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Presumptive test for modafinil?
Given a sample that is probably modafinil plus common fillers, but could be fake, what reagent tests would best determine whether it contains modafinil? Assume a known-good sample for comparison.
(Modafinil is a sulfur-containing atypical stimulant which has become common in some circles, but which, due to off-label use and the high price of the US name brand, is mostly obtained by mail order from slightly-sketchy websites in India and the UK. There are rumors that some fraction of the modafinil imported this way is fake, but this is only rumor because the effects vary to begin with, and most people don't seem willing to pay for laboratories to do proper testing.)

Why are you limited to reagent tests? Could you acquire an NMR spectrum of the compound?

@MadScientist - it sounds like the OP might want a kitchen-level test to distinguish good/bad when he opens the package after delivery.

Yep, pretty much that, with the additional wrinkle that a lot of the testing should be done by third parties who're more cost-sensitive than I am. Saturating the market with reagent test kits is feasible in a way that saturating it with gift certificates for NMR is not.

Which circles is it popular in?

@jimrandomh: Modafinil as a molecule is a sufficiently large and complex organic molecule to mean that I'd be fairly confident that there isn't really going to be a good 'add this to it and only modafinil does X' test. You could consider testing physical properties and something like solubility and pH instead; they're hardly unique, but relatively easy to test.
I don't think there's even a good easy reagent test for sulphoxides, unfortunately.

Honestly, I don't think the average kitchen is equipped to identify (with any accuracy) the presence of this molecule in pills that likely vary in impurities, fillers, etc. depending on the website you buy them from.
With that said, I don't think it's unreasonable that someone could develop an inexpensive testing kit that could at least give a "positive" or "negative".
The area of the molecule I would focus on is not the amide or sulfoxide, but the diphenyl group attached to the sulfoxide. These conjugated rings will absorb UV (literature suggests lambda max around 220-230 nm). So you have a molecule that absorbs UV and is hydrophobic, and you need to separate it from the other pill contents for analysis. One way to do this would be with a mini-TLC kit, where the Modafinil could be detected as dark spots under UV light.
Example Kit:

Small vial + cap
Organic solvent (this would be determined by your
own testing) in dropper bottle
TLC plate(s)
Small glass capillary (for spotting plates)
UV-LED Light
Cool UV-protective glasses

Example Procedure:

Crush pill, add to vial
Add a few drops of organic solvent (provided by you in dropper bottle)
Cap vial, shake well (trying to extract Modafinil into organic phase)
Spot TLC plate, elute with appropriate solvent
Compare Rf of spot to the Rf of Modafinil standard (alternatively, the presence or absence of the Modafinil spot may be enough to provide a positive or negative result).

If you are the person developing such a kit, calibration would need to be done using a known "good" sample of the Modafinil. I would be careful to watch for other UV absorbing impurities that may be found in the pills (e.g. reagents from synthesis). You could confirm these results using an HPLC with UV detector or NMR (or whatever other appropriate techniques you have access to).
If all else fails, do as the previous answer says and smell your pee!

Complex structure
Modafinil is a pretty complex molecule and judging by its structure (see below) I think it will be extremely difficult to find a single reagent test that will identify it.

Post-consumption identification
Interestingly it is much easier to find after consumption, either by the strong sulfur smell of your urine or through chromatography techniques (also here).
'Sketchy' tests
There are apparently some simple tests you can try like dissolving it in hot water and vinegar (which should happen without formation of gas), but these tests are far from fool-proof and might even cause you to think something is a fake while it is not.
Combination of physical/chemical tests as a 'solution'
The best thing to do if you really want 'kitchen-level' tests is to test for a whole range of chemical/physical properties. For example this reference has some information on the pH, water solubility and melting point which are fairly easy to test. Passing the tests will certainly not guarantee anything, but at least it will take away a significant part of the doubts that the product is fake.
Something to consider when designing these tests is that there will always be fillers (e.g. lactose monohydrate, microcrystalline cellulose, pregelatinized starch, croscarmellose sodium, povidone, and magnesium stearate) in the sample so you have to find ways to correct for that.


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How to determine the order of boiling points and solubilities for (chloro-)alkanes, ethers, aldehydes, and alcohols?
How to determine the order of boiling points and solubilities for (chloro-)alkanes, ethers, aldehydes, and alcohols? What rules can be applied to make these choices?
Take for example the following question:

Which one of the following compounds  has the highest solubility in water?
  Which one of the following compounds  has the lowest boiling point?

ethane, $\ce{CH3CH3}$
dimethylether, $\ce{CH3-O-CH3}$
acetaldehyde, $\ce{CH3CHO}$
methanol, $\ce{CH3OH}$
chloromethane, $\ce{CH3Cl}$
methane, $\ce{CH4}$

Solubility in water
The two most important (and somewhat related) factors are the polarity of the molecule and the presence of hydrogen bonding groups. As a rule of thumb, the higher the polarity the more soluble a molecule will be in water (which is highly polar as well). Also, the more hydrogen bonding groups, the more soluble in water.
With these two rules of thumb you should be able to order the compounds you listed. I'll leave it to you to actually do it.
Boiling point
The website masterorganicchemistry.com[1] gives a fairly good summary of 3 important factors that determine the boiling point of organic compounds. They are:

The relative strength of the four intermolecular forces is: ionic > hydrogen bonding > dipole dipole > van der Waals dispersion forces. The influence of each of these attractive forces will depend on the functional groups present, but generally stronger forces means a higher boiling point.
Boiling points increase as the number of carbons is increased.
Branching decreases boiling point.

With these 3 'rules' you should be able to figure out the ordering of your listed compounds.
[1]: Master Organic Chemistry: 3 Trends That Affect Boiling Points or via the Internet Archive

source: Stack Exchange

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How do you find the pH of a solution with a known pKa and Molarity?
How do you find the pH of a solution with a known pKa and Molarity?

Do you know the formula:
$$K_\text{a} = \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}$$
?
You will need to find a concentration first, however, on the equation.
$$\ce{HA <=> H+ + A-}$$
And solve for the dissociated hydrogen ion, say $x$.
$$K_\text{a} = \frac{[x][x]}{[\ce{HA}]-x} = \frac{x^2}{[\ce{HA}]-x}$$
The logic is: if $x$ of $\ce{HA}$ dissociated, then we would get $x$ of $\ce{H}$, $x$ of $\ce{A}$ and $[\ce{HA}]-x$ left of $[\ce{HA}]$.
Does this make sense?

the expresssion you wrote is for $K_a=10^{-pK_a}$. The p subscript means $-log[...]$

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At equilibrium, the mass of reactants and products are equal. Does that mean the reaction stops?
This one is troubling me, and I even got it wrong in my exam:

at equilibrium the mass of reactants and products are equal, does that mean the reaction stops?

Please explain this one.

By "mas" do you mean "mass?

No, the system is in a state of "dynamic equilibrium".  This means that the forward reaction and the reverse reaction are still taking place, but at the same rate so the concentrations of each don't change.

At equilibrium, the reaction continues, but the rate of the forward reaction is equal to the rate of the backward reaction.
Let's consider the first order reaction $\ce{A <=>[k_f][k_b] B}$
So, the rate of the forward reaction is $k_f[A]$, and the rate of the backward reaction is $k_b[B]$. Initially, $[B]=0$, and $[A]$ is high, so the forward reaction is very fast, and the backwards reaction doesn't occur at all. As time passes, $[A]$ decreases and $[B]$ increases, so the forward reaction becomes slower and slower, and the backwards reaction speeds up. At one point, both reactions have equal rates ($k_f[A]=k_b[B]$). Here, both reactions take place, but for every time when an $A$ becomes $B$, another $B$ becomes $A$. The reaction is taking place, just that you don;t notice it since the net effect is zero.
Here's a diagram for how the rates of the reaction proceed for the formation of ammonia. Note that the rates never reach 0:

A reaction at equilibrium never stops (by itself), it rather remains in a dynamic equilibrium where forwards and reverse reactions occur at the same rate. However, macroscopically you cannot differentiate between a reaction that stopped and a dynamic equilibrium.
Also another note: It is wrong to say that the masses of reactants and products be equal at equilibrium. You cannot even say that the amounts, concentrations or any other variable is equal for reactants and products. Rather, for each reaction there is a specific ratio of concentrations that is reached at equilibrium, commonly known as the equilibrium constant.
$$\ce{$a$A + $b$B <=> $c$C + $d$D}$$
$$k_\mathrm{eq} = \frac{[\ce{C}]^c [\ce{D}]^d}{[\ce{A}]^a [\ce{B}]^b}$$
The mass ratio at an equilibrium can be anything. Most notably, it can be both $1\,:\,1$ and $10^{99}\,:\,1$ and there is no rule as to which mass ratio would be present at a given equilibrium.

Because at equilibrium state forward and reverse reactions occur at same place but they occur in opposite directions so reaction don't stops..

The forward and reverse reactions always occur in opposite directions. That's why they are called forward and reverse. They also always occur in the same place - the reaction vessel. I think what you are trying to say is that the rate of the forward and reverse reaction is the same but this has already been comprehensively covered in other answers so this doesn't add anything new.

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Predicting molecular geometry from a formula
Can someone explain the following types of molecules and predict their structure and shape?
It will help my understanding if you could provide an example.
AX3E,AX5 and AX6

note that A = central atom and X = bonding atoms.

Welcome to Chemistry.SE! One way you help yourself get a better answer to your question is by giving it a more descriptive title. "Help predicting molecular geometry" is more helpful to the community than "Help me with this question". I have edited your title. If you don't like it, let me know.

The title is OK with me.

You should read this: http://en.wikipedia.org/wiki/VSEPR_theory

It will depend on the number of lone electron pairs as well.
For example:
$BCl_3$ will be triangular planar as it has no lone electron pairs.
$NH_3$ however will be triangular based-pyramidal because there are 3 bonded pairs of electrons and one lone electron pair.
Basically, it comes down to the number of electron pairs (both bonded and unbonded) around the central atom.

Two electron pairs - Linear

Three electron pairs - Triangular planar

Four electron pairs - Tetrahedral

Five electron pairs - Triangular based bi-pyramidal (it's like triangular planar, but with one on top and one below, or two triangular pyramids with their base stuck together)

Six electron pairs - Octahedral (Square based bi-pyramidal)

Those geometries include the lone electron pairs. For the molecular geometry, you will then need to exclude the electron pair.
E.g. $H_2O$ : central atom O. O has 4 electron pairs: 2 bonded, 2 lone. The shape, including lone electron pairs is Tetrahedral. Ignore the lone pairs now, you will be left with a bent molecular geometry.

Blue is oxygen;
Yellow is hydrogen;
White/Grey are lone/unbonded electron pairs.
I hope this makes sense. :)

If you convert your equation to mol format or another format supported by OpenBabel you can use its tools to add 3D structure information to the files. Also if you need quality conformations you could use tools OpenEye's Omega or as I prefer Balloon which can quickly give biologically significant structures. Balloon works on many different atom types and most smiles Omega is parameterised for a select few smiles.


