text:
How to determine the Bravais lattice and atom basis vectors from a CIF file?
Say I have a CIF file describing some material in terms of its symmetry space group, lattice parameters and in-cell atom positions. A simple example might be,
data_global
_chemical_name 'Graphene'
_cell_length_a                 2.46
_cell_length_b                 2.46
_cell_length_c                 1
_cell_angle_alpha              90
_cell_angle_beta               90
_cell_angle_gamma              120
_symmetry_space_group_name_H-M 'P 3 m 1'
loop_
_atom_site_label
_atom_site_fract_x
_atom_site_fract_y
_atom_site_fract_z
C  0.00000   0.00000   0.00000
C  0.33333   0.66667   0.00000

I would like to extract from this Bravais lattice vectors and atom basis vectors. For this specific example, this might look like,
 2.456   0.0     0.0
-1.228   2.126   0.0
C
 0.0      0.0      0.0
 0.0      1.418    0.0

where the first two lines give the two lattice vectors (in some basis), while the next two give the positions of the atoms in the unit cell.
I have tried using Open Babel for this, converting from the CIF format to a VASP one.  This works well for the example above, but fails for a material such as tin sulfide (SnS), which has a structure similar to black phosphorus---four atoms in a unit cell, only two of which are explicitly listed in the CIF file.  The positions of the other two are implied by the symmetry group, and programs such as Mercury correctly visualize the structure based on the CIF file.  However, converting the CIF into VASP only gives you two of the four unit cell atoms.  I tried all other formats in Open Babel, with no success.
Is there a reliable way to convert a CIF file into a description in terms of a Bravais lattice plus atom basis vectors?

You can use Open Babel. You just need to "fill" the unit cell (i.e., generate all the symmetric atoms).
You just need the --fillUC option, documented here:

For a crystal structure, add atoms to fill the entire unit cell based on the unique positions, the unit cell and the spacegroup. The parameter can either be strict (the default), which only keeps atoms inside the unit cell, or keepconnect, which fills the unit cell but keeps the original connectivity.

e.g.,:
obabel SnS.cif -oVASP --fillUC >POSCAR
Result is:

Herzenbergite
1.000
   4.330000000000000   0.000000000000000   0.000000000000000
   0.000000000000001  11.180000000000000   0.000000000000000
   0.000000000000000   0.000000000000000   3.980000000000000
Sn  S   Sn  S
1   1   3   3
Cartesian
     0.4979500000000001148      1.3192399999999999682      0.9949999999999999956
     2.0697400000000003573      9.5030000000000001137      0.9949999999999999956
     0.4979500000000001148      1.3192399999999999682      0.9949999999999999956
     0.4979500000000001148      1.3192399999999999682      0.9949999999999999956
     0.4979500000000001148      1.3192399999999999682      0.9949999999999999956
     2.0697400000000003573      9.5030000000000001137      0.9949999999999999956
     2.0697400000000003573      9.5030000000000001137      0.9949999999999999956
     2.0697400000000003573      9.5030000000000001137      0.9949999999999999956

I would recommend you to use vesta (http://jp-minerals.org/vesta/en/). It can allow you to convert almost all types of quantum chemical coordinate file to your desitred one. To get VASP format, just drag you CIF file on vesta and it will show you the structure and then you can save it as POSCAR format.


text:
What is the difference between moisture regain and water holding capacity?
Water retention is how much water the fibre can retain. Water holding is function of hydrophilicity and construction of the fibre (fabric). What is the difference between those two? Because to retain is also to hold, in my knowledge.

According to the FabricLink Textile Dictionary:

Moisture Regain - The amount of water a completely dry fiber will absorb from the air at a standard condition of 70 degrees F and a relative humidity of 65%. Expressed as a % of the dry fiber weight.

Water Holding Capacity is the amount of water a completely dry fiber will retain after soaking in liquid water and allowed to "drip dry".

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sort: votes


text:
What's so special about chelation?
There is another closely related post here , and I've also read the referenced wiki article including applications of chelation, but I still don't see what's so special about chelation. I understand the structure of the bond - the 'capturing' of a metal cation by an organic agent, but I still fail to see why this chemical reaction might be so special relative to other possible organic chemical reactions with a metal cation.
Why is chelation so special?

Chelation is an entropy driven reaction, i.e. complexes with polydentate ligands are more stable.
Consider what happens when you dissolve copper sulfate in water:
\begin{align}
\ce{CuSO4 + 6H2O <=> [Cu(H2O)]^2+ + SO4^2- (aq)}\\
\end{align}
This complex forms and with it comes some kind of ordering of the water molecules around it. Dissolution is often favoured because of the solvation energy. Now consider a bidentate ligand like ethylene diamine, $\ce{H2N-(CH2)2-NH2}$, short $\ce{en}$. Adding this ligand to a copper sulfate solution will result in a ligand exchange:
\begin{align}
\ce{[Cu(H2O)]^2+ + 3en <=> [Cu(en)3]^2+ + 6H2O}\\
\end{align}
In this reaction you increase the number of particles from four on the left side to seven on the right side. Taking into account that you also disturb the order of the surround water molecules, this reaction is very much favoured by entropy. In an attempt of zeroth order approximation, one could assume that entropy roughly doubles. Since the Gibbs energy is inversely correlated to the entropy, this means, that the reaction will be spontaneous. This is usually referred to as the chelate effect.$%edit$

To complement Martin's answer with regard to the applications: chelation is particularly useful in the body with regard to iron transportation in siderophores and porphyrin rings and also in medicine where excess metal ions build up such as Wilkins disease ($\ce{Cu^2+}$) etc. Typically you see drugs with connected phenoxides, hydroxamates or EDTA fragments exactly for this purpose directed to iron deposits in the liver. Upon complexation you remove the chelated product easily. Without the Gibbs drive, metal plaques up. In addition catalysis and organometallic chemistry is full of chelation chemistry.

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text:
What happens when I use cyclohexanol to prepare cyclohexene?
I've done a student experiment using cyclohexanol to prepare cyclohexene.
We mix $10.5~\mathrm{g}$ cyclohexanol and $5~\mathrm{mL}$ $85\%$ $\ce{H3PO4}$ in a $50~\mathrm{mL}$ flask, then distill while heating.
I want to know during this process what kind of side reaction happens because I notice there is something light yellow in the flask. After reaction, the liquid remaining in the flask is red and upon it there is something dark green and muddy.
I have a guess that there is some phenol in the cyclohexanol (since cyclohexanol may be prepared from phenol), and that during reaction the phenol oxidized, producing 1,2-benzoquinone (which is red) and 1,4-benzoquinone (which is yellow). Is this idea reasonable? If it is, why does the red color appear later?
And about the dark green thing: can it be produced by some kind of carbonized organic matter? I have been told that if the $\ce{H3PO4}$ and cyclohexanol were not well mixed, there would be some cyclohexanol carbonization.
(Can I ask one more question?) We are required to compare the chemical properties of cyclohexene and cyclohexane. So we reacted them with $\ce{Br2}/\ce{CCl4}$ respectively, under light or covered by aluminium foil. Should it be different when cyclohexene reacts under light or covered by aluminium foil? I notice the result is slightly different; the reactions in the light become a little opaque.

Hmm, it's strange - have done everything right?

@Mithoron I don't know... but other classmates also get something like this. However, most of them only get something light yellow. Only some get red.  I know our reagent may not be that pure - they want to save cost.

Light yellow should be ok - cyclohexene - it's not purified so it's slightly colored. Othrwise you probably did sth wrong.

Cyclohexene gets probably polymerized

The acid-catalyzed dehydration of alcohols involves a carbocation intermediate, which can act as a Brønsted acid; the loss of a proton yields the desired cyclohexene. Furthermore, the carbocation may undergo other side reactions. In particular, the carbocation may be subject to a reaction with other molecules to give substitution products; or the carbocation may undergo carbocation rearrangement.
The coloured residue of the distillation is probably the result of cationic polymerization of cyclohexene. This reaction in combination with rearrangements may yield various oligomers.


text:
Intensive and extensive properties of gases
Every chemistry textbook I have ever seen defines intensive and extensive properties the same way: Intensive properties are those which do not depend on how much matter is present, while extensive properties do. I came across a question while looking through sample FE exam questions for which my understanding of these properties by this definition failed. The gist of my question really just revolves around pressure and volume in largely compressible substances (i.e., gases).
It seems that, when considering a gaseous substance, these definitions are not robust enough:

Volume can be shown to be independent of the amount of material present by simply compressing the gas.
Density can be shown to be dependent on the amount of material present by keeping a fixed volume and simply adding more gas molecules.

While I have heard the convincing argument that we should simply consider smaller subsets of the original to determine intensiveness and extensiveness (for instance, if I consider half of the container instead, the volume halves from $V_0$ to $\frac{1}{2}V_0$ and the density remains constant at $\rho$), this doesn't help me to identify why the conventional definition remains in use with what seems like flawed implications. Can anyone help to clarify this?

To give you a more clear explanation of the definition of these properties, it's more easy to remember them this way :
-Extensive properties are those which *are additive * ( mass, volume ) for example 1 L + 1 L = 2 L. On the other hand, intensive properties are not additive ( temperature, pressure, concentration ) for example 2 mol/L + 3 mol/ L does not give you a 5 mol/L solution. These are properties which don't change at any circumstance.
Some other thing to be considered :
The ratio of two extensive properties, gives us an intensive property. For example density= mass/volume. Since it is a ratio it will always be the same regardless of how much of a sample is measured.

While I do understand your answer and think it is helpful, I don't think it helps to explain why the definition which is typically presented in chemistry textbooks is still valid. Simply saying that an extensive property "depends on the amount of matter present" would imply that density is extensive.
For this reason I am wondering less about a correct way to interpret it and more about why the definition that discusses the amount of mass is the standard one when it seems to fail in the case of density.

