text:
Fractional Subscripts?
What is the physical meaning when we have an empirical formula with fractional subscripts such as $\ce{LiMn_{1-x}Fe_{x}PO4}$ $(0 \le x \le 1)$?
This an example of the paper the notation above is used:
Bai, J.; Hong, J.; Chen, H.; Graetz, J.; Wang, F. J. Phys. Chem. C 2015, 119 (5), 2266–2276. DOI 10.1021/jp508600u.

These indicate "doping" or non-stoichiometric compounds.
In this case, there's some fraction of Fe, and some fraction of Mn in a lithium iron phosphate battery. If I remember correctly, you need to dope a different ion to increase the conductivity of $\ce{LiFePO4}$.
See also here: How are non-Stoichiometric compounds determined?

I think it is a bit more than that: the nomenclature here is used when the ($Mn$,$Fe$) site in the lattice is fully occupied, and $x$ is telling you how many have $Fe$ vs $Mn$ in the sample. This is indicating that there is a single phase across $LiMnPO_4$ to $LiFePO_4$, trading out $Mn$ for $Fe$.  Non-stoichiometry ($Mn$+$Fe$ isn't 1) would be indicated by a $\delta$.  If $Mn$ did not substitute perfectly for $Fe$, it might be shown as $LiMn_{1-x-\delta}Fe_{x}PO_4$.

Yes, that's a useful clarification. I agree that in this particular case, there's a single phase and x vs. 1-x is giving you a ratio of $\ce{Fe}$ to $\ce{Mn}$ in this compound. However, I think most people would consider this case as a part of the broader subset of non-stoichometry, since there's no simple integer to define the formula. The OP's title question referred to the general question of fractional subscripts, which does go to non-stoichiometry.


text:
Mechanism of hydrolysis of polycyclic oxetan
The reaction is this:
The only idea I have thus far is if we have a chloride acid that can form an ester, then we might be able to find a suitable mechanism. This would require a chloride reagent.

You'd think it all starts with protonation at the oxetane oxygen.

You might want to look up non-classical carbocations.

Oh, I'd also suggest making a 3D model of the species, either physically using a modeling kit or using a molecular visualization tool.

And look at Wagner-Meerwein rearrangements, too!

I have drawn a mechanism but I haven't checked for any references as I am at home now, so it's more like paper chemistry. The chlorine atoms should play some role and possibly disfavour formation of certain carbocations that you would expect in a standard Wagner - Meerwein. It is possible to draw a concerted mechanism (not shown here) but seems very unlikely
**EDIT: I have changed my initial mechanism so it explains better the methyl ester formation (see the comments below). The stereochemistry at C* has also been corrected.
As a general comment you cannot propose a non trivial mechanism without drawing one, it is handwaving. So I did.

Can you comment on your thought process? Much appreciated.

Is there a specific step or steps that you are asking about? In general I tried to use the most stable (looking to me) intermediates and ... get there. Remember also that this is only a suggestion not based on any literature. I did have a quick look today but couldn't find anything relevant, if you or anyone else have any references please post

@ K_P: I don't think step 3 is correct; you show methanol attacking the oxocarbenium ion, but the reaction takes place in water/sulfuric acid

I don't love this either but you have to end up with the methyl ester somehow, and the most reasonable source is MeOH produced by the initial hydrolysis of the dimethyl ketal. You could probably make it more "believable" by a mechanism that involves an intramolecular attack by the methoxy. I'll have a look later as I'm on my mobile now.

Considering your second proposed mechanism, I find the 3-membered ring closure unlikely. Essentially it's a cationic enol ether, which shouldn't be nucleophilic, attacking a carbocation. Put another way, why doesn't water just attack the carbocation directly instead of through an alkene?

Also, I think after the enol ether O-acylates, the intermediate will be very strained due to the bridgehead pi-bond. You could build a model to check it out.

@jerepierre : It's true the intermediate as drawn looks ridiculously strained. You need to imagine it as the "sum" of resonance forms where the pi system is stabilising the methyl carbocation thus leading to the cyclopropane ring formation. Still you possibly need to rearrange some steps eg have the cyclopropane ring forming before the lactone intermediate. The problem is that Abel's more "natural" alternative approach didn't give the desired product at least in my hands.

I haven't looked at the literature either, but it seems to go like this:

protonation at the oxetane
ring-opening of the oxetane. This gives a tertiary carbenium ion and the keto group (elimination of HCl from the $\alpha$-chloroalcohol).
skeletal rearrangement to yield the dioxocarbenium ion; the resonance stabilization is what provides the driving force
nucleophilic attack of solvent, to lose a methyl group and form the methyl ester

It would help if you post a diagram. A picture is worth a thousand words.

@ron: I know.  It's difficult to draw the polycyclic system in an aesthetically pleasing way.

@AbelFriedman please post a picture of your mechanism. I think starting at the oxetane makes sense, but I can't see out how a rearrangement (unless it's more than one) that gives a dioxocarbenium ion also gives the correct carbon skeleton.

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text:
Describe the bonding in a tribromide anion, what is its structure?
I'm thinking it will be linear with a negative charge on the middle bromine. Someone told me it is trigonal bipyramidal geometry with 3 lone pairs around the central bromine and the two other bromines above and below making it linear. Is this correct? Also is it just covalent bonding in the molecule?

Try looking up the structure of the triiodide ion.

In fact, tribromide ion has the generic formula $\ce{AX2E3}$ according to Lewis diagram. Where  $\ce{A}$ is the central atom, $\ce{X}$ an atom bonded to $\ce{A}$ and $\ce{E}$ a lone pair on $\ce{A}$.  Using Valence Shell Electron Pair Repulsion (VSEPR) theory,  the geometry  of this ion will be
 :

So, the molecular shape is linear, while the electron pair geometry is trigonal  bipyramid.
 And the bonds in the ion are covalent.

While VSEPR concludes on the correct geometry, the way it got there is just plain wrong.

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text:
Can Ice Packs Freeze Water at 21 °C?
I receive $120~\mathrm{mL}$ bags of medicine in sterile water. The medicine is refrigerated and then packed with a significant amount of ice packs in an insulated bag. Generally three times the volume of all the medicine. When I remove the ice packs, they are still solid two hours after being packed. So, they are very cold.
Is there enough energy to freeze the water directly in contact with the ice packs when packed and then melt immediately? The air temperature is around $21~\mathrm{^\circ C}$ at the time.

It really depends on the initial temperature of the ice packs and their composition. If they start at a temperature far below the freezing point of water, they could conceivably freeze the water as their container comes to thermal equilibrium. That said, they’re going to have to be quite cold.
To cool by $1°\mathrm{C}$, liquid water requires the removal of roughly double the amount of energy that ice would give up in moving up $1°\mathrm{C}$ (per gram.) If your bags of medicine started at $5°\mathrm{C}$, (stored in a refrigerator) then the ice will warm by about $3°\mathrm{C}$ as it brings the medicine down to its temperature because it has three times the mass and gives up half as much heat per degree gram.
Once the water has reached $0°\mathrm{C}$, it will begin freezing. For every gram of frozen water, about $160~\mathrm{g}$ of ice will warm by $1°\mathrm{C}$. The entire mass of ice will be enough to freeze ~$2.3~\mathrm{g}$ of water for each degree it is below freezing.
Assuming the ice packs came from a normal freezer, their starting temperature is about $−20°\mathrm{C}$. They would warm to about $−17°\mathrm{C}$ in bringing the water to its freezing point. From there the ice could then freeze about $40~\mathrm{g}$ of the medicine before it all came to equilibrium.
Note: There are significant assumptions (notably, that the heat capacity of the ice pack is equal to that of ice and that its melting point is equal to $0°\mathrm{C}$) and some healthy rounding here. If the ice packs melted at a temperature significantly lower than $0°\mathrm{C}$, their heat of fusion could be sufficient to freeze the water, given time.


text:
Dipoles and determining molecular (a)symmetry
What is a dipole? Also, how do you know if a molecule is symmetrical or asymmetrical?
For example, are the following symmetrical or asymmetrical? Can you tell just from the molecular formula? Which ones have dipoles?

$\ce{O2}$
$\ce{H2O}$
$\ce{CO2}$

Are these concepts related?

Take a look at something like http://chemwiki.ucdavis.edu/Physical_Chemistry/Quantum_Mechanics/Atomic_Theory/Intermolecular_Forces/Dipole-Dipole_Interactions, please and then edit your question to address the concepts that you still don't understand.

I'm nominating to reopen so it can be closed again as a duplicate of https://chemistry.stackexchange.com/q/1107/194.

You should easily find the answers by searching the web.
A dipole is the separation of a positive charge and a negative charge in a molecule. Any 2 different atoms bonded together will have a dipole since their electronegativities are different.
A symmetrical molecule will either have a mirror plane or a rotation that will superimpose the atoms. Benzene has both a mirror plane and a rotation of 60°.
An asymmetrical molecule does not have a mirror plane or rotation. Ethanol is an example.


text:
What makes protons give the property of acidity?
Brønsted Lowry definition of an acid says that an acid gives away protons.
What I don't understand is what makes dissolved protons grant the property of acidity? Like, why would dissolved protons eat up materials?
My question applies to bases too, I suppose. And I guess you could include Lewis-acids and bases in your argumentation...
It's a question that has been bothering me for long. By the way I am not chemist(nor do I study chemistry).

that's due to hydration energy

Does this answer help at all?

Whether something is a Brønsted acid is defined by its pH.
$$pH = -log_{10}[H^+]$$
In other words, the better a substance is at releasing protons, the higher the concentration of protons [H+], and correspondingly the lower its pH.
Whereas for Lewis acid, it is the ability for it to strip electrons from anything it attacks (electrophilicity). There is no direct measurement for the strength of Lewis acidity, as many factors have to be taken into consideration, such as enthalpy change, steric factors. In fact, many substances can act as Lewis acids can also act as Lewis bases, such as water, ammonia, and even some alkanes.
Ultimately, what makes acids and bases so corrosive is its ability act as an oxidant or reductant.
Oxidation: Lose electrons/Gain hydrogen.
Reduction: Gain electrons/Lose hydrogen.

While it is true that oxidizing strong acids like nitric and sulfuric are more corrosive than a non-oxidizing strong acid like hydrochloric, the latter still is able to dissolve materials (e.g, metals, minerals, proteins) and cause burns to the skin. A strong acid like HCl can (in the case of human skin or other biological materials) catalyze the hydrolysis of proteins, fats, and carbohydrates down to their water-soluble components. This is what happens in your stomach, where there is an abundance of hydrochloric acid, as well as enzymes to speed up the process.
The most widely accepted measure of how strong an acid is, its its acid dissociation constant, not the pH of a solution of it.


text:
Freeware that converts structure in a conformational model?
In order to draw chemical structures I use ChemBioDraw Ultra.
As an exercise I've to draw the most stable conformation.

