good point. Didn't think of that, but makes perfect sense

Are we counting organometallics?

@Klaus-DieterWarzecha don't forget the hydride. Smallest Lewis base I could think of.

By IUPAC definition of radical, the organic molecules with odd number of electrons must have at least one of them unpaired and are therefore radicals. The fundamental reason is the definition.
Bear in mind, that other spin states must be taken into account, as simplest example the oxygen. It exists as a triplet, so two unpaired electrons.
The situation gets different, when you do have a metal in your compound, but it is out of the scope of the question.


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What would explain urine that resists freezing?
A frozen urine specimen is required for an organic acids lab test, but the urine refuses to freeze after 4 days in a regular freezer turned as low as it will go. What would explain this, and how cold might it need to be to freeze?

Various not-so-legitimate sources say around -120 to -80 degrees celcius. I'd be surprised if there is a simple way to estimate the actual freezing temperature in this case.

That definitely sounds very far from the mark. A slight bit of time with a search engine afforded this source, which states that the freezing point of urine is normally between -1.3 and -2.2°C. I suspect either the freezer is not working properly, or the urine has been significantly preconcentrated, or the urine has been mixed with something else.

Add a tiny jot of ground glass, a boiling chip, or scratch the side of the container with a glass stir rod.  You need  nucleation centers.

@UncleAl: unless its sterile filtered there will be a few cells already in there (e.g. exfoliated epithelial cells) so nucleation centers should not be a problem.

I'd procude a control sample myself and put it in there to see what happens. Of course, tap water does freeze, right?

Urine is an aqueous solution of urea, uric acid, sodium salts and whatnot. These solutions usually exhibit freezing-point depression.
As Brian has correctly pointed out in a comment, the effect isn't that strong and a quick and dirty look in the literature shows that a freezing point around -1.3 to -2.2 °C.
You might want to try a cooling bath with dry ice in isopropanol or acetone or other drastic methods to achieve shock-freezing.

Freezing point depression, as caused by salts such as NaCl, is usually on a much smaller scale. I tried calculating using concentrations from wikipedia, and the freezing point depression came out to around .5 K. Why do we see this 80-120 K freezing point depression?

some additions can reduce freezing points dramatically.

According to another source, freezing point depression based on the urea content is -5.6 °C.


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Activity coefficient greater than 1
From Practical skills in Biology by Allan Jones et al.

At high concentrations of certain ionic solutes, $\gamma$ (activity
  coefficient) may increase to become greater than unity.

My questions : If activity coefficient $\gamma$ $=$ $  {a}/{[C]} $ where $a$ is activity which signifies effective concentration and $[C]$ is concentration, how can $a$  be greater than $[C]$ ?

A) Gamma < 1 means attractive forces dominate, and there is a decrease of vapour pressure of solute A relative to pure A. B) Gamma > 1 means repulsive forces dominate, and there is an increase of vapour pressure of solute A relative to pure A.  You can think of activity coefficients as a qualifier for "escaping tendency" of the solute.  With case A) above, the partial molar excess Gibbs energy of mixing is negative, meaning the solvent stabilises the solute, and energy is given away as a result of the mixing event (i.e. mixing is spontaneous). With case B) above, the partial molar excess Gibbs energy of mixing is positive meaning the solvent destabilises the solute, and energy is required in order to achieve mixing (i.e. mixing is NOT spontaneous). Think of activity as escaping tendency - a small activity means the solute remains in the solvent due to attractive forces whereas a large activity indicates a large escaping tendency of the solute because repulsive forces dominate.  a = gamma x [conc.], right? So, if gamma > 1, then a is going to be larger than [conc.]. It might help you to consider fugacity, which is the analogue of activity for gases (which employs similarly the fugacity coefficient). Also consider an aqueous solution of ammonium hydroxide.  You make it to be X M, right?  But you can always smell ammonia since the solution you made is losing quantities of ammonia.  Is the gamma for NH3 > or < than 1?

In chemical thermodynamics, activity (symbol a) is a measure of the “effective concentration” of a species in a mixture, in the sense that the species' chemical potential depends on the activity of a real solution in the same way that it would depend on concentration for an ideal solution.
(From Wikipedia.)

So the main point is that that activity is just a measure of the deviation from the ideal solutions, because the laws you are using in physical chemistry  assume that solution is ideal. But usually it is not true. Laws can work for real life with a certain precision, but you should always remember about their "ideality" and the non-ideality of our world.

Hi! Thank you for your input. But I still don't understand how gamma can be greater than 1.

What is meant by effective concentration is whatever number makes the equilibrium equation work.  To a first approximation it is true that, for example, [Na$^+$][Cl$^-$] is a constant in any solution saturated with NaCl at a given temperature, regardless of whatever else might be in the solution; but if you add lots of HCl, although [Na$^+$] does go down as [Cl$^-$] goes up, the product is not exactly constant. You can tabulate values for each concentration that will make the product constant, and these are called activities.  As @Gordon Driver mentioned, these values can be either greater or less than the actual concentration, depending on the specific system.