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Normal Spinel and Inverse Spinel
According to my notes, octahedral site stabilization energy (OSSE), sterical effects and electronic effects in stability create a confusion in this theory. However, we are able to solve this problem with few rules:

Tetrahedral splitting energy is nearly equal to 4/9 octahedral splitting energy.
As the OSSE increases, the $O_\mathrm{h}$ geometries will be more favored.
These are useful to decide to whether the structure is spinel or inverse spinel.

I got $\ce{ZnCr2O4}$ ($\ce{Zn^2+}$, $\ce{2Cr^3+}$, $\ce{4O^2-}$):

$\ce{Zn^2+}$; $\mathrm{d^{10}}$ ion with CFSE = $0$ and thus, there will be no effect on $O_\mathrm{h}$ holes.
$\ce{Cr^3+}$; $\mathrm{d^3}$ ion with CFSE = $\pu{-8 Dq}$ and OSSE = $\pu{8.45 Dq}$ so, this is thought to occupy $O_\mathrm{h}$ holes.

Since $\ce{Zn^2+}$ can only occupy $T_\mathrm{d}$ holes, then it suits the rules for being normal spinel.
On the other hand, if we got a group II metal with a non-zero CFSE, then what would the result be?

Could you state an example of a group II metal ion with a CSFE unequal to 0?

Let's take a (2-3) spinel ($\ce{A^{II}B^{III}_{2}O4}$) as an example. If it is normal or inverse can be predicted from the total crystal field stabilization energies. For a normal spinel, the total CFSE wold be: $$\Delta_{T}(A)+2\Delta_{O}(B)=\Delta_{T}(A)+\Delta_{O}(B)+\Delta_{O}(B)$$ and for an inverse spinel $$\Delta_{O}(A)+\Delta_{T}(B)+\Delta_{O}(B)$$ with $\Delta_{T}$ being the tetrahedral and $\Delta_{O}$ being the octahedral splitting energy of $\ce{A^2+}$ and $\ce{B^3+}$, respectively.
If $\Delta_{T}(A)+\Delta_{O}(B)>\Delta_{O}(A)+\Delta_{T}(B)$, the normal spinel structure should be favored, as the resulting CFSE is higher. If it is the other way around, an inverse spinel is to be expected.
To sum it up, if the CFSE gained by placing $\ce{A^2+}$ in the octahedral holes outweighs a potential destabilization caused by putting a part of $\ce{B^3+}$ in the tetrahedral holes, an inverse spinel can be formed. However, a normal spinel is favorable when $\ce{B^3+}$ clearly "prefers" the octahedral and/or $\ce{A^2+}$ the tetrahedral holes.


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How, chemically, would you remove protruding phenyl groups?
How, chemically, would you remove protruding phenyl groups? I mean for instance in the molecule phenethylamine there's a phenyl group sticking out (or protruding) how would you go about removing this phenyl group?

The only method I'm aware of for removing arbitrary alkyl substituents from benzene is reversed Friedel-Craft (treating the substance with a Lewis acid catalyst under acidic conditions at very high temperature). With a primary carbon attached, extremely harsh conditions are required, and the possibility of side reactions will likely yield a mixture of polysubstituted benzenes and, if you're lucky, some of your substituent in a recoverable form. At those extreme conditions, however, I frankly have no idea if your amine group will actually survive without autoigniting or otherwise degrading.

@GregE. I think that your example would serve as a fine answer if you wanted to convert your comments into a full response.

You might try doing artificial evolution in the lab with ribozymes, thus achieving high selectivity and high rates after only a couple of cycles. It's pretty equipment-heavy and not really suited for Chem.SE, though. Also it is pretty non-trivial to find a suitable starting molecule.

FWIW, Reaxys has a grand total of zero hits for this specific transformation.


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How would you change the chirality of a compound such that (S,R) becomes (S,S)?
How would you change the chirality of a compound such that (S,R) becomes (S,S)? That is how would you synthesise a compound from its diastereomer?

Can you make this question a bit more concrete? Both (S,R) and (S,S) forms of a molecule are diastereomers as they have more than one stereocenter. Beyond regioselective stereoinversion, this question doesn't really have a specific answer.

You mean you want me to add an example of particular significance to me?

@BrentonHorne - An example would be helpful. The methods of changing a chirality center's configuration vary significantly in ease and reagents by functional group.


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Can we process waste just by decomposing it to homogenous mass and extracting the elements from it?
I. e., what if we just burn trash completely or dissolve it in acid, get the resulting paste and extract pure chemical elements out of it by applying some physics and/or chemistry to it until no more can be extracted? Just curious.
EDIT
To be perfectly clear: my main concern is whether there are any means to extract any useful element from this bulk of burned/dissolved/blender-ed trash. For example, we burn with Thermite the whole contents of some waste disposal site, until it's only multicolored ash there. Or pour the HSbF6 over it. From there, can we get the pure carbon out of the resulting mass? Pure aluminium? Pure iron? Gases, maybe, like oxygen?

Sure, this absolutely can be done.  In particular, you cannot destroy elements using conventional means like burning, so all the initial elements are still in the mass.  The question is whether or not this is a good way to extract what you want.  There may be better chemical techniques to do the extraction that are short of burning.  The branch of chemistry which deals with the optimal conversion of elements and chemicals to other useful substances is chemical engineering, and figuring out which process is "best" (e.g. fastest, cheapest, highest yield) is crucial for industry.
There is a very funny story related to this idea on the Periodic Table of Videos.  A scientist was trusted with the entire UK supply of plutonium (10mg, a very small amount) during World War II.  He accidentally spilled it on to his lab bench.  He then took a saw, sawed a large circle out of bench, burned that wood, and recovered nearly all the plutonium!  So yes, this works.  Here's the video, and this story begins around 15:30: Plutonium Video
EDIT: As to your update, there are lots of methods available, but it depends greatly on the starting conditions and materials.  For example, burning is a great way to extract carbon: burning hydrocarbons or carbohydrates burns off the hydrogen or water (respectively) and leaves mostly pure carbon.
But extracting metals might take more work.  Often, metals exist as ionic compounds.  For example, silver nitrate, $\ce{AgNO3}$, contains siver $\ce{(Ag)}$ in the +1 state.  While it might be possible to dissociate the silver with high enough temperature, it is far easier to extract metallic silver by reducing silver in the presence of a more reactive metal.  "More reactive" simply means a metal that will more easily give up an electron to the less reactive sliver, changing the silver ion $\ce{Ag+}$ to metallic $\ce{Ag}$.  Silver nitrate is highly soluble in water, and if you simply drop in a piece of metallic copper $\ce{(Cu)}$ metallic silver will slowly come out of solution and plate the copper with silver, while copper dissolves and creates copper nitrate.  The equation is:
$\ce{2AgNO3 + Cu->2Ag + Cu(NO3)2}$
but the crucial part is the ionic equation:
$\ce{2Ag+ + Cu->2Ag + Cu^2+}$
which shows that the silver ion is reduced to silver, and the copper metal is oxidized to copper ion.
Again, Periodic Videos has the footage of this exact reaction: Silver Video
Now I picked this example precisely because of this footage, but it's not the best example because silver is so low in reactivity that it exists metallically in nature (i.e. you can find lumps of silver if you are lucky!)  But many metals are so reactive that it takes a huge amount of energy to extract them.  Examples are sodium or potassium, which are so reactive they explode in the presence of water (go search YouTube and have a field day!).  To extract them as metals commercially usually requires electrolysis, which uses electricity to force electron transfer which would not otherwise occur spontaneously.
And on and on with other methods.  There is no universal extraction method, at least not yet, and this is why we have chemcial engineers to figure out better and better procedures to use.

I've updated my question with more elaborate explanation of my idea. If there really are such methods, why they don't implement them everywhere?

Can you, please, extend your consideration about the universal extraction method? What do you think are possible choices? Radiation with proper signal? Changing structure of the substance and then doing something?

Sounds like you're referring to plasma waste conversion.  If you heat matter to 4-figure temperatures all molecular bonds are broken and you have a gaseous mass of dissociated atoms.  It is theoretically possible to extract the constituent elements from the plasma.
If you want to preserve a particular element then the cooling of the plasma has to be controlled so that the element does not come into contact with others with which it naturally reacts until it is below the threshold temperature for those reactions.
Fun bonuses: Not only can you recover specific elements from homogeneous waste streams, but plasma waste reactors can also run with an energy surplus created by the "burning" of energy-rich inputs.  They can also be designed to produce synthetic gas from organic feedstocks.

Amazing! So, it seems that such technology exists at all! Exactly what I was curious about the most.
I think it's very unprobable, though, that this method will become widespread. I suspect energy requirements for manipulations with positions of just the elements we're interested in inside the plasma stream would be enormous.

My original thought was about liquefying the waste, not torching it, however. :) Less stuff blowing up, which is safer.

The problem with liquefying, blending, or anything short of molecular disintegration of "waste" (i.e., heterogeneous compounds of indeterminate composition) is you can't decompose everything to elements.  Plasma waste conversion is just one step beyond incineration, which is already widely used on both municipal waste and weapon stockpiles, including chemical weapons!  There's no problem with things "blowing up" when that's the objective.  The thing that isn't widespread is elemental extraction from general waste plasma, probably because general waste doesn't contain enough valuable elements.


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Explanation of the Selection rule in light of IR spectroscopy
I am in mathematics, but I am writing a paper on molecular vibrations (because you can use representation theory to handle this).
I want to prove the selection-rule, but I also have to give a presentation of my intermediate results. Can someone give a little (maybe trivial) explanation (in one or maximal two small sentences) of what the selection-rule does? (for example for the IR-spectrum). This is for my presentation if someone doesn't know something about the mathematical and chemical background.

Perhaps you will get a better response on physics.SE

The operator for interaction with infrared light is the permanent dipole of the molecule, $\mu$.  I don't know anything about representation theory, but in group theory, you decompose the molecule by its symmetry elements to find the eigenmodes.  The infrared active modes are those modes in which the permanent dipole changes over the course of a vibration.  An excellent reference is Symmetry and Spectroscopy by Harris and Bertolucci which should be available in your library.


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Recreational drugs by half life
I am interested in the Biological Half Life of substances. In the previous wikipedia link, I've already found the half lives of common pharmaceutical products. I can't seem to find similar data for recreational substances, does anyone know these half lives?

Sorry, didn't realise there was a biology beta. Does a pharmacology questions belong there? I don't know, flagged it.

I think it would need some elaboration to be worthy of migrating, to be honest.  I think as it is, it's okay here, but asking about specific classes of drugs would make it more scientific and rigorous.

