text:
What does it mean when it is said that an sp³ orbital has 25% s character?
My textbook frequently mentions:

$\mathrm{sp^3}$ hybrid orbital has 25% $\mathrm{s}$-character and 75% $\mathrm{p}$-character.

What are these "characters"? And how do these characters influence the bond and the hybrid orbital?

Hybridisation is a purely mathematical concept, which makes it possible to explain experimentally found structures. The most prominent example for this is methane, where you can consider the central carbon atom to be $\mathrm{sp^3}$ hybridised. Formally, the $\mathrm{s}$ orbital and the three $\mathrm{p}$ orbitals can be linearly combined to form four equivalent orbitals. Hence $\mathrm{sp^3}$ is a contraction of $\mathrm{s^{\frac{1}{4}}p^{\frac{3}{4}}}$. Therefore you can see, that the orbital has $25\,\%$ $\mathrm{s}$ character and $75\,\%$ $\mathrm{p}$ character.
Generally, the concept of hybridisation has to be treated as a concept only, it is nothing which can be physically observed. It is a mathematical model to understand geometries better. Please also note, that hybridisation is a function of the geometry, not the other way around. Having said that, it is fairly obvious, that I am not a big fan of the concept. I would not go as far as Alexander Grushow* and preach to abandon the whole thing, but I advise caution in the use of this concept to not obtain wrong conclusions.

Despite significant experimental evidence and theoretical advances to indicate that hybrid atomic orbitals do not exist and do not appropriately describe molecular bonding, their description still permeates chemical education at many levels, and the model still finds its way into modern chemical literature.

* “Is It Time To Retire the Hybrid Atomic Orbital?” Alexander Grushow, J. Chem. Educ., 2011, 88 (7), 860–862.

Wouldn't it be better to say "hybridisation is a simplification of the actual physical orbitals" rather than saying it's not physically observable outright? The orbitals do behave as hybrid orbitals, after all - that's the whole point of why they were postulated in the first place. Unless you're just saying this to avoid confusion, of course - a lot of chemistry education takes the path of "we already know this is completely false, but it mostly works and it's a lot less work, so learn to use it" :D

@Luaan I am not sure I understand what you have written. Do you have any reference for the statement, that orbitals behave like hybrid orbitals? We might be able to look at energy differences of molecular orbitals and in that way we might be able to conclude that they are observable, but hybrid orbitals are no molecular orbitals, they are just linear combinations of atomic orbitals, which are one of the more fundamental approximations, we already include. I also always thought, that they were a by-product VB theory and the obsolete VSEPR model.

Well, I meant just the fact that the resulting orbitals are all equivallent (just with "different orientation"), not the "25%s + 75%p" part; explaining the molecular structure of methane and friends. But you're right that they don't really make any sense outside of molecules, and even there they're the result of trying to map the individual parts of the molecule rather than modelling the whole molecular orbitals. Perhaps you might want to add a paragraph on just the basics of MO? Although you're right that it wouldn't really help to answer the OP's question. My bad.

The hybridisation concept allows for thinking about individual atoms (and their orbitals) within a molecule. In a molecule, it is the molecular orbitals which are important, but within reasonable approximation, you can dissect them back to atomic orbitals, but they come out mixed (hybrydised) in a proper way.

@ssavec Please see my edit in response to your comment. I really believe that thinking in localised orbitals does not help understanding structure and reactivity of compounds. I also do not see why the hybridisation concept would be necessary to think about the individual atoms in a molecule. After all, the quantum theory of atoms in molecules does not require orbitals at all. And I do not understand, what you mean with 'mixed in a proper way'.

@Martin-マーチン: AIM theory operates only on density, that's fine. But it will never tell you "energy of the bond". On the other hand, Natural Bond Orbitals (NBO) analysis gives you exactly these hybridised atomic orbitals we are talking about. So if you want to think in bonds, it is this method which shows them to you. Both AIM and NBO are just interpretation tools for the (possibly many-electron) wavefunction. AIM is atom-centric, NBO is bond-centric. I agree that hybridisation is not an important pedagogic concept. But if you wish, you can rigorously obtain the hybridised AOs.

@ssavec NBO is a localisation technique that will also never tell you the energy of a bond. It is in itself just a mathematical transformation and you lose all of the information about the eigenvalues of the orbitals. The only theory which would make sense in that framework would be valence bond theory. AIM on the other hand can be performed on a physical observable, the electron density.

@Martin-マーチン I think the discussion got too technical for the OP. The analysis of wavefunction is obviously hot topic and better be cooled with beer.

I think a good way to understand the importance of hybridization is to start with an illustration.
You have two structures: $\ce{SnCl2}$ and $\ce{SnCl4}$, and you are asked to rationalize why $\ce{SnCl2}$ has a stronger bond than $\ce{SnCl4}$ (this information is given). How would you do that?
Now there are two ways to approach this:
(You can skip this if you are not interested in the argument, but it is fairly straightforward)

The oxidation state of Sn is +4 in $\ce{SnCl4}$, while it is +2 in $\ce{SnCl2}$. The higher the oxidation state of the atom, the valence orbitals will be more contracted and lower in energy as the effective nuclear charge increases without a compensating increase in electron-electron repulsion (no electrons are added obviously). Therefore, as the coordination number is higher for $\ce{SnCl4}$ and the orbitals are more compact, this will weaken the $\ce{Sn-Cl}$ bond in $\ce{SnCl4}$.
A simpler way though, would be to say that according to the $\ce{SnCl4}$ geometry, it has to be $sp^3$ hybridized ($\ce{Sn}$ is group IV like carbon), and in $\ce{SnCl2}$, it will be $sp^2$ hybridized (2 $sp^2$-$p$ $\sigma$ bonds, 1 lone pair of electrons and 1 empty orbital). As the $sp^3$ bond has 25% $s$ character but the $sp^2$ bond has 33% $s$ character, and as the $s$ orbital is lower in energy than the $p$ orbital due to penetration effects, this means that the $\ce{Sn-Cl}$ bond in $\ce{SnCl2}$ will be stronger than that of $\ce{SnCl4}$.

So that's one use of the hybridization argument. Hope this helps in some way.

$\ce{SnCl2}$ is not $\ce{sp^2}$ hybridized.  The $\ce{Cl-Sn-Cl}$ bond angle is close to 90° (reference) implying that the central tin atom is essentially unhybridized, it uses its p-orbitals to bond to chlorine. But your thinking is correct.  Just compare bond strengths or lengths in C(sp3)-H vs. C(sp2)-H vs C(sp)-H


text:
What happens if I make a hole in a container with superheated water?
Superheated water is achieved by heating water above 100 degree Celsius under high pressure.
Now, suppose the temperature at which the water is getting heated is around 250 Celsius and sufficient pressure is provided. If, I now make a small-hole in the container, the pressure will decrease obviously, but what will happen to the superheated water?
Will it come out in the form of vapor with lot of energy or some explosion might take place or just about something else?

Be very certain, that if you have superheated water with a lot of pressure and you make a hole in the container, this is very dangerous. Please be careful.

I know that, that's why I'm asking that what will exactly happen? Some websites (i guess it was wikipedia) say that superheated water is very stable and won't show any explosive sort of behaviors unless you shake it or add something like sugar.  But I doubt that previous statement is correct. Anyways, would be great if someone could tell me what exactly would happen. Thanks anyway. :-) Edit: Thanks for your concern and I can assure you that I won't do that unless I get some expert advice. :)

"Superheated" doesn't mean the liquid is under sufficient pressure to keep it from boiling.  It means that the liquid should be boiling according to the charts, but it is not.  It's a metastable state, that can occur if the container is very clean, and the liquid is free of any solid particles.  https://www.youtube.com/watch?v=SC_NtH8vWSc

If you have water at a temperature of $T=250\ \mathrm{^\circ C}$, you need a pressure of approximately $p=39.2\ \mathrm{atm}=4.0\ \mathrm{MPa}$ to keep it liquid. Liquid water at this temperature and pressure has a specific enthalpy of $h_0=1086\ \mathrm{kJ/kg}$.
If you open the container, the pressure drops to the ambient pressure of $p=1\ \mathrm{atm}=101\,325\ \mathrm{Pa}$. A part of the hot liquid water flashes to steam. The temperature drops to the boiling point ($T=99.974\ \mathrm{^\circ C}$) at the new pressure. A new equilibrium of liquid water and steam is established at the new temperature and pressure. The corresponding specific enthalpy of the steam is $h_\text{steam} = 2676\ \mathrm{kJ/kg}$; the specific enthalpy of the liquid water is $h_\text{liquid} = 419\ \mathrm{kJ/kg}$.
The flashing process is very quick, hence we may assume that there is no significant heat exchange with the environment; i.e. we may assume the following enthalpy balance in order to estimate the vaporized fraction $x$:
$$\begin{align}
h_0 &= x \cdot h_\text{steam} + (1-x) \cdot h_\text{liquid}\\
x &= \frac{h_0 - h_\text{liquid}}{h_\text{steam} - h_\text{liquid}}\\
x &= \frac{1086\ \mathrm{kJ/kg} - 419\ \mathrm{kJ/kg}}{2676\ \mathrm{kJ/kg} - 419\ \mathrm{kJ/kg}}\\
x &= 0.296
\end{align}$$
This means that approximately $29.6\ \%$ of the superheated water flashes to steam when the container is opened.

I would not be inclined to stand anywhere near that.

One thing that i guess you understand wrong i.e. I'm not going to open the whole container. Instead I'll be making a tiny hole(say 0.5 cm in diameter) or opening this tiny hole. So, will the water vapor still come out at same pressure? If yes then how much. NOTE: the water is still getting heated, even after opening this tiny hole. You can ask me for more clarification. And thanks :)

@Dahano If the opening is small, the flow rate of the escaping fluid (liquid water and steam) is limited. For the given example, a simple opening ($\zeta=1$) with a diameter of 0.5 cm causes an initial mass flow rate of about 0.3 kg/s. Nevertheless, the initial and the final state (temperature and pressure) are not affected by the size of the opening. However, the enthalpy balance is a simplified approximation since it ignores the work that is required to expand the steam against the ambient pressure.

@Loong From my understanding, will something like this would happen --> The water say, 15litre, is superheated @ 250C and now I open the tiny hole; the water will come out in form of vapor with lot of energy, which would be enough to push something (and eventually 3rd law of newton)~ Will something like this happen? If yes, then how much push that energy could create(in kg) and how long would this last?
PS: I know I've been asking a lot of questions but it would be seriously great if you can answer them all. Thanks a ton! :-)

@Dahano In principle, your idea is correct. A volume of $V_0=15\ \mathrm{l}$ of liquid water at a temperature of $T_0=250\ \mathrm{^\circ C}$ and a pressure of $p_0=39.2\ \mathrm{atm}=4.0\ \mathrm{MPa}$ corresponds to a mass of $m_0=11.98\ \mathrm{kg}$.
After relaxation to atmospheric pressure, we have a mixture of $V_\text{liquid}=8.8\ \mathrm{l}$ liquid water and $V_\text{steam}=5925\ \mathrm{l}$ steam. Therefore, a large fraction of this mixture is pushed through the opening out of the small container. The container and its remaining content are pushed into the opposite direction.

@Dahano Based on some guessed values, I estimate that the initially escaping mixture is mostly liquid ($x\approx0.05$) with a density of $\rho\approx220\ \mathrm{kg/m^3}$. The mass flow rate through the small opening with a diameter of $d=0.5\ \mathrm{cm}$ is about $\mathrm{d}m/\mathrm{d}t\approx0.3\ \mathrm{kg/s}$. Thus, the mixture is released with a velocity of $v\approx70\ \mathrm{m/s}$, which corresponds to a thrust of $F\approx21\ \mathrm{N}$.

