text:
Can compressed CO2 sprayed onto a hot mass/liquid be an effective means to cool it?
I thought the answer was obvious, but endless google searching has revealed nothing but speculation and people who have completely conflicting opinions. Some people say $\ce{CO2}$ absorbs no heat, others say it can cool things extremely rapidly. Anyone who has ever had compressed $\ce{CO2}$ sprayed on themselves or on hands knows it can cause very rapid drop in temperature on what ever it hits very fast.
So, if I have a really hot mass of liquid, would spraying compressed $\ce{CO2}$ on it be an effective means of freezing/solidying the mass? - Assuming the liquid is well above room temperature (100+) and turns to a solid after a 20-30 degree drop in temperature?
I understand why the $\ce{CO2}$ is cold coming out of the tank, as it is decompressed, but is it effective at absorbing heat?

Celebrity chef Alton Brown produces ice cream using a device called the Jet Cream.  He sprays $\ce{CO2}$ from a fire extinguisher at a stream of chocolate milk, instantly producing ice cream. The only difference in your scenario and his is the starting temperature of the liquid.
To say that $\ce{CO2}$ absorbs no heat is absurd.  If it is colder than what it touches, it should absorb heat.

Yes, blowing $\ce{CO2}$ over a hot object would cool it. The second law of thermodynamics says that heat cannot move from a colder material to a hotter material. So if you had a hot object and applied any material cooler than it, the hot object would cool down. The $\ce{CO2}$ offers some advantages because it's a gas. As the gas heats, it expands. By the ideal gas law, any expanding gas must cool down, assuming a constant pressure. However, this effect should apply to any gas, and compressed air would be cheaper than pure $\ce{CO2}$. You might also apply dry ice. This could be considered really compressed $\ce{CO2}$. The dry ice offers a further advantage because of the phase transition from solid to gas, not to mention its much colder initial temperature. As the dry ice sublimates its temperature remains constant throughout the phase transition, allowing to draw much more energy out of the hot object. However, water would also offer this advantage, as it has a high heat capacity, and is much cheaper than dry ice or compressed $\ce{CO2}$. It should also be safer. Suddenly heating a block of dry ice in an enclosed space could be explosive. Even if it doesn't explode, the large amount of $\ce{CO2}$ released could displace the air from the room and choke anyone in the room.
tl;dr, use compressed air or liquid water.

Yes, but carbon dioxide will be colder than compressed air or water so might do a better job and, since it is a gas, won't leave as much mess as water nor will it affect the warm material as badly. Air won't work as well as compressed air is a gas but carbon dioxide is a liquid and will be colder when emitted.

Theoretically it will as any mass of expanding gas will absorb heat from its surrounding.
Practically if it is feasible depends on multitude of factors (object to be cooled, its temperature, its mass, temprature drop required, amount of compressed CO2 required etc.)

I would not generalize: https://en.wikipedia.org/wiki/Joule%E2%80%93Thomson_effect

Agree! There could be lot more to consider than just work done in compressing CO2 in a cylinder and work it does while expanding when released.


text:
Explanation of certain chemical bond diagrams
Here's a "simple" bond diagram of 1,1,1,2-Tetrafluoroethane which I believe is a Lewis structure:

OK, that's easy enough - four bonds from each carbon atom, and each fluorine atom having one bond.
Here are some more "complicated" bonds which I have yet to do in my high school chemistry class, so I don't understand how to interpret them.
Here's caffeine:

And chloroform:

How would I go on interpreting these diagrams? Or is there missing information?
Image sources: Wikimedia

There are many different ways of representing molecular structures, see the Wikipedia page on structural formula(e) for a summary. The one used most frequently by organic chemists is a skeletal formula (often referred to as a "bond-line structure" or "line-angle formula"), which is what you see in that representation of caffeine.

The second structure you posted is a line angle diagram. In this diagram, methyl groups (-CH3) are implied using sticks with nothing at the end.

Carbon atoms are also implied; carbons exist at every "corner" of the polygon which doesn't have an element explicitly identified. Hydrogens attached to carbons are also implied. So at the intersection of two lines, if there are no other attachments and no charge, you can expect there to be two hydrogens (carbon is generally tetravalent).

The third structure you posted is a wedge-dash diagram. Wedges depict substituents angled toward you; dashes are substituents pointing away from you; the lines are coplanar. The advantage of this diagram is that it shows the 3D nature of the molecule. Your first picture of the fluoro-carbon compound implies 90 degree bond angles. However, you should realize that the carbon atoms are sp3 hybridized and thus should not have 90 degree bond angles.

Further reading and exploration of other depictions of molecular structure:
http://www.masterorganicchemistry.com/2010/10/15/the-many-many-ways-to-draw-butane/

Thanks for the detailed explanation! Can we always assume that a stick with nothing at the end implies $CH_3$? Also, what compound is in the first image you posted?

Yes. That's always true. The compound in the first image is 2-butene. The "ene" ending implies the presence of a double-bond (alkenes are hydrocarbons with double bonds); the 2 identifies the location of the double bond (the double bond starts on carbon number 2).

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Electric conductivity vs potential
What is the relationship between electric potential and conductivity?
For example, gold has a more negative electric potential than silver (-1.100 vs -0.800), but silver is a better conductor.

I would not expect to find any relationship.  The potential you are referring to involves breaking metallic bonds and ionizing the resulting atoms while conductivity involves electrons moving through the solid metal.  Copper's is -.34 and it is between gold and silver in conductivity.  Aluminum's is 1.66 and it is #4 behind gold in conductivity.

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Removing unreacted Mg metal in a Grignard reaction
A Grignard reagant was prepared from an alkyl halide and magnesium chips, then reacted with a ketone. The resulting solution has been stirring for a few days at low temperature (0-5°C). It's now time to transfer the reaction solution slowly to a vessel containing an acid quench. My problem is the unreacted magnesium chips (smaller than when we started, of course) have settled, and are plugging the tubing and stopper valve.
What solvents could I use to flush the magnesium through, breaking it up, and possibly dissolve it? This can't be added to the reaction solution (the Drain-O method), but only through the tubing itself. Obviously water is a bad idea. Any insights?

Plug the tubing with improvised filter made of mineral wool ?

Pure water would certainly be a bad idea, but a 1% solution of 1 M hydrochloric acid in ether would probably work without a vigorous reaction.

We had what we would call "filtersticks", they were a cannula with a metal cylinder with a bore at one end, around which we would tie glass filter paper.  Larger quantities we transfer through polyethylene tubing to a filtration column and filter through a pad of Celite.  Both methods worked well.
Since your product doesn't sound air-sensitive, you might consider quenching with wet THF.  Have an ice bath and a fire extinguisher ready.


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How much can the pH change through dilution?
Consider an acidic solution with Hydrogen ion concentration, $\ce{[H+]}$ of $10^{-5}\:\mathrm{M}$. Since $\:\mathrm{pH} = -\log \ce{[H+]}$ the $\:\mathrm{pH}$ of solution is $5$. Suppose we dilute solution 10 times with water. Now, $\ce{[H+]}$ is $10^{-6}\:\mathrm{M}$ and $\:\mathrm{pH}$ is $6$. Further dilution should increase $\:\mathrm{pH}$ from $6$ to $7$ and then from $7$ to $8$ and so on. Can this go in for ever? Does this not imply that an acidic solution can be made basic/alkaline simply by adding water? But that doesn't happen? What prevents it?
Is there anyone already found answer for this problem or it is just an unsolved basic problem of chemistry?

Water undergoes autoionization, i.e., it reacts as follows:
$$
\ce{H2O + H2O <=> H3O+ + OH-}
$$
The equilibrium constant for this reaction at standard conditions is $K_w = [\ce{H3O+}][\ce{OH-}] \approx 1.0 \cdot 10^{-14}$. In pure water, $[\ce{H3O+}] = [\ce{OH-}]$, hence $[\ce{H3O+}] = \sqrt{K_w} \approx 1.0 \cdot 10^{-7}\ \textrm{M}$.
Suppose we dilute a solution with some initial concentration of $[\ce{H3O+}]_i = \frac{n_i}{V_i}$, where $n_i$ is the initial number of moles of $\ce{H3O+}$ and $V_i$ is the initial volume. If I now add a volume $\Delta V$ of pure water (which would contain $(1.0 \cdot 10^{-7}) \cdot \Delta V$ moles of $\ce{H3O+}$) the resulting final concentration can be crudely approximated as:
$$
[\ce{H3O+}]_f \approx \frac{n_i + (1.0 \cdot 10^{-7}) \cdot \Delta V}{V_i + \Delta V}
$$
We can see what value this expression approaches as we make the solution more and more dilute by taking the limit as $\Delta V \to \infty$:
$$
\lim_{\Delta V \to \infty} \frac{n_i + (1.0 \cdot 10^{-7}) \cdot \Delta V}{V_i + \Delta V} = 1.0 \cdot 10^{-7}
$$
Therefore, $[\ce{H3O+}]$ tends towards $1.0 \cdot 10^{-7}$ with further dilution, so $pH$ will approach a value of $7.0$.

I add a question here....what will be change in ph when a buffer soln containing 0.1 MCh3COOH and 0.1 M CH3CHOONa. is diluted 10 times

You forget that pure water already contains $\ce{H_3O^{+}}$ ($\ce{Kw}=\ce{[H_3O+][OH^{-}]}=14$,). If you have an acidic solution, you will be able to reach the neutral point ($\ce{pH} = 7$) after enough dilution, but it will never be alkaline. You can see this as similar to a limit or infinite sum $\lim_{x\rightarrow \infty } \frac{x}{x+1}=1$, but when you look at finite $x$ it will never actually be 1.

I've uploaded a figure, just for illustration (red line: calculated pH without considering autoionization, blue is the actual one). You can see the difference.

It can not be alkaline because the water is not really alkaline, more like neutral. But dilution in an alkaline solvent will easily make the solution alkaline.
EDIT: seems my answer was not considered an answer. Well since the question is composed out of more than 1 question...
Using water, you won't be able to get past aprox 7 PH. Using an alkaline non reactive solvent will allow you to go as far as the PH of the solvent.
Additional clarification: If no reaction happens, the PH limit of the solution is bound between the substance with the highest PH, and the one with the lowest.
No, an acidic solution can not be made alkaline by adding water.