Also, you say that intensive properties are not additive, but pressure is an intensive property, and if I take 2 mol of a gas at a particular pressure and add another 2 mol of gas that is at the same pressure, the total pressure (i.e., the sum of the partial pressures) is the additive sum of their individual pressures. How does that fit into this definition?

I edited my answer for your density question,  moles do not define the pressure !  2 mmHg and 2 mmHg are not equal to 4 mmHg.

I am confused by your answer. After reviewing my chemistry textbook, it states very clearly that if I combine two gases, each with a partial pressure of 2 mmHg, the total pressure is unequivocally 4 mmHg. Am I misunderstanding your point?

Well it is UNequivocally, that's what I also said. You may probably have troubles because some other factors you must clearify  such as Dalton's law and Rault's law.


text:
What is the necessary volume of concentrated HCl to prepare a solution with a pH of 1.8?
I am trying to solve the following exercise:

What volume of a concentrated $\ce{HCl}$ solution, which is $36.0\%$ $\ce{HCl}$ by mass
  and has a density of $1.179~\mathrm{g~mL^{-1}}$, should be used to make $5.00~\mathrm{L}$ of an $\ce{HCl}$
  solution with a $\mathrm{pH}$ of $1.8$.

I found the respective volumes that $\ce{H2O}$ and $\ce{HCl}$ contributed.
I then found the molarity of the $\ce{HCl}$ aqueous solution given.
I used $10^{-\mathrm{pH}}$ to get the amount concentration of $\ce{HCl}$ needed, I then proceeded to find the amount of substance of $\ce{HCl}$ by multiplying the concentration needed by $5.00~\mathrm{L}$. I then took the amount of substance of $\ce{HCl}$ needed and multiplied by the density given to get an answer of $2.4~\mathrm{L}$.

Am I correct?

Unfortunately no, you're not right. As mentioned in a now deleted comment by Bruno, your answer gives a $\mathrm{pH}$ of $-0.74$, which is not correct. You didn't give detailed steps of your calculations, and your first step doesn't seem like anything we should do.
We can solve this by making the calculations step by step. Let's start with the concentration of $\ce{H+}$, which will be the same as the concentration of $\ce{HCl}$, we will achieve:
$$\ce{[H+]} = 10^{-\mathrm{pH}} = \pu{0.01584893 mol//L}$$
Since we're going to prepare 5 liters, this means the amount of substance of $\ce{H+}$ should be
$$5 \times \pu{0.01584893 mol} = \pu{0.07924466 mol}$$
To transform this into amount of $\ce{HCl}$, we need the acid's molar mass ($\pu{36.46 g//mol}$) to convert from amount of substance to mass:
$$\require{cancel}%
g_{\ce{HCl}}
= \mathrm{\frac{36.46\ g}{\cancel{mol}}}
  \times 0.07924466\ \mathrm{\cancel{mol}}
= 2.88926 \ \mathrm{g}$$
Now that we have the amount of substance of $\ce{HCl}$, we calculate the volume, which is where the density comes in:
$$V_{\mathrm{HCl}}
= \mathrm{\frac{2.88926\ \cancel{g}}{1.179\ \cancel{g}\ mL^{-1}}}
= 2.450602 \ \mathrm{mL}$$
Now the last step: Our stock solution isn't pure $\ce{HCl}$, it's diluted in water. So actually the volume we need is:
$$V_{\mathrm{HCl}}
= 2.450602 \times \frac{100}{36}
= \pu{6.807229 mL}
\approx \pu{6.8 mL}$$
This might look that it is not much, but remember $\ce{HCl}$ is a very strong acid. For example, to prepare $\pu{1000 mL}$ of a $\pu{0.001 M}$ solution you only need $\pu{0.086 mL}$ of concentrated $\ce{HCl}$!
You can check this result by imputing the given density, weight percentage, final volume and $\ce{H+}$ concentration (first step) into Sigma-Aldrich's Molarity Calculator.

1.000 L of 36.0 % HCl with the density of 1.179 g/ml will weigh 1179 g and will contain 0.36*1179 g of HCl or 0.36*1179/36.46 = 11.641 mole of HCl.
At pH = 1.8 => [H3O+] = 0.015849 (M).
To make 5.0 L solution we need 5.0 * 0.015849 = 0.079245 (mole).
Suppose we need x liter of the conc. HCl solution:
x*11.641 = 0.079245 => x = 0.006807 (L)
Answer: 6.8 ml.

It's simple. We all know that density = Mass/volume.
The density of your hcl is 1.179
Mass of hcl in one molar solution which is the molar mass is 36.5
but the percentage purity of the hcl is 35% according to you. That means, 35% of that 36.5g is in 1 molar/dm3.
Which means, your mass will be 35% of 36.5
=36.5*35/100
35.5*0.35=12.775
Therefore, 1.179=12.775/v
v=12.775/1.179
v=10.84 (approx)
This is the volume of your conc hcl needed to prepare 1M in 1 dm3
But you want to prepare 5 dm3
5 dm3=10.84*5=54.20
The simple formula that can guide you is
Molar mass times percentage purity everything over density
MM x pp
D
The volume u get using the above formula will prepare 1M in 1 dm3 so, you can now multiply by the number of molar concentration u want to prepare.
Spasiba

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text:
Why does solubility of solids sometimes decrease with temperature increase?
Recently I was researching $\ce{Ca(OH)_2}$, and I found this on Wikipedia:
\begin{array}{|c|l|}\hline
{\rm \color{blue}{Solubility~in~water}}
 & {\mathrm{0.189\,g/100\,mL}\   (0\,^\circ\mathrm{C})} \\
 & {\mathrm{0.173\,g/100\,mL}\  (20\,^\circ\mathrm{C})} \\
 & {\mathrm{0.066\,g/100\,mL} \ (100\,^\circ\mathrm{C})} \\\hline
\end{array}
Can you explain why solubility decreases when temperature increases?

In short, for this group 2 hydroxide crystal, the dissolution and solvation process is an exothermic equilibrium. Increasing the temperature drives the reaction in the reverse, favouring crystallisation!
However, it would be nice to rationalise some of these terms, so I will lay out a very brief overview of general solubility factors so you can understand the process in a little more detail.
Solubility is a complex thermodynamic process that takes many factors into account. It can be understood as the process of dissolving a crystal in a solvent, and then forming a solvent cage around the solute (solvation). It is a function of three interactions at heart;

solute-solute
solute-solvent
solvent-solvent

Large solute-solute interactions hinder solubility and are associated with high lattice enthalpies (of crystalline solutes), these must be overcome in order to break up the solid and dissolve it. Solvent-solvent interactions must also be overcome in order to solvate the dissolved solute, therefore for a highly interacting solvent the solubility is reduced.
The formation of ordered solvent structures such as clathrin cages etc orders the solvent, and therefore reduces solvent entropy significantly. Therefore solvents with high degrees of freedom hinder solubility.
Solute entropy favours solvation, this reflects the higher solubility of flexible molecules, this is because in the solid they are all locked up, while in solution their rotational and vibrational modes are populated.
In addition to these if the solute has multiple ionic forms then the solubility will be a function of pH too, so must be held constant in practice. The surface profile of the solid too, as well as the extent of cavitation and porosity are also key factors (since they affect the number of solvent-solute interactions possible).
Importantly the type of solvent into which you are trying to dissolve your crystal in is also of fundamental importance! The rule "like dissolves like" is best understood by the (semi-empirical) general solubility equation,
\begin{equation}
\log S =\frac 12-0.05 (T_m-T)-\log P
\end{equation}
A large partition coefficient (the ratio of how soluble the crystal is in organic solvent to aqueous solvent) reduces the solubility, (and mathematically since $\log x$ follows $x$ in limits).
To further complicate the situation there are several definitions of solubility that we can take. In practice we choose intrinsic aqueous solubility, mainly used for high throughput and speed in drug development.
Solubility is really an experimentalists tool for parametrising the extent of Gibbs energy for the dissolving-solvation process, which in turn is usually modelled by complex thermodynamic cycles via gas phase species or super cooled liquids. Modern research into solubility focuses on finding either molecular descriptors through random forest/QSPR or support vector models etc (which are all just number crunching techniques to find correlations between large sets of data) or through fundamental physics and chemistry of the process.
\begin{equation}
\log S=\frac{(-\Delta G_{sol}/RT)}{\ln 10}
\end{equation}
So we understand solubility as an equilibrium constant for the Gibbs change of the solvation process (whatever particular cycle you use to get that data).
Now that we know something about solubility we can ask ourselves, under what conditions does increasing the temperature reduce solubility?
In practice the temperature can effect the entropic term, but also the enthalpies of the crystal lattices, the heats given out by the solute-solvent interactions etc. It is the balance of these process that determines the solubility of something.
In your case we are observing a calcium hydroxide crystal dissolve in aqueous solvent. As we increase the temperature the solubility decreases. There is an equilibrium between crystallisation and (dissolution + solvation). Increasing the temperature is shifting that equilibrium to favour the crystallisation rather than (as we crudely expect) in the forward. This is simply an enthalpic term, due to the solvation process being an exothermic process, increasing the temperature drives the equilibrium in the reverse by Le Chatlier's principle.
Hope that is a bit clearer, and that you now understand some of the ideas behind solubility. :)

Is it possible to explain the effect through Le Chatlier's Principle as follows: since increasing the temperature causes increased dissociation of water molecules, the amount of hydroxide molecules increases and this causes the equilibrium to shift backwards, hence reforming calcium hydroxide (something in relation to the Common-Ion Effect)?

2 tennispro1213: I would not rely on that. After all, there are some ionic salts with similar behavior, and they definitely don't share common ions with water.

Like this answer very much. However, I'm not sure why you rely only on enthalpy, being that entropy changes with temperature.