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text:
Difference between Superheated steam and Superheated water?
I'm working on a project (school level) and in it I'll be dealing with the pressure generated by water when superheated.
I could understand the concept of superheated water but superheated steam is hard to understand. Are both inter-related, i mean that when water is superheated then it changes into superheated steam?
Or is it something different? And since in my project I require high-pressure-energy to be generated from water, so which one will be better, superheated steam or superheated water?

Boilers and similar steam generators usually produce wet steam, which is a mixture of saturated steam and liquid water droplets. Saturated steam is steam that is in equilibrium with liquid water. The temperature of the steam is equal to the temperature of the liquid water; it corresponds to the boiling point at the given pressure. At normal pressure ($p=101\,325\ \mathrm{Pa}=1\ \mathrm{atm}$), the temperature is $T=99.974\ \mathrm{^\circ C}$. The corresponding specific enthalpy of the steam is $h = 2676\ \mathrm{kJ/kg}$; the specific enthalpy of the liquid water is $h = 419\ \mathrm{kJ/kg}$.
If heat is supplied to the wet steam, liquid water vaporizes. At constant pressure, the temperature remains constant until all liquid water has been vaporized, i.e. until the steam is dry.
If further heat is supplied to the dry steam, the temperature and the specific enthalpy of the steam increase. Such steam at a temperature higher than its boiling point is called superheated steam. For example, superheated steam at normal pressure ($p=101\,325\ \mathrm{Pa}=1\ \mathrm{atm}$) and a temperature of $T=150\ \mathrm{^\circ C}$ has a specific enthalpy of $h = 2777\ \mathrm{kJ/kg}$.
Superheated water is liquid water under increased pressure at temperatures above the boiling point at normal pressure. For example, water at a temperature of $T=150\ \mathrm{^\circ C}$ and a pressure of $p=506\,625\ \mathrm{Pa}=5\ \mathrm{atm}$ is liquid. The specific enthalpy of the liquid water is $h = 632\ \mathrm{kJ/kg}$.
The energy content of superheated steam (e.g. $h = 2777\ \mathrm{kJ/kg}$ at $T=150\ \mathrm{^\circ C}$ and $p=101\,325\ \mathrm{Pa}=1\ \mathrm{atm}$) is much higher than the energy content of superheated water (e.g. $h = 632\ \mathrm{kJ/kg}$ at $T=150\ \mathrm{^\circ C}$ and $p=506\,625\ \mathrm{Pa}=5\ \mathrm{atm}$) because of the latent heat, i.e. the enthalpy of vaporization $\Delta H_\text{vap}$.

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text:
Are these 3 molecular structures of ammonium bicarbonate the same?
Are these 3 molecular structures of ammonium bicarbonate the same?
A:

B:

C:

As far as representations for molecular structures go, B seems wrong to me, as there is no oxygen nitrogen bond in the compound.

Agreeing with @Martin-マーチン, what are the sources of these images, just out of curiousity

This is from a chemicals company's website. http://www.mpbio.com/product.php?pid=02150107&country=96

The site says "MP Biomedicals is a global corporation, with headquarters in Santa Ana, California, USA and regional offices in Europe, Asia Pacific and Australia. "

I found one more form (C). http://www.caslab.com/Ammonium_bicarbonate_CAS_1066-33-7/

@user2986288 Why did you add structure that is definitely wrong?

Structure C is even worse! Not only is there a non-existent $\ce{N-O}$ covalent bond, but such a bond would exceed the octet on the nitrogen atom!

I found the information from a  website of an analytical lab. C is for reference.

My chemistry is bad. Why the bond in nitrogen dioxide is allowed?

A non-integer bond order can exist in resonance. Does this help?

Are these really molecular structures or ionic structures?

Like Punarbasu Roy already stated, the depicted structures are certainly not the same. To be frank, only one of them (A) is correct, one of them is not completely wrong (B - but only because it is incomplete), and one is completely, utterly wrong (C).
In lieu of an actual crystal structure of ammonium hydrogencarbonate,[1] I'll refer to the structure of the parent compound ammonium carbonate monohydrate to make my point.[2] It is actually quite difficult to assign one correct Lewis structure. In its solid form ammonium (bi)carbonate is a salt, in which ammonium and (bi)carbonate ions are interacting via ionic bonds in a lattice. In solution this certainly is broken up and we will find separate ions, according to the following equilibria:
\begin{align}
\ce{(H4N)HCO3~(s) &~->C[(H2O)] NH4+ ~(aq) + HCO3- ~(aq)}\\
\ce{NH4+ ~(aq) + H2O &~<=> H3+O~(aq) + NH3~(aq)}\\
\ce{HCO3- ~(aq) + H2O &~<=> H2CO3 ~(aq) + {}^{-}OH~(aq)}\\
\ce{HCO3- ~(aq) + H2O &~<=> CO3^2- ~(aq) + H3+O~(aq)}\\
\end{align}
The same equilibria will establish for ammonium carbonate with different concentrations.
In the crystal structure of ammonium carbonate monohydrate you can see the different molecular entities (ions) held together via hydrogen bonds, reproduced from [2]:

Wikipedia[3] therefore offers the best possible Lewis structure:

Strictly speaking a Lewis structure may only contain covalent bonds, i.e. two-centre-two-electron bonds, while ionic bonds are implied by placing the (molecular) ions in close proximity to each other. However, the hydrogen bond is a well known concept in chemistry and it was already extensively discussed by Linus Pauling. In his book "The Nature of the Chemical Bond" he uses dashed lines to represent the hydrogen bond.[4] Certainly another interesting read is the approach of a more unified theory about the hydrogen bond.[5]
Another possibility is to introduce the donor-acceptor notation with arrows, to indicate a bonding interaction. Therefore one of the following structures may also be considered correct in an extended Lewis concept.

The following article reads like this is still a work in progress: R. Brooks, and T. C. Alcock, Nature, 1950, 166, 435-436. (It does also not come with any supporting information.
A. D. Fortes, I. G. Wood, D. Alfè, E. R. Hernández, M. J. Gutmann, and H. A. Sparkes, Acta Cryst. Sec. B, 2014, 70, 948-962. This quite interesting article is open access.
Ammonium bicarbonate, Ammonium carbonate
Linus Pauling: The Nature of the Chemical Bond. Cornell University Press, 1960.
G. Gilli, P. Gilli, J. Mol. Struc., 2000, 552 (1-3), 1-15.

My chemistry is bad. Why is the bond in nitrogen dioxide allowed?

Well, that cannot be answered in one or two sentences, as it refers to the stability of certain compounds. You may try ask a new question, but it might be too broad to be acceptable for this site. maybe you should read a little on bond formation and molecules first and then ask a question with specific points of what you did not understand.

Would you allow a variant of structure B, in which the order is oxygen, hydrogen, nitrogen, hydrogens, a.k.a. $\ce{HO-CO-OHNH3}$? Just for reference and clarity.

@Jan At least I would consider this as more correct as structure B, but still incomplete. But Lewis structures hardly ever are. If you are true to the concept, then you cannot allow hydrogen bonds, which makes it quite impossible to construct any valid Lewis structure. But one can extend the concept, maybe I should include this...

They are certainly not the same. A is in dissociated form. Say in aqueous solution. B refers to not dissociated form but I have never come across non dissociated structure. There may not be actual bond between N atom and O atom.

"May not" is certainly not the right choice of words. There is no bond between oxygen and nitrogen

I was not sure sir thus I said 'may not'.


text:
Running behavior of glucose, galactose and mannose in TLC
I have separated the sugars glucose, galactose and mannose by thin-layer-chromatography. My question is why these sugar have different running behaviors and a separation is possible.
Glucose have the lowest Rf-value, then galactose and mannose have the highest Rf-value in my experiments.
I thought it might be something to do with the position of the $\ce{-OH}$ groups (axial, equatorial) but I have no firm explanation at the moment.
Thanks for every explanation concerning the topic.

Short answer: Every molecule that is not the enantiomer of a reference molecule will run slightly differently on a TLC. Glucose, mannose and galactose are all diastereomers so their physical properties differ so they will run differently.

Isn't the polarity the main reason for the different Rf-values? Is there a difference between these molecules or which other criteria has an impact on the Rf-value?

Well, yes. But different molecule structures (diastereomers) automatically mean different polarities. And there are other factors contributing as well, like the ability of the mobile phase to interact with the molecules in question. Compare cis- and trans-stilbene which can be separated by TLC (and silica chromatography).


text:
The Propinquity Effect
I am currently in the process of studying enzyme catalysis and am struggling to get to grips with this concept. As far as I understand, the propinquity effect allows substrates to bind with enzymes in the conformation which would facilitate the reaction, thus increasing the reaction rate. Could someone expand on the why and/or how enzymes are able to do this? What specific reactions benefit from this the most?

how: an enzyme is able to effectively make a reaction intramolecular by binding the reactants in the correct conformation in close proximity within the active site this can speed up the rate of a reaction dramatically, intramolecular rates avoid diffusion control and have prearranged conformations which is far more efficient than typical intermolecular reactions,
why: enzymes are outrageous, evolution has pushed them, through trial and error, towards optimal configurations that, almost magically, provide ideal conditions for a reaction to occur


text:
Why doesn't gold corrode?
As the electronic configuration of gold is [Kr] $4d^{10} 4f^{14} 5s^2 5p^6 5d^{10} 6s^1$, the valence shell is not completely full. Why doesn't gold  corrode?

http://www.quora.com/Why-does-gold-not-rust

Gold's 6s orbital is contracted due to relativistic effects making the 6s electron less available for oxidation (or for that matter, any chemical reaction).  See this earlier answer

You can look it up on James Huheeys book, gold's few properties have been qualitatively explained by relativistic theory. Due to that and high nuclear charge acting on the electron, the single electron in $6s$ is actually attracted to the nucleus strongly, thus the electron is not available to react that easily and thus gold does not corrode.


text:
Develop slow smoke for mini windtunnel
I want to create a small wind tunnel for testing of aerodynamics for Radio controlled units (Cars, planes and helicopters).
In real wind-tunnels water mists are used but water isn't a good thing for the electronics.
I have considered the following:

Smoke bombs, but the smoke develops too fast.
Burning stuff like laves or coal, but it creates odor, and i don't want my RC stuff to smell.
Incense, same as above.

My chemistry level is a College equivalent degree of chemistry that has been unused for 15 years.
And the materials to make this should be somewhat harmless, and legally available.
Any suggestions on how to do this would be highly appreciated.

The water condensed from steam is distilled and fairly safe for electronics as long as allowed to evaporate before turning on.

The DLR (German Aerospace Center) has developed a couple of DIY instructions for school education. Among these is a PDF in German for the construction of a wind tunnel.
For the smoke, they suggest to use either

incense (the sticks or cones that you usually find in asia shops or esoteric shops) or a
fog machine, like the ones used for entertainment (dance clubs, etc.)

In the latter case, a mixture of 1,2-propanediol and water is heated and pressed through a nozzle at elevated temperature. Given that properly insulated machines, build for safe use at home, are available from DIY electronics shops for roughly 30 € up, I wouldn't start to build one myself.