Wikipedia reports biological half-lives for several recreational drugs I just tried (cocaine, THC, MDMA). In each case, I went to the page for the particular substance, and Wikipedia keeps an information box at the top of the article containing the pharmacokinetic data. THC's half-life has a citation to the scientific literature, while the other two I checked do not. I think it might be slightly harder in general to get good data for the purely recreational drugs. I would expect drugs without an accepted medical benefit to be less well-studied. To find what's there, though, it looks like you will be able to look up each active ingrediate individually. Depending on what you need, Wikipedia might be okay. If it's not, follow the Wikipedia citation back to the literature if it's there, or try PubMed.

Try DrugBank. It's a comprehensive and free-to-use database for a large number of drugs. I saw it has data on Cocaine and Heroin (look in the pharmacology section).
(Yes, this question is three and a half years old, but it still may help someone...)

What is the maximum number of atoms in a DrugBank molecule? [Your answer is "three and a half years old" now ;) ]

Not sure I understand what you mean...

You suggest using DrugBank. Most datasets provide information on the number and type of atoms (e.g, QM9 dataset is consist of 134000 molecules made of C,O,N,F and H with maximum 9 atoms in a molecule). Was wondering if you have those info and can add them to the answer.

Ok, well, I don't have that info, but even if I had it, I do not understand how it would be related to the original question.


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Can I deduct heteronuclear van der waals parameters from homonuclear parameters in a forcefield?
I got a forcefield (OPLS-AA in this case) that lists lennard-jones parameters for all homonuclear van der waals interactions. Is there a way to deduct the heteronuclear vdw-interaction parameters from those?

Found the answer myself. In case someone else stumbles over this, the combining rules for Lennard-Jones parameters of van der Waals interactions are:
Distance parameter:
$r_{XY}=\frac{1}{2}(r_{XX}+r_{YY})$ or $\sigma_{XY}=\frac{1}{2}(\sigma_{XX}+\sigma_{YY})$
Depth parameter:
$\varepsilon_{XY}=\sqrt{\varepsilon_{XX}\varepsilon_{YY}}$


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How does caesium’s density change with temperature?
According to Wikipedia caesium’s density is $1.90\ \mathrm{g/cm^3}$ at $\Theta =20\ \mathrm{^\circ C}$. How does this change when $T$ changes? E.g. will it expand when melting?

The only element I know that contracts on melting is iron.

Interesting fact. However, I wonder if there is something like a chart that plots density versus temperature.

There are tables of data for elements so it should be easy to chart it.

Yes, a table would be equally fine. Have you got any sources?

Start here

And here

Hm, that's quite nice, but gives only density at room temperature.

It's just a question of googling around until you find something. I got the links just looking around...

Looking at the wikipedia page you can see that the density is $\rho=1.93$ kg/l at room temperature and $\rho=1.843 kg/l$ at it's melting point.
Additional information can be found in this paper, where they experimentally explore the density of liquid Cesium in the vicinity of its melting point. In the range of 302 to 375 K to be exact.
From their experimental data they deduce the following approximate equation for the density as a function of the temperature:
$$\rho= 1829.12- 0.61483 \left(T-T_{melt}\right) $$
where $T_{melt}=301.6$ K.
This shows that the density reduces with temperature. Plotting the information that we have from the paper and from wikipedia yields:

The jump at the melting point is from the mismatch between the paper and wikipedia. In the paper they specify specifically that the density is for the liquid case, on wikipedia it might (I'm not sure) be the density for the solid case at the melting point. If that is the case then the jump in density from solid to liquid is: $1843-1829=14 $kg/m$^3$

Since I cannot comment yet, the only other way I know to say something here is to post an answer.
To address your question about the units, the thermal expansion coefficient is a percentage. From wikipedia:
$$\alpha_L = \dfrac{1}{L}\dfrac{dL}{dT}$$
So, for a change dT of temperature, the metal will expand by a percentage of $\dfrac{dL}{L}$.

I think I found what I was looking for on Wikipedia. They give the thermal expansion coefficient for Caesium as $97 µm\cdot m^{−1}\cdot K^{−1}$.

One last thing: why is the coefficient $K^{-1}$? I would have assumed to have units $m\cdot K^{-1}$, so that I can multiply a temperature with the coefficient to get an expansion in meters.

Its dependants on the length of the piece Caesium- a longer piece of Caesium expands more. Its a standard dimensional quantity.


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Alternatives to Congo Red?
I require Congo red indicator for an experiment I wish to perform involving dipping a cellulose sponge into an acid and a base.  I first want to dye the cellulose sponge with Congo red, and observe the color change of the sponge as I place it in acids and bases.
Congo red apparently binds to cellulose and thus is suitable for this experiment.  Unfortunately the laboratory doesn't have this indicator at the moment.  Are there any other indicators that could be used for this effect instead?

Yes, congo red bind the cellulose because is a substantive (or direct) dye these dyes don't need a mordant so it binds suddenly to the cellulose.
 There are others dyes for example safflower, cochineal with this property. In biology Congo Red is frequently substitute with Sirus Red F3B and Tioflavone S. At this link you can find some alternatives to Congo Red use for demonstrating amyloid, however I don't know if you can apply these dyes to your case, because I don't find any reference about their use as pH indicator. Although is reported that many substantive dyes are used as pH indicator I can't find any list.
Take in account that acid pigment are not substantive to cellulosic fiber and most of the pH indicator are acids.
I think that maybe you can try to use a mordant to obtain the same result with a pH indicator normally used, however I'm afraid that using a mordant you risk to change the behavior of your pH indicator.


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Writing the net reaction for reaction of sodium hypochlorite with hydrochloric acid and potassium bromide
I am currently taking a general chemistry course and during a lab they had us do the following procedure:

Place $\pu{2 mL}$ of $5\%$ $ \ce{NaClO} $ solution in a test tube.
Add $10$ drops of cyclohexane.
Add ~$10$ drops $\pu{6 M}\,\ce{HCl}$
In another test tube, add a pinch of solid $\ce{KBr}$
Use a dropping pipette to withdraw the chlorine/cyclohexane layer from above and add it to the test tube containing $\ce{KBr}$

When I did this, the mixture turned dark red and bubbled with a hint of purple to it. In my lab report, my teacher wants me to write the net ionic equation(s) of this reaction, but I have no idea how to do that.
How would you proceed?

I'd like to add some things to what Tanith Rosenbaun said.
First, the point of net ionic equations is not just to write out the equations in terms of ions, but to focus on only those ions that experience a change in charge or valence.  Ions which don't change in charge don't really participate in the reaction and called spectator ions.
Here's the last equation from Rosenbaum's answer:
$$\ce{2H3O+ + 2Cl- + 2Na+ + CO3^2- -> 2H2O + 2Cl- + 2Na+ + CO2}$$
Notice that the sodium and chloride ions are unchanged in this reaction.  So we drop them, and just write out the ions that change:
$$\ce{2H3O+ + CO3^2- -> 2H2O + CO2}$$
This is the net ionic equation: it focuses precisely on what is interesting, which in this case is the neutralization reaction.
Now let's go back to your problem:  People have been asking what you know about it.  You seem to know that you are transferring a solution of chlorine dissolved in cyclohexane.  You got that chlorine from the following reaction:
$$\ce{2HCl + NaClO -> Cl2 + NaCl + H2O}$$
and the chlorine preferentially dissolves in the cyclohexane.  Now it's important to understand that the cyclohexane doesn't play any role in the second part of your experiment, it's just how you transport the chlorine.  (And it's why you need to be careful with acids around bleach: chlorine gas is poisonous!)
The key question is: what happens when you put elemental chlorine in the presence of dissolved potassium bromide?
$\ce{Cl2 + KBr -> ?}$  (Note: this is not balanced yet!)
In particular, do you know what you produced when that solution turned dark red?  Hint: this is a very important property that halogens like chlorine and bromine have towards each other, which is indeed seen best via a net ionic equation.  I'll happily provide the answer if you've tried some stuff and are still stuck.

A ionic equation of a reaction isn't much different from a regular reaction equation, except that you write the ions separately. As per usual, reactants go on the left, products go on the right.
As an example, let's let hydrochloric acid react with sodium carbonate in aqueous solution. First you figure out what ions your reactants dissociate into. For $\ce{HCl}$ that would be $\ce{H+}$ and $\ce{Cl-}$, and for $\ce{Na2CO3}$ that's $\ce{2 Na+}$ and $\ce{CO3^2-}$, and lastly you got water, $\ce{H2O}$. That's the left side of this example equation. Which in sum would look like this:
$$\ce{H+ +Cl- +2 Na+ +CO3^2- + H2O ->?}$$
Next you need to figure out what ions react and in what way. In this example, first, $\ce{H+}$ and $\ce{H2O}$ react to form $\ce{H3O+}$, and after that, $\ce{H3O+}$ and $\ce{CO3^2-}$ react to form $\ce{H2O}$ and $\ce{CO2}$. You would write the two parts of the reaction like this:
$$\ce{H+ + Cl- + 2 Na+ + CO3^2- + H2O -> Cl- +2Na+ + CO3^2- + H3O+}$$
$$\ce{Cl- + 2 Na+ + CO3^2- + H3O+ -> Cl- +2Na+ + CO2 + H2O}$$
Now for the last bit, stoichiometry. Stoichiometry simply means making sure the numbers of atoms left and right are the same, and the ratio of atoms needed to make the reaction work is right. The first line works as is, but in the second line you'll notice that we're using a single positive charge in $\ce{H3O+}$ to react with two negative charges in $\ce{CO3^2-}$. Something's fishy there, as charges need to match too. And indeed, we need two $\ce{H3O+}$ for one $\ce{CO3^2-}$. After we changed that in the second line, we need to alter the first line too to match, like so:
$$\ce{H+ + 2Cl- + 2 Na+ + CO3^2- + 2H2O -> 2Cl- +2Na+ + CO3^2- + 2H3O+}$$
$$\ce{2Cl- + 2 Na+ + CO3^2- + 2H3O+ -> 2Cl- +2Na+ + CO2 + 3H2O}$$
That's what your teacher wants you to do for the reaction you cited.

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


text:
Can high-tin bronze be made sharper than low-tin bronze?
I am not sure how sharpness is measured, but I do know that certain materials can be designated sharper than others, for example obsidian is said to be the sharpest material:

The prismatic glass blade is infinitely sharper than a honed steel edge

So my question is, will a honed edge of high-tin bronze be sharper than a honed edge of low-tin bronze, and if so why?

Unfortunately I don't have a solid reference for this, but what I remember from my undergrad courses is that the limit to sharpening is basically determined by the porosity and tensile strength of the metal you are trying to sharpen. At some point you will just run into the microscopic roughness of the metal itself and that is where sharpening ends.
Low- and high-tin bronze are pretty rough classifications so I couldn't tell you which will result in a lower porosity or higher strength, but if I would have to guess than I think low-tin bronze can be made sharper, because copper has a higher strength than tin.
Maybe if you can get your hands on this book chapter, you can find out a more exact answer.