I expect a twist-boat like structure (instead of usual chair conformation for substituted dioxanes)
Is there a freeware or an easy computer online tool to check if my propose (right side picture) is correct?
When I go to >Edit > Get 3D Model the program gives me a small picture with a 3D-Model:

By double-clicking this small image it opens ChemBio3DUltra and I can rotate this molecule:

@Martin: Is this correct (I've expected a twisted cyclic ring)?
@Martin I draw the structure at right with i-Bu and H exchanged in Chemdraw and do >Edit > Get 3D Model

The reason why I first thought it should be twist boat like was that one C-O is pointing up and the other pointing down. May be someone knows why people gives this stereoinformation inside the cyclic ring?

As I've drawn it as a twist boat I get exactly the same with ChemBioDraw. Is there a way to calculate by how many kcal/mol the first one is more stable then the second one with ChemBioDraw?
What's interesting: I searched for a structure in which the twist-boat conformation is the most favorable one and found one:

However, when there are two large substituents in the cis-1,4 arrangement on
  a cyclohexane ring, neither of which will go easily into an axial position, then
  it appears that the twist-boat conformation (Section 12-3A) is most favorable
  [http://authors.library.caltech.edu/25034/13/BPOCchapter12.pdf, page 458]

(The A-value of a tert-Butyl is in the range of 5 kcal/mol which is around the value of the energy difference between cyclohexane in twist-boat and chair conformation.
I've drawn cis-1,4-di-tert-butyl-cyclohexane with ChemDraw and used >Edit>Get 3D Model and ChemBioDraw gave me the structure which is NOT the energetical most suitable one! Why?

If you drew it with ChemBioDraw already, why don't you also check it with the Chem3D implementation?

Well the program might only give you its best estimate, which is clearly basing it on chair conformations. So you might have to edit your molecule to be a twisted conformer.// On a second view, your stereochemistry in the right hand picture is at least ambiguous, if not wrong (as it says R, S).// I also would not have expected a twisted conformer, but I can be wrong on this one.

You can use programs like Avogadro to get the molecular mechanics energy of different force fields. I believe Chem3D also supports MMFF or MM3 or similar force field evaluations. I agree with @Martin that you will need to test the twist-boat conformation and you will likely need to edit the 3D structure for that.

Is whatever forcefield you are using correctly parametrized for cyclic acetals?  If you have access to the CSD, why not use actual data?

For your recent update (i.e., for cis-t-butyl) I don't know how ChemDraw and Chem3D generate the 3D structures. That's a tech support question. I doubt they do a significant force field minimization or test multiple ring conformations. Very few programs evaluate ring conformers automatically.

Short answer: yes, the dioxane has a minimum when it's in a twist-boat form.
I use Avogadro which offers a "auto-optimize" tool. You can grab atoms and throw them around to force rings into different conformations while the force field is minimizing.
So here's what I did.

File -> Import -> Fetch by chemical name: (2R,4R)-2-isobutyl-4-methyl-1,5-dioxaspiro[5.5]undecane. This grabs the 3D geometry from the NIH Resolver (i.e., PubChem)
Auto-Optimize Tool: Force Field MMFF94

Then I started minimizing and looking at the energies. I pulled various atoms around to try to minimize steric clashes and flip rings from chair to boat, etc.
For the figures that follow, I've removed the hydrogens for clarity, but they exist as far as the energies and minimization. I just removed them to save the graphic.
Chair/Chair: 24 kJ/mol

Twist-boat/Chair: 14 kJ/mol

So yes, there's a decent stabilization in the twisted form of the twist-boat. You have to get the hydrogens right, or you'll end up with a bunch of eclipsed interactions between the dioxane oxygens and hydrogens on the cyclohexane.

This is a very useful description! When I optimize I get a value of 402.258 kJ/mol for the optimal configuration. Is there a way to directly see the 24kJ/mol or have you used a calculator? "Chair/Chair: 24 kJ/mol" means that this chair is 24 kJ/mol higher in energy as the normal cyclohexane chair?

What force field are you using? The default is UFF, but I switched to MMFF94 which is designed for organic molecules. UFF is less accurate but defined for all elements and strange geometries.

I used MMFF94. See my screenshot: https://www.dropbox.com/s/gz6jyz5raezlc4u/avogadro.PNG?dl=0

I'd be happy to move this to chat. I don't see hydrogens, and your structure is completely planar, so of course the energy is higher.

Thank you. I'd like to move this to chat too, but how? I just imported by name, clicked AutoOptimization, MMFF94 and pulled on some bonds still the value gets smaller. Is there a easy way to know which bonds I have to pull? Now I recognized that for optimization I have to display the hydrogens. Now I recognized that for optimization I have to display the hydrogens. I got a value of 12.6463 kJ/mol which makes more sense.

Let us continue this discussion in chat.

It's tough to see in your Chem3D picture (try turning off display of H's), but the dioxane is in a chair conformation. Truthfully, I don't think you need a 3D package to determine this, although using the energy minimization can confirm your hypothesis.
My approach is to assume that the dioxane is going to be in a chair conformation. I'm much better at converting six membered rings to the chair if I start with the ring completely in the plane. In this case, that means redrawing the original so that the oxygens are in the plane and the isobutyl and methyl groups are viewed as substituents off the dioxane (left side of figure). Now it becomes clear that the substituents should be in a trans-1,3 relationship. (Note that you have an error in your ChemDraw pics that these groups are cis. The R/S designation even changes at the isobutyl group.) Next, I draw a chair and place the oxygens. I like to number the atoms to avoid mistakes, although many people are better at this than me and can skip this step. In my first chair drawing, I see that the isobutyl group is placed axially, and interacts with the fused cyclohexane. Going through a ring flip puts the isobutyl equatorial and the methyl axial. There's no way to avoid 1,3-diaxial interactions completely, so I would predict that the right chair structure is the most favorable.

Where the 3D packages can really help is to see how much the chairs are distorted toward a twist conformation, but for a first guess, I think the hand drawn structures are useful.

Thank you for this very helpful answer! I guess that the equilibrium is only a very tiny shifted towards the right one because i-Bu is like a Me because of avoiding interaction of the i-Pr attached on the first carbon atom.


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Oxidation of formate ion by peroxydisulfate in water solution: Why are there "strange" kinetics?

As you can see, the first step is the slowest, but by using the rate-determining step approximation you wouldn't arrive at the correct rate law which is:
$r=k\ce{[HCOO^{−}]}^{1/2} \ce{[S2O8^{2−}]}$.
The first reaction is very slow, so most of the peroxydisulfate is consumed in the third reaction. The correct rate law can be obtained by applying the steady-state approximation to the two radical species.
WHAT does this mean?
So, why is the rate constant NOT according to the slowest step?
Why it is not just $k\ce{[S2O8^{2-}]}$?

I think you already have got an answer to your question here.

@Philipp
No but I am m=not understanding the meaning of that line

The rate of the third reaction affects the rate of the first reaction, i.e. they are competitive reactions and in that case the rds method can't be applied.
The rate of the first reaction is $r=-\mathrm{d}\ce{[S_2O_8^{2-}]}/\mathrm{d}t$, but the change of the concentration of the peroxydisulfate depends on the third reaction, too:
$$-\mathrm{d}\ce{[S_2O_8^{2-}]}/\mathrm{d}t=k_1\ce{[S_2O_8^{2-}]} + k_3 \ce{[CO_2^-]}\ce{[S_2O_8^{2-}]}$$

In this answer, I will attempt to explain using chemical principles, rather than detailed mathematics.
As has been said already, reactions 1 and 3 both consume $\ce{S2O8^{2-}}$. Now, for the rate determining step to dominate the decay of this species, there would necessarily be very little consumption of it in any other steps. IE, there would be little consumption of it in reaction 3, implying that there is very little supply of the $\ce{CO2^-}$ radical.
Now, unfortunately for the complexity of the kinetics, there is a rapid interconversion of $\ce{SO4^-}$ to $\ce{CO2^-}$ by reaction with formate, which in turn rapidly reacts with more $\ce{S2O8^{2-}}$ to reform the $\ce{SO4^-}$ which can react again with formate, you get the picture. So, although there will only be a very small concentration of $\ce{SO4^-}$ and $\ce{CO2^-}$ present (by virtue of the slow reaction in step 1), the sheer relative speed of formate induced interconversion makes these other steps important. Of course, the more formate we have, the more $\ce{CO2^-}$ we can liberate (to a point, hence the square root), and the faster we can remove $\ce{S2O8^{2-}}$. This is, in fact, a catalytic cycle within the greater reaction scheme.
This is what is meant by the steady state approximation. As the concentrations of $\ce{SO4^-}$ and $\ce{CO2^-}$ are always very small, we can assume that their total rate of change across all reactions in the scheme is (extremely close to) zero. This allows us to write a set of simultaneous equations from which we can try to determine the (often quite complex) rate law by estimating the "steady state concentrations" of the low concentration species in terms of the other species of interest.

I have updated your use of mathematical/ chemical environments. The way you tried to use it, it sometimes breaks at the end of the lines, making it very hard to understand.


text:
How to Locate the Excited State Dipole Moment in Turbomole Output (TD-DFT)
I want to calculate the dipole moment for the first excited state with a Q.C. program (TURBOMOLE: TD-DFT). The calculation lists the following output:
ground state total dipole moment:
-1.129096
-2.295685
 0.000000

I think that's $\langle\Psi_0|\vec{\mu}|\Psi_0\rangle$ with $\Psi_0$ the wavefunction for the ground state.
1st excited state electric transition dipole moment (length representation):
-2.413321
 0.367463
 0.000000

Should be $\langle\Psi_1|\vec{\mu}|\Psi_0\rangle$ with $\Psi_1$ the wavefunction for the 1st excited state.
2nd excited state electric transition dipole moment (length representation):
 0.841417
-0.329533
 0.000000

Should be  $\langle\Psi_2|\vec{\mu}|\Psi_0\rangle$.
And a summary:
             nuc           gs,e       ->  gs,tot          excit     ->  total
 ------------------------------------------------------------------------------
   x     -29.158609      28.029513      -1.129096       1.370060       0.240963
   y      48.811850     -51.107535      -2.295685       0.086529      -2.209156
   z       0.000000       0.000000       0.000000       0.000000       0.000000

I don't know how to get the values denoted by excit in the summary and they are not present anywhere else in the output. There's also no hint which excited state they apply to, so how do you calculate the total dipole moment for the 1st excited state e.g.?

I think you may need to ask the TURBOMOLE developers, e.g., in the user forum. I don't see any common open source program that parses TURBOMOLE excitation runs (e.g., cclib, Open Babel, etc.)


text:
Number of mercury atoms in a drop of mercury
Question:

A) Calculate the mass of a mercury atom in kg.
  B) How many atoms of mercury does a drop of mercury contain, knowing that the volume of a drop of mercury is $\pu{0.05mL}$ and that the density of mercury is $13.6 \times 10^3 \mathrm{kg/m^3}$

I know that $\mathrm{1 mL = 1 cm^3}=1.0\times10^{-6} \mathrm{m^3}$
And the mass of a drop of mercury is $\mathrm{m=v}\times \mathrm{density}
= 5.0\times10^{-18} \times 13.6\times{10^3}
=6.8\times10^{-14}$
What I want to know is how to proceed to find how many atoms mercury are present in its one drop? Do I simply divide the mass of a drop of mercury by the mass of a mercury atom, or do I use the Avogadro's number?  I don't want a straight-off answer, I only wish to know how to calculate it.