I don't think you sufficiently answered the why, which is the main question. That makes this not an answer.

I answered the question "how much does PH change through dilution.

@GregE and Jori's answers both give an answer using straightforward math and their answers seem to be pretty well liked by the community.

I agree with Tyberius - As the question already has a (three year old) answer, you should really only be adding another answer if you think there's important information that hasn't been covered by the existing answers, or if you think existing answers are too unclear to adequately answer the question. -- As it looks like you're new to the site, please take the tour and look at the help section if you haven't already. Stack Exchange is a Q&A site, not a forum, so we do things a bit differently here.


text:
On the process of heating a gas at constant pressure

Calculate the final temperature and the change in internal energy when $\pu{500 J}$ of energy is transferred as heat to $\pu{0.900 mol}$ $\ce{O2(g)}$ at $\pu{298 K}$ and $\pu{1.00 atm}$ at constant pressure. Treat the gas as ideal.

I am trying to understand how to approach this question without simply plugging the numbers into an equation. Can someone please explain their thought process while solving this? For example, does having constant pressure automatically mean there will be a temperature change and if so, why? What about volume? Why do we not need to know the volume for this equation? Basically, I want to understand what's going on conceptually rather than memorize a few rules. Any help would be much appreciated.

OK, let's go one question at a time.

does having constant pressure automatically mean there will be a temperature change and if so, why?

Not quite.  Adding energy means there will likely be a temperature change. What
constant pressure means is that it will be easy to calculate this change, because you can use the value of the constant-pressure molar heat capacity, $C_p$, which is the same for all diatomic ideal gases.  It is also independent of temperature, volume and pressure.  You will find it in your book.
The molar heat capacity tells you how much energy you need to transfer to one mole of the gas to raise the temperature by one degree.  You now have everything you need to know by how much the temperature changes, and therefore the final temperature.

What about volume? Why do we not need to know the volume for this equation?

Actually you do know the volume, or can easily figure it out.  And though the volume is not important for finding the temperature change, for the reason explained above, it's helpful in finding the change in internal energy. Why?
The change in internal energy is the energy you put in (500 J) minus whatever energy the gas spends expanding.  Since the pressure is constant the energy spent expanding is simply the product of the pressure by the increase in volume.  You can calculate the increase in volume because you know the pressure, number of moles and initial/final temperature.
Now a bit of meta advice.

I am trying to understand how to approach this question without simply plugging the numbers into an equation. Can someone please explain their thought process while solving this?

You don't make clear what, if anything, you know (either conceptually or equation-wise) and what, if anything, you've tried.
You'll get help faster if you show you've done your part.

Basically, I want to understand what's going on conceptually rather than memorize a few rules.

Understanding what's going on conceptually goes hand-in-hand with learning the rules.  You can't really do one without the other.  (Well, you can memorize the rules without understanding the concepts, but you won't know which ones to apply.)


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The definition of formal Charge
Ok, most people including me know that formal charge is just a book-keeping tool. I know how to calculate it, apply it to Lewis structures etc etc. But I am confused in its very basic definition which says - "Formal charge is the charge on an atom if the electronegativity differences are ignored" . If the electronegativity difference is ignored, then how can an atom have charge ? I know this is not a place to ask such a basic question , but still I am very confused about this miniscule definition.

Take a look at this previous question to see if it helps you out.

@NicolauSakerNeto Thanks but I know all that (how to calculate and use formal charges). I just want an explanation to this definition.

In terms of your definition, formal charge is the charge on an atom when electronegativity differences, regardless of relative electronegativity, are ignored. You must realise that a formal charge is not the same as a real charge. They are used simply as a bookkeeping method only. But, Polarity considers electronegativity rather than formal charge.

A good description appears in the book, "Chemistry A Molecular Approach".
In the molecule of hydrogen fluoride, we know that it has a dipole moment, and fluoride is slightly negative. Ignoring this information (electronegativity difference) and we share the bonding electrons equally, the formal charge can be calculated, and both of them are determined to be 0.
In reality, we know that fluoride has the electron from the hydrogen most of the time (due to electronegativity difference). We say the hydrogen has a partial positive charge and the fluoride has a partial negative charge.
From this, it is apparent that the formal charge is different from the charge on the atoms in a molecule.

Does the formal charge affect the overall charge seperation? I mean in carbon monoxide formal charges are -1 on carbon and +1 on oxygen but considering only electronegativity we should expect the negative charge on oxygen whereas the electrostatic map potential shows some negative charge on carbon. So can we say that the overall charge depends also on formal when and only when the proposed lewis structure have formal charges opposite to these according to electronegativity differences?

@adosar Actually, the formal charge in carbon dioxide is +4 on carbon and -2 on oxygen. Since formal charge is for bookkeeping, it has no influence on what the charge distribution in the molecule is.

I said for the carbon monoxide.

@LDC3 I think the electronegativity difference indicates which of the 2 atoms is attracting the "bonding electron density" more strongly. By assuming equal sharing, you know that the charge on oxygen is +1. The further polarisation of shared electron density towards oxygen would perhaps reduce +1 to a δ+ partial charge (since the oxygen atom can't completely attract the shared electron density; that would make the bond predominantly ionic )

The formal charge is the charge on an "ion" that results when all valence electrons participating in bonds are assumed equally shared between this "ion" and the others it is bonded to. If we consider, for example, nitrogen bonded to three carbon atoms using its 3 bonding electrons then it loses, assuming equal sharing, 1/2 an electron to each carbon but gets back a half share in an electron from each carbon. Net charge on the nitrogen: 0. Now assume that the unshared pair form a second bond with one of the carbons. The nitrogen loses a half share in each of those two electrons but does not get back a half share from the carbon in this case. Thus the nitrogen has a formal charge of +1. But this is a formal charge. A bookkeeping convenience - nothing more. In actuality the electronegativities of carbon and nitrogen will determine where the electrons spend their time and thus the apparent charge distribution. Obviously it would be quite tedious to try to figure out what those charge distributions would be. It's handy sometimes to assume, when considering covalent bonds, that they are perfectly covalent.

It's best to think about why electronegativity is used in this definition of formal charge and perhaps replace it with a concept that is familiar. One definition of electronegativity is the ability of an atom to attract electron density towards itself. So if we think of a simple diatomic series with (lewis pairs added) like: H:H,  H:Br, H:Cl and H:F, with electronegativities H < Br < Cl < F, then those two electrons ":" are closer to F than they are to Cl, which in turn are closer to the Br atom and in H:H they are shared equally. Now the formal charge is the number of electrons ascribed to the atom minus the atomic number. So the question is how do we ascribe an electron to each atom? If we try to include electronegativity arguments then we would ascribe more electrons to F then to Cl in those diatomics and the formal charges in HF and HCl would be different. However, if we ignore the electronegativity and let the atoms share the electrons equally (like in a lewis diagram) then we arrive at the familiar formal charge definition that you have noted.

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Bombarding atoms with electron gun?
Just to explain this question in a better way just think about the Rutherford's experiment(the alpha particle bombarded on Gold foil) be conducted using a electron gun in place of the Alpha one.. So what will happen when electron is targeted on a gold foil?
Either the atom rather the electrons of Ag atom become excited and therefore emit some sort of waves(if yes,then what kind of waves?), or in return electrons are ejected from the atom along with some waves,  or it may be possible that the electron pass by the gold foil without any effect...

Well, I think, in principle, depending on the energy of electrons, different events can happen, but in the classical experiment electrons will diffract from the spacing within crystal lattice of gold in the foil.
P.S. The experiment is actually very well known in the history of science due to the following reason (quoted from here).

It is interesting to recall that G.P. Thompson, who shared the 1937
  Nobel Prize with Davisson for these experiments which proved that
  electrons are waves, is the son of J.J. Thompson who received the
  Nobel Prize in 1906 for proving that cathode rays were actually
  particles - electrons! And the amazing thing is that they were both
  right.

IT may also be possible that electrons strike the atom resulting in emitting of electron present in the atom of Ag and those Ag electron may possibly have K.E. causing observation on wave nature of newly emitted electrons.

??????????????????????

So, it was the question? Yes, it possible. But how can you explain a diffraction pattern by this process?

Multiple interactions can happen.
First, lets look at the "simple" case of an electron interacting with a single atom.  There are three options.  (1) the electron does not interact with the (neutral) atom - it misses completely and keeps on going.  (2) the electron may interact with the electron cloud of the atom.  This may lead to a scattering event, kicking a bound electron into a higher energy level, or knocking it off the atom entirely (generating the secondary electrons used for imaging in an SEM).  The atom can then relax, possibly releasing an x-ray or an Auger electron, both of which are used for composition analysis of materials.  The incident electron is deflected from the original path because of the scattering event. (3) the electron may interact with the nucleus and undergo Rutherford scattering (yup, same formula applies).
The more complex case of an electron impinging on a crystal adds the periodic potential of the crystal.  All of the events listed above apply, but in addition the periodic potential allows for diffraction, where the electron scatters off the Bloch waves of the crystal.  In a TEM diffraction can be imaged from the transmitted beam, in an SEM it can be imaged from the back-scattered beam (EBSD).
So, pretty much everything you mentioned can and does happen.  All are used for various measurements in materials science and physics.


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Why do SN1 and SN2 reactions not occur at sp2 centres?
I was told in my organic chemistry course that $\text{S}_\text{N}1$ and $\text{S}_\text{N}2$ reactions did not occur at $\text{sp}^2$ centres. When I asked why, I was not given a satisfactory explanation. For $\text{S}_\text{N}2$ it was suggested that the reaction could not proceed with inversion of configuration, as this would disrupt the orbital overlap causing the $\pi$ bond. I couldn't find any explanation in Clayden's Organic Chemistry (2nd ed.), or even a mention of it.
Could anyone please explain why? I have attached some curly arrow diagrams that seem reasonable enough to me, although I realise that being able to draw a reaction mechanism does not mean it is valid.