I would also like to add another answer which may give a better understanding of the application of Le Chateliers principle over here.
Dissolution of $\ce{Ca(OH)2}$ is an exothermic process ($\Delta_\mathrm{sol} H<0$)  thus it can also be represented as:
$$\ce{Ca(OH)2(s) + H2O(l) <=> Ca^2+(aq) + 2OH- (aq) + \Delta}$$
At particular temperature a particular net amount of heat is evolved in the system during this process ,thus if the temperature was increased, the heat content of the system would increase and there would be stress on the right hand side of the equilibrium representation since heat is being evolved there,so by Le Chateliers principle the system would consume some of that heat by shifting the equilibrium to the left to reduce the stress,thereby producing more $\ce{Ca(OH)2}$ in solid state thus decreasing the solubility.
Conclusively

In general,if in a nearly saturated solution,the dissolution process is endothermic ($\Delta_\mathrm{sol} H>0$),the solubility should increase with rise in temperature and if it is exothermic ($\Delta_\mathrm{sol} H<0$) the solubility should decrease.These trends are also observed experimentally.

Please check out this post =3

If we increase temperature and then decrease it (equal to initial temperature ) will the solubility change?

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text:
What is the difference between the esterification and transesterification?
What is the difference between an esterification and a transesterification. Esterification objective is the formation of an ester by reacting an alcohol and a carboxylic acid. I am more concerned transesterification or reference to a journal article.

esterification is the formation of an ester from an acid and alcohol, transesterification is the switching of an existing esters alcohol fragment for another alcohol.

yes I know. But is there any definition for it? I want to use it in my paper-work. Btw Thx very much!

@user16060 That is the definition as AngusTheMan stated

Angus, esters can be made in many ways, your definition is the "Fischer esterification" only. Also once they switch they are no longer alcohols. Ahem.....Homework question!

You are getting too specific as though these are biological processes. They do not have to be... ANY ester made is an esterification reaction.
So any ester can be transesterified into a different alkyl alkanoate and any carboxylic acid or derivative (even nitriles) can be esterified. See the diagram:


text:
What is the value for the reverse reaction?
My question deals with the reaction $\ce{2NO(g) + O2(g) -> 2NO2(g)}$ with $K_c = 1752$.
This is what I have so far; $K_c$= products over reactants, so is the reverse reaction reactants over products to find the answer
Such as
$$1752 = \frac{[\ce{NO2}]^2}{[\ce{NO}]^2[\ce{O2}]}$$
So then does the answer become
$$1752 = \frac{[x]^2}{[x]^2[x]}$$

${K_c}$ of reverse reaction should just be the reciprocal of forward ${K_c}$. So, ${K_c}$ of reverse reaction is $\frac{1}{1752}$.
Now, I am not sure, why you used 2nd equation. Suppose you have 1 unit conc of $\ce{NO}$ and $\ce{O_2}$, then after time $t$, if the conversion is $x$, then your equation should be
$1752 = \frac{[2x]^2}{([1-x])^3}$.
Can you please check your 2nd equation again?


text:
Why/How am I calculating freezing point depression wrong?
What is the freezing point of water made by dissolving $\mathrm{10.11~g}$ of magnesium chloride in $\mathrm{86.03~g}$ of water?
The freezing-point depression constant of water is $\mathrm{1.86~^\circ C~m^{-1}}$.
To start I take the $\frac{10.11~\text{g}\ \ce{MgCl}}{86.03 \times 0.001 \times 1.86~\mathrm{^\circ C~m^{-1}})}$, which I keep getting as $\mathrm{63.18~^\circ C}$.
What am I doing wrong?

How did you manage to move the temperature from the bottom to the top? Oh wait, your equation is wrong. It should be $-1.86°/molar \times {10.11 g MgCl_2 / 95.21 g/mol \over 86.03 g \times 1kg/1000g}$.

The equation for freezing point depression is this:

$ΔTf=−K_f\cdot m \cdot i$.
Where $ΔT_f$ is the freezing point depression, $K_f$ is the freezing point depression constant $m$ is the molality of the solute, and $i$ is the van 't Hoff factor.

The van 't Hoff factor is the number of moles of ions the solute dissolves into, which in the case of $\ce{MgCl2 -> Mg^2+ + 2Cl-}$, is 3.
Your problem here is that you used the mass of $\ce{MgCl2}$ ($\ce{MgCl}$ is incorrect) instead of its molality, and did not include the van 't Hoff factor.
Molality is calculated by dividing moles of solute by kilograms of solvent. In this case,
$\frac{0.1062 \: \mathrm{mol  \:of}  \:   \ce{MgCl2}}{0.08603  \:  \mathrm{kg \: of}  \:  \ce{H2O}}$.
This shows that the molality of $\ce{MgCl2}$ is $1.234 \: \mathrm{mol~kg^{-1}}$. If you use this value instead of mass to calculate $ΔT_f$, the value comes out to be $-6.888 \:\mathrm{°C}$.


text:
finding cation of unknown salt in solution from precipitate
I should probably add that I was originally given an unknown solid mixture of 2 salts. I separated the insoluble salt out by centrifuging and then decanting. I found that the insoluble salt is $\ce{MgCO_3}$ by adding $\ce{HCl}$ (bubbles) and $\ce{NaOH}$ into the solution (jelly like white ppt). The decanted liquid was then what I observed as having 1 soluble salt. The soluble salt I know by testing for $\ce{SO4^{2-}}$ with $\ce{BaCl_2}$ and observing a violet flame in a flame test. Then I had a second bottle with 3 unknown, 2 of which I discovered above. I know that the 3rd one has 2 soluble and then my original question follows.
I have 2 soluble salts in aqueous distilled water solution. I know that one of the salts is potassium sulfate ($\ce{K_2SO_4}$) and I know that the anion of the second salt is chloride ($\ce{Cl^{-}}$). I know that the second salt is either magnesium, sodium, calcium or potassium chloride ($\ce{MgCl_2}$, $\ce{NaCl}$, $\ce{CaCl_2}$, or $\ce{KCl}$).
I tested for the anion for the second salt by adding silver nitrate ($\ce{AgNO_3}$) into the solution and observing a white precipitate. When I added sodium hydroxide ($\ce{NaOH}$) to it, I observed that the cloudy white precipitate disappeared and turned into small black precipitate which then settled to the bottom of the tube. Is there any way to get the cation of the second salt with this information, and are there any other test(s) I can do to find what is the cation for the second salt?

Among the options for the second salt in solution, one is less likely to be found in the presence of $\ce{K_2SO_4}$.
These solutions for analysis are prepared by adding decent amounts of two salts with good solubility to water. Under these conditions, a double displacement reaction may occur and in one case, a precipitate will be formed:
$\ce{K_2SO_4 + CaCl_2 -> 2KCl + CaSO_4}$
I doubt that the person preparing the solutions for analysis used calcium chloride, since
this would have required to remove the precipitate by filtration or decanting off the supernatant solution
the concentration of $\ce{Ca^{2+}}$ in the solution would be rather low. Remember that these qualitative tests are not necessary suitable for trace analysis.
The second salt therefore is most likely sodium chloride ($\ce{NaCl}$), potassium chloride ($\ce{KCl}$) or magnesium chloride ($\ce{MgCl_2}$).
I would do a flame test first. If you see a yellow coloured flame, the second cation is the sodium cation ($\ce{Na+}$). Similarly if you see a lilac coloured flame, the second cation is the potassium cation ($\ce{K+})$ and if you initially see a colourless flame and finally see an intense white coloured flame,the second cation is the magnesium cation ($\ce{Mg^{2+}}$).
If you want to test for the magnesium cation ($\ce{Mg^{2+}}$), place a drop of the solution on a spot plate, make sure it's alkaline by adding aqueous $\ce{NaOH}$ if necessary, and add a bit of a titan yellow solution. This azo dye with its misleading name (there is no titanium in it) will form a red precipitate with the magnesium cation ($\ce{Mg^{2+}}$).

Note: yellow flame is not necessarily sodium! Sodium flame coloring is so intense it is known to overwhelm all other colors. So, if you get violet (potassium) or red (calcium), it is valid, but yellow may arise from trace amounts of sodium!

Thanks for the suggestions. I should probably add that I was originally given an unknown solid mixture of 2 salts.I separated the insoluble by centrifuge and decanting. I found that the insoluble salt is MgCO3 (by adding HCl (bubbles) and NaOh into that(pcp jelly like white)). The decanted liquid was then what I observed as having 1 soluble salt. The soluble salt I know by testing for SO4 with BaCl2 and observing a violet flame in a flame test. Then I had a second bottle with 3 unknown, 2 of which I discovered above. I know that the 3rd one has 2 soluble and then my original question follows.

UPDATE: Looks like I don't have access to Titan Yellow, I will try to acquire either Azo Violet or 4-(p-nitrophenylazo)resorcinol


text:
Azimuthal Quantum Numbers
Why do azimuthal quantum numbers have a value lesser than Principal Quantum number.
example-
if n=1
it will have subshell "s" as 0

In order to understand the exact reason for this you will have to study quantum mechanics. It is simply something that evolves naturally from  the Schroedinger equation of the hydrogen atom. You should be able to find the proof for that in basically every theoretical physics textbook on the subject.

See http://chemistry.stackexchange.com/q/31639/5017 for more

@Philipp, I might guess the OP would like an explanation a little bit friendlier to the novice than your average theoretical physics textbook

@BenNorris You are certainly right, but I couldn't think of an intuitive way to show why the azimuthal quantum numbers are bounded. From your answers on this site I gather that you are pretty proficient in quantum mechanics so I guess you know about the derivation of the angular momentum operator's eigenvalue spectrum for the hydrogen atom. I feel this topic lacks approachablilty via non-mathematical means and simply cannot be explained without delving into the math but if you can think of a nice layman-explanation I would be very interested to see it.


text:
How to interpret TD-DFT results for finding λmax and amplitude
I am trying to understand how to interpret the TDDFT results ( I know there is another post for this but it does not answer my question ).   Here is part of the result from GAMESS:

I downloaded Chemissian and this is its result for this state:

I have two related questions:
1-How did it calculate this 1.41 from the amplitude of this state ?
2-Does it have anything to do with λmax ?