You've got an error in your volume in part B.  You state that $1ml = 1cm^3 = 1\times 10^{-6}m^3$, but when you use the volume of a drop, 0.05ml, you somehow change that into $5\times 10^{-18}m^3$.  You don't need to cube the number, the cube is there solely as part of the units.  It's $5\times 10^{-8}m^3$.  And yes, you just divide the mass of the drop by the mass of a single atom, assuming both masses are in the same units.

You can use dimensional analysis. It is a very useful tool to solve equations such as these.
What you have is the volume of one drop, $V(ml)$, the density of mercury, $\rho(kg\cdot{}m^{-3})$, the molar mass of mercury, $Mr(g\cdot{}mol^{-1})$, and the Avogadro's constant, $N_A(mol^{-1}$).
To find out the number of atoms in a drop of mercury, all you have to do is to manipulate those variables via dimensional analysis.
You have: $\frac{Mr(g)}{(mol)},\frac{N_A(atoms)}{(mol^{})},\frac{\rho{(kg)}}{(m^3)},\frac{V(ml)}{drop}$ and you know that $\frac{1(kg)}{1000(g)},\frac{10^6(ml)}{(m^3)}$
Now, try arranging them to form an equation such that the units cancel off,in order to get the answer you need.
A: Mass of mercury:
$$\frac{kg}{atom}=\frac{1(kg)}{1000(g)}\cdot{}
\frac{Mr(g)}{(mol)}\cdot{}\frac{(mol^{})}{N_A(atom)}
$$
Notice that this is dimensionally equivalent. Therefore,
$$\frac{kg}{atom}=\frac{Mr}{1000\cdot{}N_A}
$$
B: Atoms of mercury per drop:
$$\frac{atoms}{drop}=
\frac{N_A(atoms)}{(mol^{})}\cdot{}
\frac{(mol)}{Mr(g)}\cdot{}
\frac{1000(g)}{1(kg)}\cdot{}
\frac{\rho{(kg)}}{(m^3)}\cdot{}
\frac{(m^3)}{10^6(ml)}\cdot{}
\frac{V(ml)}{drop}$$
Notice that this is dimensionally equivalent. Therefore,
$$\frac{atoms}{drop}=
\frac{N_A\cdot{\rho}\cdot{V}}{Mr\cdot{10^3}}$$


text:
Nernst equation and electrolysis
Let's say we have a stable solution of
$\mathrm{A}_{\left(\mathrm{aq}\right)}^{2+}$
and $\mathrm{C}_{\left(\mathrm{aq}\right)}^{+}$
, with their respective counter-ions.
The associated redox reactions are:
$\mathrm{A}_{\left(\mathrm{aq}\right)}^{2+}+\mathrm{e}^{-}\rightleftharpoons\mathrm{A}_{\left(\mathrm{aq}\right)}^{+}\qquad\mathcal{E}_{\mathrm{A}_{\left(\mathrm{aq}\right)}^{2+}/\mathrm{A}_{\left(\mathrm{aq}\right)}^{+}}^{\circ}$
$\mathrm{C}_{\left(\mathrm{aq}\right)}^{2+}+\mathrm{e}^{-}\rightleftharpoons\mathrm{C}_{\left(\mathrm{aq}\right)}^{+}\qquad\mathcal{E}_{\mathrm{C}_{\left(\mathrm{aq}\right)}^{2+}/\mathrm{C}_{\left(\mathrm{aq}\right)}^{+}}^{\circ}$
where $\mathcal{E}_{\mathrm{A}_{\left(\mathrm{aq}\right)}^{2+}/\mathrm{A}_{\left(\mathrm{aq}\right)}^{+}}^{\circ}<\mathcal{E}_{\mathrm{C}_{\left(\mathrm{aq}\right)}^{2+}/\mathrm{C}_{\left(\mathrm{aq}\right)}^{+}}^{\circ}$
, since no reaction between $\mathrm{A}_{\left(\mathrm{aq}\right)}^{2+}$
and $\mathrm{C}_{\left(\mathrm{aq}\right)}^{+}$
is occurring.
The questions are:

Provided that neither $\mathrm{A}_{\left(\mathrm{aq}\right)}^{+}$
nor $\mathrm{C}_{\left(\mathrm{aq}\right)}^{2+}$
are present in the solution, how can we determine, form the Nernst equation, the $\mathcal{E}$ needed to start the electrolysis process
$\mathrm{A}_{\left(\mathrm{aq}\right)}^{2+}+\mathrm{C}_{\left(\mathrm{aq}\right)}^{+}\rightleftharpoons\mathrm{A}_{\left(\mathrm{aq}\right)}^{ +}+\mathrm{C}_{\left(\mathrm{aq}\right)}^{2+}$?
In other words, how can the Nernst equation deal with activities that are equal to zero?

Thanks in advance!

Interesting question. At first glance this seems to involve taking the natural logarithm of inf or 0, since the Nernst equation involves a ln[red]/[ox] term.

I don't know that the Nernst equation can handle this kind of starting condition.  That said, I would try to approach it from the point of view of Gibb's free energy, with $\Delta G = -nFE$, if possible.

I don't get it completely... As far as I understand, you want to "do electrolysis" on a solution that does not contain the ions you need?

@tschoppi: Only the product ions are absent. So, in the Nernst eq. their activity would be zero.

@DavideLaVardera Yes, everything starts to be clear now.

The Nernst equation only technically applies when the system is in electrochemical equilibrium—when there's no net current flow, but there's actually an exchange current. A system in which the reaction can only happen in one direction because there are no products wouldn't apply. Source:    Electrochemical Dictionary; Bard, A. J.; Inzelt, G.; Scholz, F., Eds.; Springer Berlin Heidelberg: Berlin, Heidelberg, 2008.
What I'm not sure about is the validity of ∆G values calculated from $∆G=∆G°+RT\ln Q$ at the  limits where the system is all products or all reactants. ±∞ don't seem to be meaningful values, but the system should quickly move away from these boundary values, even if only by a small amount. These chemical thermodynamics methods are well-behaved when near equilibrium, but well away from equilibrium, they can produce invalid results.

$$\mathrm{A}_{\left(\mathrm{aq}\right)}^{2+}+\mathrm{C}_{\left(\mathrm{aq}\right)}^{+}\rightleftharpoons\mathrm{A}_{\left(\mathrm{aq}\right)}^{ +}+\mathrm{C}_{\left(\mathrm{aq}\right)}^{2+}$$
If you don't have $\mathrm{A}_{\left(\mathrm{aq}\right)}^{+}$ or $\mathrm{C}_{\left(\mathrm{aq}\right)}^{2+}$ you must be having atleast  $\mathrm{A}_{\left(\mathrm{aq}\right)}^{2+}$ or $\mathrm{C}_{\left(\mathrm{aq}\right)}^{+}$, the reactants to produce the products when there is no product the case is:
$$\mathcal{E}=\mathcal{E}^\circ-\frac{\mathcal{R}T}{\mathcal{\nu F}}\ln{\frac{[\mathrm{A}_{\left(\mathrm{aq}\right)}^{ +}][\mathrm{C}_{\left(\mathrm{aq}\right)}^{2+}]}{[\mathrm{A}_{\left(\mathrm{aq}\right)}^{ 2+}][\mathrm{C}_{\left(\mathrm{aq}\right)}^{+}]}}$$
when the $\ln$ term is zero, the $\mathcal{E}\to\infty$ implying an instantaneous reaction which requires no external force, also implies by the Le'Chatelirer's principe but note that when even a slightest amount of product is created the tendency to procees forward decreases until it reaches the equilibrium where the reaction attains the equilibrium.


text:
How long must a sample be irradiated before all 59-Co atoms are converted to 60-Co?
For the interaction,  n$^0$ + $^{59}$Co $\to$ $^{60}$Co

At a constant neutron flux, this is a first-order reaction:
$ - \frac{{{\text{d}}\left[ {{}^{59}{\text{Co}}} \right]}}{{{\text{d}}t}} = k \cdot \left[ {{}^{59}{\text{Co}}} \right]$
$\left[ {{}^{59}{\text{Co}}} \right] = {\left[ {{}^{59}{\text{Co}}} \right]_0} \cdot {\operatorname{e} ^{ - k \cdot t}}$
Theoretically, $\left[ {{}^{59}{\text{Co}}} \right]$ will never become zero.


text:
Colour of Nickel thiocyanate
What is the colour of $\ce{C2N2NiS2}$ (Nickel thiocyanate)?
In most of the web sites it only molecular weight,formula and other specifications but its colour is missing every where.

In this paper, it is reported that $\ce{Ni(SCN)2}$ is yellow: http://link.springer.com/article/10.1007%2FBF00935928

@JannisAndreska Is it anhydrous? Most Ni(II) salts are yellow in anhydrous form and green in hydrate form.

@t.c I think they are referring to the anhydrous salt.

Chemical Book describes nickel(II) thiocyanate (CAS #: 13689-92-4) as "Nickel Thiocyanate Greenpowder" (see Freddy's answer elsewhere), but according to the Ref.1, it is yellow in color:

[...] This chloroform solution was green in contrast to the usual intense blue of nickel amino complexes. When the chloroform was allowed to evaporate, a viscous green scum was left. There was no evidence of crystallinity. This green amorphous matter was allowed to stand in the air for a few days and in that time the product had largely decomposed to yellow nickel thiocyanate.

This work has been published in a peer-review journal (Ref.2) so that I tends to believe this report, which is also supported by Matt Cliffe's and permeakra's answers.
References:

Don Wayne Carle, "The preparation of nickel(II) thiocyanate complex compounds with certain organic bases and the determination of their heats of formation," MS Thesis 1952, Oregon State College, OR, USA.
Albert V. Logan, Don W. Carle, "The Preparation of Nickel(II) Thiocyanate Complex Compounds with Picolines and the Determination of their Heats of Formation," J. Am. Chem. Soc. 1952, 74(20), 5224–5225 (https://doi.org/10.1021/ja01140a517).

After bit research i got finally one link where they mentioned the colour. The other name of(Synonym) of Nickel thiocyanate is "Nickel Thiocyanate Greenpowder" which suggest it is Green in colour.

Reference:ChemicalBook

It's a kind of a yellow/green brown, assuming you mean the solid. Aqueous (and similar solvents) solutions are green.
Reference

Dubler, E., Relier, A., & Oswald, H. R. (1982). Intermediates in
thermal decomposition of nickel (II) complexes: The crystal
structures of Ni(SCN)2(NH3)2 and Ni(SCN)2, Zeitschrift für
Kristallographie-Crystalline Materials, 161(1-4), 265-278.

The proper answer 'it depends'

PROPERTIES:
  Formula weight 174.88. Dark chocolate-colored powder. On
  addition of water, becomes first yellow and then dissolves with a
  green color.