For the record, the second mechanism probably does happen in limited cases: https://pubs.acs.org/doi/10.1021/ja972267c

Not a bad question, but a poor answer. The person who mentioned $\mathrm{S_N2}$' is on a much better track.
Nucleophilic attack on a carbonyl or 1,4-Michael acceptor is not an $\mathrm{S_N2}$ reaction. It is also very misleading. It's like saying a chicken and a human are the same thing because they both have two legs.
$\mathrm{S_N2}$ reactions involve nucleophilic attack on the sigma-antibonding orbital. In those two examples, (carbonyl & 1,4-Michael acceptor), the nucleophile is attacking the $\pi$-antibonding orbital. Moreover, there is no inversion of configuration, and it is not a concerted bond-breaking/bond-making reaction. So no, it is in no way an $\mathrm{S_N2}$ reaction.
Back to the original question: in the $\mathrm{S_N2}$ case, the lowest energy unoccupied molecular orbital is the $\ce{C-C}$ $\pi$-antibonding orbital. Thus, the nucleophile will overlap with that orbital. The $\ce{C-X}$ $\sigma$-antibonding orbital, which is what the nucleophile would attack in the mechanism you're drawing, is higher in energy, and the resulting transition state would be higher in energy compared to the reaction in which the $\ce{C-C}$ $\pi$-antibonding orbital is attacked. Thus, the reaction does not proceed in that manner.

Sometimes, especially in introductory courses the instructor will try to keep things "focused" in order to promote learning.  Still, it's unfortunate that the instructor couldn't respond in a more positive and stimulating way to your question.
These reactions do occur at $\ce{sp^2}$ hybridized carbon atoms, they are often just energetically more costly, and therefore somewhat less common. Consider when a nucleophile reacts with a carbonyl compound, the nucleophile attacks the carbonyl carbon atom in an $\ce{S_{N}2}$ manner.  The electrons in the C-O $\pi$–bond can be considered as the leaving group and a tetrahedral intermediate is formed with a negative charge on oxygen. It is harder to do this with a carbon-carbon double bond (energetically more costly) because you would wind up with a negative charge on carbon (instead of oxygen), which is energetically less desirable (because of the relative electronegativities of carbon and oxygen).

If you look at the Michael addition reaction, the 1,4-addition of a nucleophile to the carbon-carbon double bond in an $\ce{\alpha-\beta}$ unsaturated carbonyl system, this could be viewed as an $\ce{S_{N}2}$ attack on a carbon-carbon double bond, but again, it is favored (lower in energy) because you create an intermediate with a negative charge on oxygen.

$\ce{S_{N}1}$ reactions at $\ce{sp^2}$ carbon are well documented.  Solvolysis of vinyl halides in very acidic media is an example.  The resultant vinylic carbocations are actually stable enough to be observed using nmr spectroscopy.  The picture below helps explain why this reaction is so much more difficult (energetically more costly) than the more common solvolysis of an alkyl halide.  In the solvolysis of the alkyl halide we produce a traditional carbocation with an empty p orbital.  In the solvolysis of the vinyl halide we produce a carbocation with the positive charge residing in an $\ce{sp^2}$ orbital. Placing positive charge in an $\ce{sp^2}$ orbital is a higher energy situation compared to placing it in a p orbital (electrons prefer to be in orbitals with higher s density, it stabilizes them because the more s character in an orbital the lower its energy; conversely, in the absence of electrons, an orbital prefers to have high p character and mix the remaining s character into other bonding orbitals that do contain electrons in order to lower their energy).

What are the curved line segments between the central carbon and the oxygen atom in your first diagram?

Good question, I should've explained it.  The curved lines above and below the C-O sigma bond are meant to represent the pi electrons.  The arrow below the bottom curved line represents the electron flow of these pi electrons away from carbon and onto oxygen.

Ron's explanation is very good. One more reason why these reactions are not explored in introductory courses is that they're generally not applied in synthesis of more complex compounds, which is one of the main goals of organic chemistry.

Wouldn’t an $\mathrm{S_N2'}$ mechanism be even better to show $\mathrm{S_N2}$ at $\mathrm{sp^2}$ carbons?

Yes, that would be a nice example, somewhat analogous to the Michael addition example.

The other answer seems to be challenging your stand. Anything to say to what the user Schrodinger mentioned?


text:
Finding the chemical name of SnMg₂O₄
Can somebody help me in identifying the name of $\ce{SnMg2O4}$?
Other examples are $\ce{ZnAl2O4}$ (zinc aluminate), $\ce{ZnIn2S4}$ (zinc thioindate).
Please kindly provide the reference if possible.

$\ce{Mg2SnO4}$ Magnesium Orthostannate
$\ce{MgSnO3}$ Magnesium Metastannate

$\ce{SnMg2O4}$ and $\ce{ZnAl2O4}$ belong to spinel group ($\ce{A^4+B^2+2O4}$ and $\ce{A^2+B^3+2O4}$, respectively) and should be treated as solid solutions of ternary oxides:

$\ce{SnMg2O4}$: tin(IV) dimagnesium oxide or dimagnesium tetraoxostannate(IV);
$\ce{ZnAl2O4}$: zinc dialuminium oxide or zinc tetraoxodialuminate.

$\ce{ZnIn2S4}$ is a thiospinel ($\ce{A^2+B^3+2S4}$):

$\ce{ZnIn2S4}$: zinc diindium sulfide or diindium tetrathiozincate.

In general, IUPAC Red Book [1, p. 237] discourages from using ortho- and meta- prefixes with only a few exceptions (for more details refer to What does the prefix meta mean in inorganic chemistry?), and suggests leaving the formula unit with a structural type notation as, for example, $\ce{SnMg2O4}$ (spinel type):

A mineral name may, however, be used to indicate the structure type. Where possible, a name that refers to a general group should replace a more speciﬁc name. For example, large numbers of minerals have been named that are all spinels, but which have widely differing atomic constituents. In this case, the generic name ‘spinel type’ should be used rather than the more speciﬁc names chromite, magnetite, etc. The generic mineral name, printed in italics, should be accompanied by a representative chemical formula. This is particularly important for zeolite types.[...]
Examples:
   1. $\ce{FeCr2O4}$ (spinel type)
   2. $\ce{BaTiO3}$ (perovskite type)

References:

IUPAC “Red Book” Nomenclature of Inorganic Chemistry; Connelly, N. G., Ed.; IUPAC Recommendations; Royal Society of Chemistry Publishing/IUPAC: Cambridge, UK, 2005. ISBN 978-0-85404-438-2.


text:
Why does a Fluoride ion only have a -1 charge and not a -2 charge or more?
Transition elements can form ions with different charges. Why can't elements other than transition elements form ions with different charges? If it is a Fluoride ion, why does it have to be an anion of charge -1? Why can't it have a -2 charge, or more?

The octet rule is usually used to explain the charges that atoms in groups I, II, VI, and VII have in ionic compounds, by saying something like "noble gases are stable, therefore having 8 valence electrons must be stable for some reason." This is true, but understanding why noble gas configurations are stable is important, and without knowing that, it does seem like magic that some atoms don't adopt a noble gas configuration in ionic compounds.
The octet rule is a useful way to quickly predict ionic charges and write Lewis structures, but it's not a magical law that atoms follow. They don't "want" or "try" to obtain specific electronic structures, although we often talk like they do. What is really important is the total energy of the electron configuration. The electron configuration with the lowest energy is the most stable one.

"Stability" really means "lowest energy"

The total energy of the electron configuration is governed by two things:

The energy of the electron orbitals
The effective nuclear charge that is "seen" by the electrons

The orbital energies determine the order in which atomic orbitals will be populated by electrons in the ground state - in other words, which orbitals the electrons will exist in for a given atom. We know from empirical evidence and from quantum mechanical calculations that the order goes like this for most elements (not for some transition metals and heavier elements):

Electrons will occupy these orbitals from the "ground up" (called the Aufbau principle). We know, therefore, that for 4th-period elements, the $3d$ sub-shell will fill before the $4p$ sub-shell will.
The atomic orbital diagram indirectly tells us something else. The principle quantum number, $n$ (which for the valence shell is equal to the row or period number), roughly corresponds to the distance from the nucleus of the orbital's highest electron density region. In other words, any sub-shell with $n=4$ will have most of its electron density further away from the nucleus than a sub-shell with $n=3$. This means that $3d$ electrons are closer to the nucleus than $4s$ electrons, even though $3d$ electrons initially have  higher energy.

The distance of the electron from the nucleus is roughly proportional to $n$

Since the electrons have a negative charge, and the nucleus is positively charged, there is an electrostatic potential energy between the electrons and the nucleus. If the electrons were uniformly distributed throughout the atom, we would expect that the nuclear charge "felt" by each electron would be equal to the number of protons - the atomic number. However, because of the "rules" of quantum mechanics (the allowed solutions to Schrodinger's equation), electrons are not uniformly distributed - they occupy orbitals in a specific arrangement. Since some electrons are closer to the nucleus (lower $n$) than others, this means that the nuclear charge is partially screened by the inner electrons, and the outer electrons experience an effective nuclear charge that is somewhat less than the full nuclear charge. This diagram from wikipedia demonstrates the idea:

And this one shows the idea using the Bohr model (note that this is a schematic - not an accurate picture of what electrons "look" like in orbitals - only the relative distances from the nucleus is right)