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Chemical equilibria and changes in concentration
I learned that if concentration of reactants or products is changed, the position of equilibrium changes but value of $K_c$ remains the same.
This just does not make intuitive sense to me... I mean whatever concentration of reactants I start with does not matter, $K_c$ remains same....?
Also, can someone please elaborate the term position  of equilibrium

Assuming the following reaction:
$$\ce{A + B <=> C + D}$$
Where:
$$K_\text{c}=1.23\times10^{-2}$$
We say that:
$$K_\text{c}=\frac{\ce{[C]*[D]}}{\ce{[A]*[B]}}=1.23\times10^{-2}$$
And the reaction will be at equilibrium for any concentrations (actually, it should be activities, which are in essence effective concentrations) of A, B, C and D that satisfy the above.
I can understand your confusion. What they mean is... If you have a reaction at equilibrium, and remove half the reactants, then the reaction will proceed towards reactants, until equilibrium is once again reached.

Their way of saying it:

You remove reactants, and the position of equilibrium is shifted towards the left, because the reaction will now proceed towards the left hand side to restore equilibrium.
It's confusing, because equilibrium occurs at the same conditions as before, so you haven't really shifted the equilibrium position in the way it sounds like.

In general, $K_c=1$ does not imply that you have equal amounts of reactants and products. Consider the reactions $\ce{A + B -> C}$ or $\ce{A + B -> 2C}$ or $\ce{A -> B + C}$.

You're right of course


text:
What is the name of this molecule? (C₈O₃H₁₃N)
@tschoppi
Tried taking a stab at this for me and got this much:
1-(1'-oxoethyl)-2-(something-something ester)-pyrolidine

@Singh Please do not use markup in the title, compare http://meta.chemistry.stackexchange.com/q/261/4945

If I managed to see the structure correctly it should be Methyl 1-acetylprolinate

The natural (S)-isomer.


text:
Is it possible to create a salt with multiple anions with only one cation(e.g. CaBrCl)?
I know that usually more electronegative anions would displace less electronegative anions but is there a way to get different anions to "share" a cation?

A cursory search showed some interesting cases.
The oldest reported examples seemingly are mercury halides $\ce{HgXY}$.

$\ce{HgClI}$ was prepared by iodination of solid $\ce{HgCl2}$
$\ce{HgBrCl}$ was obtained from the chlorination of an ethanolic suspension of $\ce{HgBr2}$
$\ce{HgBrI}$ was prepared by mixing ethanolic solutions of $\ce{HgBr2}$ and $\ce{HgI2}$

The crystal structure of $\ce{Cs6Cl3[Fe(CN)6].H2O}$ has been reported.
Passing $\ce{H2S}$ through solutions of oxidosulfido molybdates and tungstates let to various mixed salts with halide, hydrogen sulfide and thiosulfate as additional anions beside the trisulfidometalates $\ce{[MoOS3]^{2−}}$ and $\ce{[WOS3]^{2−}}$.
But the concept isn't limited to inorganic chemistry.
The crystal structures of mixed salts of amino acids with different anions were reported.
From the field of organic conductor research, salts of BEDT-TTF (bis(ethylenedithio)tetrathiafulvalene) with mixed anions of $\ce{[CuBr2]-}$ and $\ce{[CuCl2]-}$ are known.

Mixed anion salts are more interesting than I thought! I like how the Hg salts have different properties depending on the order that anions are switched.

Sure, why not?  In the bad old days geology claiming isomorphous substitution had a hissy fit with chemistry claiming exact stoichiometry.  Rochelle salt is potassium sodium tartrate. Potassium aluminum sulfate (alum) is often doped with colored transition metal ions substituting for some of the aluminum (chrome alum).  Ruby, where Cr(III) substitutes for some (Al(III).
If size is about the same for identical charge, the alien ion goes into the crystal lattice.

This answer discusses salts with varying cations, while the question asks for anions.

It works the same way either way.

One example that I stumbled upon: Mixed halide compounds, $\ce{MFX}$ where $\ce{M=Ca,Sr,Eu,Ba}$ and $\ce{X = Cl,Br,I}$.

$\ce{MFX (M = Ca, Sr, Eu, Ba; X = Cl, Br, I)}$ compounds have been prepared by solid-state reaction. Lattice parameters and X-ray diffraction patterns are presented for these compounds, which are all isostructural with tetragonal $\ce{PbFCl}$. Attempts to synthesize solid solutions of $\ce{Sr(Eu)FCl}$ and of $\ce{MFCl}$ compounds with several rare earth oxychlorides are reported.

Reference

H.P. Beck, A study on mixed halide compounds MFX (M = Ca, Sr, Eu, Ba; X = Cl, Br, I), Journal of Solid State Chemistry, Volume 17, Issue 3, 1976, DOI: 10.1016/0022-4596(76)90132-8.