Which reaction are you talking about? The question is vague. Energy can be transferred through collisions even in liquids. And such collisions can lead to emission of electrons. The Helium-Neon laser operates like this http://en.wikipedia.org/wiki/Helium%E2%80%93neon_laser

that should just be the transitions of the orbitals and your $\lambda_\mathrm{max}$ should be somewhere around $9.3~\mathrm{eV}$ judging from the spectrum. Note that the state you have selected has 0.000 oscillator strength.

@Martin-マーチン this is the output for Acetone and related to my other question about Acetone behavior. I am confused what is going wrong and how this 1.41 is calculated

These are terrible tags! Reaction mechanism? Heat? Transition state theory? They don't make any sense.

You are looking at the wrong numbers.
You should look for excitation  that has non-zero oscillator strength. In the text file "State # 1,2 .." and in Chemissian on the left, list with columns "# Energy Strength" The transitions with zero or very small oscillator strength has a very low intensity in the spectrum, so you don't see them.
In your case the relevant excitation are #2 (Energy: 6.691 eV, Strength: 0.027) and #4 (Energy: 8.353 eV, Strength: 0.042), and you can see them as small peaks on the graph down.
The "amplitude .. 1.41" has nothing to do with the oscillator strength. It tells you which orbital contributions are the most important ones for the given excitation.


text:
Chemiluminescence vs. Bioluminescence
Luminescence in general, is the term of emission of light from specific substances/objects. Chemiluminescence is the emission of light due to chemical reactions. Bioluminescence on the other hand is the emission light from living/biological organisms such as insects ( this is of course the way I understood them ), but I'm a little bit confused because what is the cause that makes insects emmit light? Doesn't that also have to do with chemical reactions but inside living organisms ?
Just a clear explanation of the difference  between these two terms  would be helpful.

Yeah, you're right.

Simply searching the title of your question into Google yields a multitude of sources that can explain the difference between the two, including this one. Try Google first before asking.

Doesn't that also have to do with chemical reactions but inside living organisms?

Exactly! That's why there is one scientific society for both phenomena: ISBC - International Society for Bioluminescence and Chemiluminescence ;-)
The thermal reactions, that lead to an excited species which eventually emits a photon are quite similar. From a reactive compound, typically called a luciferine, a four-membered ring, often a dioxetanone is formed. This decarboxylates and gives rise to a molecule in an electronically excited state. In reality, the decarboxylation is often not just a spontaneous process but requires an activator. W. J. Baader at IQUSP in Sao Paulo, Brazil has examined that in great detail.
The luciferines can be quite different compounds, such as

diaryloxalates (these are used in glow sticks, where the diferent colours come from diferent fluorphores that absorb the initial photon and emit at different wavelength)
benzothiazole derivatives (firefly)
triazine derivatives (marine organisms)


text:
Entropy change in an adiabatic expansion

Question
  One mole of ideal gas initially at a pressure of 1 atmosphere and $T = 298\ \mathrm K$, is
  expanded into a volume 50 % larger adiabatically.

In this adiabatic process, no work is done on the surroundings and $\mathrm dQ=0$. And entropy is defined as $\mathrm dS=\frac{\mathrm dQ_\text{rev}}{\mathrm dT}$. However, there must be a change in entropy surely?
Is the definition $\mathrm dS=\frac{\mathrm dQ_\text{rev}}{\mathrm dT}$ only at a constant volume? Because I have also seen entropy defined as $\mathrm dS=\frac{C_V}{T}\,\mathrm dT+\frac{R}{V}\,\mathrm dV$ and there would be a change in both temperature and volume. However, how do I find what these changes are? Would it make a difference if the expansion was reversible or irreversible (the question didn't make it explicitly clear)

I am sorry to be posting this exact same comment in every single question about entropy but I believe that proper notation is important. The definition of entropy is $\mathrm{d}S = \mathrm{d}q_{\text{rev}}/T$ and not $\mathrm{d}S = \mathrm{d}q_{\text{rev}}/\mathrm{d}T$. The latter is a ratio of two infinitesimals (i.e. it is a partial derivative) and is not an infinitesimal; in fact it is the definition of heat capacity (whether it is the heat capacity at constant pressure or at constant volume depends on the process for which the quantity $\mathrm{d}q$ is measured).

Is it possible to look at the problem from the other definition of entryopy $S=klnW$?  Clearly the number of possible microstates accessible to the gas is changing so S has to change.

"One mole of ideal gas initially at a pressure of 1 atmosphere and T = 298 K, is expanded into a volume 50% larger adiabatically."

The question doesn't have sufficient information for a solution, because we don't know if the adiabatic expansion is reversible or irreversible.
If reversible:
If the expansion is done reversibly, then we know entropy of the universe can't be changing (because reversible), and that heat flow from the gas to the surroundings is zero (because adiabatic).  Without heat flow, the only way the entropy of the universe can stay constant is if both the entropy of the universe and of the gas stay constant.  In this case you can apply both of your equations.  But keep in mind that your equation for $dS$ is only true for ideal gases.
$dS = \frac{\delta q_{rev}}{dT} = 0$
$dS = 0 = \frac{C_V}{T}dT + \frac{R}{V}dV$
$\frac{dT}{T}=(-\frac{R}{C_V})\frac{dV}{V}$
This is now a differential equation you can solve for temperature if you know volume.
As the other answers have indicated, a reversible expansion will involve the gas doing work on the surroundings.  One way to think about why this is so is that for the process to be reversible, the potential of the gas to do work must not be lost, and so if this potential for doing work doesn't remain in the gas, it must go to the surroundings.  A loss in the potential to do work (of the universe) is an increase in entropy.  And in a reversible process that can't happen.
Think of gas in a piston, where the piston is held down by a pile of sand.  The reversible expansion involves slowly removing sand, grain by grain, from the piston.  As the grains are removed, the piston rises and the gas lifts the remaining sand, doing work against gravity.
If irreversible:
...then we still don't have enough information because there are many ways the process can be irreversible.  But one particular, extreme case of irreversibility is if the adiabatic expansion occurs without transfer of any energy, then we can solve the problem.  In this extreme scenario, neither heat nor work are not transferred.  You can thus infer that $dU=0$ for the gas, so
$dU=-TdS + p dV=0$
$TdS = p dV$
Using your definition for $dS$ for an ideal gas,
$T \frac{C_V}{T}dT + T \frac{R}{V}dV = p dV$
Keep in mind that $\frac{RT}{V}$ for an ideal gas is $p$, so you should get that
$C_V dT = 0$
Which is also a differential equation that you can integrate to get the temperature change.  (The obvious solution is that the temperature does not change in this case.)
In this second case, now that we know that $T$ is constant, we know $dT$ is zero, and thus using your equation for $dS$, we get that $dS = \frac{R}{V}dV > 0$, i.e., the entropy of the gas increases, which makes sense because the process was irreversible.  The potential to do work has been lost.
Think of gas confined to one half of a cylinder, and the other half of the cylinder is empty (i.e. vacuum).  If the divider between the filled and unfilled halves is suddenly removed, that would be a case of adiabatic, isoergic expansion.

In calculating the change in entropy of a closed system that has undergone an irreversible process, you need to first focus exclusively on the initial and final equilibrium states.  You then need to identify a reversible process path between the exact same pair of equilibrium states.  The reversible process path does not necessarily  have to bear any resemblance to the actual irreversible process path.  Therefore, even if the irreversible process path is adiabatic, the reversible path does not have to be.  You then calculate the integral of dq/T for the reversible process path.  This is equal to the entropy change.
Pretty simple, huh.  I don't know why it isn't taught properly to students.
Regarding the original question of why the equation for dS in terms of dT and dV for an ideal gas gives the correct answer, it is because the equation applies to the changes in dS, dT, and dV over a continuous sequence of thermodynamic equilibrium states. This is the same thing as a reversible path. So, all one needs to do is to integrate this equation from the initial thermodynamic equilibrium state to the final thermodynamic equilibrium state determined obtained with the irreversible process. Note that the two terms on the right hand side of the equation are exact differentials.

An important thing to know is that on the adiabatic processes, the system is isolated, due to that no heat can be given or taken, but that doesn't mean that the volume can't change. Because we can't take heat, we can lower/rise the temperature from T1 to T2 . We don't have any work done only when the volume is constant. Entropy will be at a constant volume or pressure only when you see the index v ** or **p in the formula. If the process would be reversible ( which i think it is) only the change of temperature should be in infinitezimal amounts. Hope I understood the question right

So does entropy change in this situation?

Yes it does, everytime there is work done, entropy will also change.

But there is no work done?

The gass expand 50% more, of  course there is work done, but probably not in the sorroundings !

If it is adiabatic, surely it's expanding into a vacuum. Therefore, no work is done. Or is this not the case?

No it isn't, because when the heat is zero ( Q=0), from the first law of thermodynamics we know that work will be equal to the inside energy (A=-U),  if you still have it unclear I can edit my answer writing the formulas.

Please may you edit your answers with the formulas. I don't understand

Sorry I don't know why its not allowed to put a photo ( probably because Im on my phone ), however i searched on google there are plenty explanations just write ' work done in adiabatic expansion' and you will find the formulas.


text:
Comparing the Lewis acidity of carbon tetrafluoride and silicon tetrafluoride
I've been asked to compare the relative acidity of $\ce{CF4}$ and $\ce{SiF4}$. As both atoms have noble gas configurations, neither can act as a lewis acid and so I assume the question is inferring that they are in solution.
My guess is that under aqueous conditions they could react via some sort of hydrolysis to give $\ce{H+}$ and $\ce{F-}$ alongside an alcohol in the case of carbon or $\ce{Si(OH)_{(4-x)}F_{x}}$ with silicon.
As the energy match between the $\ce{Si}$ and $\ce{F}$ HOMO and LUMO is worse this means that this is likely to be a weaker bond and so $\ce{SiF4}$ is more acidic? Is there an argument that could be made about the relative stability of the hydrolysis products?