When the valence electrons experience a large effective nuclear charge, they are harder to remove (the energy cost for removing them increases), and when the effective nuclear charge is lower, they are easier to remove. The actual effective nuclear charge is very difficult to accurately calculate, but it is very easy to estimate - all you need to do is take the number of protons, then subtract the number of "screening" electrons. Basically, we are assuming that each electron between the valence shell and the nucleus cancels out one proton:
$$
Z_{eff} = Z - S
$$
where $Z_{eff}$ is the effective nuclear charge, $Z$ is the actual nuclear charge, and $S$ is the screening constant, here assumed to equal the number of "core" electrons.
To find $S$, all you have to do is count the number of electrons with $n$ less than the valence shell (highest $n$).
Taking fluorine as an example, the electron configuration is:
$$
1s^22s^22p^5
$$
We see that $Z=9$ and $S=2$, giving an effective nuclear charge of +7. Compare that to lithium: $3-2=+1$. The valence electrons on fluorine are roughly 7 times harder to remove than the valence electron on lithium. In fact, any element to the left of fluorine has a lower effective nuclear charge for the valence shell, and will give up electrons more easily than fluorine will. In other words, it's not so much that fluorine won't give up electrons, it's that nothing can take them from it.
What if we add an electron? For $\ce{F-}$, the electron configuration becomes:
$$
1s^22s^22p^6
$$
$Z_{eff}$ is still +7, since the number of core screening electrons hasn't changed. This means that it is energetically favorable for fluorine to pick up an electron, since the additional electron would still "see" a large positive nuclear charge.
What if we add another, to make $\ce{F^{2-}}$? Now the configuration looks like this:
$$
1s^22s^22p^63s^1
$$
$Z_{eff}$ is now $9 - 10 = -1$ - note that the number of screening electrons jumped to 10 because the "new" valence shell has $n=3$. In other words, because of the allowed orbital configurations the new electron would "see" a negative charge on the nucleus - it would require energy to keep it.
If you work through this process for all of the "fixed charge" elements, you find similar things - the effective nuclear charge is large for elements that gain electrons, until they gain too many (an octet), at which point it becomes negative. For fixed-charge metals, something else happens. Let's look at magnesium.
$\ce{Mg}$: $12-10=+2$
$\ce{Mg^1+}$: $12-10=+2$
$\ce{Mg^2+}$: $12-2=+10$
Once you remove two electrons, the effective nuclear charge jumps to +10 for the remaining valence electrons. The same is true for all the group I and II metals - once you get to the noble gas "core" - the octect - the effective nuclear charge seen by the valence electrons jumps way up.
This explains why "fixed charge" elements always have the same charge in ionic compounds, but what about transition metals?
Let's look at chromium. The electron configuration looks like this:
$$
1s^22s^22p^63s^23p^64s^23d^4
$$
Which electrons are the valence electrons? The ones with the highest $n$ - the 4s electrons. The effective nuclear charge they "see" is $24 - 22 = +2$ - note that the 4 electrons in the $3d$ sub-shell are counted as screening, since they have $n=3$ which is less than 4, putting them closer to the nucleus than the valence shell, despite their initially higher energy.
If we remove an electron, we get:
$$
1s^22s^22p^63s^23p^64s^13d^4
$$
(This isn't exactly correct, there would really be no $4s$ electron and 5 $3d$ electrons, but the reason is complicated and it doesn't matter to this explanation)
Note that we removed a valence electron - a $4s$ electron. Now $Z_{eff} = 24 - 22 = +2$ - it hasn't changed. We can remove one more to make $\ce{Cr^{2+}}$ and get:
$$
1s^22s^22p^63s^23p^63d^4
$$
Now the definition of "valence" becomes a little less clear - we know that the highest $n$ value corresponds to the valence shell, but what about $l$, the orbital shape? It turns out that d orbitals are larger than s and p orbitals, and so the s and p electrons end up screening the nucleus from the d electrons. This means that d electrons would be removed next, and the effective nuclear charge "seen" by them is $24 - 18 = +6$. This is a larger charge, but not so large that other elements (notably oxygen and the halogens) can't still pull the d electrons away. This gives us a range of chromium charges (oxidation states) from +1 to +6. For $Cr^{6+}$, the electron configuration would be:
$$
1s^22s^22p^63s^23p^6
$$
Now the effective nuclear charge is $24 - 10 =  +14$ - much too strong for anything else to "win" against.
As you can see, the rules of quantum mechanics determine the relative size of atomic orbitals and the order in which they are filled, which in turn determines the effective nuclear charge that valence electrons "see" for any given neutral atom or ion. The combination of the two leads to constraints on the charges that atomic ions can have, with the practical result being that group I, II, VI, and VII elements adopt a "filled octet" or "noble gas" configuration, while the other groups have more flexibility.

Nice usage of nuclear effective charge, and a good answer in general. I never considered that in extreme situations the calculation could actually result in a negative value, strongly suggesting at least one valence electron is unbound from the nucleus. However, I should point out to other readers that your calculations are very qualitative, much more so than even those based on Slater's rules. Your argument still stands, of course, but I would not recommend anyone take the magnitude of the calculated effective charges too seriously.

@Nicolau - that's why I tried to make it clear that it was an estimate, not the actual effective charge.

@thomij  I'm wondering about your statement, "This means that 3d  electrons are closer to the nucleus than 4s  electrons, even though 3d electrons have higher energy."  Is that true, why would an electron closer to the nucleus have higher energy?

@ron - The argument commonly used is that the local electron density of the d shells is higher than the next higher s shell, both because of "crowding" in the d sub-shell and because of the proximity to the 3s and 3p orbitals. The real reason is more complicated, but in short, the 4s electrons have density peaks closer to the nucleus, even though the outermost peak is further out. The energy levels actually move around as the orbitals fill, and by the time the 3d orbitals have started to fill, they may or may not have higher energy. I will edit my answer to make it more correct.

The electronic structure of a fluorine atom is $\ce{1s^2 2s^2 2p^5}$.  There is a strong driving force for atoms to attain an octet (achieve an inert gas configuration) due to the extra stability associated with a filled shell of electrons.  In order to complete its octet and achieve the neon inert gas configuration ($\ce{1s^2 2s^2 2p^6}$), fluorine must gain 1 electron and become the fluoride anion ($\ce{F^{-}}$).  If it gains two electrons and becomes $\ce{F^{-2}}$, or loses 1 electron and becomes $\ce{F^{+}}$, it won't have an octet in its outer shell - it won't be anywhere near as stable as if it just gains one electron.  The same reasoning explains why sodium, for example, prefers to lose only one electron. Transition elements have larger electronic configurations and may need to gain or lose 3 or 4 electrons to achieve an inert gas configuration.  It requires a lot of energy to add or remove 3 or 4 electrons, so adding or removing 1 or 2 electrons, while not achieving an inert gas configuration, becomes an acceptable alternative because of the lower energy required to remove fewer electrons.  In these cases, oxidation states of (plus or minus) 1, 2, 3 or 4 become feasible.

Some main group elements do form multiple ions, however, analogues to the transition elements are less common.
Examples:

Hydrogen - Hydrogen forms both the $\ce{H+}$ ion (the proton) and the $\ce{H-}$ ion (the hydride)
Tin forms the $\ce{Sn^{2+}}$ and $\ce{Sn^{4+}}$ cations

The fluoride anion can have a negative two charge. It's not impossible to add two electrons to the fluoride anion. But at that point it would be rather unstable as the fluoride anion with a negative 1 formal charge is isoelectronic with a noble gas - neon.
In addition, consider the high charge density of the fluoride anion. Tacking on an extra electron to fluorine is OK, but not as favorable relative to the less charge dense chlorine atom. Fluorine actually has a lower electron affinity than chlorine despite fluorine's higher electronegativity. Tacking on two extra electrons would be unfavorable.

Here's an easy argument to complement the others, which doesn't directly involve octets and electron shells. Compare the isoelectronic species  $\ce{Na^0}$, $\ce{Ne^{-}}$ and $\ce{F^{2-}}$, which all have the same $1s^22s^22p^63s^1$ fundamental configuration. Notice that going from left to right in the list, you are gradually removing a proton from the nucleus, making the valence electrons more poorly bound to the atom (less positive charge in the nucleus to pull the electrons inwards). Elemental metallic sodium is already very reactive because it has a comparatively loose valence electron (a low ionization energy). Removing two protons from the nucleus to generate $\ce{F^{2-}}$ would create a species much more reactive than even metallic sodium(!), and it would be expected to immediately lose at least one electron in any conditions.
In truth, $\ce{F^{2-}}$ is too unstable to be formed in the first place. Again, comparing the isoelectronic species in the list above, we look to $\ce{Ne^{-}}$. Neon actually already has an electron affinity that is practically zero or even endergonic. In other words, the reaction $\ce{Ne^0_{(g)} + e^{-} -> Ne^{-}_{(g)}}$ already has very little drive to the right, if not actually being driven to the left. This suggests the reaction $\ce{F^{-}_{(g)} + e^{-} -> F^{2-}_{(g)}}$ (fluorine's second electron affinity) is strongly endergonic, and will not ever happen to a significant extent.


text:
Neutralization reaction between ammonia and nitric acid

Complete and balance the following molecular equation (in aqueous solution); include phase labels. Then, write the net ionic equation.
$~\ce{NH_3 + HNO_3 -> ?}$

I thought that the acid $\ce{HNO3}$ would just give its hydrogen to $\ce{NH3}$ and make the resulting reaction:
$$\ce{NH_3 + HNO_3 -> HNH_3 + NO_3}$$
However the correct answer is $\ce{NH_3 + HNO_3 -> NH_4NO_3}$. Why is this?

Your answer is very close to the answer given, except for the following two tidbits (the first being more significant).

An acid-base reaction is not the exchange of a hydrogen atom $\ce{H}$. It is the exchange of a hydrogen ion (or proton) $\ce{H+}$. Thus your answer should be: $$\ce{NH3(aq) +HNO3(aq) -> NH4+(aq) + NO3-(aq)}$$
The given answer combines the two ions produced into a single compound. $$\ce{NH4+(aq) + NO3-(aq) ->NH4NO3(aq)}$$

The result of #2 would be reasonable if you had not been told that the reaction was occurring in aqueous solution. In aqueous solution, the products should have been those of #1, since ammonium nitrate is freely soluble in water.

Molecular equation:
$$\ce{NH3(aq) + HNO3(aq) -> NH4NO3(aq)}$$
Net ionic equation:
$$\ce{NH3(aq) + H^+(aq)-> NH4^+(aq)}$$
Note: $\ce{NH4NO3}$ is not solid in aqueous solution. It is quite soluble.

All this is already in previous answer.

@Mithoron:The only previous answer that I see has NH4NO3 as a solid, and does not show the net ionic equation with NH3(aq) + H^+(aq), so that is why I submitted the current answer. If it is redundant, can it be removed?


text:
Molecular connectivity index
I would like to know what is valence molecular connectivity index of 3rd order cluster. Please provide an example?
Thanks for your help.