(c) Brauer
In general you should be extremely cautious guessing colors of $d$-metal salts when the metal ion is not hydrated. The color may range widely. The only reasonably safe assumption is that $d^0$-salts are usually colorless, but then agains $\ce{MnO4-}$ is violet, so the colors must be memorized separately for each case.

As @permeakra said, the color 'depends'. The color various in different forms. From atomistry.com:

Nickel Thiocyanate, $\ce{Ni(CNS)2}$, has been obtained in the
anhydrous condition as a chocolate-coloured amorphous powder. The
hydrated salt, $\ce{2Ni(SCN)2.3H2O}$, obtained by dissolving nickel
carbonate in thiocyanic acid is a yellowish brown crystalline powder.

Most hydrated Ni(II) salts are green.
For transition metals, its salts are often colored because of transitions of electrons between d-orbitals of different energy, and such transitions release protons of a specific wavelength (which happen to be within the visible range of light for most d block elements).
Therefore, for a transition metal with the same oxidation state, its hydrated salts often exhibit a characteristic color.
In fact for most transition metal salts, the color can be predicted simply by knowing the cation of the salt.
http://en.wikipedia.org/wiki/Color_of_chemicals#Ions_in_aqueous_solution
For example, Cu(II) salts are blue, Ni(II) salts are green.
As thiocyanate has the charge of -1, the nickel in $\ce{Ni(SCN)2}$ has the charge of +2, and $\ce{Ni(SCN)2}$ is therefore green, as confirmed in the answer given by @hey.

authors: شیر محمد فدا حسین

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


text:
What are the pros and cons of using Gilman reagent over Grignard reagent to synthesize a hydrocarbon?
As far as I know, in order to synthesize a hydrocarbon $\ce{R-R'}$, Grignard reagent or Gilman reagent could be used.
$\ce{RMgX + R'X -> RR' + MgX2}$
$\ce{R2CuLi + R'X -> RR' + LiX + RCu} $
What are the pros and cons of using Gilman reagent over Grignard reagent? Gilman reagent is significantly harder to prepare and involves a reactive organolithium compound. I'm not looking for the pros and cons in terms of cost and yield, but rather specificity:

Are there any $\ce{RR'}$ groups that must be synthesized via Gilman reagent and not Grignard reagent?
Are there any $\ce{RR'}$  groups that must be synthesized via Grignard reagent and not Gilman reagent?

Someone (outside of SE) pointed out to me how pointless it is for chemists to synthesize hydrocarbons using these methods (as they are readily available to purchase). While I agree with him, I'm still interested to find out "theoretically" if Gilman/Grignard reagent is specific to certain R/R' groups.

Not all hydrocarbons are available for purchase though, so you might want to know how to make them.

I am not sure if Gilman reagents are harder to make than Grignard but they are softer nucleophiles and so they are used to perform 1,4-(conjugate) additions on $\ce\alpha,\beta-$unsaturated electrophiles. Grignard reagents will generally go for 1,2-attack instead. See this Wikipedia article for more info:

There is a flaw in your first reaction. $\ce{RMgX + R'X}$ is in equilibrium with $\ce{R'MgX + RX}$. This is a type of halogen metal exchange.
Since the reaction is in equilibrium you will get 3 products instead of one $\ce{RR', RR, R'R'}$ and therefore sheer wastage of Grignard reagent. On the other hand, Gilmann reagent is pretty good at it.
Only symmetrical alkane could be prepared by Grignard reagent (Here I am talking only about synthesis of alkane). So, as you say, Gilmann reagent is harder to care so its better you use Grignard reagent there, but I am not sure of that. Though I am pretty much sure that preparation of unsymmetrical alkane leads to complete wastage if Grignard reagent is used.

@t.c thanks for pointing out the flaw . I think I have NOT used congusing language by the way I was talking for purpose of making alkanes only

Yes, I realised, in my question I said R-R', and I did not mention about C=O

@t.c By using "SPECIFIC" . Are there any special features among alkanes that one reaction which is happening with some reagent might not react with other . I think the reason usually is steric hinderance which is not the case here.

Are there any reasons why wouldn't PhMgBr react with CH3Br to form toluene, for example?

@t.c I think the reaction you mention do occur by Sn2 mechanism


text:
How do you determine how many electrons occupy the anti bonding orbitals?
For molecular orbital theory:
How do you determine how many electrons occupy the anti bonding orbitals?
Also, why does hydrogen gas have no anti bonding orbitals- if each atom's electrons form a molecular orbital, then their energy is reduced so where does the extra energy go ?

Hydrogen has antibonding orbitals - just that it is unfilled.

Why are they unfilled? Wouldnt the conservation of energy require an antibonding orbital and a molecular orbital to be both filled up?

Forming a bond makes the molecule more stable. So the "extra energy" is released to the system. That's the definition of an energetically favorable reaction ($\ce{H + H -> H2}$).

Not necessarily @LDC3. The bond may just be weakened.

@Dissenter OK, I'll clarify. If you promote an electron into an anti-bonding orbital, the bond weakens and may undergo a reaction. If the molecule obtains an electron which must go into the anti-bonding orbital, the bond will likely break, resulting in a radical.

@LDC3 Okay, that's something more agreeable.

This is the MO diagram for Hydrogen. As you can see, for hydrogen atom, there is only 1 electron in the 1s orbital. Two hydrogens could come together to form a sigma bond, resulting in lowered energy. This doesn't mean that there is no sigma antiorbital, but rather due to the fact that hydrogen only contributes one electron each leading to a sum of two electrons, there are not enough electrons to fill the antiorbital, hence it is unfilled.
Now, imagine if it was helium instead. Helium has two 1s electrons, therefore, if two He atoms form a bond, 4 electrons has to be placed into the molecular orbitals. Since every orbital can hold a maximum of 2 electrons, two electrons would be in the sigma orbital, and the other two would be in the sigma antiorbital. An orbital and antiorbital would effectively cancel out, therefore, you see Helium exist in the monoatomic state.

So each molecule can have up to only one molecular orbital/antiorbital?

It depends. For diatomic molecules, yes. For anything other than those, the electron distribution may not be that simple. For example acetylene http://www.chemgapedia.de/vsengine/media/vsc/en/ch/2/oc/stoffklassen/systematik_struktur/acyclische_verbindungen/ungesaettigte_kohlenwasserstoffe/monoene_monoine/alkine/mo_modell_ethin.gif

@user58953 For every bonding orbital, there's an antibonding orbital. As you can see in the lower diagram, there are four bonding orbitals and four antibonding orbitals.

Note that these MO diagrams only work for simple diatomics. More complicated diatomics or molecules will require symmetry and energy and overlap considerations.

You have tried to implement @Dissenter's comment, but how you wrote it it is wrong. MO Theory can be and is applied to any molecule - you just cannot put it in a two dimensional diagram anymore. Also note, that hybridisation is only a mathematical tool, hence hybrid orbitals are not a necessity to form any bonds.

@Martin Thanks. You're right. Anyway I've deleted the part which I didn't explain properly, since it does not really address OP's question.

"An orbital and antiorbital would effectively cancel out, therefore, you see Helium exist in the monoatomic state." Does this mean the reason an orbital with complete electron pairs cannot form a new bond is because the electron pair would fill both a bonding and anti-bonding orbital, canceling each other out?

It is many years since I did bonding theory, more than I care to remember. In the 1s orbtal there is only one electron and the volume occupied fills a large region of space. The anti bonding orbital may be very small. In the 2s orbital with 2 electrons the region of space exhibited by the bonding orbital is so large that the nodal plane of the anti bonding orbital lies within the domain of space occupied by the bonding orbital. See T W G Solomons, Fundamentals of Organic Chemistry, 1982. The illustration of the molecular orbitals in this way gives an insight into the relative size and reinforcement of orbitals in a way line diagrams do not. I hope this is helpful. Chris


text:
Structure of benzene
How does molecular orbital or valence bond theory explain the delocalised structure of benzene? I.e. when the carbon forms 3 bonds with other carbons/hydrogen, what happens to the remaining hybridised orbital?

You would like to see this

The short answer is that it's not hybridized.
In benzene, the carbon atoms are sp2 hybridized. That means that each carbon's set of atomic orbitals is three sp2 orbitals and one p orbital. The sp2 orbitals are utilized to make sigma bonds to the adjacent carbons and hydrogens. The unhybridized p orbital contributes to the pi system.


text:
Hybridised orbitals
What determines which type of hybridisation (sp3/sp2/sp) a molecule will take? Methane/ethylene/acetylene all have the same electron configuration but undergo different different types of hybridisation- why?

Why did I get voted down? O_o

It's a pretty long topic and you would like to have a look at this article. In short I can present this from the same link, also some explanation at the last:

Methane $\ce{CH4}$, Ethylene $\ce{H2C=CH2}$ and acetylene $\ce{HC#CH}$ all have same electronic configuration but the number of  unhybridised orbitals( and of course the hybridised orbitals) are different in each case due to formation of $\pi$ bonds.


text:
How is Carboxymethyllysine formed from the heating of olive oil?
Olive oil has significant levels of Carboxymethyllysine.
But how does it form from the heating of olive oil, chemically speaking? Olive oil barely has any protein in it to begin with (though it still has some), so where would all the lysine come from?

What unit is kU/mL? I'm guessing kilounits per milliliter. Without knowing what a kU is, it's hard to say how much there actually is. That amount might be "barely any protein".

@jerepierre This was my question as well.  The best I could find was that they defined the smallest amount their immunoassay could detect as 0.1 U/mL.

Given that the levels are expressed in arbitrary units (based on the results of an immunoassay), it's difficult to say how much CML there actually is and accordingly, how much lysine is needed to produce it. It may be that only a small quantity of lysine is required) In fact, a different study found no CML in raw olive oil.
CML is produced during cooking from glycation of the side chain amine, starting with the Maillard reaction between the amine and a reducing sugar's carbonyl at high temperature. However, in the paper you cited, there was already significant CML in raw olive oil that was only somewhat increased by cooking. This paper suggests that CML formation is related to the oxidation of fatty acids and found that oils with increased concentrations of unsaturated fatty acids (olive oil is in the middle of the 5 oils they tested) were more prone to forming CML, but they deliberately added lysine to test it. It could be that this oxidation is something that happens in storage and the study you linked was using old oil or something. Also, just as with any other plant derivative, there could be significant variation in lysine content from different olives. This site indicates a lysine content of 3 mg/8.3 g whole olive, so there is some there. How much actually makes it into oil, I don't know.


text:
How to interpret the Maxwell-Boltzmann distribution to find the activation energy?
I understand activation energy to be the relative difference in energy between that of the activated complex/transition state (at some temperature) and the average energy of the reactants like shown here:

(Source: Wikimedia, User Bkell)
If this is correct, I have questions about graph like this one:

(Source: webchem.net (not available anymore, Internet Archive), retrieved via the Internet Archive)

Is the location marked activation energy supposed to be the energy level of the transition state provided the energy axis is absolute?
Can the Maxwell-Boltzmann distribution be done for heterogeneous systems (i.e., two chemical reactants)?
Unrelated to the graph, why does negative activation energy result in a decrease in reaction rate (per Arrhenius equation)?