So why do so many articles and sources claim that higher tin bronze can be made sharper, higher tin bronze is used in the edges of the blade in bimettalic swords. the edges have more tin content, making them harder and capable of retaining a sharper edge

Ok, so apparently my GUESS was wrong and apparently high-tin bronze does not behave 'more like tin' as I assumed in my guess. Maybe adding more tin will lower the porosity and will therefore allow higher sharpness. I am still convinced that the explanation of porosity and tensile strength of the metal is correct, it is just hard to apply these 'rules' if you don't know the exact details of a metal alloy.


text:
What does under-stoichiometric or stoichiometric mean?
When people talk about oxides like $\ce{SiO2}$ or $\ce{Al2O3}$, they use expressions like stoichiometric or under-stoichiometric. I understand that this refers to the relative composition of materials, but I don't really know what it exactly means.
Say for $\ce{SiO2}$, does under-stoichiometric mean that it isn't exactly $\ce{SiO2}$, but more like $\ce{SiO_{1.8}}$? Does the term usually refer to the oxygen composition?

Yes, the convention is to refer to the oxygen atoms when talking about over- and under-stoichiometric in metaloxides. Normally they also indicate this explicitly by writing $\ce {SiO}_x$ (like in the article here for example) or $\ce {Al_2O}_x$. Being explicit about it is a good habit because it avoids ambiguity.


text:
Is there a specific term for reactions of the form A + B → C + D?
I'm working with an abstract model of chemical reaction networks. In this model, some of the reactions have the form $\ce{A + B <=> C}$ and others have the form $\ce{A + B<=>C + D}$. I'm wondering if there is some existing terminology I can use to refer to these two cases.
I know that the terms "unimolecular" and "bimolecular" refer to the number of reactants, but in this case the important thing is the number of products. The ideal thing would be if there's a general term describing whether a reaction has the same number of products as it does reactants.
Alternatively, since the species I'm modelling are abstract polymers, I can use the terms "cleavage" and "ligation" to refer to the $\ce{A + B <=> C}$ case, depending on the direction in which the reaction proceeds. If there is a corresponding term for $\ce{A + B<=>C + D}$ type reactions in polymers, that would also be useful to know.
I suppose at a stretch I could call it an exchange or a substitution reaction, but (as far as I know) that tends to imply that the thing being substituted or exchanged is a single atom, rather than a whole section of polymer. (In my case the reaction might be something along the lines of $\ce{aaabb + ccdddd -> aaadddd + ccbb}$, where $\ce{a}$, $\ce{b}$, $\ce{c}$ and $\ce{d}$ are different types of monomer.)

There are 4 'base-types' of chemical reactions:

Synthesis reactions ($\ce{A + B -> AB}$)
Decomposition reactions ($\ce{AB -> A + B}$)
Substitution reactions ($\ce{A + BC -> B + AC}$)
Metathesis reactions ($\ce{AB + CD -> AD + CB}$)

Given that you talk about polymers that in one case combine to form a new (single) polymer: $\ce{A + B <=> C}$ you could call this a synthesis reaction.
You other example, $\ce{A + B <=> C + D}$, is an exchange of parts of the polymers so that would be a metathesis reaction.
For polymers there are a couple of additional options to name the types of reactions, but these depend a lot on the specific chemistry (e.g. hydrolysis if water is one of the reactants) so that is probably difficult if you have 'abstract' polymers.

Thanks, this is helpful. One slight problem though is that some of my $\ce{A + B<=>C + D}$ type reactions would be classed as substitution reactions according to this scheme and others would be metathesis reactions - ideally I'd like a term that encompasses both possibilities. Do you think it would be reasonable to use the word "metathesis" to include substitution reactions as well? I suppose I could just say "substitution/metathesis" throughout the paper, but it's a bit of a mouthful.

Both substitution and metathesis reactions are also often called single resp. double replacement reactions, so I would call them replacement reactions then.


text:
How does the reaction Mg + O₂ → MgO occur?
$\ce{Mg + O2 → MgO}$
Magnesium, oxygen and magnesium oxide form the above equation.

How?
What do I need to learn to do this? atomic numbers/ valencies?

Always try to balance your equations in order to comply with the Law of Conservation of Mass. It makes things simpler to understand.

First of all, do you know that chemical equations are means to express chemical reactions?
Second, this equation represents the reaction of magnesium with oxygen. Magnesium and Oxygen are the reactants and Magnesium Oxide is the product.
The arrow indicates the direction of the reaction.
To write basic chemical equations, you only need to know your reactants, your products and how to write the symbols for them. Usually, you can 'guess' what the products can be from generic patterns or mechanism, but I don't think you will be there yet (or required to remember them) soon if you just got started on them.
To get the molecular formula or ionic formula in this case, you need to know the valencies of the elements constituting that compound. For instance here, Magnesium has valency 2, Oxygen has valency 2, so $\ce{MgO}$ is the formula for Magnesium Oxide.
For example, Sodium Oxide, Sodium has valency 1, Oxygen has valency 2, so $\ce{Na_2O}$ is the formula for Sodium Oxide.
Iron (III) Oxide becomes: $\ce{Fe_2O_3}$ (Iron valency = 3; Oxygen valency = 2)
Magnesium Chloride becomes $\ce{MgCl_2}$ (Magnesium valency = 2; Chlorine valency = 1)
Does this answer your question or were you looking for something different?


text:
Homemade preservative for natural aloe-vera gel?
I suppose this question can be extrapolated to food in general, but it is the case that I would like to know if there are any methods that would help me to keep the aloe gel fresh, i.e. avoiding it to rot.
I mostly squeeze the gel and pour it into a small jar. So far I have tried the following methods, being some of them obvious, but would appreciate corroboration anyway:

Sterilize the container and the tools, e.g. boiling or steaming.
Keep it in the fridge.
Stir with C and E vitamin powder to avoid oxidation.

I know little about chemistry, but have heard about pasteurization and some other methods that I couldn't afford at home so, even if you mention those, I will be mainly interested in homemade methods.
PS: I am forced to choose the tags for the post but I am not sure if they are the ones that best suit the question. Please, let me know if I should change them.

Freezing should work, but why don't you just use it right off the plant?

@Nile Yes I could freeze it in chunks... I don't use it right off the plant because I cannot harvest every day. And it would be inconvenient to process it every day aswell, e.g. leave the cut-off leaves into an upright position for a day to let the resin drain out, stir in the blender, etc.

Just read about some more preservatives: rosemary extract, borax powder, baking soda, citric acid... What do you think?

It also depends on what you want to use the gel for. If you want to eat it, I really wouldn't advise using borax powder or baking soda... :P

I see @kaliaden, thanks for the advise. I forgot to mention that I read honey is a good preservative for eating purposes, probably safer than borax powder.

I suffer from chronic back pain and have tried just about everything on the market.Now I grow and harvest my own Aloe Vera gel,
put the gel in a 50ml roll-on bottle 1/2 full with gel, a 1/2 tsp vitamin C powder
and 10 drops vitamin E oil then add 98.9% DMSO by syringe
10ml at a time till the strength is good enough for me. I personally use 25ml
DMSO with the Aloe gel for remarkable pain relief

Thank you for your answer. Does it help with your back pain?

@clapas Be advised, that anything that is dissolved in DMSO will enter your body through your skin. You need to be very, very clean and careful. Consider that when choosing packaging, too, because DMSO dissolves quite a lot, especially rubbers and plastics. This answer is probably not mentioning the potential dangers of the used substances enough.

High concentrations of ethanol might work well if you just want to put it on skin. Ethanol should be safe, it's the main ingredient in those hand sanitizers.

Blend only gel, then freeze in ice cube trays: slip tray into a large freezer bag and expel air. Use a tie to seal the bag. When frozen place cubes in an air tight container and return to freezer.
Very cooling for the skin and put a cube in filtered water to drink.
Don't ad anything to it (except a squeese of fresh lemon juice to drink).
Don't use ANY chemical/ethanol/crap....borax/DMSO WTF!

Lemon juice is just a mix of chemicals. I think you're providing contradictory advice here...


text:
Spontaneous explosion of a bottle containing alcohol and milk
A friend of mine gave me a bottle of home-made liquor which is a concoction of pure alcohol and sweetened milk. I accepted his gift happily and then I put the bottle in my cocktail cabinet. After an hour, we happened to hear a terrible noise of a some sort of explosion. We were indeed right - the bottle had exploded with actually no reason! Could someone please explain this strange phenomenon?
I am not sure whether this is important but he claims he had been keeping it in his fridge, although the difference of temperatures (in the fridge and in the cocktail cabinet) does not seem terribly significant to me. Also, the bottle was almost full meaning there was little room for the air in the bottle.
Please find attached the full list of ingredients:

0.5 litre of condensed, unsweetened milk,
0.5 litre of condensed, sweetened milk,
0.25 litre of pure alcohol,
few sweets.

I'm guessing that the mixture generated carbon dioxide gas, and that storing in the fridge retarded the formation of the $\ce{CO2}$. Alcohol is a weak acid, and in proximity to calcium carbonate, which may be in your milk, you could obtain $\ce{CO2}$.  But I can't say this for certain, which is why I'm posting this as a comment.

Please don't cross post your questions.  The proper protocol is to post it one place and then ask for it to be moved if it hasn't gotten much attention.  I think it's probably better off topic-wise here, to be honest.

@jonsca: Wait, are you calling it off topic here or on Physics? (IMO it's on topic on both, but better here).

@ManishEarth Sorry, my words blended together, it was "better off", "topic-wise".  No, I think it's fine here! :)

@jonsca: Ah, that makes more sense :). I've closed the one on Physics and left a note; I'll open that one if there aren't any answers here (by the normal policy of only cross posting/migrating overlap questions when they don't get answers)

Just out of curiosity, are we talking about a "the cap shot off" type explosion, or a "the bottle physically broke" type of explosion?

The bottle can't have exploded for "no actual reason"; there may have been no apparent reason, but everything does indeed happen for a reason, at least at this level. But this isn't English.SE...
My bet is on a combination of things. As with many things in everyday chemistry, the substances involved are never pure enough to pinpoint one exact cause. Case in point, milk is an emulsion containing a significant percentage of fats and proteins. So, possible causes include:

Hygroscopy and acidity of the alcohol causing the milk to curdle, by taking the water from the milk to form an alcohol solution that the complex milk proteins and fats precipitate out of due to lower solubility. This may produce a "greater than the sum" volume of product by leaving these proteins and fats in a porous gel (basically cottage cheese, though usually a stronger acid than ethanol is used in cheesemaking).
Esterification. Milk, as stated, contains fats including fatty acids. Add alcohol, and the fatty acids react with the alcohol to produce esters (neutral fats). The byproduct is water, formed from the acid stripping the alcohol's hydroxyl group. Again, the products of this reaction can take up more volume than the reactants.
Biological reduction. I actually do not think the concentration of alcohol in this solution would be antiseptic; if the alcohol is pure then we're only talking about a 20% solution by volume, and that would have to be a lab-distilled product like Everclear; "natural" distillation in open air only gets you to about 150 proof, and using even 100-proof vodka gives this mixture only a 10% concentration by volume of ethanol. According to at least one source, ethanol needs to be concentrated to at least a 65% solution by volume to be effective as an antiseptic. So, to a bacteria's point of view, the alcohol is an overworked police officer; it'll slow down bacterial growth, but there's way too much fat and sugar here for the alcohol to be much of a deterrent. Consumption of organic fuel by cellular metabolism, as a rule, produces CO2 gas, and in a closed container at room temperature, well, you get the picture.