I just checked your suggestion, a smoke machine like the one you talked about is available for €40 at my local store, that is within my limits, thanks!


text:
How is it possible to distinguish antimony from lead?
I just bought some (pure) antimony stones from a local store. However, I've heard that some stores deceive their customers by selling lead instead of antimony. I was wondering if there is any simple test that I can perform at home to make sure these stones are antimony and not lead. Since I don't know much about chemistry I thought of asking here...

Lead has much bigger density as melts much easier.

@Mithoron I actually thought of melting it in the oven but wikipedia says the melting point of lead is 327.4 and antimony is almost double of that. The maximum temperature of my oven is around 250 degrees Celsius. But thanks for the suggestion about density.

Yeah, you could do Archimedes' experiment :)

Definitely agree with @Mithoron. Lead is almost twice as dense as antimony and Archimedes' experiment is easy and safe to do so and should give a clear result.

Eureka! @Mithoron

As mentioned in the comments, you could perform Archimedes' Experiment.

Firstly, take the mass ($m$) (in grams) of your sample.
Then, measure the volume $V$ displaced (according to the first link), in a method summarised in the image below:

You could use any kitchen graduated container. Even though these have graduations for mL (millilitres) and even though the water density changes with temperature (source:USGS), for the temperatures in a home experiment, the following conversion could be used 1 mL = 1 cm3 of water displaced.

Then determine the density by $\rho = \frac{m}{V}$ and compare it to the known values for lead at 11.3 g.cm-3 and antimony at 6.7 g.cm-3.

Another observed feature is that according to the LennTech site for lead is that as opposed to antimony, lead will tarnish on exposure to air.

Thanks. The density turned out to be 6.06 (with mL) but I didn't have a graduated container so I used a ruler and a regular container which did not have a 100% flat bottom (and probably it resulted in some error). It looks like the stone is antimony I guess :).

Another method that doesn't require calculations, but does involve sacrificing only a very small portion of your sample is to do a flame test. By heating a pure sliver of antimony in a hot flame you should observe a pale green color as opposed to a blue-white color if the sample is lead. If you are a poor judge of color then its helpful to have known pure samples of the elements to compare with your unknown sample.
This method, as well as the gravimetric method suggested by Santiago can fail if your sample is an alloy (mix) of the two elements (and perhaps other ingredients!). But here the flame test can help by using a spectroscope for a more quantitative accounting of the constituents.

Very good point - a spectroscope is not necessarily expensive nor hard to use.


text:
What does the 129Xe NMR spectrum of XeOF4 look like?
What will the $\ce{^129Xe}$ NMR spectrum of $\ce{XeOF4}$ look like?
I know that its electron pair geometry is octahedral, and the molecular pair geometry is square pyramidal.
I believe that the lone pair on $\ce{Xe}$ will take the equatorial position (less sterically crowded), alongside the oxygen and 2 other equatorial fluorines in the same plane, to minimize electron pair repulsion.
So there will be 2 equatorial $\ce{F}$ and 2 axial $\ce{F}$. Therefore, am I right to say that the $\ce{^129Xe}$ NMR will be a triplet of triplets?

The $\ce{^{129}Xe}$ nmr spectrum of $\ce{XeOF4}$ is a sharp, quintuplet of peaks (reference, p. 445).  As you note, $\ce{XeOF4}$ adopts a square pyramidal geometry (reference), but the lone pair occupies an axial position.

Therefore, the 4 equatorial fluorines are equivalent and since they are spin 1/2 nuclei they will couple with the $\ce{^{129}Xe}$ nucleus (also spin 1/2) producing the observed quintuplet. $\ce{XeOF4}$ is an IUPAC suggested standard for $\ce{^{129}Xe}$ nmr spectroscopy (but xenon gas is more commonly used because $\ce{XeOF4}$ can decompose in the presence of moisture to produce $\ce{HF}$).

Hi, I am curious as to why the lone pair chooses to adopt an axial position. Doesn't this lead to the lone pair being at right angles to 4 other substituents vs being at the equatorial position? And also why is the oxygen at the axial position for the same reason?

You might want to ask that as a separate question, the answer requires more than a few sentences.  If this answer was helpful please consider accepting and\or up-voting it, thanks.


text:
Bond length in ICl2- and ICl2+
Why is the $\ce{I-Cl}$ bond length in $\ce{ICl2-}$ (charge of -1) is longer than in $\ce{ICl2+}$ (charge of +1) ?
Here are a few explanations that I have in mind. Please critique.

I in $\ce{ICl2-}$ is $sp^3d$ hybridized. $\ce{I}$ in $\ce{ICl+}$ is $sp^3$ hybridized. Since additional promotion energy is required to form the $sp^3d$ hybrid compared to the $sp^3$ hybrid, the bond lengths in $\ce{ICl2-}$ is longer.
For $\ce{ICl2-}$, if we let $\ce{I}$ be $sp^2$ hybridized such that the 3 lone pairs lie 120 apart in a plane, and then let the 2 $\ce{Cl}$ atoms and the $\ce{I}$ atom lie in a straight line, then we can form a 3-centred, 4 electron system. This is weaker than the 2-centred, 2 electron system in the $\ce{I-Cl}$ bond in $\ce{ICl+}$.
As $\ce{I}$ in $\ce{ICl2+}$ has a higher effective nuclear charge, its valence orbitals are smaller and more compact, hence the $\ce{I-Cl}$ covalent bond is stronger as the electrostatic attraction between the nuclei and the overlap density is stronger.

The iodine atom in $\ce{ICl2}$ is what is termed hypercoordinated (or hypervalent). Basically, this means that the 3 atoms combine to produce 3 molecular orbitals (the 3-center 4-electron model). There is no need to invoke d-orbitals.

In neutral $\ce{ICl2}$ there are three electrons spread across the two $\ce{I-Cl}$ bondss. Placing those electrons in our molecular orbitals, 2 go into the bonding orbital and one into the nonbonding orbital.
In $\ce{ICl2^{+}}$ we have removed the one nonbonding electron.  As a consequence there will be less electron-electron repulsion and only the bonding molecular orbital is occupied, because of this we would expect shorter bond lengths in $\ce{ICl2^{+}}$ compared to $\ce{ICl2}$.
In $\ce{ICl2^{-}}$ we have added another electron and it will go into the nonbonding orbital.  This will increase the electron-electron repulsion in the molecule without increasing bonding, because of this we would expect longer bond lengths in $\ce{ICl2^{-}}$ compared to $\ce{ICl2}$.
These arguments tell us that we would have shorter bond lengths in $\ce{ICl2^{+}}$ compared to $\ce{ICl2^{-}}$.

This paper suggests a $C_{2\mathrm{v}}$ symmetry for $\ce{ICl2+}$ with a bond angle of 98° contradicting the assumption that any kind of 4e-3c bond be present. (F. Bailly, P. Barthen, H.-J. Frohn, M. Köckerling, Z. anorg. allgem. Chem. 2000, 626, 2419.)

@Jan  Thanks for pointing out the theoretical geometry for the cation.  What level of computation did they use?  +1 for your nice answer based on that geometry.

It’s in the linked paper, table 1, saying LANL2DZ/RHF. It was ‘only’ the first paper I could find entering $\ce{ICl2+}$ into SciFinder and gathering references with crystal structure although there is no crystal structure easily found. I would love to find one though to verify either of our claims

The article is behind a pay wall, I can only see the abstract.  I'm not convinced that a 15 year old calculation has predicted the correct geometry.  Why have the chlorines so close together (steric and electronic repulsion) when it is not necessary?  The 3-center 3-electron bond model (for the cation) would have the chlorines further apart and still have the lone pairs in low energy sp or (s and p) orbitals.  Perhaps you or @Martin-マーチン could perform a better calculation?

Unfortunately I’m not in a group that does calculations any more. I could ask an ex-colleague, though. However, I remember our first inorganic chemistry lecture talk about how to predict interhalogen compounds and their ions with the explicit step of giving the central atom (iodine) the charge, drawing 2-electron 2-centre bonds until octet and filling in the remaining atoms via 4-electron 3-centre bonds — using that technique, I arrive at the same structure. Additionally, the same professor presented us with the structure of the related cation …

… $\ce{ClF4+}$, whose structure can be seen as the combination of a 90° $\ce{F-Cl-F}$ fragment of 2e-2c bonds with a perpendicular $\ce{F-Cl-F}$ fragment composed of a 4e-3c bond. I wished I had found an actual crystal structure for $\ce{ICl2+}$ to finalise my argument, though.

I’ll agree with ron concerning the non-hybridisation of d-orbitals and the description of $\ce{ICl2-}$. However, I disagree with ron’s description of $\ce{ICl2+}$, which was calculated to have an $\ce{Cl-I-Cl}$ bond angle of 98°.[1]
I also believe that the contribution of the s-orbital to any bonding, non-bonding or anti-bonding orbitals in the iodine atom is neglegible, the reasoning being that the energy difference between iodine’s s- and p-orbitals is larger than for the atoms of the second period.
The $\ce{ICl2+}$ ion can thus be explained by assuming one double-filled s- and one double filled p-orbital plus two half-filled p-orbitals on iodine. Each of the two half-filled orbitals then create a normal $\sigma$-bonds to a corresponding p-orbital of one of the chlorine atoms, predicting a bond angle of 90° (closer to 98° than if we assumed sp³-hybridisation which would suggest 109.5°).
This means the following:

In $\ce{ICl2+}$, we have to bonds with a bond order of 1 to each of the chlorines. Compared to a bond order of ½ for $\ce{ICl2-}$ this means a shorter bond length.
The ion’s charge, which is located mostly on the iodine (electronegativity!) means that the iodine atom is contracted and the two chlorines can move closer. This effect is not present in $\ce{ICl2-}$, where the charge is distributed evenly across the two chlorine atoms.

For the reference and on the topic of predictions, I would predict the bond length of $\ce{ICl}$ to be comparable but slightly longer than that of $\ce{ICl2+}$, while still being notably shorter than that of $\ce{ICl2-}$.

[1]: F. Bailly, P. Barthen, H.-J. Frohn, M. Köckerling, Z. anorg. allgem. Chem. 2000, 626, 2419.


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Linear PEI Solution Making
I'm making a solution with linear poly(ethyleneimine) $\ce{[CH2CH2NH]_{n}}$.  The bottle indicates that the substance is hygroscopic and must be dealt with under inert gas.  Is this really necessary?  I don't currently have access to such a space - just the PEI and a balance.

Well, depends if water in your solution makes problem.

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


text:
steam distillation of 4-bromoaniline , why render it basic prior
In the multistage synthesis of 4-bromobenzenamine, why must the solution be rendered alkaline prior to the steam distillation after the hydrolysis of 4-bromo-N-ethanoylbenzenamine?

Which compound are you distilling, the 4-bromoaniline or the 4-bromo-N-ethanoylphenylamine?

the 4-ba might be a zwitterion and in the acidic medium it may be attracted to the cation forming a salt of whatever acid was used.This would raise the boiling point of the 4-ba enough to not come over to the receiving flask. When made basic with a reagent like naoh the sodium ion will react with the cation from the 4-ba salt and the 4-ba will be left as the free base which is more volatile and can be distilled .

In principle, it is difficult to distill salts (ionic compounds). Therefore you first make sure that you have a free base in the solution. This is guaranteed in alkaline solution. For the steam distillation the compounds should be immiscible, which is another reason for having a free base. Let’s compare boiling points and solubility for aniline and its salt:

Aniline b.p. 184 °C, solubility in water: 36 g/L
Anilinium chloride b.p. 245 °C, solubility in water: 1070 g/L


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Molten vs Liquid
When I start studying electrochemistry, I learn the words “molten” and “aqueous”. I don't have a problem for “aqueous”, but I'm a little bit confused about “molten”.
For me, “molten” means melt, which means to become liquid from solid. But why do we say molten lead(II) bromide instead of liquid lead(II) bromide?
I mean both are the same, but liquid is arguably easier to understand.