It sounds like you're needing help with chemical graph theory.
I'm no expert in this topic, but it is one that has interested me in the past. Hence I can recommend 'Topological Methods in Chemistry' by R. E. Merrifield and H. E. Simmons—a book which explicitly deals with chemical graph theory in chapters 3 and 4; various topological indices feature throughout.
As a complement to this book, I would recommend the book on Topology by Munkres. The above book isn't rigorous enough to approach this topic perhaps, and so a mathematical treatise is also needed. I'm currently working through these topics myself for my own research.
It doesn't quite answer your question, but I hope it helps in your lofty quest!


text:
Safety when handling cyclohexylamine in non-industrial amounts
Today I was pipetting some cyclohexylamine into smaller bottles so that I don't have to use the liter flask every time I need a little bit of catalyst or plasticizer. I pipetted several 10-20 mL portions in the open air on a regular table.
Because I lack experience I'm extra precautious with what I do. From the MSDS's of cyclohexylamine I found that 10 mL will kill an adult male, which if you look at is isn't very much, less than a swig of water.
Of course I'm wearing a lab coat, safety goggles and chemical resistant gloves, but a few things concerned me (or at least made me wonder if it was the right/safe way to handle). According to the MSDS of ScienceLab the odor threshold is 26 ppm, while according to CAMEO, the PAC-3 for cyclohexylamine is 30 ppm, which Wikipedia defines as:

PAC-3 : Life-threatening health effects.

When I opened the bottle the amine fish odor was very strongly noticeable, I however found no irritating effect on my eyes or lungs, although according to the documents provided above I was inhaling a lethal concentration. I'm quite confused here. More easily put as a question: as long as I don't feel irritation of my lungs or eyes, am I safe? And if I do, is there enough time to clean up (i.e. the most essential things, like shutting off electricity and closing open jars of chemicals) and leave?
Finally I accidentally touched a drop of cyclohexylamine (from the pipette) when I was cleaning up, with no gloves on. Although cyclohexylamine is listed as highly corrosive, I did not feel any irritation at all. Then I rinsed my pipette 3x with distilled water and put the waste in a jerrycan, but I still haven't figured out if I can pour the diluted mixture of water and cyclohexylamine down the drain, without poisoning myself or the neighborhood. The MSDS's keep quiet about that.
Can someone clear some confusions up for me. I will be greatly appreciated!

It is possible to use nitrite in acidic solution to get rid of alkyl-amines. Idea is, that they form diasonium salt, that is quickly destroyed into alcohol and nitrogen.

Yes, but I need the alkylamines for the reaction. Or are you saying this as a way of disposal? In that case, your comment is appreciated and I'll look into it.

I once did some work with some small chain alkyl amines and they were bad news.  They would actually burn you if some got on your skin.  I always used a respirator and worked in a hood when using them.

as long as I don't feel irritation of my lungs or eyes, am I safe?

If this is a one time occurrence, given the small quantities, I'd say the odds are way in your favor, but they call it "odds" for a reason.  My advice would be not to do it again without a respirator and hood where the exhaust is treated.

And if I do, is there enough time to clean up (i.e. the most essential
  things, like shutting off electricity and closing open jars of
  chemicals) and leave?

If you ask an industrial hygiene professional, I'm sure the answer would be "no".  If it were me and I just began to notice some very mild sensation, I think I'd hold my breath and try to shut things down.  I know it doesn't make sense, but that's what I'd do.  If I noticed anything beyond the onset of very minor discomfort, I'd clear out fast and call the fire department.

Finally I accidentally touched a drop of cyclohexylamine (from the
  pipette) when I was cleaning up, with no gloves on.

I'd keep washing it every so often during the day and I'd put some burn cream on it in between the washings.  I'd also take a good shower.  No harm in overreacting.

still haven't figured out if I can pour the diluted mixture of water
  and cyclohexylamine down the drain

I'm almost positive that the answer is no.  Besides you and the neighborhood, it might react with the pipes or what's stuck in the pipes and create longer term toxicity problems when you're in that room at some later date.   Where I live, we have hazardous chemical drop-off centers (mainly for paint, oil and household chemicals, but used for everything chemical).  I'd pay the relatively small amount they charge and drop it off there.
One final note, even when using the right protective gear, some people become "sensitized" to amines.  This means that future exposure at even lower levels to a wide array of amines can be more dangerous.  Also, I'd try to open up the exposed room to as much outside air as possible, use a room fan too if you have one.

Wow, and I haven't even started heating it up for refluxing. I don't have a professional lab, I did install a fumehood, but it remains amateur work. Notice I'm talking about liter quantity (one liter to be precise and 20 mL in situ), not industrial quantities. When reading reports about reactions involving alkylamines I never say people explicitly warning about them. I always thought that methylamine and n-buthylamine for example are widely used chemicals, that require safety measures, but not something like a respirator  and professional filtering fumehoods

How are people using these compounds at home?! :) Pretty sure no one I ever saw mentioning these compounds has such safety measures installed.

n-butylamine (nBa) is what I worked with, really bad stuff.  They are widely used in industry, but so is cyanide.  I talked with guys who worked with drums of nBa in industrial manufacturing applications, that's where I heard the sensitization stories.  1 liter sounds like a lot to me.

Not in situ, 10-20 mL in situ. 1 liter in storage. Or do you mean that? I couldn't even get less, else I would just have ordered 100 mL or so. The chemical supplier I got it from seemed it not dangerous enough to not supply it to individuals. I mean, I knew it was bad stuff, but always considered it something similarly dangerous to sulfuric acid or concentrated lye (NaOH), which can be relatively safely handled by amateurs.

Yep, I understood you, I just wouldn't want a liter around for someone to knock off the shelf or whatever.  The industrial guys took precautions with the small alkyl amines that they didn't take with acids and bases - like respirators and only using it in hoods marked for amine use that led to rooftop scrubbing before release to the atmosphere.

Hmm okay. So let me get this clear: amateurs shouldn't be using alkylamines (cyclohexylamine in this case)? I don't have that kind of equipment installed as you can might imagine, nor do you at home I guess :P I might be setting a bounty for this, because it's very important for me to know the hazardous of the compounds I'm working with, and as always like multiple opinions. No offense of course.

No offense taken. Multiple opinions is the best way to go. Maybe the industrial folks who prepared me to work with nba were being overly cautious.

Good Ron. You know I respect you a lot :) Do you a lot for the Chemistry community.

There are some general points that need to be made about this question, and some will not directly answer your question. They may, hopefully be useful for future generations of home-chemists seeking answers on this site.
Firstly, all of these very practical questions should have been asked well before you carried out this process. Hindsight is a wonderful gift. Prior to any chemical handling process, all industrial and research chemists will carry out an extensive Hazard Identification and Risk Assessment. Prior to taking possession of this chemical you should be aware of how to store, handle and dispose of your chemicals. Disposal management includes spill control and safe disposal of unwanted waste materials.
Secondly, spilling some material on yourself during cleanup indicates that you were not sufficiently protected. Gloves, glasses, labcoats and other protective equipment need to be used during cleanup as well. Accidents can happen at any stage, and you need to be prepared for this.
Thirdly, if you are using a hazardous chemical in an environment where you can so readily smell it suggests to me that you have inadequate ventilation in your work area. You should address this immediately.
Finally, the first Risk Control measure the home chemist (any chemist!) needs to consider is subsitution. Is this really the only suitable reagent for you to use, or can you substitute it for another, less harmful, chemical?
Cyclohexylamine is a nasty chemical. It has a GHS signal word DANGER, which should immediately make any handler seek further information prior to using it. It is classed as

Category 3 flammable liquid,
Category 2 Oral Acute Toxicity,
Cat 3 Dermal Acute Toxicity,
Cat 1 Skin corrosion,
Cat 1 Serious Eye Damage and
Cat 2 Reproductive Toxicity.

Yes, Cyclohexylamine is a nasty chemical. Now, on to your questions.......

"as long as I don't feel irritation of my lungs or eyes, am I safe?"

No, no, no, good god, no. You will be subject to the dangerous effects of any chemical as long as you are exposed to them. Many effects will not be instantaneous; many can take hours or days (or even weeks) to manifest symptoms. I recall a case in the mid-90's, as an example, where a postgrad student died after cleaning up a spill of acrylonitrile in a lab. Symptoms of distress only became apparent several hours after the incident, and the person was hospitalized and died within 24 hours. Acrylonitrile essentially dissolves the tissues of the lungs. Not a nice way to die. Your lungs and eyes are the most sensitive areas of your body, and are likely to show irritation first. Consider, however, that your skin (especially your face) may have a much greater area of exposure, and this may be a pathway for contamination.
The PAC guidelines you ask about relate to extended periods of exposure. It does not mean that if you inhale 30ppm of a particular chemical you will experience life-threatening health effects immediately. Perhaps a more relevant and practical exposure guide is the Acute Exposure Guideline Levels (listed in your first reference), which shows that extended exposure at 30ppm for an hour could lead to life-threatening health effects or death. Or 10 minutes at 38ppm. The take home message from this is that if you can smell it, you are working in a dangerous environment.

"I accidentally touched a drop of cyclohexylamine (from the pipette) when I was cleaning up, with no gloves on".

Clearly, your previous statement "Of course I'm wearing a lab coat, safety goggles and chemical resistant gloves" was not quite correct. Personal safety equipment should be the first thing to go on, and the last thing to come off. I hope you treated the exposed area according to the MSDS first aid advice. Like other adverse chemical affects, corrosive properties do not need to be instantaneous. They can take a long time to manifest symptoms; HF, for instance, does not produce inflammation or burns until many hours after exposure.

"I still haven't figured out if I can pour the diluted mixture of water and cyclohexylamine down the drain, without poisoning myself or the neighborhood."