The marked location corresponds to a level of kinetic energy in the reactants sufficient to result in a successful collision (energy wise, it says nothing about orientation).  The energy required for a successful collision is the gap in potential energy between the reactants and transition state.
A heterogeneous system is a system where the reactants are in different phases.  A common example is a gaseous reactant that collides with a solid catalyst.  In that case the Maxwell Boltzmann distribution can be applied to the gaseous reactant.
Can you clarify this?  What reactions have negative activation energies?  The reactions you'll commonly encounter have positive activation energies.

For Q1: is that basically the same as the energy level of the transition state?                                    For Q2: what about for combustion reactions (two gases), in this case is it possible to represent the Maxwell Boltzmann Distribution for both gases combined? Is that what the graph I linked is doing?
For Q3: I don't have a specific reaction in mind and I was reading this page: http://en.wikipedia.org/wiki/Activation_energy#Negative_activation_energy. Is it theoretically possible to heat any substance hot enough such that activation energy is negative?

Q3: The original wiki is referring to empirical activation energy, i.e. a value that you would got if you calculate it from temperature dependent kinetic data. For that kind of temperature dependence you need formally a negative energy value if you use this simple model. Therefore, it is more like of an artifact.

@brinnb For Q2, if a heterogeneous Maxwell Boltzmann Distribution is possible (say reactants A and B), is it correct to interpret it as the sum of the energies of A and B molecules which collide and the distribution is the representation of the sum?

In question 3, it is never possible in a unimolecular or bimolecular reaction to have an negative activation energy. No matter how you draw two potential energies as intersecting parabolas, with the transition state at the intersection, there is always a positive activation energy.
In reactions with a complex mechanism, one reaction following on from another, then the measured activation energy for however many reactions are present can be negative.
The Boltzmann distribution applies to all reactants and products, but on average we may take the difference of average energy of reactants and the average at the transition state to be the activation energy. As the Boltzmann distribution applies this is the reason that reactions with large activation energies (>> kBT) are slow; only very occasionally do the reactants transiently acquire enough energy from the solution, or from a collision if in the gas phase, to react. (There are other effect that may need to be considered in particular cases, such as translational diffusion limiting reaction or reactants having such complex structures that only certain orientations permit reaction.)


text:
How can I know when I have added an excess of magnesium oxide?
The following instructions were used to prepare magnesium sulfate crystals, $\ce{MgSO4 . 7H2O}$.

Measure $50~ \mathrm{cm^3}$ of dilute sulfuric acid into a beaker and warm the solution.
Using a spatula, add some magnesium oxide and stir the mixture. Continue
adding the magnesium oxide until excess is present.
Separate the excess magnesium oxide from the solution of magnesium
sulfate.
Heat the solution until crystals form. Obtain the crystals and dry them.

The question I have an issue with is this: How would you know when excess magnesium oxide is present in step 2?
The answer key says that I would know when no more solid (that is, no more magnesium oxide dissolves). However, since we are talking about dilute sulfuric acid, there should be water, and I read on the internet that $\ce{MgO}$ is soluble in water.

Check the solubility of magnesium oxide.

In what sense? As in how much can dissolve in water or if it can dissolve in water at all? Or are you talking about acids

$\ce{MgO}$ is insoluble in water (solubility: 0.0086 g/100 mL). On the other hand, $\ce{MgSO4}$ is soluble in water (solubility: 25.5 g/100 mL).
The reaction that is taking place is an acid-base reaction:
$$\ce{MgO + H2SO4 -> MgSO4 + H2O}$$
Now for your question, how would you know when excess $\ce{MgO}$ is present? Here you are trying to make $\ce{H2SO4}$ the limiting reagent. When all the $\ce{H2SO4}$ has reacted, no more $\ce{MgO}$ can further react. Therefore, you will see solid in your reaction mixture that cannot be further dissolved.
At the end of your reaction,
$$\ce{MgO_{excess} + H2SO4 -> MgSO4 + H2O + MgO_{remaining}}$$

Thank you! But I am a little confused as to why other sites ( BBC Bitesize, for one) say it is soluble in water to form MgOH

@dadadok Both MgO and Mg(OH)2 are insoluble - you can easily filter them. Note that for MgSO4, if the solution is too basic it will react with OH- to form Mg(OH)2,  leading to low yield.

Thanks!This is pretty much the kind of response I was looking for... and goes some way towards relieving exam stress. Thanks again.

@dadadok Mg(OH)2 is just barely soluble in water, that's why you're finding some sources saying it's insoluble and others saying it is soluble.  The same is true for MgO, which will dissolve to a very small extent in water.  "Very small" and "barely" here mean only a few thousandths of a gram per 100mL of water.


text:
Kinetics of the conversion from metal oxide to metal hydroxide in water
How long can a metal oxide such as magnesium oxide remain in solution before being converted to the metal hydroxide?

Rate is dependent on many factors such as temperature and concentration.

This will happen very fast. What do you need this info for?

Satisfaction of curiosity.


text:
Actual meaning of equilibrium
What is actually meant by that concentration of reactants and products remain constant in equilibrium?

t.c is right, but to expand on it, equilibrium means that a reaction is still occurring, i.e. reactants are converting to products and products are converting to reactants, but they're doing so at the same rate so there's no net change in the concentrations of any of the components. Contrast this with a metastable system, in which concentrations of components may be remaining the same, but the two aren't interconverting because of a large kinetic barrier.

Does it mean that same amount of quantity is converted to product as the quantity as is converted to reactants after attaining equilibrium ?

Over a given period of time, yes. That's what same rate means. Rate = change in concentration/change in time.

For an equation
$\ce{Reactant <=> Product}$, at equilibrium, the rate of reactants forming products is the same as the rate of products converting back to reactants. Therefore, the concentration of products and reactants remain constant.


text:
Why is equilibrium achieved at different stages of a reaction?
In other words, I want to know why some reactions attain equilibrium early in the reaction while some reactions obtain equilibrium at the end of the reaction.
Why is this the case?

If you define early or end of reaction by the how much the concentration of reactants change from initial reaction to once equilibrium is reached, it is because the equilibrium constant itself is essentially a ratio of the forward and reverse rate constants (can be approximated by the Arrhenius equation posted by t.c.). A high ratio means "more" products (at least a higher concentration) are present at equilibrium than reactants (so in a sense near the "end of the reaction"), since the forward rate is much higher than the reverse, a lower concentration of reactants compared to products is needed to maintain equilibrium. A low ratio means the opposite.

The rate of reaction is dependent on the rate constant, as well as the order of reaction.
The rate constant, which is approximated by the Arrhenius equation, is dependent on the Activation energy, Ea, and temperature, T. Activation energy can be lowered by using a catalyst (if any).
 $$k = A e^{-E_a/(R T)}$$
From this equation, you can see that the higher the temperature and the lower the Ea, the faster the rate.

I reinterpreted the question a bit, but I think your answer still fits the edit, anyway.

I want to know why some reactions attain equilibrium early in the reaction while some reactions obtain equilibrium at the end of the reaction.

@pcforgeek Okay, so you were looking for an explanation on rate and not thermodynamic equilibrium. I've edited my answer.

The way I read it the asker wants to know why some equilibrium constants are very large (essentially all reactants turn into products at equilibrium) and others are very small (almost no reactants turn into products at equilibrium).


text:
How does aminoguanidine interfere with the formation of advanced glycation endproducts?
This says that aminoguanidine interferes with the formation of advanced glycation endproducts (AGEs) through interacting with 3-deoxyglucosone.
But how does it interact with 3-deoxyglucosone that prevents AGE formation?

What is AGE? Please define your acronyms for the benefit of those who are not familiar with the jargon.

okay i've added it

I would expect the N-amino group of aminoguanidine to react with 3-deoxyglucosone at either of the adjacent, reactive carbonyl groups to form a stable guanidinylimine derivative (analogous to a hydrazone). This derivative, unlike 3-deoxyglucosone itself, would not be able to react with free amino groups on proteins and continue on the pathway to AGEs, but rather would be excreted as is.


text:
What does being an "elementary substance" mean?

Which of the following isn't an elementary substance?

Ozone
Sulfur
diamond
quartz

I don't get what elementary substance means.
By seeing the options and the word elementary, I guess it means something like pure or obtained in its original form.

From LATINTOS:

An elementary substance is a pure chemical substance that consists of atoms belonging to a single chemical element.

Now you should be able to solve your question.

ok so it's quartz.


text:
How is bonding in the trihydrogen cation and triatomic hydrogen possible?
According to wikipedia $\ce{H^+3}$ and $\ce{H3}$ are possible and have been synthesized. It says that $\ce{H^+3}$ is stable too. Can anyone please tell me how is the bonding possible?

Hydrogen has a valency of 1 but it still bonds like Ozone. How is this possible?
Links:

$\ce{H^+3}$ - Trihydrogen Cation
$\ce{H3}$ - Triatomic Hydrogen

Because the cation is aromatic?  (Bonding analogous to pi bonds in the cyclopropenyl cation.)

Here are the vital parts from the links about $\ce{H3}$ in the question:

The molecule can only exist in an excited state [...] the electronic state for a trihydrogen cation with an electron delocalized around it is a Rydberg state.

and from second-level link

Rydberg states of the neutral triatomic hydrogen molecule can be viewed as
  an electron attached to a tightly bound $\ce{H3+}$ core

Basically, at least some states can be viewed as an $\ce{H3+}$ ion working as a nucleus in a bigger hydrogen-like atom. If said 'nucleus' catches the electron, the molecule immediately dissociates. So, let's focus for a bit on $\ce{H3+}$. In this ion a two-electron three-center bond is formed. Such bonding is actually pretty common, even if (almost never) mentioned in school. As long as orbitals of several atoms can overlap effectively and have fitting symmetry, they can form a single highly symmetrical bonding orbital, that can accept two electrons. In this case the bonding order between atoms is fractional, so in $D_\mathrm{3h}$ (triangular) $\ce{H3+}$, each of the three $\ce{H-H}$ bonds has bond order $1/3$. This kind of bonding plays a critical role in the chemistry of boranes (and Group 3 elements in general, maybe except thallium) and some families of compounds of d-block elements.
To obtain more intimate and correct understanding you should invest in molecular orbital theory. Your link for $\ce{H3+}$ contains a comprehensive description of bonding in terms of said theory.


text:
Radioactivity and Kinetics of disintegration
Well I've just started on kinetics and there are a few things I am absolutely confused about!

First of all, is a neutron actually made up of a proton and electron? If yes, then how is this possible?
In K-electron capture where the electron of the K-Shell is captured by the nucleus, doesn't the species become positively charged or ionized as it is losing an electron?
When metastable technetium disintegrates into normal technetium and a gamma particle, what is the change brought about? Isn't the technetium on both sides of the equation basically the same?