As you said, the meaning is exactly the same. Molten reduces the ambiguity, because you emphasize that you know, that it is solid substance at laboratory conditions and you heat it to become liquid (while staying pure substance).
As an example, the "electrolysis of liquid sodium chloride" is in principle enough to tell you all about the process, but as we often like to dissolve the salt in water, you might be slightly unsure, whether the meaning is 100% clear. Therefore "electrolysis of molten sodium chloride" is better choice.

Molten is melted from a solid. Liquid means it exists as a liquid in room temperature. Molten substances exist as solids in room temperature. Thus the difference between molten and liquid.

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


text:
Factors controlling enzymatic phosphorylation of D-ribulose 5-phosphate
Considering the Metabolic Pathway:
http://www.genome.jp/kegg-bin/show_module?M00166
Consider the reaction R01523(the very first reaction). How is it possible to know that the compound C00199(D-Ribulose 5-phosphate) will get transformed to C01182(D-Ribulose 1,5-bisphosphate)? What are the factors on which this transformation depends? One factor I know is structure but how does structure effect this transformation? Does two similar structured compound means that one of the compound can get transformed to the other one?
Note: It ia a part of biology but a chemical reaction or transformation is basically a part of chemistry, so asked this here

Cross posted http://biology.stackexchange.com/questions/34545/factors-on-which-one-compound-transforming-to-another-compound-depend

After looking at the website I found the reaction is based off of an enzyme called: phosphoribulokinase. In biological systems many reactions occur under the auspice of enzymes or proteins. These are incredibly large molecules made of an extremely long chain of amino acids which are a set of 26 or so small molecules that can 'hook' into the chain structure of a protein. Each amino acid has different properties and thus experiences different forces pulling it towards and away from other amino acids on the chain. What this means is the whole chain folds up into an extremely complex structure. For a protein to allow a reaction to occur it has a small region that the reacting molecule can hook into. There are then sets of atoms and molecules that act as scaffolds that allow atoms in the reacting molecule to be moved without the need for nearly as much energy as would be required without the scaffold. Now this is how it changes structure. Now whilst the protein can just as easily allow the molecule produced to be converted back to the original, this may be less energetically favourable which brings us to equilibrium chemistry involving dynamic forward and backward reactions. What makes this more complicated is that proteins often have separate 'hooks' for molecules to disable them. That is if a lot of substance A is around then the protein will be disabled and no longer convert B into A until more A is consumed and thus stops inhibiting the protein.

So just the folding rules the transformation of one compound to another?

In biological systems the structure of the protein's that regulate the reaction are indeed the main factor in deciding how the structure is modified in an energetically likely way (the energy needed for a particular reaction is likely to be found at the temperature the reaction is taking place at). That's not to say all reactions occur with proteins, just a lot of them and specifically the one you linked in the question.

Temperature as in the temperature of the environment in which the reaction takes place , right?

yes that's correct


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Angular dependence of x-ray and neutron scattering factors
Why are neutron scattering factors essentially independent of $2\theta$, whereas X-ray scattering factors drop off with increasing $2\theta$ ?

Well, the interactions that lead to the scattering are different. Do you understand why the X-ray scattering factor drops off with increasing $2\theta$? If so, do those fields/potentials/forces apply to neutron scattering?

In X-ray scattering you have a dipolar electronic interaction, in neutron scattering presumably it is colour force. I don't really understand why there is a drop off in XRD.

Assume neutron scattering is sold hard spheres. Calculate the differential scattering cross section. You will find that it is flat vs either energy transferred or angle scattered (they are equivalent). For the x-Ray case there are many derivations online...

What search terms did you use ? I couldn't find anything which helped me to understand it

'X-Ray scattering cross section' worked just fine for me.

Scattering factors describe scattering of an atom
To predict how an object will diffract, you essentially have to do a 3D Fourier transform of the scattering density (https://chemistry.stackexchange.com/a/115815/72973). The scattering factor is derived from the scattering of a single atom (placed at the origin). For xray diffraction, the relevant density is that of the electrons. For neutron diffraction, it is that of the nucleus. See slide 11, https://www.embl-hamburg.de/biosaxs/courses/embo2012/slides/neutron-scattering-basics-trewhella.pdf
The fuzzier the scattering density, the higher the angular dependency
The Fourier transform of a Gaussian is a Gaussian again; a sharp Gaussian results in a broad transform, and vice versa. For X-ray diffraction, the electron density is fuzzy (same magnitude as the wavelength), so there is a large angular dependence. For neutron diffraction, the scattering density is very compact, leading to almost no angular dependence of the scattering factors.

I would also point out that in a neutron scattering experiment there is little to no backscattering of neutrons simply due to the fact that a radioactive decay of the nuclei is needed to produce additional neutrons. This is not the case for x-ray photons as these can be produced by excited electrons.

authors: Валерий Заподовников


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Predicting incomplete combustion?
I understand the concept of incomplete combustion and that it occurs when oxygen is not a large enough supply to allow for complete combustion. My question is: how can we model combustion properly in terms of predicting the ratio of complete to incomplete combustion? Is it possible to use basic kinetic with reaction rate constants to model the rate of each reaction and then obviously model how the changing concentration from both affects each individual rate etc.

Very interesting question - I would be very interested if there has been a study based on this.

This sounds possible in principle, but quite hard in practice. Wonder if anyone has some good input on it.

Before entering the kinetics realm, it is advisable to check the equilibrium. Besides the straightforward combustion reactions, you have two more, Boudouard reaction, which favors formation of $\ce{CO}$ from $\ce{CO_2}$ and Water-gas shift reaction.
The actual reactions which contribute to the equilibrium depend on the fuel. In case of pure carbon, you at least get rid of the Water-gas shift reaction.
You might need to solve the system iteratively, as the temperature of the equilibrium mixture is result of the combustion reactions, but also affects the position of equilibrium.

so what you're saying is that I could simply model complete combustion and then use these equilibriums to model conversion of original products into those of incomplete combustion?

What exactly do you mean by incomplete combustion? For me, simply speaking it is combustion, where there is not enough oxygen.

Incomplete combustion is the process of combustion in which carbon dioxide and water isn't produced but rather carbon monoxide and water

@user2353082: yes, indeed. Is the reason for this that there is not enough oxygen? Or because the fuel leaves the combustion chamber so fast, that it cannot burn fully?

I'm looking at instances mainly where the fuel is stationary but the surrounding oxygen levels are not high enough to allow 100% complete combustion or at least it becomes energetically more favourable to take incomplete combustion paths instead of 'waiting' for complete combustion. For some fuels such an environment is the normal environment.

@user2353082 in this case kinetics is not important for you, solve the equilibrium conditions. It is simpler, but definitely not trivial. Collecet reactions, their $\ce{\Delta G}$ as function of temperature and solve for different partial pressures of oxygen.


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Why are HgO and CdS compounds colourful?
We know that the electron configuration in $3d$ orbitals of $\ce{Zn^2+}$, in $4d$ orbitals of $\ce{Cd^2+}$ and in $5d$ orbitals of $\ce{Hg^2+}$ ions are completely same.
Then why the compounds of $\ce{Hg(II)}$ and $\ce{Cd}$, such as $\ce{HgO}$, $\ce{CdS}$ radiate versatile colour where $\ce{ZnO}$, $\ce{ZnS}$ are white?

related http://chemistry.stackexchange.com/questions/2547/why-are-some-salt-solutions-coloured

$\ce{HgCl2}$ and $\ce{CdCl2}$ are colourless; just for reference.

We see a similar situation with oxides of alkali metals:    https://chemistry.stackexchange.com/questions/72243/why-color-of-alkali-metal-peroxides-superoxides-and-ozonides-deepens-down-the-g?r=SearchResults

The band gap decreases as either electronegativity of the anion decreases (electronegativity of anion in CdS < CdTe , ZnS < ZnSe) or the overlap decreases (overlap in ZnS > CdS).
Also, note that colored compounds only result when 1.8 < Band gap < 3.0. Larger gaps give white/transparent & smaller gaps give a black colour to compounds.
(The band gap in ZnS is 3.54, the band gap in CdS is 2.45.)

The electronegativity decreases from S, over Se to Te.


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Predicting relative solubilites of salts without Ksp values
Is there a way of comparing which salt is more soluble without knowing their $ K_{sp}$ data or performing experiments?
For example, comparing $\ce{Hg2Cl2}$ and $\ce{Cr2(SO4)3}$.
I am not sure but can I compare lattice and hydration energy relatively or the ion size and estimate?
Experiments will give concrete results but I was wondering if there is another method just to gauge the solubility.
(The relevant chat discussion)

As I mentioned in chat, I feel one of more thorough and reliable ways to answer this question is actually start off by calculating the solubilities for a large amount of binary salts from $\mathrm{K_{sp}}$ data, then compare how each cation/anion correlates to solubility individually. The calculations themselves are trivial, the trick is in plotting the data without it being a mess!

I believe someone has to have done an experiment at one time. Even if you look at lattice energies and hydration energies, somebody must have measured these... yet the question remains, what is the theoretical (mathematical) connection between these values.

Maybe an application of Pearson's hard and soft acids and bases concept would help here?

If you were comparing salts like AgCl , AgBr , AgI then you could use polarization and Fajan's Rules to qualitatively compare their solubility

To a certain extent, you can predict the relative solubilities of two salts, especially when there is a large difference in solubility. The existing answer mentions Fajans' rules and obviously it is quite easy to predict which of $\ce{NaCl}$ and $\ce{AgI}$ is more soluble.
But I'm inclined to be pessimistic and say no, this is not generally applicable.
You're perhaps on the right track. The dissolution of the salt arises from a couple of measurable quantities: the lattice enthalpy of the salt, and hydration enthalpies of the ions. You could measure both, and try to come up with a trend.
However that's often far easier said than done. The trends are sometimes quite subtle, and even if you have all the relevant data and limit yourself to a "simple" system like the alkali metal halides, you'll find that the interplay between two factors leads to a very delicate balance, and the resulting trend is quite complex. I don't think it's something that we can predict solely in our heads without doing any measurements.
For example, if you simply compare $\ce{NaF}$ and $\ce{NaI}$, $\ce{NaI}$ has a smaller lattice enthalpy (favours dissolution) but the $\ce{I-}$ ion also has a smaller hydration enthalpy (disfavours dissolution). So which factor wins out? There's no simple heuristic or rule of thumb that applies here, not any that I know of, at least.
How about two obviously more different salts? Let's try $\ce{NaF}$ and $\ce{MgF2}$. $\ce{NaF}$ has a smaller lattice enthalpy (favours dissolution) but also the sum of the hydration enthalpies is smaller than $\ce{MgF2}$ (more ions and $\ce{Mg^2+}$ is more charged than $\ce{Na+}$ - which disfavours dissolution of $\ce{NaF}$). Again we have the same dilemma.
In fact, that's just the enthalpic changes, so only half the story. There's also entropy to care about, which can either be positive or negative (and it's extremely difficult to predict this!).