SiF4 can act as a lewis acid and they both can't hydrolyse.

So CF4 is not acidic at all and SiF4 would be slightly acidic?

Sorry, SiF4 does hydrolyse to form hexafluorosilicic acid - nzic.org.nz/ChemProcesses/production/1C.pdf

SiF4 can hydrolyse because of the presence of vacant d-orbitals.

$\ce{SiF4}$ does not hydrolyse due to the presence or absence of hypothetic d-orbitals. Rather, the $\ce{Si-F}$ bond can be cleaved more easily by nucleophilic oxygens.

Both molecules can theoretically act as a Lewis acid. This would involve the approach of a Lewis base (i.e. a lone pair-carrying group), its interaction with the central atom and the formation of a four-electron-three-centre bond. These bonds with a bond order of $0.5$ can be understood by invoking the following two resonance structures:
$$\ce{\overset{-}{X}\bond{...}AF3-F <-> X-AF3\bond{...}\overset{-}{F}}$$
Where $\ce{A}$ is carbon or silicon in the context of this question. The structure that this arrangement would require is a trigonal bipyramid; the central atom is $\mathrm{sp^2}$ hybridised with its unhybridised p orbital forming the 4e3c bond to the electronegative groups.
You may immediately recognise this as the transition state of an $\mathrm{S_N2}$ reaction which is possible both with carbon and silicon centres.
These four-electron-three-centre bonds are typically more stable if the central atom has a lower and the terminal atoms have a higher electronegativity. If you compare interhalogen compounds, only iodine is able to form such bonds with chlorine ($\ce{(ICl3)2}$) while all other halogens can only form these bonds with fluorine (e.g. $\ce{ClF3}$). This already should allow us to conclude that $\ce{SiF4}$ is the stronger Lewis acid. But the discussion is not finished here.
We also need to discuss steric effects. Carbon tetrahalides are very crowded molecules — even with fluorine even though it is the smallest halogen. Giving up the tetrahedral structure of $\ce{CF4}$ to arrive at trigonal bipyramidal $\ce{CF4X-}$ would require the three fluorines that do not take part in the 4e3c bond to move closer together, increasing steric crowing even more. Indeed, the required interaction would be the Lewis base’s lone pair with $\sigma_{\ce{C-F}}^*$ — an orbital which is already well hidden by the three other fluorines. Thus, steric effects further reduce $\ce{CF4}$’s ability to be a Lewis acid to the point where we shouldn’t even be saying it be a weak Lewis acid: we should just call it Lewis inactive.
Silicon is larger than carbon and better able to stabilise 4e3c bonds due to its lower electronegativity. Thus, 4e3c bonds are sufficiently accessable and stable for silicon in $\ce{SiF4}$ to act as a Lewis acid. Depending on the nature of $\ce{X-}$, the pentacoordinated $\ce{SiF4X-}$ species may either break down under liberation of either $\ce{F-}$ ($\mathrm{S_NSi}$ substitution) or $\ce{X-}$ (reverse reaction) or — especially if $\ce{X} = \ce{F}$ — it may add another to give $\ce{[SiF4X2]^2-}$ like the well-known hexafluoridosilicate anion $\ce{[SiF6]^2-}$.


text:
Does anisole react with concentrated nitric or sulfuric acid
Can we add concentrated $\ce{H2SO4}$ and concentrated $\ce{HNO3}$ to anisole?
I heard that we can't add acids into phenol and aniline which make them blast.

Anisole (methoxybenzene) is an electron-rich arene that will react with sulfuric acid or nitric acid in an electrophilic aromatic substitution.
I suggest that you look the reaction up. Then grab a pencil, some sheets of paper and start to draw. Draw the arenium ions (Wheland intermediates) for the addition of the electrophiles at the different positions of anisole, have a look at the possible resonance structures and enjoy chemistry.

UPDATE
Btw, you can do the same reaction with aniline or phenol without causing an explosion.
My comment on aniline and phenol only works on paper but not in real life and Mithoron's comment on aniline and phenol is correct!
Nitroanilines and nitrophenols are synthesized in a different way.
In principle, nitration of aniline is possible with a mixture of nitric and nitrous acid, or after protecting the aniline as its formanilide.
Technically, 4-nitroaniline is produced by reacting 4-chloronitrobenzene with aqueous ammonia at elevated temperatures in an autoclave.
Details can be found in Ullmann's Encyclopedia of Industrial Chemistry:

In the alternative continuous process, molten 4-chloronitrobenzene and 40 % ammonia liquor in a molar ratio of 1 : 17 are pumped at 20 MPa through a steel reaction coil held at 237-240 °C by oil bath heating. Inflow is controlled
  to give a contact time of ca. 10 min, and outflow is directed into an expansion chamber with a tangential flow of hot water to give, in quantitative yield, a suspension of product that is readily isolated and washed.

The same is true for the formation of 4-nitrophenol. Again, the starting material is 4-chloronitrobenzene, which is reacted with strong aqueous $\ce{NaOH}$ to yield the phenol.
These two reactions are examples for nucleophilic aromatic substitutions!

aniline or phenol won't explode with conc. H2SO4 and conc. HNO3 but nitration won't occure but oxidation

Aniline can definitely be nitrated under these sorts of conditions. With phenol you might be better off using milder conditions e.g. just dilute nitric acid.

@J.LS I thought so too, but according to the Ullmann, the direct nitrations don't seem to play a role nowadays. That is, at least on a technical scale.

I'd be interested to know what reactions/synthetic approaches have replaced them in industry !

@J.LS I've added a procedure from Ullmann's Encyclopedia of Industrial Chemistry to my answer.


text:
Chemical kinetics of a reversible reaction
I am having trouble with the following question:

Consider the following reversible reaction in which the reaction is first order in both
  directions: $$\ce{[A] <=> [B]}$$ $k_\mathrm a$ is the rate constant for the forward reaction and $k_\mathrm b$ is the rate constant for the reverse reaction.
If only reagent $\ce{A}$ is present at $t = 0$, such that $[\ce{A}](t = 0)$ = $[\ce{A}]_0$ (and likewise for $[\ce{B}]_0$), then at all subsequent times $[\ce{A}]+[\ce{B}] = [\ce{A}]_0$. Write down a differential equation for $\frac{\mathrm d[\ce{A}]}{\mathrm dt}$ where $[\ce{A}] \equiv [\ce{A}](t)$ is the concentration of the reagent $\ce{A}$, and hence show that:
$$[\ce{A}]=[\ce{A}]_0\frac{k_\mathrm b+k_\mathrm a \exp[-(k_\mathrm a+k_\mathrm b)t]}{k_\mathrm a+k_\mathrm b}$$

Here's my attempt:
$$\frac{\mathrm d[\ce{A}]}{\mathrm dt}=-k_\mathrm a[\ce{A}]+k_\mathrm b[\ce{A}]\tag{1}$$
Then using the identity in the question:
$$\frac{\mathrm d[\ce{A}]}{\mathrm dt}=-k_\mathrm a[\ce{A}]+k_\mathrm b([\ce{A}]_0-[\ce{A}])\tag{2}$$
Rearranging...
$$\frac{\mathrm d[\ce{A}]}{\mathrm dt}+(k_\mathrm a+k_\mathrm b)[\ce{A}]=k_\mathrm b[\ce{A}]_0\tag{3}$$
I believe that this is a first order linear differential equation with the integrating factor: $\mathrm e^{\int (k_\mathrm a+k_\mathrm b)\,\mathrm dt}=\mathrm e^{(k_\mathrm a+k_\mathrm b)t}$ Thus:
$$\frac{\mathrm d}{\mathrm dt}\left([\ce{A}]\mathrm e^{(k_\mathrm a+k_\mathrm b)t}\right)=k_\mathrm b[\ce{A}]_0\mathrm e^{(k_\mathrm a+k_\mathrm b)t}\tag{4}$$
$$[\ce{A}]\mathrm e^{(k_\mathrm a+k_\mathrm b)t}=\int k_\mathrm b[\ce{A}]_0\mathrm e^{(k_\mathrm a+k_\mathrm b)t}\,\mathrm dt\tag{5}$$
This gives me:
$$[\ce{A}]\mathrm e^{(k_\mathrm a+k_\mathrm b)t}=k_\mathrm b[\ce{A}]_0\frac{\mathrm e^{(k_\mathrm a+k_\mathrm b)t}}{k_\mathrm a+k_\mathrm b}+c\tag{6}$$
But this doesn't give me the final result if I put in the boundary conditions. Where have I gone wrong?