No, as hey answered, neutrons are made of two down quarks and an up quark.  Protons are made of two up quarks and a down quark.  It is the conversion of a down quark into an up quark that results in the neutron --> proton conversion.  Read here for more information about this reaction.
The confusion with this probably comes from the fact that net reaction  in a neutron to proton conversion is $n^{0} \rightarrow p^{+} + e^{-} + \overline{\nu_e}$, so it looks like a neutron is falling apart into three sub-particles.
No, the atom will not become ionized from this event itself.  Consider what is happening in this reaction:
$p^{+} + e^{-} \rightarrow n^{0} + \nu_e$
Yes, an electron is removed from the atom, but at the same time a proton is as well.  We should know that the atoms remains neutral because of conservation of charge. If the atom has no net charge at the beginning of the reaction, and no charged particles are ejected, it must be neutral at the end of the reaction as well.
Jon Custer's response is spot on here.  Gamma ray emission represents the movement of the nucleus from a metastable state to its ground state via the release of electromagnetic radiation.

For 3., this is nuclear physics (or chemistry - Glenn Seaborg was a chemist). You clearly understand electron structure of an atom - there are various energy levels that can be occupied, and the ground state has all the electrons at the lowest states.  An excited (neutral) atom has an electron occupying a higher electron state, and it can decay back to the ground state by emitting a photon.  Well, the nucleus has various energy states for the protons and neutrons to occupy (the forces and math aren't the same, but the general idea is similar).  A metastable nuclear configuration is not in the ground state.  However, it can drop into the ground state, and has to release the energy somehow, and just like the electron it does so by emitting a photon, just a high energy one. Gammas are from nuclear transitions, x-rays from electronic transitions.

Is neutron actually made up of a proton and electron?

Answer is no. Neutron is neutral(no charge) doesn't mean it is made up Electron and Proton. The neutron is made of two Down and one Up quark.


text:
How is the bond length calculated from the total electronic energy?
In the quantum chemistry course that I currently attend, it was said several times that one of the key quantities derived from molecules by means of computation is the total electronic energy $E_\text{el}$, because from this many derived quantities can be calculated, such as bond length or $J$-coupling values for NMR experiments.
However, I've never gotten a good explanation as to how to proceed when I've calculated a value for $E_\text{el}$ and want to find out the bond lengths in the molecule in question.
What are the formulae and/or computation steps necessary for this?

The traditional way of getting the right geometry for your system is by taking a fixed atomic structure, perform an electronic structure minimization, calculate the forces acting on the atoms from that, use them to alter the atomic structure according to the forces and then use the altered structure to start over again. This is repeated until you converge to the equilibrium geometry.
However there is an alternative (quite elegant) way for DFT calculations using pseudopotentials or the PAW method with plane-wave basis sets: the ab-initio molecular dynamics method introduced by Roberto Car and Michele Parrinello. The "killer feature" of the Car-Parrinello method is that it allows the simultaneous solution of the electronic structure problem and the equations of motion for the nuclei which allows for the relaxation of the nuclei to find stable structures, i.e. the equilibrium geometry, as well as for thermal simulations of solids and liquids - so this method does offer much more than a simple geometry optimization. How does it achieve that? In the Car-Parrinello approach, the total Kohn-Sham (KS) energy is the potential energy as a function of the positions of the nuclei. The forces from this energy then determine the molecular dynamics (MD) for the nuclei. The special feature of the Car-Parrinello algorithm is that it also solves the quantum electronic problem using MD. This is accomplished by deriving the equations of motion from a fictitious Lagrangian which contains a fictitious kinetic energy of the electronic (KS) wave functions and some constaints to ensure that the wave functions remain orthonormal:
\begin{align}
  \mathcal{L} &= \sum_{n} f_{n} m_{\psi} \langle \dot{\psi}_{n} | \dot{\psi}_{n} \rangle + \frac{1}{2} \sum_{i} M_{i} \dot{\vec{R}}_{i} \! {}^{2} + E_{\mathrm{DFT}}[\psi_{n}, \vec{R}_{i}] - \sum_{m, n} \Lambda_{n,m} \left( \langle \psi_{m} | \psi_{n} \rangle - \delta_{n,m} \right)
\end{align}
where $f_{n}$ is the occupation of the $n^{\mathrm{th}}$ Kohn-Sham eigenstate $\psi_{n}$, $M_{i}$ and $\vec{R}_{i}$ are the mass and the position of the $i^{\mathrm{th}}$ nucleus, $\delta_{n,m} = \begin{cases}1 & n=m \\ 0 & n \neq m \end{cases}$ is the Kronecker delta, $\dot{\psi}_{n} = \frac{\mathrm{d} \psi_{n}}{\mathrm{d}t}$ and $\dot{\vec{R}}_{i} = \frac{\mathrm{d} \vec{R}_{i}}{\mathrm{d}t}$ are the first derivatives with respect to time and $\Lambda_{n,m}$ is a Lagrange multiplier ensuring the constraint of wave function orthonormality.
The first term in this Lagrangian represents the fictitious kinetic energy of the wave functions. There is no physical correspondence for this term. Ideally one would choose the fictitious mass $m_{\psi}$ of the wave functions equal to zero. This introduction of this unphysical quantity also lead to naming the Lagrangian fictitious.
The second term describes the classical kinetic energy of the nuclei.
The third term $E_{\mathrm{DFT}}[\psi_{n}, \vec{R}_{i}]$ is the density functional total energy, which is a functional of the electronic wave functions and the atomic positions.
The last term is the constraint of orthonormal wave functions $| \psi_{n} \rangle$.
Applying the principle of least action to this Lagrangian leads to the following Euler-Lagrange equations (equations of motion for the ab initio MD simulation):
\begin{align}
  M_{i} \ddot{\vec{R}}_{i} &= \underbrace{- \nabla_{\vec{R}_{i}} E_{\mathrm{DFT}}}_{= \, \vec{F}_{i}} \\
  m_{\psi} | \ddot{\psi}_{n} \rangle &= - \underbrace{ \frac{1}{f_{n}} \frac{\delta E_{\mathrm{DFT}}}{\delta \langle \psi_{n} |} }_{= \, \hat{H} | \psi_{n} \rangle } + \underbrace{ \sum_{m} | \psi_{m} \rangle \frac{\Lambda_{n,m}}{f_{n}} }_{\tilde \, | \psi_{n} \rangle \epsilon_{n}}
\end{align}
where $\ddot{\psi}_{n} = \frac{\mathrm{d}^{2} \psi_{n}}{\mathrm{d}t^{2}}$ and $\ddot{\vec{R}}_{i} = \frac{\mathrm{d}^{2} \vec{R}_{i}}{\mathrm{d}t^{2}}$ are the second derivatives with respect to time, $\hat{H}$ is the Kohn-Sham hamiltonian, $\epsilon_{n}$ is $n^{\mathrm{th}}$ Kohn-Sham eigenvalue and $\vec{F}_{i}$ is the force acting on the $i^{\mathrm{th}}$ nucleus.
The first set of equations is just Newton's equations of motion for the nuclei moving under the forces derived from $E_{\mathrm{DFT}}$.
The stationary solution of the second set of equations is equivalent to the Kohn-Sham equations, since for a steady state all time derivatives vanish so that:
\begin{align}
   \hat{H} | \psi_{n} \rangle &=  \sum_{m} | \psi_{m} \rangle \frac{\Lambda_{n,m}}{f_{n}}
\end{align}
Constructing a matrix $\mathbf{\Lambda}$ from the Lagrangian multipliers $\Lambda_{n,m}$ shows that $\mathbf{\Lambda}$ is proportional to the transpose of the matrix representation of $\hat{H}$, i.e. $\Lambda_{n,m} \propto H_{m,n}$. Diagonalizing $\mathbf{\Lambda}$ leads to the eigenvalues $\epsilon_{n}$ of the Kohn-Sham equations.
So, this Lagrangian does not lead to the time-dependent Schroedinger equation for the electronic wave functions. Rather it creates a dynamic that (in the ideal case) keeps the electrons always in their ground state.

Under which conditions does the Car-Parrinello method break down? Excited state analysis already seems to be out of the question...

@tschoppi I've only seen it being used with plane-wave DFT methods. But it is more generally applicable than this. I think you can even use it with a CI calculation but the CP-method is rather expensive since it makes extensive use of forces which are usually quite costly to calculate. As for excited states: There are papers indicating it can be done (see for example here and here).

The geometry of a system is what is known (or given) and a corresponding energy is computed.  If you have an energy then you should have the geometry (or in your case, bond length).
The wave function, $\Psi$, in the Schrodiner equation is simply a function of all the positions of the particles within the system (nuclei and electrons).  Typically chemists invoke the Born-Oppenheimer approximation which assumes that the nuclei are stationary (frozen in space) with respect to the very 'fast-moving' electrons.  Therefore, we freeze the nuclei in space and compute the energy of this stationary state.  Move the nuclei in your molecule a little bit and you will get a new energy.
As Wildcat was explaining, optimization procedures essentially moves the nuclei in your molecule around in space, computing an energy for each configuration. The goal of this type of routine is to find an energy that is lower than all the rest.  Take, for example, the $\ce{H2}$ molecule.  It contains one bond length and therefore the only coordinate we need to consider in an 'optimization procedure' is the internuclear separation of the two $\ce{H}$ atoms.  It is the chemist which will choose the initial positions of these atoms.  Say we initially set these atoms to be 200 pm apart.  We compute an energy.  Now, can we do better?  (i.e. can we perturb this geometry in such a way that gives rise to a lower energy?)  Well we can certainly try other bond lengths so we now set our bond length to 150 pm and compute a new energy.  Lo and behold we see that this new energy is lower than the first!  We keep trying different bond lengths until we reach a point where the energy increases no matter what we do with the separation (whether we increase it or decrease it).  Once we are at this point, we have found the lowest energy configuration of $\ce{H2}$.  Of course optimization routines are smarter than doing a brute-force approach to sampling all the possible quantities each geometrical parameter can take on.  Furthermore, second-derivatives are essential for characterizing the nature of the stationary state that was determined (i.e. is the stationary state a true minimum, a transition state or some higher order saddle point?).

So if I got this correctly: It is not from the calculated quantity, but by way of calculating it, that we find the bond length?

The bond length is what 'you' make it.  From that you compute an energy.  Of course optimization procedures will take what you give it and manipulate the geometry as it tries to find the lowest energy.  The geometry information is printed to the output file for every iteration the procedure takes.

There is no formula. In short, for an electronic state of interest you do what is known as the geometry optimisation, i.e. for the chosen electronic state you solve the electronic Schrödinger equation for a number of fixed (or clamped) nuclear configuration chosen in a certain way by the employed minimisation procedure with the aim to find the minimum of the electronic energy. And the nuclear configuration that corresponds to minimum of the electronic energy for the chosen electronic state describes the equilibrium geometry of a molecule being in the chosen electronic state. Then from such equilibrium nuclear configuration you can easily extract structural parameters, such as, for instance, bond lengths.

NMR shielding is a bit more complicated story. You need to become familiar with the basics of the so-called response theory, to understand how some electric and magnetic properties, such as NMR shielding, are calculated.


text:
Intermolecular forces in Carbon Nanotubes?
What are the intermolecular forces that can be found in Carbon Nanotubes?

I am unfortunately not really getting what you are asking. Are you asking about the bonds between the carbons, or the forces that are inside a nanotube, or are you asking about the molecular forces between nanotubes? It would not hurt if you could expand your question a little.