Like dissolve like. Thus, a more ionic compound would be more soluble in water and a covalent one would be more soluble in organic solvents. Speaking in context of your example, and assuming solvent to be water,
The nucleus of Cr3+ ion attracts the bond pair electrons more strongly for its greater charge & smaller diameter than $\ce{Hg+}$ ion.
According to Fajans' rules, Cr2(SO4)3 is of more covalent character than Hg2Cl2. Again the factor of anion is less significant here as oxygen & chlorine are quite closer  in their electronegativity.
So, from that point of view Hg2Cl2 has better solubility than Cr2(SO4)3.

Certainly an interesting starting point, but I believe the OP just named those two compounds as an example and is actually looking for a conceptual approach to the underlying problem.


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How do I predict the solubility of a compound in a mixture of water and an organic co-solvent?
I'm trying to predict the solubility of long alcohols (e.g. 1-octanol, 1-decanol or 1-dodecanol) in a mixture of water and an organic co-solvent such as DMSO, acetonitrile or ethyl acetate.
For example, decanol in water with 25% DMSO or saturated with ethyl acetate (8.3%). I of course know the solubility of alcohols in water, but not in organic solvents or in mixtures.
I found a server to predict the solubility of any compound in any solvent, but I'm not sure how accurate it is and it only does predictions for pure solvents, not mixtures: http://showme.physics.drexel.edu/onsc/models/multisolvent.php
Any input on this would be greatly appreciated!

I think this is very much an open problem. As you may see from the Drexel pages, even solubility prediction in one solvent is still fairly hard, partly because of the lack of experimental data.
The only software that I know capable of making predictions about solubility in mixed solvents is COSMOtherm, based on the COSMO solvation model.
I have never used it, but the capability is mentioned on the website.

I had a workshop on COSMO once, iirc the theory is sound, but the predicted values are often far off the experimental results. (But this was several years ago and Klamt was still improving it.) This is probably due to computational limitations. Note that the model used here is COSMO-RS.

@Martin-マーチン I think it's an extremely hard problem and there's obviously not a lot of experimental data to calibrate and test models. My guess is that only in the next 5-10 years will this be possible, and only if there's a lot of open solvation data.

I would be surprised if you found anything predicting solubility in mixtures as there is poor modelling due to non-linear effects. Taking solubilities of two solvents and trying to interpolate does not work. As little as a few hundred ppm to 4% of one solvent in another is enough to confound standard predictive equations without empirical data.


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Deformation of the nematic phase of liquid crystals surviving in the solid state
Does anyone know if I were to cool a liquid crystal with a deformed nematic phase quickly, if it will preserve the deformation in the crystal lattice?
I didn't find any information in classical books on liquid crystals.


text:
Sugar dissolution test
Is there a way to detect whether sugar is completely dissolved in a cup of water other than looking at the bottom of the cup?
I was wondering if there is a test or something that can be carried out to determine whether sugar is fully dissolved.

You could filter it to remove any undissolved sugar.

The sugar will keep dissolving until it reaches it's saturation concentration of around $\pu{900 g/L}$, which is of course horribly sweet, thick and undrinkable as a beverage.  Assuming you don't completely saturate the water with sugar in this way, the undissolved sugar will continue to dissolve so long as there is still any remaining in the glass.
A common way of measuring the dissolved sugar concentration is by refractometry, as described below (note that home refractometers can be purchased for as little as $30 US):

A refractometer is a laboratory or field device for the measurement of
  an index of refraction (refractometry). The index of refraction is
  calculated from Snell's law while for mixtures, the index of
  refraction can be calculated from the composition of the material
  using several mixing rules such as the Gladstone–Dale relation and
  Lorentz–Lorenz equation.
  ...
  In homebrewing, a refractometer is
  used to measure the specific gravity before fermentation to determine
  the amount of fermentable sugars which will potentially be converted
  to alcohol.

So, in your case a "homebrewing" type of refractometer is probably your best bet. As long as sugar remains in solution, it will continue to dissolve until it reaches saturation.  This also means it's refractive index and specific gravity will continue to increase.  Once this increase is no longer observed, then the sugar is no longer dissolving, meaning you've either reached saturation or all of the sugar has dissolved.  Again, saturation would mean a thick, honey-like solution. Thorough stirring throughout the measurement process is a must.


text:
Chemical equilibrium and Le Chatelier's principle
$$\ce{energy + 2KClO3 (s) <=> 2KCl(s) + 3O2 (g)}$$
The reaction will be more active in the reverse direction if:

$\ce{O_2}$ is added
more $\ce{KCl}$ salt is added.
More $\ce{KClO3}$ is added
Increasing the temperature.

My thoughts:
I think (1) and (2) are both right. But the book states that only (1) works and I can't find reason for that searching google was useless.
$K_c = [\ce{O2}]^3$, so if we increased $\ce{KCl}$ or decreased $\ce{KClO3}$ it won't affect the equilibrium?

$$\ce{Fe(OH)(s) <=> Fe^{2+}  + 2OH^-}$$ The reaction will be more active in the reverse direction if : ... is added

$\ce{KOH}$
$\ce{Na_2S}   $
$\ce{Fe(OH)_2}  $
$\ce{Fe(NO3)2}$

Won't it be (1) and (4)? The book says it's (4) only.

Why do you think (1) and (2) are correct?

Can you write out the formula for the equilibrium constant?

Because if I added KCL le chattele ( never written his name right)  says the reaction will go in the direction which makes it get back to equilibrium and so will happen if we added o2.  maybe it's misunderstanding from me.

Ok ,    Kc = [0_2]^3

You're on the right track there, please consider adding your thoughts presented in your comments into your question.

Now I think I understand. When I add o2 I will increase the number of moles  and as a result the concentration will increase but the KCl is out of the Kc so it won't affect the equilibrium , right ? The problem is that my teacher told me that is wrong ...

I think your reasoning is sound,  (have upvoted as well)

Why do you think  KCl is out of the Kc? You're wrong...

Because it's solid (long explanation). H2O liquid is not counted too

@Mithoron perhaps expand on that into an answer?

Help , now he confused me more.

Bro, I think I am right check this : http://www.chemguide.co.uk/physical/equilibria/kc.html       (Writing an expression for Kc for a heterogeneous equilibrium)  You will see "The important difference this time is that you don't include any term for a solid in the equilibrium expression." this too http://bouman.chem.georgetown.edu/S02/lect8/lect8.htm

You are correct @HoytVolker - and the answer below provides a good explanation emphasising what you found.

But I don't understand what he does mean by oxygen pressure I think we are not playing with pressure , are we ?

@mamun's comment below explains what is happening

@HoytVolker I would suggest taking the 2nd question you edited in and writing that as a separate question, linked to this one.

Ok , thanks for the suggestion , I will. Ops , seems that I can't I have reached the limited 2 questions.

1) Increasing Oxygen pressure will increase the reverse rate because to reestablish equilibrium some of the oxygen should be consumed.
2) Out of system due to precipitation.
3) Out of system due to precipitation.
4) This is endothermic reaction. So increasing temperature will increase the forward rate to consume the added heat.

"Oxygen pressure" ??

Adding gas phase oxygen means increasing the partial pressure of oxygen.

Because we are increasing the number of atoms in Unit of volume ?  p(ro) = m/Vol

Yeah. Think about it from mathematical point of view: You have a constant K which has numerical value of 100. K is also equal to ${O_2^3}$. So that means ${O_2}$ is 4.64 bar roughly. Now you increase it to 5 bar and the equilibrium wants it to stay at 4.64 bar. So the equilibrium will consume 0.35 bar worth of oxygen molecule to reestablish it to 100.

A reference that helps explain what is happening with the oxygen is at http://chemwiki.ucdavis.edu/Physical_Chemistry/Equilibria/Chemical_Equilibria/The_Equilibrium_Constant/Effect_of_Pressure_on_Gas-Phase_Equilibria

Thanks I am checking it atm @santiago. Thanks too mamun maths explanation is better , but please check the new example it confused me.

authors: เพชรลัดดา บุดดี


text:
Clarifications about the mechanism of the Wittig reaction
Ok so this is the proposed mechanism for the Wittig Reaction as stated by my textbook, LG Wade. I have some doubts regarding this mechanism:

1) Why is only triphenyl phosphine used here? Why not some stronger
  nucleophile like say trimethyl phosphine?

  2) Why would the lone pair on P attack the C of the alkyl halide in the
  first place? Isn't the lone pair in resonance with the benzene ring?
3) Why is butyllithium used particularly.Can't any other reagent be
  used to abstract H ?
4) Why is the four membered ring formed in step 2?
5) Why does the ring collapse in step 3?

Note: I'm just a beginner in organic chemistry (Org 2) so it would be helpful if you can keep adjust your answer to suit my level of understanding.

It is preferred if you can post separate questions instead of combining your questions into one. That way, it helps the people answering your question and also others hunting for at least one of your questions. Thanks!

That would be making 5 posts on the same topic ! @Loong

related: http://chemistry.stackexchange.com/questions/5313/why-is-the-wittig-reaction-syn-stereoselective

Yes, but splitting the questions helps keep the answer chains clean.  It's perfectly all right to hyperlink among all the various questions to make it clear that they're part of the same broader search for understanding.

@SanchayanDutta - Multiple question posts is preferred. Think about it this way: What if I know the answer to questions 1) and 2) from your post, but  I do not know how to answer questions 3), 4), and 5)? I might not feel that I can post anything since I cannot answer all five parts. If you post them separately, then I can answer the ones I know how to answer and leave the others for people who know more than I do.

@SanchayanDutta - You can also think of it this way - More posts means more chances for everyone to earn rep, badges, privileges, etc.

Ok I understand your point @BenNorris.But think from the asker's perspective also.Chem SE allows one post per 30 mins.So it will take 30*5 mins for asking this question part by part.That means almost 2 and a half hours!!Thats really a taxing job.But,well as you are saying I'll try that from my next post.

1) Why is only triphenyl phosphine used here. Why not some stronger
  nucleophile like say trimethyl phosphine?

Trialkyl phosphines can particpate in the Wittig reaction just fine, but they create some other problems.  Let's say you were going to create the phosphonium salt from ethyl iodide and trimethyl phosphine.  It would form in high yield, but in the next step where you deprotonate it and form the ylide you would have a problem.  You could form the ylide from deprotonation of the desired ethyl group, or you could form an ylide by deprotonating any of the 3 methyl groups.  This would give rise to a mixture of ylides and consequently a mixture of products.  On the other hand, there are no protons on triphenyl phosphine that can be removed once the phosphonium salt is formed.
The following additional reasons also favor usage of triphenyl phosphine:

it is cheap
it is very cheap
it is a solid and easy to handle and weigh out, many of the smaller trialkyl phosphines are liquids
triphenyl phosphine is less toxic and doesn't smell as bad as many of the smaller trialkyl phosphines

2) Why would the lone pair on P attack the C of the alkyl halide in the
  first place? Isn't the lone pair in resonance with the benzene ring?

Triphenyl phosphine has the aromatic rings arranged in a non-planar, propeller-like geometry which makes delocalization less likely.  Also $\ce{2p-3p}$ overlap would be involved which is not as effective as $\ce{2p-2p}$ overlap.

3) Why is butyllithium used particularly. Can't any other reagent be
  used to abstract H?

In the phosphonium salt, the protons alpha to phosphorous are not very acidic (pKa ~ 20-35), a strong base is required to remove them.  Other strong bases could be used, but BuLi is commercially available in high purity at a reasonable cost.  When BuLi abstracts the proton butane is formed and escapes as a gas - one less side product to worry about separating out.

4) Why is the four membered ring formed in step 2?

The 4-membered ring may not be as strained as you imagine.  Remember, phosphorous is in the third row of the Periodic Table.  It has longer bond lengths and can tolerate smaller bond angles.