Third equation looks right to me.  That's a non-homogeneous (some would say heterogeneous) first-order differential equation.
I think the heterogeneity is what is giving you problems.   Those equations have a general solution which is a sum of a "complementary" solution and a "particular" solution.  Don't try to use any boundary conditions until after you have combined particular and complementary solutions.
$$\frac{d[\ce{A}]}{dt}+(k_\mathrm a+k_\mathrm b)[\ce{A}]=k_\mathrm b[\ce{A}]_0 \tag{3}$$
The "complementary solution" is the solution to the corresponding homogeneous equation $\frac{d[\ce{A}]}{dt}+(k_\mathrm a+k_\mathrm b)[\ce{A}]=0$, which is $[\ce{A}]_\mathrm c=K e^{-(k_\mathrm a+k_\mathrm b)t}$, where $K$ is an unknown constant of integration.
The "particular solution" is a polynomial in $t$ of the same degree as the hetereogeneous term, which in this case is a polynomial of degree 0, $[\ce{A}]_\mathrm p=C$, where $C$ is an undetermined constant.  We solve for this constant by substitution into the original equation:
$$0 + (k_\mathrm a+k_\mathrm b)C=k_\mathrm b[\ce{A}]_0  \tag{4}$$
This provides the value of $C$ as $C=\frac{k_\mathrm b[\ce{A}]_\mathrm 0}{k_\mathrm b+k_\mathrm a}$.
The general solution is thus the sum of the particular and complementary solutions:
$$[\ce{A}]=[\ce{A}]_\mathrm c + [\ce{A}]_\mathrm p=K e^{-(k_\mathrm a+k_\mathrm b)t} + \frac{k_\mathrm b[\ce{A}]_0}{k_\mathrm b+k_\mathrm a}  \tag{5}$$
Now we can apply boundary conditions.  At time 0, $[\ce{A}]=[\ce{A}]_0$, which gives an algebraic equation we can solve for $K$
$$[\ce{A}]_0=K + \frac{k_\mathrm b[\ce{A}]_0}{k_\mathrm b+k_\mathrm a} \tag{6}$$
If I did my algebra right, then the solution is $K=[\ce{A}]_\mathrm 0 \left (1-\frac{k_b}{k_\mathrm b+k_\mathrm a}  \right ) $.  Substituting this value back into the general solution from my equation 5 gives:
$$[\ce{A}]=[\ce{A}]_0 \left (1-\frac{k_\mathrm b}{k_\mathrm b+k_\mathrm a}  \right )e^{-(k_a+k_b)t} + \frac{k_\mathrm b}{k_\mathrm b+k_\mathrm a}[\ce{A}]_0 \tag{7}$$
This solution satisfies the boundary condition and also the original equation.  It can be simplified a bit by algebraic manipulation to:
$$[\ce{A}]=[\ce{A}]_0\frac{k_\mathrm a e^{-(k_\mathrm a+k_\mathrm b)t}+k_\mathrm b}{k_\mathrm a+k_\mathrm b} \tag{8}$$
From this expression it is easy to check that (i) the condition at $t=0$ is satisfied, that (ii) the expected exponential dependence in time is obtained, and (iii) that as $t \rightarrow \infty $, the reaction should go to equilibrium.  The last point about equilibrium can be checked by applying the two chemical definitions for the equilibrium constant (the ratio of product $B$ to reactant $A$ vs. the ratio of rate constants $k_a/k_b$), i.e. $\frac{\ce{B}_{\infty}}{\ce{A}_{\infty}}=\frac{\ce{A}_0-\ce{A}_{\infty}}{\ce{A}_{\infty}}=\frac{k_\mathrm a}{k_\mathrm b}$ and using equation 8 to check consistency between our obtained expression for $\ce{A}_{\infty}$ and the chemically intuitive definition we just made.

The simplest and most general way to solve this type of scheme is to calculate the effect of an amount x that reacts with initial amounts $A_0$ and $B_0$, thus
$$\ce{A~~~ <=>[k_a][k_b] ~~~B}  \\  A_0-x  ~~~~  B_0 +x $$
which produces the rate equation
$$ \frac{dx}{dt} = k_a(A_0-x)-k_b(B_0+x)$$
the variables can be split to give a straightforward integration instead of having to solve a complicated pair of differential equations.
$$\int _0^x \frac{dx}{k_aA_0-k_bB_0-(k_a+k_b)x} = \int _0^t dt$$
which, after some rearrangement, gives
$$\ln\left(1-\frac{(k_a+k_b)}{k_aA_0-k_bB_0}\right) = -(k_a+k_b)t$$
and this can be rearranged to
$$x= \left( 1-e^{-(k_a+k_b)t}  \right)\frac{(k_aA_0-k_bB_0)}{k_a+k_b}$$
with the initial condition that at $t=0$, $B_0=\ce{[B]=0 }$ this becomes
$$x= \left( 1-e^{-(k_a+k_b)t}  \right)\frac{k_aA_0}{k_a+k_b}$$
The concentration of A is $A_0-x$ which is
$$ \ce{[A]}_t =A_0 \frac{k_b+k_ae^{-(k_a+k_b)t}}{k_a+k_b}$$
This expression has the correct form at all times. $\ce{[B]}_t$ can be calculated similarly from $B_0+x$.

I know its been a long time, but shouldn't is be B(0)+x in the first bracket..

Yes. Thank you, well spotted!  I have corrected it.

Nice thank you.

We know, that at time $t = 0$, $[A] = A_0$,
utting these values in the equation you derived,
we will get $$c=\frac{K_a A_0}{K_a + K_b}. \tag{Constant of Integration}$$
And voilà your answer is in front of you, you just need to transpose the exponential term in the left hand side to the right hand side.


text:
How to explain the dipole moment anomaly of methyl fluoride and methyl chloride with respect to their hydracids?
I was checking dipole moment orders, and came across a very peculiar result:
Methyl fluoride $(\ce{CH3F})$ has lesser dipole moment than methyl chloride ($\ce{CH3Cl}$), but hydrofluoric acid ($\ce{HF}$) has more dipole moment than hydrochloric acid ($\ce{HCl}$).
At first I thought it was about distance, but upon seeing the second one, it led me to think about polarizability. So my question shortens to this:
Which factor dominates while checking dipole moments and when?

Summary
A systematic examination of the $\ce{CH3F}$ and $\ce{CH3Cl}$ complexes with density functional techniques and high-level wave function theory indicates that the charge-separation within these complexes influences the relative corresponding dipole moments for these compounds.  Though fluorine is indeed more electronegative than chlorine, $\ce{CH3Cl}$ had a larger dipole moment due to the longer C-Cl bond with respect to the C-F separation.

Introduction
Recall that a dipole moment can be explained relatively simply by a separation of two charges.  The dipole moment, $\mu$, is dependent on two things: the magnitude of the charge ($q$) AND the separation between the two charges ($r$).  This can be expressed as:
$\mu = q \cdot r$
We know from general chemistry that fluorine is the most electronegative atom on the periodic table.  Therefore, for simple molecules such as $\ce{HF}$ or $\ce{LiF}$, we would expect a rather large dipole moment than if we considered $\ce{HCl}$ or $\ce{LiCl}$.  However, a dipole moment depends on separation as well.  This can lead to some discrepancies in our general trends that we chemists love to use.
Consider chlorine.  The atomic radius for chlorine is larger than it is for fluorine.  A C-Cl bond is going to be longer than a C-F bond.  In the case of methyl fluoride ($\ce{CH3F}$) and methyl chloride ($\ce{CH3Cl}$), we will have a difference in separation of the methyl group and the halogen (larger for the latter).  Looking back at our electric dipole equation, an increase in $r$ translates into an increase in $\mu$.  Perhaps it is this larger separation between the carbon and chlorine atoms that gives rise to these relatively similar dipole moments between $\ce{CH3F}$ and $\ce{CH3Cl}$.
Super Mini Research Project Time
Let us apply electronic structure theory to the problem.  Dave has already provided a link to NIST which is a superb resource for these types of simple computations. However, to be thorough, we will perform our own computations.  The experimental dipole moments for $\ce{CH3F}$ and $\ce{CH3Cl}$ (provided by DavePhD) are $\pu{-1.858 D} \text{ and } \pu{-1.897 D}$, respectively ([1] and [2]).  Dipole moments from electronic structure theory for these complexes can also be found on the NIST website (source3 and source4)
Methodology
Full optimizations and corresponding harmonic vibrational frequency computations are performed using the B3LYP density functional technique (sorry Martin) using the Gaussian 09 software package.  The 6-31+G(d,p) basis set will be employed along with a series of Dunning's correlation consistent basis sets: heavy-aug-cc-pVXZ (where X = D, T, Q, 5).  This basis set denotes that cc-pVXZ was employed for hydrogen atoms and the aug-cc-pVXZ basis set was employed for all heavy (i.e. non-hydrogen) atoms.  A numerical integration grid consisting of 99 radial shels and 590 angular points per shell were employed for all DFT computations.  In addition, full optimizations have been performed using the CCSD(T) level of theory using the CFOUR software package.  The haDZ, haTZ, and haQZ basis sets have been employed for these computations.  All geometries were constrained to $C_{3v}$ symmetry.  Everything is converged to 'overkill'.  Frozen-core approximation applied... 5D 7F spherical basis functions used, etc. etc.
Results
The dipole moments (in Debye) as well as select geometrical parameters ($R$ in Angstroms) for the $\ce{CH3F}$ and $\ce{CH3Cl}$ complexes at various levels of theory are given in the table below.  The deviation of the computed electric dipole moment from experiment is denoted as $\delta_{\mathrm{theory}}^{\mathrm{expt.}}$.  Using this deviation as our main measurement of error, we can see that for the $\ce{CH3F}$ complex, the B3LYP/ha5Z level of theory deviates from experiment by only $\pu{0.01 D}$.  A larger deviation from experiment is observed for the CH$_3$Cl complex with a magnitude of $\pu{0.1 D}$ at the B3LYP/ha5Z level of theory.  These magnitudes are qualitatively similar at the CCSD(T) level.  Therefore, with respect to the dipole moment, the B3LYP method with a triple-$\zeta$ quality basis set is necessary to describe this first-order property.

Clearly $\ce{CH3Cl}$ contains a larger separation between the Cl atom and the methyl group with respect to $\ce{CH3F}$ by about 0.4 Angstroms.  If you wanted to back-calculate the charge for these complexes by assuming $R$(CX) (where X is the halogen) is equivalent to $r$ in the first equation, you will see that CH$_3$F has a larger charge (by about 0.25).  However, we still see that CH$_3$Cl has a larger dipole moment.
The final column in the table above ($\mu_{\mathrm{geom}}^{\ce{CH3Cl}}$) denotes the dipole moment of $\ce{CH3F}$ when using the optimized geometry of $\ce{CH3Cl}$ at the corresponding level of theory.  The dipole moment is significantly larger, by about $\pu{1.5 D}$, which is expected due to the larger C-F separation.
Conclusions
Though we as chemists like to resort to using the general trends provided by the periodic table, we must take into account other factors to explain deviations from these trends.  Here we consider the separation of the halogen from the methyl group as an explanation for $\ce{CH3Cl}$ having a larger dipole moment than $\ce{CH3F}$, even though F is more electronegative.  Our computational results qualitatively agree with experiment.  In addition, B3LYP with a triple-$\zeta$ quality basis set is adequate in describing this first-order property according to CCSD(T) results.  Finally, these results indicate that it is the charge-separation in these small molecules that greatly impacts the qualitative ordering of these dipole moments.