The intermolecular forces between the molecules.

There is known, but poorly understood $\pi-\pi$ stacking. While weak compared to covalent, dipole or ionic interactions, it still beats usual van-der-vaals interactions.
Carbon nanotubes are, essentially, polycondensed aromatical systems, so they can and would interact with each other as aromatic systems via $\pi-\pi$ stacking

To be fair, $\pi-\pi$ stacking are often considered one type of dispersion interactions. Since large conjugated systems like nanotubes have very high polarizabilities, the dispersion interactions are fairly large.

Carbon nanotubes are not polar or ionic.  Therefore I do not expect any ion or dipole intermolecular interactions.  Since carbon has electrons, that only leaves us with dispersion forces (induced-dipole).  Therefore carbon nanotubes exhibit dispersion as the only intermolecular force.


text:
What functional groups are subsets of the Alkoxy group?
In a quiz, I was asked to find all of the combinations of organic molecule of $\ce{C4H6O2}$. The combination excludes -COOH, -OH, double bonds and the alkoxy group.
The only problem, as far as I am concerned, is the definition of the term "alkoxy group". According to this Wikipedia article, a Peroxy or Methelenedioxy should be one of the possible combinations allowed. However, the names do have the "-oxy" suffix in them. So I am uncertain whether to include Peroxy's and Methelenedioxy's in the list.
P.S. Below is a picture of the Methelenedioxy group and the peroxy group, respectively,  from the already mentioned Wikipedia article.

There is another Wkipedia article entitled "Alkoxy group".  In it they show the following molecules as containing alkoxy groups.

They include methylenedioxy, but not peroxide in their list.

Alkoxy groups are ethers, so I would consider methylenedioxy as a subset of alkoxy groups (although the reactivity is pretty different).
Peroxy is not an ether, so I would consider it distinct from an alkoxy group.
Given those functional group restrictions, it seems like the quiz was trying to force you into a particular line of thinking. Without alcohols and ethers, I think the only thing left would be peroxides, but perhaps I'm missing something.


text:
Can butane-2,2,3,3-tetraol be oxidized by potassium permanganate?
Can the compound butane-2,2,3,3-tetraol, $\ce{C4H10O4}$, be oxidized by $\ce{KMnO4}$? If so, in what conditions? As I have learnt in class, normally the hydroxy group would oxidize to form a ketone, but what about this particular case?

You are talking about the dihydrate of biacetyl.  gem-diols are generally not stable, the equilibrium is on the side of the aldehyde/ketone, except when the keto group is next to an electronegative atom.  Glyoxal in aqueous solution is all hydrate and oligomers, as is chloral.
Biacetyl is more hindered than glyoxal, but in aqueous solution one would think they there is still some hydrate in the equilibrium.  What you'd think would happen is that on reaction with permanganate you'd get the cyclic diol and under sufficiently forcing conditions (read: acidic medium) you'd get cleavage to acetic acid.
Biacetyl is awful.  Run away!

Unless you like artificial butter flavoring. Then run toward.


text:
Will propane liquefy under these conditions?
Consider a 1:1 mixture of propane and nitrogen at $60°\mathrm{F}$ and $75\:\mathrm{psi}$. Will the propane liquefy if pressure is increased to $200\:\mathrm{psi}$? I am experimenting in the oilfield with a shallow heavy oil lease.

Is this homework?  If so, what materials have you been given to help answer the question?

First of all, you might want to consider using proper SI units.
If you know the saturation temperature (boiling point) of propane at a given pressure:
At a pressure of p = 75 psi (0.52 MPa), the saturation temperature of pure propane is 37.13 °F (2.85 °C); thus, at a temperature of 60 °F (16 °C), propane would be gaseous.
At a pressure of p = 200 psi (1.38 MPa), the saturation temperature of pure propane is 104.54 °F (40.30 °C); thus, at a temperature of 60 °F (16 °C), propane would be liquid.
However, a 1:1 mixture of propane and nitrogen contains only 50 % propane; i.e., the mass fraction of propane is wpropane = 0.5. The corresponding amount-of-substance fraction (mole fraction) of propane is xpropane = 0.39.
For ideal gas mixtures, the ratio of partial pressures equals the ratio of the number of molecules:
ppropane = xpropane • p
Therefore, the partial pressure of propane is approximately ppropane = 78 psi (0.54 MPa).
At a pressure of p = 78 psi (1.38 MPa), the saturation temperature of pure propane is 39 °F (4 °C); thus, at a temperature of 60 °F (16 °C), propane would be gaseous.
If you know the vapour pressure of propane at a given temperature:
At a temperature of 60 °F (16 °C), the vapour pressure of propane is 107.71 psi (0.74 MPa).
Thus, at a pressure of p = 75 psi (0.52 MPa), pure propane would be gaseous. At a pressure of p = 200 psi (1.38 MPa), it would be liquid.
However, once again, a 1:1 mixture of propane and nitrogen contains only 50 % propane; i.e., the mass fraction of propane is wpropane = 0.5. The corresponding amount-of-substance fraction (mole fraction) of propane is xpropane = 0.39.
For ideal gas mixtures, the ratio of partial pressures equals the ratio of the number of molecules:
ppropane = xpropane • p
Therefore, the partial pressure of propane is approximately ppropane = 78 psi (0.54 MPa). Since this partial pressure is below the vapour pressure of 107.71 psi (0.74 MPa) at a temperature of 60 °F (16 °C), propane would be gaseous.


text:
Why is C4 of butynone electrophilic?
A question asks whether C4 in 3-butyne-2-one can act as a nucleophile, electrophile, or acid (it can be more than one). The answer is that it can be both electrophilic and acidic.
The acidity makes sense, it's a terminal alkyne, sp orbitals stabilize the negative charge on the conjugate base.
The electrophilicity I don't quite understand. I can write a resonance structure that puts some positive charge on C4, analogous to an alpha-beta unsaturated carbonyl compound, but it looks bizarre. I know ketenes are a real thing, but is  this resonance structure a real contributor? Is there another explanation for the electrophilicity?
Google searches for alkynone, ynones, etc. yielded mostly peer-reviewed literature that's above my understanding.

Yes, that second resonance structure, the one on the right, is a significant contributor to the overall description of the molecule.  That resonance looks a little different because the p orbitals on the alpha and beta carbons are on $\ce{sp}$ hybridized carbons, rather than the $\ce{sp^2}$ hybridized carbon more commonly seen in enones.   As this second resonance structure explains, alpha-beta unsaturated enones and ynones are both subject to attack by nucleophiles at the electrophilic beta carbon.
Resonance structures, like the one pictured below, explain the reactivity seen in enamines and yneamines (the triple-bond analogue); however, now the positive charge is on the nitrogen and a negative charge on the beta carbon.  Therefore these nitrogen compounds are subject to attack by an electrophile at the nucleophilic beta carbon.


text:
Dissociation rates: trans-effect rule in square planar complexes
In lecture, the trans-effect was described.
A ligand $L^t$ with a higher trans-effect as $L$ (cis to $L^t$) leads to a faster dissociation of ligand $L^d$ (trans to $L^t$). I would expect that the attacking ligand $B$ is needed to make a stronger bond as the leaving ligand $L^d$ did.

In order to see if these assumptions are always valid in planar quadratic systems I've set up following problem according to the given trans-effect order:

$\ce{CO}$ has a stronger trans-effect as $\ce{CH3-}$ and therefore the ligand trans to $\ce{CO}$ will be exchanged faster as trans to the $\ce{CH3-}$. But the leaving group $\ce{I-}$ would also have a higher trans-effect as $\ce{Cl-}$ so I wouldn't expect a substitution in b). Are these statements correct?

First off, there are two nice articles on the trans effect on Wikipedia and LibreTexts that are worth reading.
You need to remember that there are two factors leading to the trans effect:

Weakening the $\ce{Pt-X}$ bond trans to another group (i.e., a structural effect)
Stabilizing the 5-coordinate transition state, since substitution reactions in square planar complexes are generally associative. (i.e., a kinetic effect)

(a) I don't think the $\ce{I-}$ will substitute for the $\ce{NH3}$ group. Instead, I think the $\ce{CO}$ "wins" the trans effect "competition" and so the trans $\ce{CH3-}$ will be replaced by the $\ce{I-}$.
Next, substitution with the $\ce{Cl-}$ should replace the $\ce{I-}$.

Now your second question seems to be if you'll actually get a substitution with $\ce{Cl-}$ since $\ce{I-}$ has a greater trans effect. Yes, you will. All of these are in equilibrium, so if you generate a lot of $\ce{I-}$ that can substitute back (and get starting material). But you'll definitely get the Cl-substituted product too if you separate the reaction mixture.
Indeed, the prototypical example for the trans effect is synthesis of trans-platin or cis-platin, starting from either $\ce{[PtCl4]^{-2}}$ or $\ce{[Pt(NH3)4]^{+2}}$. Even though $\ce{NH3}$ has a smaller trans effect, you can synthesize cis-platin from $\ce{[PtCl4]^{-2}}$ and substituting two $\ce{Cl-}$ with $\ce{NH3}$.

The trans effect on Wikipedia is quite easy to understand because there are only two different groups involved. Did I correctly understood; for step a) you've only chosen the ligand which has the highest trans effect? I'd argue that in the starting material $\ce{CO}$ trans to $\ce{CH3-}$ is more stable as $\ce{CO}$ trans to $\ce{I-}$ because of the better donor properties of $\ce{CH3-}$ there is a better push-pull-stabilization. But I think my argument is wrong because it is a thermodynamic argument and not a kinetic argument.

The unfortunate problem with trans effect is that there is a thermodynamic structural effect and a kinetic effect. Yes, you pick the leaving group trans to the strongest substituent. I'm not sure I follow the rest of your argument here.

I'd argue that in the starting material $\ce{CO}$ trans to $\ce{CH3-}$ is more stable as $\ce{CO}$ trans to $\ce{I−}$ because of the better donor properties of $\ce{CH3-}$ there is a better push-pull-stabilization that means $\ce{CO}$ favors electrons from a strong trans placed donor for a good $\Pi$ Backbonding. The methyl group acts already as the better donor as iodine does.

That's true, but besides the point. The $\ce{CO}$ group won't substitute in either case.

I'm not sure if I pointed it out clearly: It's no question, $\ce{CO}$ will not get substituted. Although the trans placed methyl is more stable as the iodine so I would not propose that the methyl will get substituted by an iodine.

@laminin As I said, all these ligands are labile, so it's a question of where the $\ce{I-}$ will go, not whether it will or not. Based on the trans effects, the methyl will be replaced.

all right. (The question, where the $\ce{I-}$ will go: I believe the negative charge is mire stable on soft iodine instead of hard methylanion, but may be this is wrong.) I guess according to trans effect I understood it in this way: 1.) In normal cases always the ligand with highest trans-effect dictates the trans placed substituent to leave. 2.) The trans-effect strength of the attacking nucleophile is not important to expect which ligand gets substituted.