Many SDS do not explicitly describe safe disposal considerations, and you should probably seek a more comprehensive source. Chemical suppliers should now be moving toward the GHS system for chemical management, and the SDS from these companies should include (albeit brief and generic) disposal considerations. However, this is a chemical which is toxic to the environment and should be disposed of correctly. Most local Council or Municipal Waste Management Facilities will have chemical collection days, or specific chemical drop-off centres to receive these types of items. Speak to your local government authority, and seek advice from them. Occasionally, for some chemicals, you can return any unused portions to the place of sale. Normally, these types of chemicals are burnt in a chemical incinerator with afterburner and scrubber. You should collect waste mixtures in an appropriate container and store safely until ready for disposal. For small volumes of waste, you may consider using the spill control guidelines, which may offer a more practical approach to providing a safer chemical to store and dispose of.

Now I must note that I only smelled the fish odor after opening the flask. It went away after that. The pipetting was done outside. Yes while cleaning up I should still wear everything, stupid mistake, but I still don't have any symptoms. Cleaning up is in issue too indeed, but the MSDS's mostly are quiet about that one too for non-industrial spill. Thanks for your answer.


text:
Glass with a low melting point?
What kind of glass can I purchase that will melt in a Bunsen burner flame?
Is there any way to treat glass such that the melting point is lowered? I tried adding some sodium hydroxide but the glass shattered when heated.

I know that soda lime glass has a low melting point, but I don't know what that is. Glass shatters when the stress is great enough, like when there is a large temperature difference in the glass. You could continue to heat what you have until it melts. If it hasn't melted within 15 minutes, I would think your temperature is not high enough.

I've tried heating a soda lime glass lab slide in a flame before--it just throws off slivers of glass and eventually shatters. Ironically, the glasses I have seen successfully heated and melted are borosilicates, which have a much higher melting point, but a lower thermal CoE, so they can more easily be heated in a flame. TL;DR low melting point alone might not be a good candidate for melting in a bunsen burner; you also need it to be tough enough to withstand the uneven heating of a flame.

What is your end goal here? Some crafts stores sell frit glass which melts below most flame temperatures, but it's sold in a powdered form

I want to make Prince Rupert's drops.

I found this youtube video that shows how to make the drops. Natural gas burners (Bunsen burner) don't have enough heat to melt the glass; you need to use a propane torch. https://www.youtube.com/watch?v=5zxZkK2aJig

There are a few options for low-melting glasses that will easily melt at the temperatures of an air-gas torch (i.e. a Bunsen burner).  I suspect that either soda-lime or lead glasses would be the soft glasses of choice for your application (making Prince Rupert's drops).
Soda lime glass should be an adequate choice.  For some applications, it can be prone to cracking if not properly annealed, however.
Lead-based glasses are among those with the lowest melting points of any glass, and probably the lowest applicable to your application.  Annealing is simple and can just be done with the torch itself.  Lead glasses also have a large "soft" temperature range (the range of temperatures below the melting point at which they are still somewhat pliable).  This is seen as a disadvantage in many use-cases, where sagging prior to sufficient cooling/hardening is a problem.  For your work, I would think this would either be a non-issue or possibly could even expand the possibilities for making cool looking drops ;)
Given the availability and affordability of these solutions, further treatment to reduce the melting point just seems like a waste of good glass-working time.

I guess, the types of glass usually prone to cracking are the ones suitable for Prince Rupert's drops. Thermal expansion is responsible for both.


text:
How to choose the level of theory for modelling reactions of polymers?
My Main purpose is using quantum chemistry for simulation of some reactions. Actually we are working on making some new polymers and I need to simulate some molecular properties of these polymers after adding some molecules to them. These include bond breaking and bond making as well as evaluation of optical properties.
That's why I started studying quantum chemistry models. I started with the Schödinger Mechanical Model and then Hartree Fock Theory, but as everyone knows these models are not that efficient for molecules larger than lithium. Then I started learning DFT and got interested in semi-empirical models, but could not learn them yet. Finally I am absolutely confused about which method should I select.
This is my question: As I have not enough time to learn all those quantum chemistry models/techniques, which method is currently best option if I just want to learn ONE method. (I have access to some super-computers in our university.)

To start a meaningful discussion you need to provide at least the following information. 1) How big your polymers are, i.e. how many atoms do they consist of? 2) What kind of properties do you want to predict? 3) How much computational resources do you have (how many CPUs, how much memory, and how much time)?

Roughly speaking (i.e. without any information provided by the request above), you could go with DFT if the polymers consist of hundreds of atoms. For larger molecules (i.e. thousands of atoms) you are limited to semi-empirical and QM/MM approaches.

@Wildcat My polymer is a ring , consisting of around 65 atoms. Mostly we are interested in ring opening reaction feasibility and optical properties of the polymer before and after that ring opening. We have access to super-computers of University of Colorado (USA) but I don't know how powerful they are. I just know they are some of the best available.

The question is by far not an elementary one. Short answer is: unfortunately, there is no silver bullet.
For a molecule of around 65 atoms I would go with DFT as the methodology which provides probably the best quality/cost ratio for molecules of this size.
Now, you need to choose functional and basis set, and this is a tricky business. But to start from:

Do not blindly use B3LYP.
Try to search literature for the same types of calculations (energetics of ring opening reactions, optical properties) for similar compounds. Maybe someone already did the so-called benchmark calculations, i.e. compared on how good different functionals in combination with different basis sets are on reproducing the properties (say, specific rotation) either measured experimentally or calculated at some high level of theory (say, CCSD(T) with reasonably sized basis set).
Do your own benchmark calculations with different functionals and basis sets. First, benchmark different functionals with the same basis set (say, Dunning correlation-consistent basis sets, cc-pVTZ or better cc-pVQZ), vary the functional, compare results with experimental values, choose the best one. Once method is chosen, try to decrease and also increase (if possible) the basis set to see the effect of basis set size (with the above mentioned bases it is quite simple: if cc-pVTZ was used to benchmark different functionals, try the best functional in combination with cc-pVDZ and cc-pVQZ). Choose the smallest basis set that reproduces experimental results with desired accuracy.

P.S. You could also ask Google, there are some resources of information here and there on how to choose method and basis set, though, as I said, this question is not elementary, so only general advice can be given.

Thank you !   I am going to dig into that and will write about here.  Thank you again !

Just as another question: How do you prefer to work with DFT ? do you write your own code or use available softwares?

@Aug writing your own DFT code would require enormous amounts of time and intellectual efforts, which, taking into account the existence of many extensively debugged effectively parallel DFT codes, is pointless. Just use available software packages.

The choice of software package is another nonelementary question. Basically, people use one which they are mostly familiar with and which is, of course, available on a supercomputer. But since you just start and since (according to your questions) your background in quantum chemistry is rather limited, you would better use a program which someone can help you to use properly and effectively. You need an assistance, at lest at the very beginning.

Strike blindly from number one. There are much more sophisticated and efficient hybrids nowadays - the wonder punsch was a nice addition in the early years...


text:
Shape Of Cetyl Alcohol
Why is cetyl alcohol bent at the 9th position?

It seems like a good space saving conformation, perfect for an image. Notice that the space filling model is drawn 'straight'.

This is simply how the chemist chose to draw the molecule in that particular instance, and cetyl alcohol is formally regarded as a straight-chain molecule. Cetyl alcohol has all single bonds and no rings, so it is not structurally locked into any particular conformation or geometry, and every bond is free to rotate.
For very simple saturated molecules like this, where the conformation is not in any way fixed (under typical conditions), the chemist is free to represent the molecule as one of any of its possible conformers.
In reality, the most stable (i.e., lowest energy) conformer should statistically predominate, and this is the one with all groups staggered and the largest groups antiperiplanar (q.v., the IUPAC Gold Book entry on torsion angles for details), the conventional explanation being that this minimizes steric hindrance/strain. The skeletal formula corresponding to this minimum-energy conformer is the simple straight-chain representation (seen here, for example).

It is not.  This flat representation is schematic.  In fact the molecule can wobble a lot, and no C-C bond is more likely to be the site of a sharp bend than any other.


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Why do different alcohols have different effects on the brain?
I've been trying to understand why different alcohols have different effects on the brain.
Methanol is obviously poisonous, apparently because it's converted to formic acid. Because the body seems to cut organic substances in two's, any odd-numbered alcohol would seem to yield methanol, and therefore be poisonous. But why is 1-butanol (according to certain sites) more toxic than ethanol? Does it have a different effect on the receptors in the brain? And if so, why? Same with other simple even-numbered alcohols: why are their effects different from ethanol?

Different molecules have different properties and metabolic pathways, whether the function group is the same or not.

As the chain length of the alcohol increases, it becomes more hydrophobic. This would make it easier for them to insert into membranes and hydrophobic pockets of proteins. This could interfere with the protein function. Hydrophobic molecules also tend to be harder to eliminate than shorter molecules and can accumulate in fatty tissues, including the brain. Longer chains can also be metabolized into a greater number of products simply because there is more to metabolize. And sometimes seemingly innocuous compounds can be metabolized into very toxic substances, such as n-Hexane, which can be converted to a molecule with an epoxide on each end. The epoxides can cross-link proteins, particularly in neuron axons.


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High/low density water solution values
I would like to know what min and max values of densities are achievable with water solutions.
There are some conditions:
1. Normal or close to normal temperature;
2. Non-radioactive solutions/components;
3. Acids are good as long as they can be safely contained.
Thanks for your help!

Highest known is about 4.25 (all
densities in g/cm${}^3$) at room temperature:
http://en.wikipedia.org/wiki/Clerici_solution
If you're willing to allow heating it comes close to 5 because the solubility of the salts increases with temperature:
http://books.google.com/books?id=kaa2qeFRXmUC&pg=PA99
My favorite runner-up is the Thoulet solution mentioned at
http://en.wikipedia.org/wiki/Potassium_tetraiodomercurate(II) but it
only reaches around 3.20.
Lowest would depend on what you mean by "water solution". If you mix a
drop of water with a gallon of a very light liquid such as diethyl
ether (which is not very miscible with water but does dissolve some
-- hence the need to dry it in the lab over sodium or molecular
sieves), then the answer is the density of the liquid (ether = 0.73).
If water has to be the majority component I'd expect the
minimum to be very close to the density of water, since mixtures tend to decrease a bit in volume. (The density of water is 1.0 at 4${}^\circ$, close to 1.0 at room temperature, 0.94 near the boiling point).