5) Why does the ring collapse in step 3?

The bond between phosphorous and oxygen is an extremely strong bond, in other words its formation is very exothermic.  Therefore, this last step is being driven by thermodynamics.  It is the exothermicity of this step that drives the reaction to completion.

Actually, $\ce{KO}t\ce{Bu}$ is often used for Wittig reactions in my group (phosphane usually being $\ce{PBu3}$) and my supervisor even said that it be way to strong — prompting me to try diazabicyclo[5.4.1]undecene (DBU) which worked like a charm on the test reaction. (Not on my product because my product is ugly, though.)

This is indeed a very good answer! Have you been gone a while? I don't remember reading something from you for at least two weeks... For your enjoyment, I found a quite comprehensive article about the Wittig reaction: Bruce E. Maryanoff , Allen B. Reitz, Chem. Rev., 1989, 89 (4), pp 863–927.. Especially the part about betaines and oxophosphetanes is interesting.


text:
Dieckmann condensation's final protonation step
This is the Wikipedia image for the reaction mechanism:

In the final protonation step, why does the enol tautomerize to the keto form?
In the enol form the carbon-carbon double bond is in resonance with the ester's carbonyl double bond. Also the enol's $\ce{-OH}$ group would form a H-bond.
Shouldn't the enol form be more stable/predominant?

They are in equilibrium - are you asking which tautomer is predominant?

http://chemistry.stackexchange.com/questions/18904/which-is-the-more-stable-enol-form

why does the enol tautomerizes to the keto form?

In 1,3-dicarbonyl systems, like the one pictured above, we would generally expect to have a measurable amount of the enol form present.  The keto-enol interconversion is an equilibrium and ratio of keto and enol will be guided by thermodynamic factors.  Let's take a look.
If we compare the structural elements in the keto and enol forms, we see that the differences between the two tautomers are as follows:

The keto form has an extra carbonyl (178), a $\ce{C-C}$ single bond (83) and a $\ce{C-H}$ bond (99)
The enol form has a $\ce{C-O}$ bond (86), a $\ce{O-H}$ bond (111) and a $\ce{C=C}$ double bond (146)

The numbers in parenthesis are standard bond energies in kcal/mol.  If we add up the keto values we get 360 kcal/mol; while for the enol it totals to 343 kcal/mol.  The numbers strongly favor the keto form (by 17 kcal/mol).  That is why is most simple systems the keto form strongly predominates.
However as you point out, both resonance and hydrogen bonding can further stabilize the enol form in the current example.
The hydrogen bond strength in $\ce{O..H-O}$ systems is typically 5-6 kcal/mol. In this system, with an intramolecular bond and a 6-membered transition state the hydrogen bond strength is likely closer to 9-10 kcal/mol. Resonance stabilization in 1,3-pentadiene is approximately 6.2 kcal/mol. Due to increased polarization in the enol from the carbonyl replacing a carbon-carbon double bond, the resonance stabilization probably increases to something around 8 kcal/mol.
A 9 kcal/mol effect due to hydrogen bonding and a 8 kcal/mol push due to resonance, shifts the 17 kcal/mol stabilization of the keto form to something very close to where the keto and enol forms have comparable energy. Therefore, in  1,3-dicarbonyl compounds we would expect a delicate balance between keto and enol forms, where both forms may be detectable.


text:
Are the bonding orbitals in methane equivalent - photoelectron spectrum
The low energy portion (the part dealing with the $\ce{2s}$ and $\ce{2p}$ electrons) of the photoelectron (PE) spectrum of methane is reproduced below.

(image source)
The reaction being examined is the following one-photon process
$$\ce{CH4 + ~h\nu~ -> [CH4]^{+} + ~e^-}$$
The fact that 2 peaks in roughly 1:3 intensity are observed is often used to argue against the $\ce{sp^3}$ hybridization model for methane. Simply put, how can a model with 4 equivalent $\ce{sp^3}$ bonds produce 2 peaks in the PE spectrum.  The argument then goes on to point out that, according to MO theory, methane has one triply degenerate bonding MO and a second single bonding MO and these two nonequivalent orbitals would explain the 2 peaks in the PE spectrum with their 1:3 ratio (see p. 827 here for an example of such an MO-based explanation).
The PE spectrum of methane shows 2 transitions with 1:3 intensity.  How does one know that it is the methane ground state that is responsible for producing two peaks with a 1:3 ratio in the PE spectrum?  Why couldn't one argue that it is $\ce{[CH4]^+}$, where we clearly have 3 $\ce{C-H}$ bonds containing a pair of electrons and one $\ce{C-H}$ bond with one electron - two different types of $\ce{C-H}$ bonds in a 1:3 ratio - that is responsible for the two peaks in the PE spectrum?
Can the PE spectrum of methane really be used as evidence to argue against the hybridization model?
EDIT:
Poking around on the internet, I came across this Wikipedia article on Orbital Hybridization - Photoelectron Spectroscopy.
It says,

"There is a popular misconception that the concept of hybrid orbitals
  incorrectly predicts the ultraviolet photoelectron spectra of many
  molecules. While this is true if Koopmans' theorem is applied to
  localized hybrids, quantum mechanics requires that the (in this case
  ionized) wavefunction obey the symmetry of the molecule which implies
  resonance in valence bond theory. For example, in methane, the ionized
  states (CH4+) can be constructed out of four resonance structures
  attributing the ejected electron to each of the four sp3 orbitals. A
  linear combination of these four structures, conserving the number of
  structures, leads to a triply degenerate T2 state and a A1 state.[16]
  The difference in energy between each ionized state and the ground
  state would be an ionization energy, which yields two values in
  agreement with experiment." (emphasis mine)

The bold face portion seems to support my contention (if I'm understanding it correctly) that it is the energy differences between the ground state and the excited state that is responsible for the spectrum.  Therefore, orbital splitting and orbital degeneracies in the excited state are likely an explanation for the observed spectrum.

That's not how I'd interpret the VB discussion in the book. If you want to preserve the idea of hybrid orbitals, then ssavec's answer is the one you want - cast the wavefunction into a linear combination of hybrid orbitals. But I disagree completely that $\ce{[CH4]+}$ is involved - it's an entirely different species with a completely different PES.

As the question gradually evolves, I would need more time to medidate on this topic. Although you do not measure spectrum of $\ce{[CH_4]^+}$ (which is different, as you pointed out in your comment), $\ce{[CH_4]^+}$  is the final state of the process (non particle conserving, alas). To be honest, I did not yet came across of rigorous enough QM description of this process (but I did not actually tried that much).

This question goes along the line of what does it mean when it is said that an sp3 orbital has 25% s-character. It also intrigued me so I have tried to find answer, which would not break my hybridised orbital view.
What you clearly see from the spectrum is that there are two bands, corresponding to different energy levels in the molecule and their intensity is 3:1. You can say, it is $3\times 2p : 1 \times 2s$ and you are done. Not at all. The most important thing to deal with is the $T_{d}$ symmetry of the molecule. That means, all four $\ce{C-H}$ bonds are equivalent.
But it is this symmetry which has the answer. Leaving out the $1s$ orbital of carbon, the occupied orbitals must have the following symmetries:
$1\times A_1$ and $3\times T_2$. And this is what you see in the spectrum.
Now how it comes with the hybridisation? Upon simple HF/STO-3G calculation, the orbital populations are following:

$A_1$: 0.62 C 2s + 4x 0.18 H 1s  ... fully symmetric, great
$T_2$: 0.57 C pz + 0.3 1H 1s + 0.3 2H 1s - 0.3 3H 1s - 0.3 4H 1s
$T_2$: 0.57 C py + 0.3 1H 1s - 0.3 2H 1s - 0.3 3H 1s + 0.3 4H 1s
$T_2$: 0.57 C px + 0.3 1H 1s - 0.3 2H 1s + 0.3 3H 1s - 0.3 4H 1s

Well, as if you would copy the symmetry table ;) The downside of this picture is that all the bonds are completely delocalised (as you would expect from canonical MOs). Although observable through PES, they do not allow you to draw line between any C and H atom.
Where do we get the hybridised orbitals? If we want to localize the orbitals, whatever scheme you choose, you just form linear combinations of the existing MOs so that you have least number of atoms involved in any bond. One of such methods is NBO and as a result you obtain:
C - H1  ( 52.55%)   0.7249* C    s( 25.00%)p 3.00( 75.00%)
        ( 47.45%)   0.6888* H  1 s(100.00%)
C - H2  ( 52.55%)   0.7249* C    s( 25.00%)p 3.00( 75.00%)
        ( 47.45%)   0.6888* H  2 s(100.00%)
C - H3  ( 52.55%)   0.7249* C    s( 25.00%)p 3.00( 75.00%)
        ( 47.45%)   0.6888* H  3 s(100.00%)
C - H4  ( 52.55%)   0.7249* C    s( 25.00%)p 3.00( 75.00%)
        ( 47.45%)   0.6888* H  4 s(100.00%)

All with 2 electron occupation and energy -0.63479 Eh.
So where is the truth? ... $\Psi$ is the answer and you can massage it to your will, depending what you want to see. You want localized bonds? You will get them. You want orbitals with the proper symmetry required by spectroscopy? As you wish.
Does the hybridisation concept have sense? In localized bond picture yes. Is it observable? No, and was never meant to be.

The short answer is that's not how photoelectron spectroscopy works. It's a one-photon spectroscopy.
You have a sample of $\ce{CH4}$ in this case, and you shoot different energy photons at the sample. At certain energies, the photon is absorbed and you get an electron off. Note, that's a single electron from one photon.
For you to have any significant amount of $\ce{[CH4]+}$ around, you'd have to hit it with two photons. The first would do:
$$\ce{CH4 + ~energy -> [CH4]^{+} + ~e^-}$$
Only then would you be able to hit that one single $\ce{[CH4]+}$ ion with another photon.
That's a two-photon process. Assuming the incident light intensity is small, that's extremely unlikely. It's much more likely that that second photon hits a different $\ce{CH4}$.
In other words, normal photoelectron spectroscopy gives a set of intensities when the species absorbs a single photon with enough energy to remove a single electron (i.e., orbital energies). This process happens on the femtosecond timescale, so it's essentially a "frozen orbital" technique. There's no time to sample the electronic structure of the resulting cation.
Also, the PES of $\ce{[CH4]+}$ is completely different (seen here), because of the Jahn-Teller distortion. The geometry is $C_{2v}$ and thus there are many more electronic transitions - it actually gets quite complicated.

Geoff, thanks for your answer.  I'm probably not being clear because I'm not very clear on my own question. I understand that PES is a one photon process.  What I'm asking is, doesn't the energetics or MO description of the product ($\ce{[CH4]^+}$ in this case) play some role in determining the energetics of the ejected electron?

@ron Nope. PES is a fast process. While it's possible to do time-resolved PES you need femtosecond pulses. The best description is the "frozen orbital" approximation. The electron is ejected before it can sample the ionized product.

@ron There are techniques where the ion matters. For example, in electrochemistry, such as cyclic voltammetry, the ionization process matters in determining the shape and peak potential.

I dug up a PES of $\ce{[CH4]+}$ which shows in Figure 11 the spectra is quite different.