References:

Marshall, M. D.; Muenter, J. The electric dipole moment of methyl fluoride. J. Mol. Spectrosc. 1980, 83 (2), 279–282. DOI: 10.1016/0022-2852(80)90051-X.
Wlodarczak, G.; Segard, B.; Legrand, J.; Demaison, J. The dipole moment of CH335Cl. J. Mol. Spectrosc. 1985, 111 (1), 204–206. DOI: 10.1016/0022-2852(85)90083-9.

I found a better experimental value for dipole moment of methyl chloride: Journal of Molecular Spectroscopy vol. 111, pages 204-206.  Has 1.8959 D  for the chlorine 35 isotope.  http://www.sciencedirect.com/science/article/pii/0022285285900839 So I'll delete my answer and enjoy your calculations!

And since I deleted my answer, I'll add the methyl fluoride reference as a comment here: Journal of Molecular Spectroscopy vol. 83, pages 279–282:  1.85840 D http://www.sciencedirect.com/science/article/pii/002228528090051X

@GeoffHutchison because once I found the more accurate value of the methyl chloride dipole moment, the experimental values were significantly different to high confidence. The difference increased to 0.04 D, which might be explainable (although not by me).


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Can you melt wood?
We talked about it in our chemistry class but we couldn't get to a conclusion, any help?

Have you tried to find it on google or some books? If yes kindly show your research efforts :)

Is wood a pure chemical, which might have a single melting point, or a mixture of chemicals with different melting points?  2. Do you understand the difference between decomposition and melting? Look them up.

@DrMoishe Pippik Decomposition is one thing breaking up into several things by chemical means, melting is a solid changing form into a liquid by physical means.

Lots of things can't be melted.  Try to melt diamond (changes allotropic form) or limestone (decomposes).

I remember in some of first moderately serious science lessons, I got a data book with melting and boiling points of various substances.  However, apart from water, not many were easily available at home.  The things at home were not in the book.  I remember wondering what the melting point of curtain was.  It was hard to imagine liquid curtain and even harder to imagine it freezing back to its original form.

NO, we can't melt wood!
From primary level we have learned that solid melts to liquid at certain temperature and on increasing the temperature further it change into gaseous. But that is not the case always.

The problem with melting wood revolves around what combustion is, and what temperature the combustion of wood happens at.  Combustion,
  also known as burning, is simply a chemical reaction that takes place
  where the combustible material (in this case wood) in the presence of
  an oxidizer (usually the air around the fire) changes its chemical
  composition and decomposes the material into other chemicals. The
  process is one that’s exothermic. As such, light and heat can be
  released.
Wood is mostly made up of things like cellulose, lignin, and water. As
  wood combusts, it’s broken down into products like charcoal, water,
  methanol, and carbon dioxide. Unlike water turning back into ice, if
  you cooled down the resulting products of burning wood, it obviously
  does not change back to its original composition.
All materials that combust will have a natural temperature at which
  the process will begin taking place. The higher the temperature, the
  quicker the process becomes (usually). If that temperature is lower
  than the temperature at which the material will melt, that material
  will never (naturally) melt because it just turns into other
  chemicals.
As for wood, it will begin a process known as pyrolysis at
  temperatures around 500-600 degrees Fahrenheit. Pyrolysis is also an
  exothermic reaction that tends to be self sustaining. At these
  temperatures, wood will begin giving off up to 100 chemicals,
  including methane and methanol (the same stuff they put as additives
  in gasoline), that will begin to burn. Once those chemicals begin
  burning, they will increase the temperature and the remaining char
  (the burned black bits present after the fire goes out) left behind
  will begin to further decompose, things like calcium, potassium and
  magnesium.

Source: todayifoundout

Would it be possible to do it in an oxygen free space? Like a chamber filled with CO2?

Wood itself is not an oxygen free environment. The carbohydrates and proteins in wood contain oxygen atoms. At sufficient temperatures, the compounds can break down without extra oxygen. It's not the same as burning, because there wouldn't be enough oxygen, but it wouldn't be liquid wood.

http://en.wikipedia.org/wiki/Pyrolysis_oil

@Cyberson Combustion is not possible without oxygen till what I know. Though I don't know if there is any new technique available.

@user137 yes it may contain oxygen atoms, but not diatomic oxygen molecules, AKA, what's required for fire. And thanks to the H2O vs H+2O lab we did in school, I now know that being a molecule vs an atom changes properties.

@Curt F. I scanned the wiki page about Pyrolysis oil and, correct me if I'm wrong, isn't heating something in a vacuum space until it becomes a liquid still melting?

No because it is not the same chemical substance and won't change back when heat is removed

Burning wood without oxygen makes charcoal.

@JosephHirch You probably mean heating or thermolysis (m.s. pyrolysis). I would also argue that production of charcoal is actually incomplete burning. As an simpler example for hydrocarbons, $\ce{C6H12 + 5O2 -> 4C + CO + CO2 + 7H2O}$. So not total absence of oxygen, just that there isn't enough for complete combustion.

Theoretically, it may be possible, but hasn't been proven.
According to this article, wood cannot even be melted in a vacuum, but may be able to melt under high pressure.

The lignin, and some of the hemicelluloses, in wood can melt under certain circumstances, such as friction welding.  For a dramatic video, see:  https://www.facebook.com/interestingengineering/videos/1891004754302553/
See also:
http://web.utk.edu/~mtaylo29/pages/Wood%20welding.htm?fbclid=IwAR1MLgBtkfESlYiaP0iEaXbv36AtLy8yXEj0iCqFaaVYBcDRitxqeZJZYtM
Here they say, about welds created between wood pieces during friction welding:  "Examination of the bondline suggests that the friction between the pieces heats and melts components of the wood (mainly lignin) and loosens fibers on the surface. These fibers intertwine in a matrix with the molten lignin and solidify to form a bond that is strong enough for structural applications."
For more details, see:  "Wood Bonding by Mechanically-Induced in Situ Welding of Polymeric Structural Wood Constituents" B. Gfeller M. Properzi M. Zanetti A. Pizzi F. Pichelin M. Lehmann L. Delmotte, Journal of Applied Polymer Science 92(1):243 - 251, 2004
Abstract from above reference: "Mechanically induced wood fusion welding, without any adhesive, is shown here to yield rapidly bonding wood joints satisfying the relevant requirements for structural application. The mechanism of mechanically induced vibrational wood fusion welding is shown to be due mostly to the melting and flowing of amorphous cells–interconnecting polymer material in the structure of wood, mainly lignin, but also some hemicelluloses. This causes a partial detachment, the “ungluing,” of long wood cells and wood fibers and the formation of an entanglement network drowned in a matrix of melted material which then solidifies, thus forming a wood cell/fiber entanglement network composite with a molten lignin polymer matrix. During the welding period some of the detached wood fibers which are no longer being held by the interconnecting material are pushed out of the joint as excess fiber. Crosslinking chemical reactions of lignin and carbohydrate-derived furfural also occur. Their presence has been identified by CP-MAS 13C-NMR. These reactions, however, are relatively minor contributors during the very short welding period. Their contribution increases after welding has finished, which explains why relatively longer holding times under pressure after the end of welding contribute strongly to obtaining a good bond." © 2004 Wiley Periodicals, Inc. J Appl Polym Sci 92: 243–251, 2004


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Is there a characteristic that characterizes all, and only, metals?
The Wikipedia article on metals describes them as elements whose polyatomic configurations* exhibit any one of a number of patterns. I could not discern any property that characterized all the patterns.
To the best of my knowledge, there are two main categories of elements: metals, non-metals. There's a third category, metalloids, whose members are those that satisfy some of the criteria for inclusion in both main categories.
It would seem very odd if there were nothing that characterized all, and only, metals. In fact it would seem that metals are better conceived of as an arbitrary grouping of what is actually a number of other categories, each of which would include all, and only, the elements that exhibit a single type of polyatomic configuration*.
Is there a characteristic that characterizes all, and only, metals?

*(I'm using the phrase 'polyatomic configuration' in order to express 'the way that atoms of a particular element stick together when you have a number of atoms of that element connected together'. Is there a standard term that expresses that idea?)

Metals will crystallize in a wide variety of Bravais lattices, much less point groups.  And, the elements are divided into many finer shades of grey than metals and non-metals. I believe that your question rests on unfounded assumptions.

@Hal Have you looked at conductivities? That is the way I would classify metals, semiconductors, insulators and superconductors... If you would like to know more about this I would be happy to write an answer but I'm not sure if you are aware of this already?

@AngusTheMan I've been reading about the periodic table and metalicness (?) on Wikipedia, looking for the answer. It did seem that conductivity was the underlying property, and that the metallic-nonmetallic dichotomy isn't best thought of as a pair of categories, but rather as opposite extremes of a continuum. I also learned that there are some more fundamental properties of elements that explain why certain patterns appear in the table. Is there a reason why conductivity correlates with the other characteristics of metals (opacity, hardness, etc.)?

One quality shared by metals is a "sea" of valence electrons, though as you state, this valence band is shared to some extent by metaloids (aka semiconductors). Many of the properties of metals are dependent on the fact that electrons, to a greater or lesser extent, are not localized.

I'm afraid there's no such characteristics and even such characteristics can't exist as there is no sharp boundary between metallic and other substances.

https://en.wikipedia.org/wiki/Conductive_polymer

The Universe is full of continuous and smoothly-transitioning properties, yet which we often stubbornly try to assign a discrete yes/no answer in order to make the world easier to cope with. Metallicity is one of them. The Van Arkel-Ketelaar triangle is a diagram which attempts to show the transition between covalent, ionic and metallic types of bonding. Here is one populated by a variety of compounds:
The closer a substance is to one of the vertices, the more that vertex is important in describing the nature of bonding in the material. As you can see, the entire triangle is populated quite uniformly. There are certainly many more materials which would fit in the gaps between the materials shown, and even simply varying the ambient conditions (pressure, temperature, etc), a same substance can be put in different regions of the triangle.
Given this behaviour, it should be clear that any attempt to rigorously define a line dividing substances between metals and non-metals will be fraught with difficulty and likely not useful in the first place. No matter where the line is drawn, there will be some compounds that will look strange for "barely being a metal", while others will be strange for "barely being a non-metal".