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What happens when we heat an atom?
Question is simple.. If we take an atom of any element and then supply heat energy to it then what will happen?
What I thought is that in the beginning, energy (quanta; due to excitement of electron and then moving back to ground state) should be  evolved and then electron should be emitted... Am I right?

If we take an atom of any element and then supply heat energy to it
  then what will happen?

A single atom can move in three dimensions (x, y, z) through space, it cannot rotate about an external axis like a diatomic molecule can, nor can it vibrate.  As you supply heat (or any form of energy) to the system containing the atom, it's kinetic energy will increase and it will move faster through space.  If you supply enough energy, then the electrons surrounding the atom will absorb that energy and undergo electronic transitions from the ground state to various excited states.  This process can be diagrammed using a Morse potential diagram , I've drawn a simple one below.  The diagram is meant to show promotion of an electron from the ground electronic state to higher and higher electronic levels as more energy is applied to the system.  Note that as the electron is promoted to higher energy levels, the average separation distance between it and the nucleus increases.  If enough energy is supplied, an electron will dissociate from the atom and we will be left with an ion.  Continue to supply even more energy and more electrons will be removed from the atom. Pumping even more energy into the system will ultimately lead to transformations in the nucleus of the atom.

Are you sure that the use of a Morse potential diagram is permissible for the description of ionization?

What is your concern, the shape of the left wall?  I was trying to use the x-axis to represent some time of 10-90 percentile of the electron's radial distribution function.  I thought if I made the left wall vertical or tilted to the right it would raise questions.  But to answer your question, I think that the Morse concept can be applied to electron dissociation just like bond dissociation.

I've just never seen it being used in this context. I thought it would be only valid for the vibrations of diatomic molecules.

I've seen it used in spectroscopy books and courses to diagram electronic transitions in a molecule.

@Philipp I agree, a Morse potential is weird for ionization. The Morse potential picture only comes if you have well defined position for both the nucleus and the electron. It's fine for diatomics in the Born-Oppenheimer picture (classical nuclei), but if you have quantum nuclei or quantum electrons... well, their position is not well defined! Energy diagrams for a particle in a Coulomb potential (with a continuum limit being ionization) would be more fitting.

Well, strictly speaking, you can not heat an individual atom, because heat is a macroscopic phenomenon: it is a mechanism of energy transfer from one macroscopic body to another one. And the same is true for the related notion of temperature.
But, both temperature and heat can be explained at microscopic level as follows.
Particles that constitute bulk matter are in constant random motion and temperature is nothing but a measure of the mean kinetic energy of particles random motion. Heat then can be understood as macroscopic manifestation of transfers of this kinetic energy of random motion from a statistically significant number of particles of one body to particles of another body.
Thus, you can heat a body (not an individual atom), i.e. you can transfer energy to it by means of heat, and while the body is heated, the mean kinetic energy of all particles (atoms, nuclei, electrons) will rise, i.e. it temperature will rise.
The kinetic energy of an electron acquired then can be partly redistributed with its potential energy when electron ``jumps'' into higher orbital. After excitation the atom may return to the ground state or a lower excited state, by emitting a photon with a characteristic energy.
And, of course, in principle, we could add so much energy to a body be heating, that electrons will indeed leave atoms.


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Height of potential barrier
Could anyone please help to find the height of the potential barrier between silver and copper or $\ce{Ag-Cu}$. I am working on a piezo-resistive sensor based on copper fabric and silver electrodes. I was wondering if I can validate my experimental results with the model given by Zhang, X.-W., et al. [1]
I am not a chemistry expert and have a little knowledge on the height of the potential barrier. It would be highly appreciated if anyone could help me with providing the value for the height of the requested potential barrier.
[1] Zhang, X.-W., et al., Time dependence of piezoresistance for the conductor-filled polymer composites. Journal of Polymer Science Part B: Polymer Physics, 2000. 38(21): p. 2739-2749.


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How to test for presence of Tobacco/nicotine in a product?
I found this paper that says nicotine was found in major toothpaste brands in India. Rather than playing blame game, wouldn't it be easier to test ourselves ?
How to test for presence of Tobacco/nicotine in a product ?

The paper states how they did it (extraction followed by GC/MS), so that method should work.

I was hoping for something that does not required a GC/MS..but if there was a cheaper/easier way, they wouldn't have gone for a costlier method .. hmmm :(

The problem is that you need to isolate the nicotine from the mess.  GC does this automagically, and when piped into the MS, you can know the identity of the product because the 70eV ei fingerprint for nicotine is characteristic of that compound (well any compound).  Other tests would be ether be less reliable or much more laborious.


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Electrochemical cell polarity and electrodes
In an electrochemical cell, the anode undergoes oxidation and the cathode undergoes reduction. This means that the anode loses electrons while the cathode gains electrons.
However, the anode is negatively-charged while the cathode is positively-charged. Why is this so?

The electrons which are "given up" by the oxidized species at the anode flow from the anode through the electric circuit to the cathode, where they are "needed" for the reduction reaction. The oxidation creates an excess of electrons (negative charge) at the anode, which is compensated by the consumption of electrons at the electron-deficient cathode (positive charge) for the reduction reaction. In summary, the flow of electrons in the electrochemical cell is from anode (electron source) to cathode (electron drain).

So when the anode is said to be 'negatively-charged'  it implies that the anode has an excess of electrons which are available for donation to the cathode?

Yes, the excess electrons from the oxidation reaction are available for the reduction reaction at the cathode.

Oxidized species give up electrons to an electron pool which prior to their flow to the other half cell create a large negative charge on the electrode where oxidation takes place.

Welcome to Chemistry.SE. If you have questions about how our site works, please consider taking the [tour] and visiting our [help]. This answer does not provide any additional information over the previous answer. Do you have more that you could add to make your answer distinct?


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Kinetic and Thermodynamic Stability
My prof noted that if the two components of any ionic compound undergo a large extent reaction with each other, then the substance CANNOT exist as a pure substance.
For example, consider ammonium hydroxide. This is simply ammonia (at least in the solution phase). Can we have a solid, crystalline ammonium hydroxide lattice?
I think this is thermodynamically valid. What about kinetically? Isn't it theoretically possible to have two substances that would otherwise undergo a large extent reaction with each other but don't due to an activation energy barrier?
Are there any such examples?
Also what do you think of the below question?

Chemistry is a field with far too many exceptions and edge cases to be making sweeping categorical statements. There are numerous examples of ionic compounds that are isolable and stable, the components of which will spontaneously decompose under certain conditions (this is where reaction kinetics is relevant). Two that readily come to mind are ammonium nitrate ($\ce{NH4NO3}$) and ammonium bicarbonate ($\ce{NH4HCO3}$). Both of those can be isolated as pure salts, and both also can and will chemically decompose when circumstances are favorable (the former by multiple possible pathways depending on conditions).

Isn't it theoretically possible to have two substances that would otherwise undergo a large extent reaction with each other but don't due to an activation energy barrier?

This isn't just theoretically possible, it's ubiquitous, it's the norm. If it weren't, human life couldn't exist. Combustion reactions are probably the most obvious and mundane example; just about every organic compound imaginable is usually more thermodynamically stable as a mixture of $\ce{CO2}$, $\ce{H2O}$, $\ce{N2}$, etc., but the kinetic barrier to the reaction (along with physical hindrances and the availability of an oxidizer, e.g., $\ce{O2}$) prevents this from occurring spontaneously without some trigger and the appropriate environmental conditions.

But ammonium is a weak acid and nitrate anion a base weaker than water. The two don't undergo a "large extent" acid base reaction. So my prof would agree with you that yes, that salt can exist.

@Dissenter, if he was talking about this strictly in the context of acid-base reactions, then yes, his point probably has broad validity.

@Dissenter, having said that, I should also mention that the situation becomes more complex if we introduce ionic compounds with mixed cations. Ammonium phosphate, for example, is unstable, but ammonium magnesium phosphate forms an extremely stable solid hexahydrate (mineral name struvite).

Ah. I'll be sure to hit him over the head with that example. But even in the context of acid-base reactions, can you think of any exceptions? Preferably ones without mixed cations? The ammonium magnesium phosphate one is one for sure. Do you know why exactly that mineral is stable? Does the magnesium cation have anything to do with it?

@Dissenter, well, the decomposition of ammonium phosphate is via an acid-base reaction liberating ammonia. In the case of struvite, it forms a hexahydrate, and if I had to hazard a guess, speculatively I'd imagine that ammonium as a hydrogen-bond donor interacts strongly with the oxygen lone pairs. That may be one factor, but I don't know with any certainty.


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Counting Nodal Planes in cyclopropane
The energy of molecule orbitals increases with more nodal planes. W1 (in the attached picture) has no nodal plane. I'd like to know how to draw the nodal planes in cyclopropane molecule orbitals but do not know how. I found no picture in the web with the nodal planes of cyclopropane.
How many nodal planes have W3, W4 and W5?

Not all nodes are planar. Some may be points, radial, angular, etc.

Could you please provide the source of this picture? I would like to see the context, as I think they assigned W4 as the LUMO, which is not true at all.

@Martin the source appears to be Rauk's book: http://books.google.com/books?id=Kn1YXlMlbPMC&pg=PA85&lpg=PA85

@SilvioLevy Thank you. I was not mistaken, he assigned W4 as the LUMO.

It seems he's not the only one to think so.  See for example page 2 of  http://www.arkat-usa.org/get-file/19033/ (and if true, this just goes to show what I wrote in my answer: the number of nodal surfaces is not an absolute predictor of the energetic order).

@SilvioLevy In a very general sense, this is true. But in the long run this assessment is wrong. I am currently preparing an answer, where I will go in more detail.