I am guessing there was only a small gap between the cap and the liquid.
It was not a fermentation with that much alcohol.
It could be a question of gas being less soluble in liquid at a higher temperature (with the gas being CO2) causing a pressure rise.


text:
How can antibonding orbitals be more antibonding than bonding orbitals are bonding?
In molecular orbital theory, the fact that a bonding and antibonding molecular orbital pair have different energies is accompanied by the fact that the energy by which the bonding is lowered is less than the energy by which antibonding is raised, i.e. the stabilizing energy of each bonding interaction is less than the destabilising energy of antibonding. How is that possible if their sum has to equal the energies of the combining atomic orbitals and conservation of energy has to hold true?

"Antibonding is more antibonding than bonding is bonding."

For example, the fact that $\ce{He2}$ molecule is not formed can be explained from its MO diagram, which shows that the number of electrons in antibonding and bonding molecular orbitals is the same, and since the destabilizing energy of the antibonding MO is greater than the stabilising energy of bonding MO, the molecule is not formed. This is the common line of reasoning you find at most places.

Usually, when atoms form bounds they release energy (heating, radiation etc). So, the energy conservation law holds, but this kind of sum rule does not work. Vice versa, one has to spend energy to get dissociation of the molecule.

stabilizing energy of each bonding is less than the destabilising energy of antibonding. Now how is that possible if their sum has to equal the energies of the combining atomic orbitals has it?

The reason is because there is a term for AO overlap that appears in the normalization constant: $\frac{1}{\sqrt{2-2S}}$ vs. $\frac{1}{\sqrt{2+2S}}$, or something. It will take some time to write it up, as it is a mathematical derivation.

@EricBrown Non mathematical but an intuitive reason would do.

related http://chemistry.stackexchange.com/questions/39431/

Mathematical Explanation
When examining the linear combination of atomic orbitals (LCAO) for the $\ce{H2+}$ molecular ion, we get two different energy levels, $E_+$ and $E_-$ depending on the coefficients of the atomic orbitals. The energies of the two different MO's are:
$$\begin{align}
E_+ &= E_\text{1s} + \frac{j_0}{R} - \frac{j' + k'}{1+S} \\
E_- &= E_\text{1s} + \frac{j_0}{R} - \frac{j' - k'}{1-S}
\end{align} $$
Note that $j_0 = \frac{e^2}{4\pi\varepsilon_0}$, $R$ is the internuclear distance, $S=\int \chi_\text{A}^* \chi_\text{B}\,\text{d}V$ the overlap integral, $j'$ is a coulombic contribution to the energy and $k'$ is a contribution to the resonance integral, and it does not have a classical analogue. $j'$ and $k'$ are both positive and $j' \gt k'$. You'll note that $j'-k' > 0$.
This is why the energy levels of $E_+$ and $E_-$ are not symmetrical with respect to the energy level of $E_\text{1s}$.
Intuitive Explanation
The intuitive explanation goes along the following line: Imagine two hydrogen nuclei that slowly get closer to each other, and at some point start mixing their orbitals. Now, one very important interaction is the coulomb force between those two nuclei, which gets larger the closer the nuclei come together. As a consequence of this, the energies of the molecular orbitals get shifted upwards, which is what creates the asymmetric image that we have for these energy levels.
Basically, you have two positively charged nuclei getting closer to each other. Now you have two options:

Stick some electrons between them.
Don't stick some electrons between them.

If you follow through with option 1, you'll diminish the coulomb forces between the two nuclei somewhat in favor of electron-nucleus attraction. If you go with method 2 (remember that the $\sigma^*_\text{1s}$ MO has a node between the two nuclei), the nuclei feel each other's repulsive forces more strongly.
Further Information
I highly recommend the following book, from which most of the information above stems:

Peter Atkins and Ronald Friedman, In Molecular Quantum Mechanics; $5^\text{th}$ ed., Oxford University Press: Oxford, United Kingdom, 2011 (ISBN-13: 978-0199541423).

I highly recommend using a different textbook (maybe any other - If you would like a list of good reads, I am happy to compile it). The energy of the orbitals has nothing to do with nuclear repulsion. Hence your given equations describe the ground and the exciting state. The generality of this example should also be disputed very much. It might hold for simple diatomic molecules, but in general MO are formed from more than only two AO.

@Martin I tried to taylor this answer specifically for diatomic molecules. And maybe I should have left it more fuzzy in terms of coulomb interactions (because those between the electrons will have a much larger impact). However, an intuitive explanation for the asymmetry was asked for, so I tried to provide one.
The way this site works, you should try to provide a better answer, which then gets upvoted. In the best case, mine gets un- and yours accepted.

@Martin-マーチン I would be interested by two or three (or more if you want) items from your list of good reads.

The sum of the energies over all orbitals is constant. The lower energy in a bonded state arises from the fact that not all MOs are occupied by electrons. Because the energy of the electrons is lowered, the total system energy is lowered by bonding. See http://www.dlt.ncssm.edu/tiger/diagrams/bonding/BondingEnergy.gif for an illustration.
Edit:
This phenomenon is due to the nuclear-nuclear repulsion of the two (or more) atoms that are binding. See Ball, D.W. Physical Chemistry p. 408

There is no such thing as conservation of orbital energies at different fixed values of $R$, the internuclear distance. Obviously, at $R \rightarrow \infty$ the energies are the same regardless of the linear combinations you use (there is no interaction and the "molecular" energy is exactly the same as the sum of the "atomic" energies); obviously too, the energies raise a lot if you make $R$ very small: you have the fusion barrier that doesn't allow you to convert your initial atomic orbitals of atoms A and B into the atomic orbitals of atom AB. Energy is obviously conserved in a dynamical process where two atoms approach and you have kinetic energy in the collisional coordinate and so on and so forth (you can immediately see that two atoms cannot form a diatomic molecule is something else does not take away energy from the system).

Another possible (and slightly more simplistic) explanation given by Fleming (2009) is based on inter-electronic repulsion, when describing the molecular orbitals of the $\ce {H2}$ molecule:

In addition, putting two electrons into a bonding orbital does not achieve exactly twice the energy-lowering of putting one electron into it. We are allowed to put two electrons into the one orbital if they have opposite spin, but they still repel each other, because they have the same sign and have to share the same space; consequently, in forcing a second electron into the $\sigma$ orbital, we lose some of the bonding we might otherwise have gained.

However, note that the author cited this as an add-on. The primary reason, as discussed in greater detail by the author, is still what was mentioned by orthocresol here. Also, this does not explain the case for the interaction of two unfilled orbitals since the absence of any electrons would mean that there is no inter-electronic repulsion.
Reference
Fleming, I. (2009). Molecular Orbitals and Organic Chemical Reactions. United Kingdom: John Wiley & Sons, Ltd.

Here is an explanation that, I believe, solves most of the problems raised by others on this thread.
TL;DR
When we bring the atoms closer from infinity, electrostatic repulsions come into play between the electrons of the two atoms (as
Martin - マーチン pointed out under tschoppi's answer, the energy of the orbitals has nothing to do with nuclear repulsion). These inter-electronic repulsions increase the potential energy of the system.
Since the energy increases, the average energy of the bonding and anti-bonding molecular orbitals also increases. Thus, the bonding molecular is not lowered in energy (on the molecular orbital diagram) to the extent expected, and the anti-bonding molecular orbital is raised in energy to a higher extent than expected. That is why "antibonding is more antibonding than bonding is bonding."
(Somewhat) Mathematical Explanation
For simplicity, let us assume a symmetrical, diatomic molecule $X_2$ (we do not know how many electrons are present in it). Let us also consider only the $1s$ atomic and molecular orbitals.
When we talk of the energies of atomic orbitals, we talk of atoms ($X$'s) at infinite separation. Let the energy of the $1s$ atomic orbital of $X$ be $E_X$. Then the total energy of the $1s$ atomic orbitals in two $X$ atoms is
$$E_i = 2E_X$$
However, when we talk of the energies of molecular orbitals, we must remember that the atoms are now a finite distance away. Let us consider the bonding and anti-bonding molecular orbitals separately.
Let the ideal/expected energy (without taking into account the inter-electronic repulsion) of the $\sigma 1s$ molecular orbital be $E_1$. However, there will be inter-electronic repulsions present in the bonding molecular orbital; at least, more so than at infinite separation of atoms. Let the extra energy due to repulsions be $E_{e_1}$. Then the energy of the $\sigma 1s$ molecular orbital will be
$$E_\sigma = E_1 + E_{e_1}$$
Again, let the ideal/expected energy of the $σ^* 1s$ molecular orbital be $E_2$. However, there will still be inter-electronic repulsions present in the bonding molecular orbital; at least, more so than at infinite separation of atoms (we do not need to know whether the repulsions are greater in the bonding or anti-bonding molecular orbital; just that they exist). Let the extra energy due to repulsions be $E_{e_2}$. Then the energy of the $σ^* 1s$ molecular orbital will be
$$E_{\sigma ^*} = E_2 + E_{e_2}$$
Therefore, the total energy of the $1s$ molecular orbitals is
$$E_f = E_\sigma + E_{\sigma ^*} = (E_1 + E_{e_1}) + (E_2 + E_{e_2}) = (E_1 + E_2) + (E_{e_1} + E_{e_2})$$
The term $(E_{e_1} + E_{e_2})$ arises due to the change in distance between atoms, i.e. from infinite to finite separation. Therefore it is the change in potential energy of the system, or the work done in bringing the two atoms to the final finite distance.
$$W = E_{e_1} + E_{e_2}$$
Now, we can apply the law of conservation of energy:
$$E_i + W = E_f$$
$$2E_X + (E_{e_1} + E_{e_2}) = E_\sigma + E_{\sigma ^*}$$
$$E_\sigma + E_{\sigma ^*} - 2E_X = E_{e_1} + E_{e_2} > 0$$
$$E_\sigma + E_{\sigma ^*} > 2E_X$$
$$\frac{E_\sigma + E_{\sigma ^*}}2 > E_X$$
Therefore, the average energy of the bonding and anti-bonding molecular orbitals is greater than that of the atomic orbitals.
To put it another way,
$$E_\sigma + E_{\sigma ^*} > 2E_X$$
$$E_{\sigma ^*} - E_X = E_X - E_\sigma$$
Therefore, "antibonding is more antibonding than bonding is bonding."