Possibly there is such a single characteristic.
I say "possibly" because in periodic table terms there is no universal agreement as to where metals stop and nonmetals (or metalloids) start.
That aside, Johnson (2007, p. 15) defines a metal as any element having an electrical conductivity at or below room temperature of at least 3 x 105 S m-1 along any direction in a single crystal of any known form of the element.
In practical terms, most authors agree that boron, silicon, germanium, and antimony are metalloids, so that all elements to the left of these are metals. There is some uncertainty as to the status of polonium and astatine. Both allotropes of polonium are metallic conductors; and astatine has been predicted to be a metal. So it is probably reasonable to regard polonium and astatine as metals.
Johnson's 2007 definition appears to be in agreement with these observations but for astatine, which he shows as a semi-metal or metalloid. However the prediction of astatine as a metal was made in 2013, so there is no real contradiction here.
In summary, as earlier Users noted, electrical conductivity may fit the bill.

Johnson D (ed) 2007, Metals and chemical change, RSC Publishing, Cambridge

I can't think of anything better. +1. You probably know that Si and Ge actually become metallic when melted (they contract as they do so), and just about anything becomes metallic under enough pressure.


text:
What fraction of hydrogen in air results in an explosive mixture?
I would like to know if there is documented evidence of the quantity of hydrogen to air that must be present in the mixture to risk explosion or flame.

Hydrogen is explosive in mixtures in air at room temperature and ambient pressure when it is between 4% and 75% by volume.  These limits are respectively called the "lower explosive limit" (LEL) and the upper explosive limit (UEL).
Values for hydrogen (and many other gases) are provided in many locations on the internet.  Engineering toolbox is one such site.
More info on LEL and UEL is available from Wikipedia.


text:
Torquoselectivity: Disrotatory ring opening
In organic chemistry lecture we discussed the beginning of the Woodward Hoffmann Rules with a key step of the vitamin B12 synthesis.
(The initial paper from their work can be found here:
http://pubs.acs.org/doi/pdf/10.1021/ja01080a054)
Let's first see the forward reaction (the electrocyclic ring closure) as Woodward and Hoffmann found (which goes by disrotatory cyclization; 6 electrons, thermal).

I'm interested in the reverse reaction, the ring opening (6 electrons, and therefore disrotatory:
I'm thinking now about the torquoselectivity of the ring-opening  (because in an exercise we had an example with 1-Ethyl-1-ethoxy(EDG)-cyclobut-2-ene).
Is there a rule how to see why the E (E: Electron withdrawing group) is in trans position when I start with the product?

EDIT:
I'm thinking if they are some stereoelectronic effects here which explain the obtained selectivity:

See also: http://pubs.acs.org/doi/abs/10.1021/jo00244a058
(Electronic control of the stereoselectivities of electrocyclic reactions of cyclobutenes against incredible steric odds)
Well here we have a conrotatory ring-opening. I don't know if they are still valid in our 6e case for disrotatory ring-opening of the reaction from W. and H..

Just to be clear, the reaction shown is the electrocyclic ring closure which goes by disrotatory cyclization (6 electron, thermal). But the reaction you are interested in is the reverse process, correct?

Are you asking, if I start with the cyclohexadiene on the top-right, why does the E group wind up trans to the chain on the other end of the double bond, in the hexatriene on the top-left?

Exactly (see update)

No, there is no general orbital symmetry rule for predicting which of the two possible disrotatory motions will be favored.  However, by examining steric effects in the two possible transition states we can often make a good guess as to which product will predominate.  There are always a few "special cases" like the one you referenced in your post, where with certain substituents under certain conditions certain things will happen.  But to my knowledge there is no general rule for this situation.
In the following diagram there are 3 thermal reactions controlled by orbital symmetry (6 pi electrons, therefore disrotatory motion).

The top line shows the symmetry allowed disrotatory closure of trans, trans-1,6-dimethylhexatriene.  There are two possible disrotatory motions - either the tops or the bottoms of the p-orbitals on carbons 1 and 6 can move towards each other in a disrotatory fashion.  One motion produces the dimethylcyclohexadiene with the two methyl groups above the plane of the screen, the other motion produces the mirror image molecule with the methyl groups below the plane of the screen.
Similarly, cis, cis-1,6-dimethylhexatriene can undergo two disrotatory motions to produce the same two products (Note: the cis, cis isomer will react much slower and probably in lower yield; steric interactions between the methyl groups will disfavor the s-trans conformation about the two single bonds that is required for the ring closure to be able to occur).
The bottom line shows the reverse reaction, thermally converting a cyclohexadiene to a hexatriene.  As shown, two possible isomers can be formed.  The ratio of these two isomers will depend on steric interactions that develop during the two possible disrotatory motions.  In this case, we will primarily obtain the trans, trans isomer; the cis, cis isomer has steric problems in its transition state from the two methyl groups bumping into each other (build a model).
In this last reaction, if we replaced the methyl groups with deuterium, we would obtain a roughly equal mixture of the two possible products, cis, cis- and trans, trans-1,6-dideuteriohexatriene.  So again, there is no orbital symmetry "rule" for predicting whether a group would wind up cis or trans on the double bond.  But if we examine the steric effects in the two possible disrotatory transition states, we can often make a good guess about which isomer should predominate.

In order to deal with the specific question I've updated my post. Your answer helped me a lot to remind the basic concepts. For small cycles we mostly consider only Hückel overlap (the methyl groups are syn): I guess a möbius overlap resulting in the two methyl groups in anti relationship is sterically disfavored, isn't it?

"I guess a möbius overlap resulting in the two methyl groups in anti relationship is sterically disfavored, isn't it?"  Mobius type overlap occurs during conrotatory motion, when the top of one p orbital overlaps with the bottom of the other p orbital to which it is forming a bond.  Conrotatory (Mobius) motion is allowed thermally in 4 pi systems.  Disrotatory motion is allowed thermally in 4n+2 pi systems.  It is orbital symmetry (not steric factors) that prevent formation the methyls in the anti arrangement in the present 6 pi electron example.

I really like your question, but I'm not completely happy with my chemical understanding of this reaction. Can we predict from stereoelectronic arguments (by assuming E=COOH) why in the initial question the thermally allowed disotatory ring closure gives this preferedly the above mentioned product (Me and H pointing towards us) and not the one in which the methyl group and hydrogen atom of the newly formed bond are pointing at the back?

Yes, at least we could make a good guess.  There are 2 possible disrotatory motions.  The one that would place the "E" down, has the "E" group and the tertiary hydrogen at the ring juncture approaching each other (sterically preferred). Disrotation to place the "E" up, has the E group and the other rings approaching each other (sterically less preferred).


text:
Optional determination techniques in Atomic Absorption Spectroscopy
I have tried to find information about this case but I couldn't find anything specific.
The key point of instrumental methods of analysis, is for the student to be able to know which one of the possible methods to use for a specific determination. As for me, that is not an easy thing to do because every method has its pros and cons. For the moment I am analysing AAS (Atomic Absorption Spectroscopy), and I simply cannot tell:
What's the reason that $\ce{As}$, $\ce{Te}$, $\ce{Se}$, $\ce{Ge}$, $\ce{Sn}$, $\ce{Pb}$ is preferred to be determined only by the hydrid method? $\ce{Pb}$ for example can be determined also with the other methods in AAS
Does anyone have any tip, how to choose between methods if it's required for any element?
I replied with a similar question earlier about the mercury determination.

Well mercury is very different...

Well I think it is the hybrid (not hydrid) method.

Remember this is "atomic" spectroscopy so it is expected that the "sample" is being totally ripped apart. So one "best technique" per element.
$\ce{As}$, $\ce{Te}$, $\ce{Se}$, $\ce{Ge}$, $\ce{Sn}$, $\ce{Pb}$ are analyzed by the graphite tube technique rather than in a flame because the technique provides more sensitivity.
$\ce{Hg}$ is an odd one since you use a cold vapor technique.
Of course depending on the particular sample matrix what you might need to change.

Inductively coupled plasma Optical emission spectroscopy (ICP-OES) is best know technique for the determination of different elemental concentration in a sample even at ppb level simultaneously. you don't need to change the lamp every time and don't need to bother for calibration before running the actual sample.
But when it's come to these elements ( As, Te, Se, Ge, Sn, Pb) the best result you can only obtained by ASS using hydride generation in a graphite furnace. Using hydride generation technique the measurement can be done at ppb level in a dilute sample. This technique is used to provide higher sensitivity compared to other techniques.

You are right, every method has its pros and cons and if you were to choose one, the choice depends on what are you trying to analyse.
The important things to keep in mind are:

With AAS you will rarely be looking for every single element in the sample, but only the ones you expect to find or the ones you're specifically searching for

For example, you have an organic product and during its synthesis you used the metal catalyst (Pt is a common one). You want to make sure the catalyst is not present in your product. You will then want to search only for that particular metal. Based on what your analyte is, you choose on a method.

The other important thing to consider is how sensitive your technique needs to be. You don't need ultra high sensitivity in all analyses and this will depend on your research (again, what are you trying to find with your measurement)

For example, if some product that's sent to you for analysis (e.g
 toys, pharmaceuticals) is regulated to have <5 ppb lead concentration, it is enough to use a technique which has sensitivity on ppb level. The sample may still contain traces of lead on ppt level and there are techniques which are that sensitive, but they are in this case unnecessary
I hope this answer is satisfactory to you because I don't believe any analyst can give you a singular answer. You may use different kinds of techniques for the same type of analyte.