The assessment of nodal planes is not as trivial as it might seem at first, because they are usually not planes at all.
Orbitals can be explained in terms of symmetry operations. There are three different operations and each of them has a unique element (in cartesian coordinates):

Rotation: Axis
Mirror: Plane
Inversion: Point

When we talk about hydrogen type atomic orbitals, we usually refer to their cartesian form. It is already fairly obvious for the $\ce{2s}$, that there is no nodal plane, while there are certainly nodes.
The representations of $\ce{s}$ orbitals are always trivial with respect to the total symmetry of the external field (= Point group). That means any symmetry operation, which is given through the external field will be matched by the $\ce{s}$ orbital. Another term for this is totally symmetric.
The representation of $\ce{p}$ orbitals are always antisymmetric with respect to inversion in any given external field.
The representation of $\ce{d}$ orbitals is always antisymmetric to two mirror planes. That excludes of course the $\ce{d_{z^2}}$ orbital, which is not a cartesian, but a spherical representation. (Note: In cartesian space there are 6 $\ce{d}$ orbitals. As quantum numbers only allow 5 spherical orbitals, they have to be transformed. See for example at chemissian.com or Schlegel and Frisch.) Early basis sets for quantum chemical calculations used the cartesian representation, because it wa easier to code and calculate.

Due to the geometric arrangement of nuclei in a molecule, the external field is given through the point group of the molecule. Every molecular orbital can be expressed through symmetry operations of that point group.
Cyclopropane has $D_\mathrm{3h}$ symmetry. And every molecular orbital has to respect that. I have already written about orbitals these orbitals in Why does cyclopropane give bromine water test?
However, I would like to shed some more light on this, because I believe the picture you found is not quite correct.
The following picture shows the depicted orbitals from your source, as well as the LUMO.

Where W1 corresponds to HOMO-2, W2 and W3 correspond to both HOMO, W4 corresponds to LUMO+3, finally W5 and W6 correspond to LUMO+2.
The ordering of these orbitals is taken from a BP86/cc-pVTZ calculation of Gaussian09. A methodology that I believe gives fairly accurate results for molecular orbitals. (It is however noteworthy, that unoccupied orbitals are constructed from occupied orbitals. That means, that their physical meaning is limited.)
Let's have a closer look at the orbitals. The HOMO-2 is of $A_1'$ symmetry and therefore behaves like an $\ce{s}$ orbital. There is no antibonding nodal surface in this orbital. That means at any bonding axis of the ring structure the sign of the wavefunction does not change. There is however a nodal surface that does not affect bonding, as was already pointed out in some comments.
The HOMOs are of $E'$ symmetry and therefore behave like $\ce{p_x}$ and $\ce{p_y}$ orbitals. These orbitals have one antibonding nodal surface. In a result of symmetry restrictions these nodal surfaces are perpendicular to each other.
This is also true for the LUMO+2s. In addition to this there is another antibonding nodal surfaces with respect to the $\ce{C-H}$ bonds.
The LUMO again is of $A_1'$ symmetry and behaves like a $\ce{s}$ orbital. There is again no antibonding nodal surface with respect to the $\ce{C-C}$ bonds, but the main electron density would be outside of any bond and there is an antibonding nodal surfaces with respect to the $\ce{C-H}$ bonds.
The LUMO+3 is of $A_2'$ symmetry and does therefore not behave like any other atomic orbital. Here we find three antibonding nodal surfaces with respect to the $\ce{C-C}$ bonds.
The bonding situation in cyclopropane is very complicated and can be explained in many different ways. It is not as obvious, that the molecule is a $\sigma$ aromate, having in plane $\pi$ orbitals (HOMO). The explanation of its molecular orbitals were first composed in terms of a Walsh like behaviour on the basis of an extended Hückel calculation with a minimum basis.
When you disregard the $\ce{C-H}$ bonds and only allow three carbon $\ce{sp^2}$ and three carbon $\ce{p}$ orbitals, then you can only arrive at the boning picture that was depicted in your source (and many other publications), as you can only have six molecular orbitals formed from these. There are, however, many more atomic orbitals that mix into the final wavefunction. First of all it is necessary to state, that the used $\ce{sp^2}$ orbitals are not of the ideal $\frac13\ce{s}+\frac23\ce{p}$ composition (see more about that in the linked question and the answer of ron therein).

Now finally for the discussion about the correlation of nodal planes and increasing energy. It is a very common statement, that is missing it's most important limitation. It is only completely true for orbitals of the same symmetry. (This is also a necessary requirement.)
For example, the energy of the $\ce{s}$ orbital series increase with the main quantum number, as you add one more nodal surface per one increment in $n$: $\ce{1s->0; 2s->1; 3s->2;...}$. The same applies to the $\ce{p}$ orbital series: $\ce{2p->1; 3p->2; ...}$.
Now this statement became popular, when talking about aromatic systems. In the framework of Hückel's molecular orbital method, this must be true, since all regarded orbitals are of the same symmetry.
However, it is usually true, that an orbital with fewer nodal planes is more stable than another with more, but this is more a gut feeling, than actual, factual science.

+1 for an excellent answer overall, but especially for your point about the necessity of like symmetry when correlating nodes to energy. It's a subtle and important point that gets glossed over routinely in most qualitative MO texts (as I know to my own detriment).

Nice answer. There is nothing better for checking the validity of a qualitative MO scheme than a real calculation.

@Martin Thanks for taking the time to write a long answer and providing the results of your calculations. It's nice that the rule of thumb about energetic ordering is not being broken in this example. But I'm still uneasy about its status as a rigorous rule, even with the symmetry caveat. Take a molecule lacking all symmetry, which is the commonest case once you leave the realm of textbook examples. Here all molecular orbitals belong the the same (trivial) symmetry class. Does it follow that more nodal planes implies higher energy?

@SilvioLevy Yes it still is valid. Every nodal surface adds a sign change to the orbital wave function and therefore the energy of this orbital has to go up as it gets larger (less dense). For the asymmetric case it is not trivial to see the sign changes, as orbitals mix with a lot of functions and it is not easy see it as clearly. Sometimes then it helps to break it down to local symmetry and look there. However, there are more factors at play than only the sign change, overlap integral, etc. This is why a computation will always be superior to the rule of thumb.

Thanks. I agree that the sign changes. But why is it that "therefore the energy has to go up"? We're talking about solutions of the Schrodinger equation in 3D. In a complicated molecule, why can't some of the many solutions that have 47 sign changes interleave in the spectrum with some of the many solutions have 48? Perhaps we can move this to chat...

@SilvioLevy Join me here: http://chat.stackexchange.com/rooms/15903/counting-nodal-planes-in-cyclopropane

@Martin: Thanks for your very helpful explanation!

Instead of nodal planes think of nodal surfaces.  For W1 there is one nodal surface, a cylinder that looks like a circle seen from above (the view in the book).
Continuing with the hints.  All the combinations that come from orbitals that have radial symmetry (the ones where the bulges point toward the center and away from the center) have the same cylinder as a nodal surface.  If you draw the circle representing this cylinder seen from above, you'll basically have completed the nodal surface diagram for W1, W5, W6.
The use of nodal surfaces helps build intuition, but it only goes so far.  For example, it doesn't have to be true that more nodal surfaces equals higher energy.


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How to crystallize out reaction mixture from DMF?
Reaction was carried out in DMF, and TLC shows that the reaction has completed. But how to separate reaction mixture i.e. product from DMF?
Edit
Steglich Esterification was carried out and the product is soluble in DMF. –  user 24 (Jul 22 at 7:13)

More details would be helpful. In the standard approach, (assuming your product is something you can't precipitate or extract by acid/base workup) you'll need to select the right organic extraction solvent (less polar than DMF but sufficiently polar to dissolve your desired product, possibly $\ce{Et2O}$, $\ce{EtOAc}$, DCM, MTBE, etc.). Add it to your reaction mixture, then wash multiple times with copious water. Ideally, much of the DMF should end up in the aqueous phase. If your product is water-soluble, the situation is more difficult.


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Does vinegar neutralize the strongly basic compounds in cement?
I have heard that washing one's hands with vinegar is effective in neutralizing the strong bases in cement and is more effective than water.
I want to know how cement burns and what bases are involved in this reaction.
I have tried googling "cement burns" and found websites explaining the importance of washing one's hands with vinegar as a preventative measure. I have tried searching PubMed but I've found mostly case reports that do not address how chemical burns actually happen.
If anyone knows of some good papers or books that cover the topic of chemical burns in greater detail than what I've found; please let me know.

If you look for Portland Cement safety issues, it is stated that cement is highly alkaline so it can cause alkaline burn and presumably since vinegar is acetic acid, it neutralises the alkaline effects of cement as tested here and the burns you suffered are probably related to alkaline burn. You can look at this article section 6.5 for some details on alkaline burns, which was suggested by @Cornelius or perhaps the mentioned chapter in this book (Harchelroad FP, Rottinghaus DM. Chemical burns. In: Tintinalli JE, Kelen GD, Stapczynski JS, Ma OJ, Cline DM, eds. Emergency Medicine: A Comprehensive Study Guide. 6th ed. New York, NY: McGraw-Hill; 2004: chap 200.).

As a specialist in poison information (SPI), I would never recommend using anything other than water to decontaminate the skin or the eyes. An acid mixed with a strong alkali can cause an exothermic reaction (release of heat) which would then cause thermal burns. Neutralizing a corrosive substance (acid or alkali) that has been ingested is CONTRA-INDICATED under all circumstances, due to the potential for exothermic reactions and the release of gas. Although I could find no specific references for skin exposures, logic would indicate the same would be true in these cases as well.
Water is always the best, most easily available and cheapest means of decontamination. AND it will do no harm.
In case of exposure, always call your local Poison Control Centre for the most appropriate recommendations.

+1 The use of much water is exactly what is recommended in the OSHA Guide for Concrete Manufacturing, the First Aid Guide of the DGUV (Deutsche Gesetzliche Unfallversicherungen) and in standard books on experimental chemistry, such as Arbeitsmethoden in der Organischen Chemie. Don't waste time to search for the vinegar! Use water and use plenty of it!

Injury to your skin can occur while the skin is in contact with the cement--the more prolonged the contact, the greater the degree of skin damage. As soon as you wash off the cement (using either vinegar or water--they will be equally effective) you will stop additional damage, but the vinegar will not reverse the skin damage that has already occurred.

authors: יוחאי בן-זיכרי