Note that had we not considered inter-electronic repulsions, i.e. the work done to bring the atoms closer, we would find the anti-bonding molecular orbital to be "just as anti-bonding" as the bonding molecular orbital is "bonding". This is the mistake made in applying the law of conservation of energy directly.

As for the case of unfilled orbitals (pointed out by Tan Yong Boon in their answer), unfilled orbitals don't really "exist" until they are filled by electrons. Orbitals have energy only due to the electrons in them, and thus we cannot say anything about the energy of unfilled orbitals (the inter-electronic repulsion idea also cannot be used here).


text:
Spontaneous inequilibrium in a gas in an isolated system?
Can an isolated system(of several phases) in equilibrium by itself move out of equilibrium?
if a system contains for example an inflammable gas dispersed with another combustible gas such that the molecular speed are insufficient to cause the reaction. but as the molecular speeds keep fluctuating, at some point, it may cross the energy barrier and may start the reaction which will eventually spread evenly but till then the temperature there would be high and thus the system in inequilibrium.
isn't this also an example that although the free energy is minimum and entropy maximum, some useful work would've been done if the temperature gradient had been utilised?

I'm slightly confused as to why you've specified multiple gases here -- an inflammable gas is, by definition, also combustible, and I'm not sure how that changes your question. Could you clarify?

it is just to make sure that at the point in space where the reaction begins there is quite a large difference of temperature created and hence the condition of inequilibrium is unanimously accepted. you can ignore that considering only the thought experiment

The short answer to the question "Can an isolated system(of several phases) in equilibrium by itself move out of equilibrium?" is NO.  Once the system is in equilibrium it will remain there.
If it is possible, as is asked in the second part of the question, to have some reaction be triggered at a higher temperature reached by fluctuation, then as has been commented, the answer is that the system was never in equilibrium.
We should never forget that there is no "absolute" starting temperature for a reaction.  If the reaction is physically possible, it will take place, but the rate of that reaction can be horribly slow.

Technically it can move out of an equilibrium situation spontaneously, since a spontaneous increase in free energy is only statistically disfavoured. If the potential energy landscape has a shallow minimum, reasonable fluctuations might happen. Though now I have two questions. Firstly, do we know of some equilibrium that fluctuates appreciably, and more interestingly, do we know one that does so around room temperature? Or does the mere free energy contribution of mixing overwhelm perceptible fluctuations? Secondly, are there any reaction equilibria with more than one free energy minimum?

@NicolauSakerNeto The presence of two minima in free energy is present in regular solutions wherein the presence of two minima necessitates the splitting of the mixture into two phases of the compositions corresponding to the minima. But I couldn't find examples of such conditions in proper chemical 'reactions'


text:
Number of bonds in a Lewis Structure
I'm asked to find the Lewis Structure  of sulfuric acid, $SO_4H^-$. After struggling for a while, I sumply looked it up.

How can this be? Sulfur has 6 electrons in its valence shell, so it should form bonds so as to collect 2 more electrons and satisfy the octet rule. So, surely it should form 2 bonds?
But it's involved in no less than six bonds here. Huh?

Welcome to Chemistry.SE! This question has already been answered before (not explicitly for sulfuric acid, but expanded valence on sulfur is covered).  Check out the answer to this question: http://chemistry.stackexchange.com/questions/444/can-an-atom-have-more-than-8-valence-electrons-if-not-why-is-8-the-limit. I had trouble search for it using the title of your question, so its understandable how you might have missed it.

The octet rule only applies firmly to elements below main energy level 3 (in fact, there are rare instances in which even that isn't entirely accurate, but molecular orbtal theory needs to be invoked to explain those situations). At main energy level 3, which Sulfur occupies, unfilled d orbitals become available, and those d orbitals serve to accommodate the additional electrons in Sulfur's bonds. In reality, Sulfur is a much larger atom than Oxygen, and the π orbital overlap between the two elements is fairly poor, giving the bonds significant single-bond character. You might therefore see resonance structures in which single bonds are drawn instead, with formal positive charges assigned to Sulfur and formal negatives assigned to Oxygen.

I notice that in this molecule, the Oxygens and Hydrogens are still following the octet rule. Will this be true in general? That is, an element below energy level 3 will always follow the octet rule, even when in a compound with elements that don't?

@JackM, yes, that's generally the case. There are, however, specific instances in which valence bond theory breaks down and cannot explain the observed bond order and geometry of certain atoms even below the third main energy level (e.g., diborane). Molecular orbital theory is necessary to adequately rationalize those situations.

But usually, a good strategy for drawing a Lewis Structure would be to make sure all the octet-rule-following atoms satisfy the octet rule, and if that means that the other atoms end up with "too many" electrons, then just so be it?

@JackM, the normal strategy for drawing a LEDS given no info other than the formula is to select a central atom and create single bonds to the other atoms, aiming for a maximally symmetrical skeletal molecule. Then, fill in lone pairs to the outer atoms (using octet rule), and then to the central atoms. Then, convert lone pairs to double and/or triple bonds as necessary with the aim of satisfying all octets and minimizing charge distribution. Exceeding the octet of centrals atoms is acceptable, assuming the atom is at the third main energy level or higher and doing so minimizes charges.

@JackM, for H2SO4 (32 valence electrons total), the procedure would be: 1) for maximum symmetry, select S as the central atom. 2) draw single bonds from S to each of the four O atoms (which satisfies S octet for the moment) and to two of the O add single bonds to H. 3) add remaining electrons as lone pairs to each O, stopping when octet is full. This leaves you with a structure in which all octets are satisfied, but two of the O atoms have -1 formal charge, while the S has +2 formal charge, so eliminate the formal charges by using one lone pair from each O to create a double bond to S.

@JackM, keep in mind that the approach I've described above is a fairly crude and naive one. It's what was taught to me as a beginning chemistry student, but it's far from perfect. Even relatively simple molecular formulas can correspond to multiple valid structures, and for complex formulas it's all but completely useless. I've yet to encounter a situation in laboratory practice where I've had to guess at a structure given a molecular formula and no other information (e.g., IR and/or NMR spectra, etc.). Even then, knowledge of functional group structures would serve as a guide.


text:
How am I to interpret the phrase "a solution of acid"?
The question title is a bit odd, I couldn't find a plainer way of saying it.
I'm given the concentration and conductivity of a solution of a particular acid, let's just call it $\ce{AH}$. I then have to deduce that acid's acidity constant.
My first instinct is to write out the expression for the solution's conductivity... but then I realize I don't know how to do that. I have the coefficients for all the relevant ions (I'm not sure what the technical term is, but essentially I mean the lambdas in the second equation here), in particular, the ions $\ce{H_3O^+}$, $\ce{OH^-}$ and the conjugate base $\ce{A^-}$.
So if I had the concentrations of those ions in the solution, I could solve the problem! But hang on... I have no idea how to figure them out. I know the concentration of the acid in the solution. Suppose the acid has just been added to the water. Then it won't have started to disassociate yet, so the water will be exactly as it was before we added the acid: $[\ce{H_3O^+}] = [\ce{OH^-}] = 10^{-7}$. But what if originally there was a lot more acid, but it's had time to react with the water? Then the water will have been filled with an amount of $\ce{A^-}$ dependent on how long ago the acid was dissolved in it.
Do I assume the acid is at equilibrium? The question doesn't specify that.

A solution of an acid usually refers to it's aqueous solution -- also written as $\ce{AH(aq)}$. And as described in Wikipedia - a solution is a homogeneous mixture. So yes, the acid will be dissolved, i.e. in equilibrium.

That's not strictly true for acids and bases though. What makes an acid an acid is the fact that it gives off protons, especially in water, which react with water molecules to form $H_{3}O^{+}$. So what you got when you dissolve an acid $AH$ in water is really a mixture of $AH$, $H_{2}O$, $A^{-}$ and $H_{3}O^{+}$. The concentrations of $AH$ and $A^{-}$ relative to each other at a given concentration in water is what the specific acidity constant of an acid is, and that's what the OP is supposed to find out.

The acidity constant is defined as the equilibrium constant for the reaction:
$$\ce{HA <=> H^+ + A^-}$$
$$K_\text{a} = \dfrac{[\ce{H^+}][\ce{A^-}]}{[\ce{HA}]}$$
Note: $\ce{H3O+}$ can be considered equivalent to $\ce{H+}$
Since you have to find an equilibrium constant, (and no further info is given) you can safely assume the solution to be in equilibrium.
Plug in the various conductivities in the second equation (Kohlrausch's Law) you mentioned getting the limiting conductivity ($\lambda^o$) of the acid. This is the conductivity of the acid if it had completely dissociated into its constituent ions.

EDIT: To find the limiting conductivity of the acid using the equation, you must have the values of $\lambda^o_{\ce{H^+}}$ and $\lambda^o_{\ce{A^-}}$.
$$\lambda^o_{\ce{HA}} = \lambda^o_{\ce{H+}} + \lambda^o_{\ce{A-}}$$

The degree of dissociation of the acid $\alpha$ (The number of acid molecules which have dissociated at equilibrium per each acid molecule present originally -- it is fractional for weak acids and bases) would be the ratio of the measured conductivity of the solution ($\lambda$, which is given) to the limiting conductivity ($\lambda^o$).
$$\alpha = \dfrac{\lambda_{\ce{HA}}}{\lambda^o_{\ce{HA}}}$$
This alpha would be the same as the ratio of concentration of dissociated acid molecules at equilibrium (since each dissociated acid molecule gives one $\ce{H^+}$ ion, this is equal to the number of $\ce{H^+}$ ions at equilibrium) to the initial concentration of acid taken (which you have mentioned as given).
Therefore,
$$\dfrac{\lambda_{\ce{HA}}}{\lambda^o_{\ce{HA}}} = \dfrac{[\ce{H^+}]}{[\ce{HA}]_\text{initial}}$$
You can find $[\ce{H^+}]$ at equilibrium using this equation and then find the other concentrations using that.
Note that all this assumes that the acid is monoprotic (i.e. there is only one proton released per acid molecule). You would have to change all the equations accordingly in case more than one proton is released.


text:
Minimizing Gibbs free energy with Helmholtz free energy expressions
If I have an expression for the Helmholtz free energy (from statistical associating fluid theory for a polymer system), can I still find the Gibbs free energy minimum under constant $T,P$ conditions?
Seems that since
$\mu_i = \left(\frac{\partial A}{\partial n_i}\right)_{T,V,n_j} = \left(\frac{\partial G}{\partial n_i}\right)_{T,P,n_j}$
where $A$ is the Helmholtz free energy and $G$ is the Gibbs free energy and $\mu_i$ is the chemical potential of component $i$.
If I can just find the expressions for $\mu_i=\left(\frac{\partial A}{\partial n_i}\right)_{T,V,n_j}$ I should be able to use this in the expression for $G = \sum_i \mu_i n_i$ and find its minimum, without having to worry about an equation of state to relate change in volume ($dV$) with respect to constant pressure ($P$)?
Edit I should clarify that I wonder about the Gibbs free energy because I'm interested in the constant $T,P$ (isothermal, isobaric case), and in the case especially for liquids. If you needed to introduce an equation of state to parameterize the change in V as a function of $T,P$ this would not be too difficult for an ideal gas, but for liquids it would be a challenge. So in this case, I wonder how to handle the $PV$ term (for constant $P$) in the Helmholtz free energy which does not appear in the Gibbs.