Did I understood your rules correctly: In normal cases always the ligand with highest trans-effect (here in A) and B) it's the $\ce{CO}$) dictates that the trans played substituent has to leave? Do you know an example where this is not the case, do you know an exception where the 5-coordinate transitionstate overrules the dissociation of the ligand trans to the ligand with the highest trans-effect?

@laminin That's the point. Unfortunately, people call "trans effect" for two different things, but the key in your question is the kinetic effect on the 5-coordinate transition state. So there's no "exception" because these are one and the same thing. The substituent trans to the highest "kinetic trans" effect will leave.

@GeoffHutchison thank you for the answer! Not completely related to this question, but I wonder: does trans-effect (structural) exist for complexes dominated by electrostatic interactions (i.e. "hard") and not orbital mixing?

@substitutedbutadiene - Please ask a new question rather than leave it buried in the questions

@GeoffHutchison How "Weakening the Pt−X bond trans to another group" leads to a lower activation energy? In many textbooks is explained by saying that it destabilizes the ground state of the molecule by weakening the bond Pt-X. But one could also argue that the Y-Pt bond is now stronger (Y is greater σ donor) so these two effects should cancel. Could we say better than this destabilization is depicted in the transition state because now the Pt-X bond breaks easier?


text:
What is the official file extension for SMILES and SMARTS files
I want to know what is the officially used file extension for files containing SMILES and SMARTS strings. Somewhere I see .SMI/.SMA and somewhere its the full .SMILES/.SMARTS.

It is just like .htm or .html, both are same

DOS\Win systems traditionally accept only three character file extensions, while Unix/Linux systems can handle longer ones. It is just a technicality, depends on software what they accept/recognize.

I don't think I've ever seen anything proposed as "official," though I admit I don't use Daylight's official toolkit.
Open Babel accepts:

.smi
.smiles
.can (ideally canonical SMILES, but that matters more for writing, not reading)

I haven't seen many SMARTS files, but I would suspect both .sma and .smarts are out there "in the wild."
Incidentally, this is probably something to bring up with the OpenSMILES effort. I know there's an OpenSMARTS specification started as well.

Thanks for the info. (Can't up-vote because it requires at least 15 reps :) )

Slightly beside the point, but here is what Daylight says:

The "chemical" MIME type, championed by H. Rzepa of Imperial College,
  London, is (hopefully) nearing ratification. Should this happen,
  SMILES will almost certainly appear as MIME subtype, i.e., the MIME
  type: chemical/smiles. Until that blessed day, the IETF recommendation
  is to use something like application/x-smiles or x-chemical/x-smiles.

Yes, that's something we use internally for SMILES format, clipboard operations and while requesting SMILES data from our components.

Sadly, if you look at the date, that was 1998. MIME types are also somewhat passé.

@GeoffHutchison, why do you say MIME types are passé?? They are still heavily used in the modern web, in the Linked Data web even more!

@EgonWillighagen I see far, far too much data on the web as text, generic XML or other generic types. Software is then left to infer that this text file is really SMILES or PDB, because of source or formatting.


text:
How does an atom's electronegativity change as it becomes more negatively charged?
I had a question about the carbon atom specifically. Carbon has 4 valence electrons, with electronegativity 2.55 (approx). If I was to fire electrons at the carbon atom, each carbon will occupy up to 4 electrons (8 in total in outer shell) to complete the octet. Now, let's assume that this happens slowly such that we have time to observe each carbon atom as it changes its charge, from $\ce{C-, C^2-, C^3-, C^4-}$
My question is: how would the electronegativity of the carbon ions will vary from each other? Will it increase as electrons are added and becomes zero after obtaining full octet? What would happen upon taking away electrons from carbons through help of highly electronegative elements?

Electronegativity is a measure of the ability of an atom to attract electrons to itself (usually in a bonding framework; outside of a bonding framework the ability to attract electrons is electron affinity).
So from simple Coulombic considerations, we can surmise that as your carbon becomes more and more negatively charged, the less it will be able to attract electrons to itself.

I'm not sure how this makes sense. You yourself say that electronegativity is defined in a bonded framework. But, the carbon anions in the question are not covalently bonded. How do they then have any electronegativity at all?


text:
How can one rationalize Avogadro's hypothesis using the equation of kinetic molecular theory?
How can one rationalize Avogadro's hypothesis using the equation of kinetic molecular theory?
$$pV = \frac{1}{3} \cdot N \cdot m\overline{u^2} = \frac{2}{3} \cdot N \cdot \frac{1}{2}m\overline{u^2}$$
$N$ is the number of molecules and $u$ is the average speed.

Pressure is the number of collisions with the container per unit area. Imagine a particle in a box of length $L$. Assuming for the moment that it only moves in the $x$-direction, whenever it collides with a wall of the container the wall will gain momentum from the particle:
\begin{align}
\Delta p &= p_{\mathrm{final}, x} - p_{\mathrm{initial}, x} \\
  &= p_{\mathrm{final}, x} - (- p_{\mathrm{final}, x})\\
  &= 2p_{\mathrm{final}, x}\\
  &= 2mu_x,
\end{align}
where $u_x$ is the $x$-component of velocity.
The particle hits a specific wall every $\Delta t = \frac{2L}{u_x}$, since it travels the length of the box twice. Force can be defined as momentum per unit time, therefore
$$ F = \frac{\Delta p}{\Delta t} = \frac{mu_x^2}{L}$$
For $N$ number of particles, the expression becomes
$$ F = \frac{Nm\overline {u_x^2}}{L},$$
where $\overline{u_x^2}$ is the mean square velocity of the $N$ particles.
Now, we can extend the situation so that our particle can move in three directions. Since the box is cubic, the $x$-, $y$-, and $z$-components are equivalent:
$$ \overline{u_x^2} = \frac{\overline{u^2} }{ 3}$$
Therefore,
$$ F = \frac{Nm\overline {u^2}}{3L}.$$
The area of any wall is $L^2$, so
\begin{align}
p &= \frac{F}{L^2}\\
  &= \frac{Nm\overline {u^2}}{3L^3}\\
  &= \frac{Nm\overline {u^2}}{3V}
\end{align}
This leads to the expression
$$pV = \frac{1}{3}Nm\overline{u^2}$$
See also Wikipedia.


text:
What would chemistry seniors like to hear about industrial hygiene?
I have a master degree in physical chemistry with close to 10 years experience as a chromatographer in pharma.  About two years ago I made a career change into industrial hygiene.
Right now I am prepping to give a senior seminar talk about my professional experiences as an industrial hygienist at the college where I earned my BS in chemistry.  The talk will be half lecture and half lab - I'll be bringing in some equipment to demonstrate.
What would chemistry students be particularly interested in hearing about from someone in industry?

Asking in Chemistry instead of Careers due to the specialized nature of the question.

To keep this answer as far from opinionated as possible, I'd suggest some of the W's:

What is industrial hygiene? (I've never heard of the term before this post and actually had to look it up myself.)
How do you do industrial hygiene? Procedures, risk assessment...
Why is industrial hygiene important? You might give examples where disaster was averted thanks to IH, and an example where it would have helped if done right.
Where is industrial hygiene applied? (I guess the office building will have less rigorous guidelines than the production facility.)
When did industrial hygiene start to be a thing? A small historical overview could be given, you could even give a who to this one if it was some distinct person who came up with the idea.

Always good is a personal anecdote, something you witnessed or someone (whom you trust as a source) told you, highlighting the bullet points above.


text:
Miller indices. What determines the appropriate axis?

So, first of all, what's up with this "plane" I see two planes!
The basis of my question is that to determine miller indices the plane cannot lie on your origin, so you have to make a translation. If I move my origin to the right then I get the miller indices that are indicated in the figure. But if I move backwards then I don't get the miller indices in the figure.
So my question then is: what determines the appropriate translation?

You want integer numbers, preferably possible smallest (except he 0,0,0 :). The physicists have a more proper definition for Miller indexes with reciprocal spaces etc.

There is no "appropriate" translation. You must remember that everything includes translational symmetry, including the Miller planes.
The Wikipedia page on crystal structures and Miller indices go into much more detail, but some key points:

The unit cells possess translational symmetry, so there are a set of Miller planes that also differ by translational symmetry.
The Miller indices (hkl) (which yes should be expressed in smallest possible integers) define the normal of the Miller plane.

That is, (hkℓ) simply indicates a normal to the planes in the basis of the primitive reciprocal lattice vectors. Because the coordinates are integers, this normal is itself always a reciprocal lattice vector. The requirement of lowest terms means that it is the shortest reciprocal lattice vector in the given direction.

So in your case, you have to remember if you "move the origin", you have to also consider the symmetrically identical unit cells around the one in your picture. As I said above, there are an infinite set of identical Miller planes that simply differ by the translational symmetry of the lattice.


text:
Why is the para product major in the nitrosation of phenol?
When phenol reacts with nitrous acid ($\ce{NaNO2 + \mathrm{conc}\ H2SO4}$), 4-nitrosophenol and not 2-nitrosophenol is formed. I cannot understand why the para-isomer should be preferred over the ortho-isomer.
I understand the reaction mechanism proceeds by electrophilic substitution by the $\ce{NO+}$ cation, and that the $\ce{OH}$ group activates the ring, which leads to ortho and para substitution.
I don't understand why the para-isomer should be preferred over ortho?

Ortho and para positions are activated due to the +R effect of the $\ce{-OH}$ but the preference of the para product can be explained by the mechanism.

The mechanism proceeds through a dienone. The ortho dienone intermediate will experience strain as  $\ce{=O}$ annd$\ce{=NOH}$ lie in the same plane. Para nitrosated intermediate on the other hand is free from strain.

Reference: González-Mancebo S., Lacadena J., García-Alonso Y., Hernández-Benito J., Calle E., Casado J. Monatsh. Chem. 2002, 133 (2), 157-166. DOI: 10.1007/s706-002-8245-0.

I feel this answer would be improved a lot by a drawing of the strained ortho-intermediate.

Due to steric hinderance and also there is some electronegativity of alcohol group due to induction it takes electron from sigma bond though not dominating therefore
These are the 2 reasons for more p- substitution

Sources, please.


text:
In water, will sucrose be hydrolysed by water to form glucose and fructose molecules?
If I dissolve sucrose sugar ($\ce{C_12H_22O_11}$) in water, would the water readily hydrolyse the sucrose sugar to form glucose and fructose sugars?

@M.A.R. Yes, but this question was asked before the other one, shouldn't it be the other way round?

Also I think if we are comparing these two questions, I can easily imagine two different people with different intents, each wanting to see a version of this similar question. This believe this is a valid case of duplication, if you want to view it that way. See https://stackoverflow.blog/2010/11/16/dr-strangedupe-or-how-i-learned-to-stop-worrying-and-love-duplication/.

It doesn't really matter. Being marked as a duplicate does not mean anything bad for your question, just that a question with a more comprehensive answer is linked at the top for future readers. Curious that you're linking Dr. Strangedupe, because I'm using precisely the same mentality here. Love the duplication, and stop worrying about it!

Not without a catalyst present. The process is usually performed using invertase, but some other enzymes with similar activity exist and addition of acid also can catalyse the reaction.