To begin with, I am not really well-versed in this area, so would appreciate any input/corrections.
Wikipedia is correct on the matter; the photoelectron spectrum of methane is not a proof that hybrid orbitals are incorrect. The same can be said of water, ammonia, and so on. When one says that two peaks in the PES are observed and that it must correspond to two different levels, Koopmans' theorem is being implicitly invoked:
$$\mathrm{IP} = -\varepsilon_i$$
Here, $\mathrm{IP}$ refers to the experimental ionisation potential, and  $\varepsilon_i$ refers to the energy of a canonical orbital $|\chi_i\rangle$, which are obtained by solving the Hartree–Fock equations:
$$\hat{f}|\chi_i\rangle = \varepsilon_i|\chi_i\rangle$$
The problem with applying this theorem to hybrid orbitals is that hybrid orbitals are not eigenfunctions of the Fock operator; the hybrid orbitals $\phi_i$ are obtained by some unitary transformation of the canonical orbitals $\chi_i$.
$$\begin{pmatrix}
\phi_1 \\ \phi_2 \\ \vdots \\ \phi_i
\end{pmatrix} = \mathbf{U}\begin{pmatrix}
\chi_1 \\ \chi_2 \\ \vdots \\ \chi_i
\end{pmatrix}; \qquad \mathbf{U}^{-1} = \mathbf{U}^\mathrm{T}$$
and Koopmans' theorem does not apply after a unitary transformation, as has been explained in this question. So, the PES cannot be used as proof for the non-existence of hybrid orbitals, since the peaks in the PES cannot be linked to energies of hybrid orbitals.

How can hybrid orbitals explain the PES? Can they even do so? It is instructive to consider the "incorrect" way of doing it, first. This is adapted from Section 2.9 of The Chemical Bond – Fundamental Aspects of Chemical Bonding, edited by Gernot Frenking and Sason Shaik. Let's say we have four equivalent hybrid orbitals in methane, $\{\phi_i\}$ $(i = 1,2,3,4)$. Then the wavefunction of methane, ignoring core orbitals, can be written as a Slater determinant
$$\Psi(\ce{CH4}) = |\phi_1 \phi_2 \phi_3 \phi_4 \bar{\phi}_1 \bar{\phi}_2 \bar{\phi}_3 \bar{\phi}_4|$$
(no bar indicates spin up, a bar indicates spin down). Now, according to the wrong interpretation, one can remove an electron from any of these four orbitals. Without loss of generality we will remove a spin down electron. This supposedly would lead to four possible new wavefunctions:
$$\begin{align}
\Psi_1(\ce{CH4+}) &= |\phi_1 \phi_2 \phi_3 \phi_4 \bar{\phi}_2 \bar{\phi}_3 \bar{\phi}_4| \\
\Psi_2(\ce{CH4+}) &= |\phi_1 \phi_2 \phi_3 \phi_4 \bar{\phi}_1 \bar{\phi}_3 \bar{\phi}_4| \\
\Psi_3(\ce{CH4+}) &= |\phi_1 \phi_2 \phi_3 \phi_4 \bar{\phi}_1 \bar{\phi}_2 \bar{\phi}_4| \\
\Psi_4(\ce{CH4+}) &= |\phi_1 \phi_2 \phi_3 \phi_4 \bar{\phi}_1 \bar{\phi}_2 \bar{\phi}_3| \\
\end{align}$$
In Koopmans' theorem, the ionisation potential is identified as the difference in energy between the ionised wavefunction and the unionised wavefunction. The theorem states that this is simply the negative of the energy of the orbital from which the electron was removed.
The problem is that these ionised wavefunctions, $\Psi_i(\ce{CH4+})$ $(i = 1,2,3,4)$, are not acceptable wavefunctions for the molecular ion, as they do not respect the $T_\mathrm{d}$ symmetry of the molecular ion (the equilibrium geometry of the molecular ion is not $T_\mathrm{d}$, but PES is a "fast" technique and the geometry must remain the same during ionisation - in other words, Franck–Condon principle).
I can make an educated guess about what exactly "respecting the symmetry" entails, but I'm not 100% sure, so it will be the subject of a separate question, and I will edit it in here if/when it gets answered. Anyway, according to Hiberty and Shaik (who wrote the chapter), a group theoretical analysis will show that the acceptable linear combinations transform as $\mathrm{A_1} + \mathrm{T_2}$, which have different energies. Hence, the PES can still be rationalised using a hybrid orbital approach.
A refutation of the "failures" of VB theory is given in much more detail by Hiberty and Shaik, in Chapter 5 of their other book A Chemist's Guide to Valence Bond Theory. But I'll have to leave that to somebody else to describe, as I don't fully understand it myself.

Sounds like the sort of argument Professor Klapötke would make (that’s the guy with the tens of resonance structures for $\ce{H2}$). $+1$


text:
Burning formaldehyde gas for disposal?
Since formaldehyde gas is flammable, is there a safe way to burn it, and can this be done to dispose of the gas or neutralize its harmful effects (tear production, respiratory problems, etc.)?
This is a more detailed form of my earlier question, including the context of "gas disposal".


text:
How to simulate surface tension?
I am trying to create a water drop simulation for measuring hydrophobicity of surface. I don't know how to simulate the contact angle which is related by younges equation to the surface tensions for each of the solid-liquid-gas interfaces. I have come across some models for molecular dynamics, which for chemistry novice like me (I'm a EE) are bit advanced. I am looking for simpler method in which I could simulate the interfacial tensions by some sort of derived mathematical model which would be simple enough to act as an objective function to an evolutionary optimizer.

I'm not sure I understand. You know about Young's relation between the surface tensions. Do you want something that predicts the surface energies, or something that solves for the contact angle and the shape of the drop?

well I need the contact angle so I guess the latter would be what i'm looking for. However based on my understanding of the problem I will have materials which will create three interfaces (solid-liquid, solid-gas, liquid-gas) which I need to generate for my simulation. I am at a loss of how to generate these inter-facial tensions for calculating the contact angle. I realize in some situations the values for these tensions are from lookup tables, however I am not sure whether it would be wise to assume static tension values in a dynamic system (water drops on surface).

Unless there's something to perturb the system (e.g., rolling the drop down a surface at an angle) the drop will quickly come into equilibrium. If you don't have the surface tensions for a particular material, you'll need some sort of particle simulation.

well one of the objectives of the simulation is to generate data to feed to an evolutionary optimizer for a design program. The two main objective functions I have seemingly isolated for are 1.) Contact angle using the cassie-baxter model, and two 2.) the wetting angle (as you mentioned above). Below I have provided the a sample simulator for the hydrophillic case, and the paper which provides the mathematical model for the objective functions.
(Hydrophillic) Simulator|
Paper

If I understand your question correctly, you want to relate a droplet's geometry to it's environment. The curvature of a liquid droplet is described by the Laplace equation of capillarity, also known as the Young-Laplace equation. This is valid for both a sessile (sitting on a surface) and a pendant (hanging) droplet. I am assuming a static experiment in which the droplet is axisymmetric, with $z$ defined as the axis of symmetry. Such methods of solving for interfacial tension and contact angles are called Axisymmetric Drop Shape Analysis (ADSA) methods. ADSA methods are based on fitting the shape of an experimental droplet to the Laplace equation. This differential equation has been numerically solved in 1883, but thanks to computers and camera's this method is easier and more accurately to apply.
The Laplace equation is given by:
$$\gamma\left(\frac{1}{R_1}+\frac{1}{R_2}\right)=\Delta P\label{eq:laplace}\tag{1}$$
in which $\gamma$ is the surface tension and $\Delta P$ is the pressure difference over the interface. After some work, this may be written as a system of differential equations.
$R_1$ and $R_2$ are the principal radii of curvature.
$R_1$ is called the meridional curvature, which turns in the plane of the screen you are reading from.
Perpendicular to this is the azimuthal curvature $R_2$.
The meridional curvature is defined as:
$$\frac{1}{R_1}=\frac{\partial \phi}{\partial s}$$
The azimuthal curvature is defined as:
$$\frac{1}{R_2}=\frac{\sin{\phi}}{x}$$
`In the absence of external forces, other than gravity, the pressure difference is a linear function of the elevation' (Del Rio and Neumann, 1996):
$$\Delta P= \Delta P_0+\left(\Delta\rho\right)gz\label{eq:rio}\tag{2}$$
$\Delta P_0$ is the pressure at the origin, at which the droplet is not perturbed and $R_1=R_2=R_0$.
Using \eqref{eq:laplace} this leads to the following expression:
$$\Delta P_0 =\gamma\left(\frac{2}{R_0}\right)\label{eq:p0}$$
Combining \eqref{eq:laplace} and \eqref{eq:rio} and filling in the expression for $\Delta P_0$ and the curvatures leads to:
\begin{align}
\gamma\left(\frac{\partial \phi}{\partial s}+\frac{\sin{\phi}}{x}\right)&=\frac{2\gamma}{R_0}+\left(\Delta\rho\right)gz\\
\frac{\partial\phi}{\partial s}&=\frac{2}{R_0}+\frac{\left(\Delta\rho\right)gz}{\gamma}-\frac{\sin{\phi}}{x}
\end{align}
Now the coordinates are transformed to dimensionless coordinates:
\begin{align}
\bar{x}=\frac{x}{R_0}, \bar{z}=\frac{z}{R_0}, \bar{s}=\frac{s}{R_0}
\end{align}
This changes the differential:
\begin{align}
\frac{\partial\phi}{\partial s}=\frac{\partial\phi}{\partial\left(R_0\bar{s}\right)}=\frac{\partial\phi}{R_0\partial\bar s}
\end{align}
Multiplying both sides with $R_0$ gives:
\begin{align}
\frac{\partial\phi}{\bar s}&=2+\frac{\left(\Delta\rho\right)gR_0^2}{\gamma}\bar z-\frac{\sin\phi}{\bar x}\label{eq:phi}
\end{align}
Furthermore the following geometric interpretation leads to the remaining differential equations.
The arc length $s$ follows the curvature of the interface of the drop. $\phi$ is the tangential contact angle, so a change in $x$ over a change in $s$ is equal to the cosine of $\phi$:
\begin{align}
\frac{\partial x}{\partial s} &= \frac{\partial \bar{x}}{\partial \bar{s}} = \cos(\phi)\label{eq:cos}
\end{align}
 Similarly, a change in $z$ over a change in $s$ is equal to the sine of $\phi$:
\begin{align}
\frac{\partial z}{\partial s} &= \frac{\partial \bar{z}}{\partial \bar{s}} =\sin(\phi)\label{eq:sin}
\end{align}
The same holds for the dimensionless coordinates, resulting in the given system of differential equations.
Solving the curvature of the droplet is an initial value problem, with the initial values being $x_0$, $z_0$ and $\phi_0$. You can feed this to your numerical differential equation solver, e.g. Matlab's ODE45 function. With this method you will be able to solve the contact angle and surface tension from a given curvature, or draw a curvature for a given contact angle and surface tension.
Suggested reading on which I have based my answer:
Y. Rotenberg, L. Boruvka, and A. W. Neumann, “Determination of surface tension and contact angle from the shapes of axisymmetric fluid interfaces,” Journal of Colloid and Interface Science, vol. 93, pp. 169–183, May 1983.
Rı́o, O. I. del, and A. W. Neumann. “Axisymmetric Drop Shape Analysis: Computational Methods for the Measurement of Interfacial Properties from the Shape and Dimensions of Pendant and Sessile Drops.” Journal of Colloid and Interface Science 196.2 (1997): 136–147. ScienceDirect. Web.


text:
Thermodynamics of photochemical reactions
Suppose I have a photochemical reaction in gas phase, such as
$$
\ce{CO2 + {$h\nu$} -> CO + O}.
$$
I would like to work through the thermodynamics of such a reaction and understand the meaning of every term.
For a non-photochemical reaction, such as $\ce{CO + O -> CO2}$, I have a good understanding of how the change in Gibbs energy breaks down into enthalpy, internal entropy and a log-concentration term corresponding to the entropy of mixing. If I ignore the photon in the first reaction above, I can regard it as simply the reverse of this reaction, which due to its photochemical driving force is able to reduce the Gibbs energy by the same amount, $\Delta G$. All of this is unproblematic.
However, $h\nu$ itself is an energy change, which I can calculate, assuming I know the frequency $\nu$ of the absorbed light. I would like to know how to think about the thermodynamics of the whole system, including the coupling of the chemistry to the radiation field. In particular, my questions are:

How does the energy $h\nu$ relate to the other terms with energy units, such as $\Delta H$ and $\Delta G^0$?
I know that while a single photon has no internal entropy, a beam of light of a particular frequency can be thought of as having a temperature and an entropy. How should I think about the role of radiation entropy in the thermodynamics of photochemistry, and in particular, how can I do second-law calculations for photochemical reactions?