In short, I agree with Curt F.'s answer above. In general, you need a suitable equation of state (EoS) to handle this calculation.
From the differential form of the fundamental equation of thermodynamics (i.e. the differential form of the internal energy $U(S,V,N)$)
$$\mathrm{d}U = T\mathrm{d}S - p\mathrm{d}V + \sum_{i=1}^{m} \mu_i \mathrm{d}n_i$$
other relevant thermodynamic potentials may be found using the Legendre transform. Helmholtz free energy is defined as a transformation from $(S\rightarrow T)$ and Gibbs free energy by a transformation from $(S\rightarrow T)$ and $(V\rightarrow p)$
$$ A(T,V,N) \triangleq U - \left(\frac{\partial U}{\partial S}\right)_{V,N} = U - TS \\
G(T,p,N) \triangleq U - \left(\frac{\partial U}{\partial V}\right)_{S,N} - \left(\frac{\partial U}{\partial S}\right)_{V,N} = U + pV - TS = A + pV $$
If an expression for the Helmholtz free energy (that is, $A(T,V,N)$) is known, along with a suitable equation of state (EoS) (e.g. $p = p(T,V,N)$), the Gibbs free energy may be calculated. Without the EoS, it is impossible to evaluate the $pV$-term.
If both the volume and pressure may be assumed to be constant, they will not affect the phase equilibrium calculations. An alternative may be to assume that the liquid phase volume is constant or that it follows some (simple) empirical relation. Thus, the last term may be evaluated and the Gibbs free energy may be minimized.

I don't think you can use the approach you are outlining without an equation of state.
$G = \sum_i \mu_i n_i$ is valid only at constant $T$ and $P$.
$A = \sum_i \mu_i n_i$ is valid only at constant $T$ and $V$.
If $P$ and $V$ are both constant, then you can equate the two and use $G=A= \sum_i \mu_i n_i$.  Actually, that assumption is probably not horrible for most liquids at "normal" pressures: I think it amounts to assuming the liquid is incompressible.

I agree with the first part of the answer, but if P, V and T are constants (in this context). What kind of process we will be talking about? Just changing mass? What can we expect to do with those results? Respect of what variables will be minimeze G?
crippledlambda : I think that we need more details for having a chance of being helpful.

Thanks. It may be ambiguous but it is for studying phase equilibria under T, P conditions using chemical potentials derived from SAFT (statistical associating fluid theory) which provides expressions for the Helmholtz free energy (from which we can derive the chemical potential under constant T, V).


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Why do frying pans stick less with heat?
I know this is simple, compared to most questions asked here, but I notice this several times a week and I'm curious.
I notice if I break an egg and drop it into the frying pan, and the pan is still warming up, the egg sticks to the pan.  But if I wait until the pan is quite hot, and I break an egg and drop it in, the egg barely sticks to the surface at all.
How does the heat make a non-stick surface work better then when it is merely warm or is still warming up?

Heat creates steam bubbles, which hold the egg away from the pan until it is somewhat cooked - see Leidenfrost effect.

Does egg yolk really have sufficient water to do this?

The solid egg is not sticky like the liquid. If the pan is hotter it is solid on reaching the pan surface, so it doesn't stick.

I don't mean as the egg gets hotter.  If I drop an egg on the pan when it's not hot, it sticks as it cooks.  If I drop an egg on the pan when it's hot, it doesn't stick.
So the variable is not how cooked the egg is, but how hot the pan is.  (I point that out in the 2nd paragraph.)

Or how cooked the surface of the egg is when it touches the pan. Egg white is used to stick biological specimens sometimes (for microscope).

That wouldn't explain it at all.  You're grasping at straws.

@Tango : I think that COULD explain it (but I immediately note that it is very unlikely), because if the bottom layer of the liquid egg is already solidified upon touching the pan (because of heat radiated upwards) it will indeed not stick.

Coming back to the unlikeliness: the time for the egg white to heat up and solidify is almost certainly far greater than the time it takes for the liquid egg to fall into the pan due to the really low heat capacity of air which stretches out the heating time.

@michielm: Also, I'm there and watching it.  None of the egg is solid at that point.

It's about surface. How could it be tested? The only idea I have is to vary the amount of oil and or the height the egg comes from and see if that modifies the result.


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Is standard electrode potential of a galvanic half-cell is zero at equilibrium under standard conditions?
At standard conditions (at 1 atm pressure and unit activity (1 molal or 1 molar concentration) of all dissolved compounds), the electrode potential is equal to the standard electrode potential by the Nernst equation: $\Delta E = \Delta E^\circ.$
For the electrochemical reaction at equilibrium the electrode potential of the cell is zero: $\Delta E = 0.$
This implies that for any reaction at equilibrium under the standard conditions $\Delta E = \Delta E^\circ = 0,$ which is quite surprising. Is this correct?

Why would you expect the standard electrode potential of a reaction in equilibrium not to be 0?

given that the standard electrode potential determines the tendency of the system to follow the forward reaction,does the value of $$ E^0 $$ change as the reaction proceeds towards equilibrium?

There is a difference in the meaning of $\Delta E^\circ$ and $E^\circ$. $\Delta E^\circ$ actually denotes the change in standard electrode potential while $E^\circ$ denotes standard electrode potential.
$E^\circ$ is always constant for a given redox couple (eg. $\ce{Zn^2+|Zn}$). So no matter what  concentration of $\ce{Zn^2+}$ you take in a certain electrode, the standard electrode potential of $\ce{Zn^2+|Zn}$ is always $-0.76\ \mathrm{V}$.
Since $E^\circ$ is always constant, it never changes. So the value of $\Delta E^\circ$ (change in $E^\circ$) would always be zero and so it is quite a useless quantity to consider.
For $\Delta E$ however, consider the Nernst equation,
$$E = E^\circ - (RT/nF)\ln Q$$
Change in $E$
$$\Delta E = \Delta E^\circ - \Delta(RT/nF)\ln Q$$
therefore
$$\Delta E = -(RT/nF)\Delta\ln Q$$
At equilibrium $\Delta E$ is zero because there is no change in the reaction quotient $Q$ of the reaction (therefore $\Delta \ln Q$ is zero).

But then we do have a list of standard electrode potentials of different half cells with respect to the standard hydrogen electrode,right?how can $$ \Delta E^0 $$ be zero?

That is quite different... $E^o$ itself is the standard electrode potential of the electrode with respect to the Standard Hydrogen Electrode. So the value of $E^o$ for a given redox couple would never change unless you changed the reference electrode (you find the electrode potential of the electrode with respect to some other electrode instead of the SHE).

I am confused.I have solved some problems,where I had to calculate $ \Delta E^0 $ for calculating emf of the cell,or $ \Delta G^0 $ where I simply put the Nernst equation at equilibrium.All questions consider the state of equilibrium,standard conditions,and given half cell connected to SHE.is it correct to get non zero $ \Delta E^0 $ ?


text:
What is a "period" of the periodic table?
I'm familiar with the periodic table / periodic system, but I wonder why it's called "periodic" since there seems not much periodic about (there seems to be little or no predictability of which elements are stable) and from physics I know Z-value of elements and that is not periodic either. So what is a "period" in this context?

I agree that the "periods" in the periodic table are not mathematically regular.  The simplest definition is that a period begins when a new s-subshell starts to fill.  Recall that in the s, p, d, f subshells there are 2, 8, 10, and 14 electrons, respectively, so the periods have to get bigger over time and thus cannot be regular.  As to the lack of predictability of stability, that may be true, but the periodic table is pretty good at predicting the chemistry of elements by collecting together similar elements into groups (the columns).  This similarity is mostly expressed by the s and p subshells, which is where a lot of the reactivity occurs.

When it started it was periodic and the name never changed.

You could also consider a more general definition. The period in a very general sense is a group of elements after which the chemical properties begin to repeat i.e. the elements coming after each period can be arranged in a way that elements in the same vertical column (group) have similar chemical properties. In the periodic table, the periods change in a regular manner (a new block is added after every two periods, etc)

So what is a "period" in this context?
  To answer this directly, a period is a row of the table.

As user467 indicates, the repeating periods reflect the filling-up of atomic orbitals. Because the valence electrons (those in the outer shell of an atom) determine many of the chemical properties, the periodic structure places atoms with similar valences in similar groups (vertical columns.)


text:
which chemical element has the lowest boiling point?
which chemical element has the lowest boiling point?
Is it Helium(He) or is it Hydrogen(H)?
Hydrogen is lighter than Helium.

Regardless of it is an element or a molecule,, helium has the lowest boiling point $(\pu{-269 ^\circ C})$. In comparison, hydrogen, which exists as a diatomic molecule, has the boiling point of $\pu{-253 ^\circ C}$.

At its present form, this question is quite bad. I think you should've had a look at Wiki before asking this question. The boiling point of Hydrogen is around 20 K, while for Helium, it's around 4 K. I don't memorize these. But, I had a dirty look at Wiki. If you want a reason for why it has...
Helium has a stablest $1s$ orbital, which has somewhat lower energy. And, this makes it chemically inert towards other elements and to other Helium atoms, which is also the reason for its lowest boiling point.

The low boiling point of Helium is due to its very weak interatomic forces. I don't think the reactivity of helium with other helium atoms has much of a role here. Apparently quantum effects also play an important role in the case of Helium due to its small size.

@kaliaden: I don't know much. That was just a guess though ;-)


text:
What is the substituent on this benzaldehyde? (MS attached)

I have no clue what the substituent may be.
The molar mass of the unsubstituted benzaldehyde is 105 g/mol (subtracting one H, where the substituent will be). So the substituent is around 30–33 g/mol. Also I'm having trouble identifying the molecular ion. Is it the really small 138 peak or the 137 peak?