Not sure what you mean by second-law calculations.  Can you clarify?

@JanJensen sure - I mean either of the following two closely related things: (i) how can I calculate the total rate of increase of entropy in the universe, including the entropy of the incoming and outgoing radiation fields, per mole reacted; or (ii) for a reversible photochemical reaction, what are the conditions for the chemical system to be in thermodynamic equilibrium with the radiation field?

Photochemistry doesn't work by thermally exciting the reacting molecules.  You can't simply convert $h\nu$ to energy or entropy and treat it as a thermodynamic variable

Your first statement is true but the second is a non sequitur. $h\nu$ is an energy. $E=h\nu$ is the equation for the energy of a photon, that's why we write it that way. It's completely uncontroversial that the heat given off by a photochemical reaction is $h\nu-\Delta H$. But that's only the first-law energy balance. I'm asking how to do the corresponding second-law calculations, that's all.

What I am saying is that the $\Delta H^\circ$ standard value you will measure for the photochemical reaction will be different in magnitude from that measured for the thermal reaction, and will not correspond to a thermodynamic enthalpy change.

PS.  Having said that, in your hypothetical example I think the answer you are looking for is this: If $h\nu-\Delta H^\circ$ worth of energy is transferred reversibly to the surroundings at temperature $T$ then the corresponding energy change is $\Delta S = \frac{h\nu-\Delta H^\circ}{T}$

$\Delta H^\circ$ is calculated as $\Delta_f H^\circ_\text{products}-\Delta_f H^\circ_\text{reactants}$. It doesn't depend on the reaction mechanism - this is what it means to be a state function - so the enthalpy change is the same for the photochemical and the "thermal" reaction by definition. The formula in your other comment does not take account of the radiative entropy. It is a correct calculation for the entropy increase due to heat given off, but it does not include the (presumably small) entropy decrease due to absorbing the light.

I have only just seen this post so by now you have probably worked it out. If not look at G. Porter Journal of Chemical Society, Faraday Transactions 2, 1983, vol 79 , p 473-482 which discusses the thermodynamics of photochemical reactions. You can also find the article in the book 'Chemistry in Microtime' by the same author.

See also https://chemistry.stackexchange.com/questions/144496/is-light-a-reactant-in-photochemical-reactions/144509#144509

$h\nu$ is not a thermal energy so thermodynamic considerations do not apply. For example, you cannot predict the probability of being in an electronic excited state from equilibrium thermodynamics.  The main reason is that the excitation is not a result of (classical) thermal energy transfer, but rather the adsorption of a photon which changes the quantum state.
With a constant light source
$$\ce{CO2} + h\nu \ce{-> CO + C}$$
will reach a steady state with constant concentrations that will be different from the thermal equilibrium concentrations.  However, you can not extract a standard free energy difference by
$$ \Delta G^\circ=-RT\ln \left( \frac{p_{\ce{CO}}p_{\ce{O}}}{p_{\ce{CO2}}} \right) $$
because the products and reactant are not in thermal equilibrium.

Thermodynamic considerations do apply to electromagnetic radiation though, the most basic results being the Stefan-Boltzmann law and the Planck formula for radiation entropy. These are for black body radiation, but formulas also exist for the entropy and temperature of light of a specific frequency, intensity and polarisation.

yes, but they still refer to the heat equivalent of the energy of light. Electronic excitation is fundamentally a quantum phenomenon, which is why only certain wavelengths work for certain systems.


text:
Calculating the concentration of copper(II) from a non-standard cell potential

A galvanic cell consists of a standard hydrogen electrode and a copper electrode.
  \begin{align}
  \ce{2 H+ (aq) + 2e- &-> H2 (g)}&
  E^\circ &= 0.00\ \mathrm{V}\\
  \ce{Cu^2+ (aq) + 2 e- &-> Cu (s)}&
  E^\circ &= 0.34\ \mathrm{V}
\end{align}
  Suppose the copper electrode is placed in a solution of unknown copper (II) concentration. The cell potential is measured to be $\pu{0.195 V}$. Determine the concentration of $\ce{Cu^2+}$.

Since there's no indication that the $\ce{Cu^2+}$ is in the standard state, I started off by using the formula for non-standard cell potential:
$$E=E^\circ-\frac{RT}{nF}\ln Q$$
However, I can’t seem to get the right answer. I was wondering if I set up the problem incorrectly or how I go about getting the concentration of copper.

You are on the right track with that equation. There is a problem with the question, though: you do not know which direction the reaction is proceeding in. Therefore, you get two possible answers.
One of the two possible ways to write the reaction that's occurring is
$$\ce{Cu^2+(aq) + H2(g) -> Cu(s) + 2H+ (aq)}$$
For this reaction, the value of $E_\mathrm{cell}^\circ$ is given by $E^\circ(\ce{Cu^2+}/\ce{Cu}) - E^\circ(\ce{H+}/\ce{H2}) = \pu{+0.34 V}$. This seems to be the more logical direction for the reaction to proceed in, as we have a positive value of $E_\mathrm{cell}^\circ$. Consequently, to achieve the smaller potential difference of $\pu{0.195 V}$, all we need to do is to reduce the concentration of $\ce{Cu^2+}$ from the standard of $\pu{1 mol dm-3}$, so that the forward reaction becomes slightly less favourable. The corresponding value of $n$ is $2$. The reaction quotient can be approximated by
$$Q = \frac{([\ce{H+}]/c^\circ)^2}{([\ce{Cu^2+}]/c^\circ) \cdot (p(\ce{H2})/p^\circ)} = \frac{c^\circ}{[\ce{Cu^2+}]}$$
since the $\ce{H+}/\ce{H2}$ electrode is operating under standard conditions (and where $c^\circ = \pu{1 mol dm-3}$). The temperature is not specified, but we could assume it to be $\pu{298 K}$.
Therefore:
$$\begin{align}
E_\mathrm{cell} &= E_\mathrm{cell}^\circ - \frac{RT}{nF}\ln Q \\
\pu{0.195 V} &= \pu{0.34 V} - \frac{(\pu{8.314 J K-1 mol-1})(\pu{298 K})}{(2)(\pu{9.649 \times 10^4 C mol-1})} \ln\left(\frac{c^\circ}{[\ce{Cu^2+}]}\right) \\
[\ce{Cu^2+}] &= \pu{1.246 \times 10^-5 mol dm-3}
\end{align}$$

If the reaction is actually running in the reverse direction, then you have the reaction
$$\ce{Cu(s) + 2H+(aq) -> Cu^2+(aq) + H2(g)}$$
with $E^\circ_\mathrm{cell} = \pu{-0.34 V}$. Going through exactly the same steps as above (but taking care to invert the expression for $Q$, since the reactants and products are swapped), you get to
$$\begin{align}
E_\mathrm{cell} &= E_\mathrm{cell}^\circ - \frac{RT}{nF}\ln Q \\
\pu{0.195 V} &= \pu{-0.34 V} - \frac{(\pu{8.314 J K-1 mol-1})(\pu{298 K})}{(2)(\pu{9.649 \times 10^4 C mol-1})} \ln\left(\frac{[\ce{Cu^2+}]}{c^\circ}\right) \\
[\ce{Cu^2+}] &= \pu{8.019 \times 10^-19 mol dm-3}
\end{align}$$
What's happening here? It turns out that as you reduce the concentration of $\ce{Cu^2+}$ further, $E_\mathrm{cell}$ drops more and more, until it actually becomes thermodynamically favourable for the reverse reaction to occur. This concentration isn't particularly sensible, but you can certainly achieve it by serial dilution.

How do we resolve this dilemma? The most straightforward way is to find out which electrode is the anode and which is the cathode, which you can do with the voltmeter. For voltmeters to display a positive reading, as is suggested by the question, you need to have electrons flowing from the black wire to the red wire through the voltmeter. This means that the black wire must be connected to the electrode which provides it with electrons, i.e. the anode, since the anode is the site of oxidation (the chemical species gives up its electrons to the electrode and is itself oxidised in the process).
So, if your voltmeter displays a positive reading and the black wire is connected to the standard hydrogen electrode, it means that $\ce{H2}$ is being oxidised to $\ce{H+}$. That means that we are in the first situation above, where the forward reaction $\ce{Cu^2+(aq) + H2(g) -> Cu(s) + 2H+ (aq)}$ is occurring.


text:
Naming double bond sidechain

In IUPAC naming double bonds take preference over an alkane chain of any length, right? So the above should be named as 2-ethylnon-1-ene.
However, when I verify with chemdraw it gives the answer as 3-methylenedecane. I cannot seem to find any mention of double bond side chains as alkylene anywhere on the internet... Is this common nomenclature, and if not why would chemdraw name it as this?

Chemdraw is almost deliberately wrong for just this reason. "Cheating"...it is robust with trivial names and other unorthodox naming systems. They are technically correct however but you would be better off learning the rules you learned in class.

According to IUPAC rules first thing you need to see is longest carbon chain which is 10(deca) over here.
Then while numbering the Carbon double bond is given preference.
So above compound is 10 carbon chain in which 3rd carbon is double bonded.
Thus, 3-methylenedecane is correct name.

Few more such examples to make it more clear:

2,6,10-trimethyldodec-1-ene

8-ethyldec-1-ene

Example Source: ChemSink

Basic IUPAC rules:

Find the longest chain in the molecule and name it.
Name all groups attached to the chain as alkyl substituents.
Number the carbons of the longest chain beginning with the end that is closest to a substituent.
Write the name of the alkane by first arranging all of the substituents in numerical order (each preceded by the number of the carbon to which it is attached and a hyphen), and then add the name of the stem.
Haloalkanes.
The halogen is treated as an alkyl substituent, without any priority.
Ethers are named as alkoxyalkanes. The first carbon of the alkoxy substituent is numbered as C-1 of that substituent.

Note that these rules are changed  as of 2008. Older HW and exams have slightly different answers.
Rules Source: http://valhalla.chem.udel.edu/nomenclature.html

It looks like you're correct.
This molecule has been given 2-ethyl-1-hexene or 2-ethylhex-1-ene

Another name for your molecule would be 3-methylidenedecane.
Added from Wikipedia:
The trivial name carbene is the preferred IUPAC name.
The systematic names methylidene and dihydridocarbon are constructed according to the substitutive and additive nomenclatures, respectively.

So do double bonds do not precedence over the longest chain length?

I'm sure somebody will dig out the IUPAC guidelines to give you the definitive answer, but I just wanted to point out another possible name.

UC davis wiki still uses the old recommendations. The longest chain is now the first criterium.