The molecular ion $\ce{[M]^{\bullet +}}$ peak is likely at 136 m/z. The peak at 137 m/z is due to the 1.1% natural abundance of $\ce{^{13}C}$.
The peak at 135 m/z is the $\ce{[M-1]^{+}}$ peak typical of aldehydes:
$$\ce{ [R-C(=O)-H]^{\bullet +} -> [R-C+=O <-> R-C\equiv O+] +H ^{\bullet} }$$
Another typical fragmentation of aldehydes is the loss of the CHO for $\ce{[M-29]^{+}}$, which you see at 107 m/z.
$$\ce{ [R-C(=O)-H]^{\bullet +} -> R+ + [HC=O] ^{\bullet} }$$
Since the main contributor to the $\ce{[M+1]^{\bullet +}}$ peak is $^{13}\ce{C}$ we can estimate the number of carbon atoms ($C$) in the molecule from the ratio of intensities of the $\ce{[M]^{\bullet +}}$ to the $\ce{[M +1]^{\bullet +}}$ peak:
$$\frac{I_{M+1}}{I_{M}}=\frac{1.1}{98.9}C$$
$$C=\frac{98.9I_{M+1}}{1.1I_{M}}=89.9\frac{I_{M+1}}{I_{M}}$$
The intensities appear to be $I_M \approx75$ and $I_{M+1} \approx 7$.
$$\therefore C=89.9\frac{7}{75}=8.4$$
Since the number of carbon atoms needs to be an integer, round down to $C=8$. We can now work backward to get a formula. Eight carbon atoms have a mass of $8*12=96$ g/mol, leaving 40 g/mol to be the rest of the formula. We know one is an oxygen (aldehyde) (16 g/mol), leaving 24 g/mol. The most number of hydrogen atoms that can be associated with eight carbon atoms is 18 (2n+2), so there has to be some other atom, maybe an oxygen or nitrogen?
I will leave rest to you:

Maybe there is a second oxygen?
How many hydrogens do you have?
What is the final formula?
How is it different from benzaldehyde?
How could those extra atoms be placed on the ring? (there are many possibilities)
Do you have any other evidence that suggests which of the possibilties from #5 is correct (IR, NMR, etc.)?

Also 107-92 and 92-77 provide some big hints.


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What are the factors behind making ice faster?
Recently I saw a video in YouTube about how to make ice in seconds.
In this video, salt is added to a glass of water and then a straw is closed from one end and the other side is lit using a lighter and after which it is stirred for 5 seconds and removed.
As soon as it is removed the ice is formed.
What is the chemical reaction that forms ice so quickly?

The simplest explanation would be that it isn't (pure) water at all. It is most likely a supersaturated solution of sodium acetate - $\ce{CH3COONa}$ (or sometimes sodium thiosulfate - $\ce{Na2S2O3}$) in water.
A supersaturated solution is a solution that contains more solute than normally soluble at those conditions. So as soon as a rough surface is provided for nucleation (formation of crystals), the solute starts crystallizing and comes out of solution as a big crystal.
The rough surface used over here is most probably the rough end of the burnt straw. This produces enough disturbance to cause crystallization. Usually an already prepared crystal of pure sodium acetate is used for nucleation but only if you want high purity crystals.
This is a pretty common trick (also the original video is from trick life -- as mentioned on the youtube description --) and most people who don't know about this phenomenon are usually easily tricked (then again, the "ice" formed would actually be hot since it is an exothermic reaction but you can't feel it through a video right :) ).
See this page and this page if you want more info.

Cool! I would never have thought of this explanation, but indeed we don't know that the crystallization that we see is water.

Thanks! Actually, I tried doing this for a science fair once but it didn't turn out too well though, I didn't take the right concentration (it's hard to get it right). That's how I came up with this explanation.

I am not sure about this specific video (although it indeed appears fake), but the trick in general is NOT FAKE.
It works by undercooling (also called supercooling) the water, which essentially means that you cool the water (far) below its freezing point but avoid ice formation. A common way to do this is to pressurize the water. This will lower the freezing point so it allows you to cool to say -20$^\circ$C, if you then change the pressure back to normal the freezing will only start when triggered by a disturbance, like a tap, shake or added ice crystals (like in the movie below).
Adding salt is another way of lowering the freezing temperature and achieving a similar result. Also with beer it is possible to achieve undercooling. Just put a closed bottle in the freezer for $\approx 2$ hours then open it and tap the top to see the entire beer freeze up. In this case the freezing occurs because your tapping removes (part of) the dissolved gases from the water thus elevating the freezing point back to normal.
A decent (visual) explanation is given here

The videao is a fake! Even at best supercooled water will freeze only partly! Second problem is that supercooling water is extreme tedious business, needing extreme lab cleanliness in Water and vessels. Last not least the tiniest mechanical "bump" will start freezing, You would not have a real chance to remove the glass from Your refrigerator.

@Georg - As I mentioned: I am not sure about the specific video. I AM sure that you can do this trick in general although it is indeed tedious but not at all impossible to try at home. The trick with the beer can work EASILY, because while the cap is still on you are NOT in an undercooled situation yet, only after you remove the cap you get there and then indeed a small tap on the bottle will start the freezing. Please READ what I wrote before you start judging.

This answer is really informative.. I didn't know about this phenomenon... and @Georg , michielm has explicitly mentioned that he is not sure about the specific video. The trick is indeed possible by this method but most probably not by the conditions in the video.

No true- supercooling happens all the time- its normal- you mean extreme supercooling like -20C. Its why they use Silver Iodide to cloud seed to suppress supercooling so it freezes at -1C. Some bactaria can get that done to -0.1C A supercooled once seeded freezes completely quickly as the whole lot of water is thermodynamically unstable.


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What does the eta notation mean in the naming of a transition metal complex?
I am reading a paper (Gutel et al., J. Mater. Chem. 2007, 17, 3290-3292) and I have come across the following reaction:

In organic solvents, the decomposition under mild conditions of
  ($\eta^4$-1,5-cyclooctadiene)($\eta^6$-1,3,5-cyclooctatriene)ruthenium(0),
  Ru(COD)(COT), under a dihydrogen atmosphere gives ruthenium
  nanoparticles (RuNPs), by releasing cyclooctane (COA).
$$\text{Ru(COD)(COT)} + \text{H}_2 \rightarrow \text{RuNPs} + \text{COA}$$

What do the $\eta^4$ and $\eta^6$ notations represent?  Do they somehow specify the coordination "connectivity" of the Ru(0) and the COD and COT coordinating ligands?
Below, I have drawn the (two-dimensional) structures of COD and COT.  How would I draw the connectivity of Ru(COD)(COT), or is it not specified in the given name?

What do the $\eta^4$ and $\eta^6$ notations represent? Do they somehow specify the coordination "connectivity" of the Ru(0) and the COD and COT coordinating ligands?

They do.
The $\eta$-notation denotes the hapticity of the ligand, which is the number of atoms in the ligand bonded to the metal center.

In your image, shown above, cyclooctadiene (COD), has two alkene functional groups. A transition metal can coordinate to just one alkene, which would be $\eta^2$, or to both alkenes, which would be $\eta^4$.
Cyclooctatriene (COT) has three alkenes. A metal center can be coordinated at one ($\eta^2$), two ($\eta^4$), or all three ($\eta^6$) alkenes.


text:
What is the influence of an alkyl-hydroxy group on a benzene ring?
We all know that an $\ce{NO2}$ group attached to a benzene is a deactivator, and it directs and substituents that would be adding to the meta position to the $\ce{NO2}$.  For an alkyl group attached to a benzene it is an activator, and it directs to the positions ortho/para to the alkyl group.
I haven't found any information on how a $\ce{-CH2-OH}$ group would affect the reactivity of the benzene ring shown below

According to Hansch, Leo, and Hoekman (Exploring QSAR: Hydrophobic, Electronic and Steric Constants 1995) the Hammet sigma value for $-CH_2-OH$ is 0 (zero) in both either the meta or para position.  Positive values indicate deactivators or electron withdrawing (+.778 for nitro in para position) and negative values are activators or electron donating (-0.268 for methoxy group).  The Hammett relationship gives a way to compare electronic effects of substituents on a benzene ring.  Further info on Wikipedia.


text:
Reaction between n-butylithium and either an aromatic bromide or an alkylbromide
What exactly happens? Does the butylithium chain add to the carbon where the bromine is attached? or do the 2 undergo halogen metal exchange?

It's quite a long time that I did myself, but I used BuLi to transmetalate. I.e. $\ce{MesBr + BuLi -> MesLi + BuBr}$

Coupling BuLi with an alkylbromide to extend an alkyl chain is an extremely minor produt.  Transmetalation or elimination products will dominate.


text:
Can iproniazid be prepared by reacting isoniazid with 2-chloropropane?
I'm asking as I fear that the 2-chloropropane might react with the pyridine ring to form k-isopropylpridine-4-carbohydrazidine where k=2,3,5. Or that no reaction will take place.

Figure 1: Isoniazid

Figure 2: Iproniazid

Figure 3: 2-chloropropane

Could you elaborate on why you think that the reaction with the pyridine ring will be preferential?

Because the only difference between a pyridine and a benzene is a C instead of N and aromatic compounds react with alkyl halides like 2-chloropropane to create products in of which the benzene and alkyl halides have reacted to give the corresponding alkylbenzene.

@BrentonHorne - Pyridine is deactivated relative to benzene, and isoniazid is likely more so. When I have time, this will become an answer, unless someone else beats me too it.

Wouldn't the chloropropane react with both the nitrogens in the side chain of isoniazid?

@kaliaden The amide NH is much less active, so no, it won't.  However, it will readily react with $NH2$ group twice unless there are sterical issues, what I doubt.

Alkylation of the pyridine ring under normal conditions is very unlikely, so that's not a major concern. The nitrogen is electron withdrawing, hence electrophilic aromatic substitution reactions are much more difficult to achieve on pyridine than benzene, particurarly absent catalysis. Additionally, the α-amino ketone substituent already on the ring is further deactivating (since it contains a π bond conjugated to the ring and a highly electronegative atom, i.e., oxygen).
One issue is that pyridine is liable to form quaternary ammonium salts when reacted with alkyl halides. See for example the substances paraquat, diquat, and the following paper on pyridine chemistry. While the quaternization of H-bearing ammonium cations is easily reversible (given that they can be deprotonated), the same can't be said for the nitrogen of the pyridine ring. Generally, the only methods I'm aware of for dealkylation of quaternary ammonium salts involve hydride donors and strong reducing agents, such as lithium aluminum hydride. Certainly, the ketone group would need to be protected in that situation.
There is also the issue of avoiding polyalkylation, as well as selecting which of the two amines in your substituent react, in view of the fact that the basicity and nucleophilicty of amines increases with the number of alkyl substituents. Sterics, however, might preclude that from becoming a problem in this instance.

The $\ce{NH_2}$ group in given compounds is very similar to regular amine fragment and is more active than acylated $\ce{NH}$. However, alkylated amines are more active than non-substituted ones. So, I'd expect a very significant amount of double-alkylated product.
To avoid purification issues I recommend to consider synthesis of acetone hydrazone followed by reduction with $\ce{Na[BH_3(CN)]}$ or similiar low-active hydrogen donor (hi-active ones will reduce amide fragment)