text:
How much water evaporates during a given time?
I read some questions such this  for example and I asked me another question. Instead of only calculate how much time does it takes to have a cup of tea at room temperature, I'm asking how much time does is take to evaporate a part of the water in a cup of tea.

Just to be clear
For this question I am considering that the cup is full of water at 100°C and the cup too. The cup is consider to have no effect on heat transfert and has a high of $12 \mathrm{\text{ cm}}$. During all the experience the temperature and the pressure in the room are constant and respectively are $T_0 =20^\circ \mathrm{C}$ and $P_0 = 101325 \mathrm{\text{ Pa}}$.
From the beginning of the experiment to the time where the temperature of water and the cup will be the same than the room it is not difficult to imagine that some water will evaportate and escape the cup and goes in air. But don't forget that their is an equilibrum between water in the liquid water and the water in the air. We will considering during all the time of the experiment the hygrometry of the air is zero. So then by the diffusion phenomena very very slowly the water in the cup will escape.
At the end we assume the pressure of water in the air is the saturation pressure of water in the air and there is only $5 \text{ cm}$ of water inside the cup.

My Question
We know how to calculate the time for water to cool from $T_0$ and room temperature. How should I know then during this time how much water goes out of the cup in gas phase ?
"this time" refers to the time for the water to goes from $\text{100}^\circ\text{C}$ to room temperature. The time during the water at room temperature will diffuse in the air is given in the next part with details for you.

Some details for you
I'm asking this question because I know how to calculate the time to evaporate water using law of diffusion but this time depends on the high of liquids there is in the cup ...
My calculus give :
$$t=\frac{\left(z_f^2-z_i^2\right) \rho RT_0}{2M_{H_2O} P_0 \mathcal{D}_{w/a} \ln\left(\frac{1-y_i}{1-y_f}\right)}$$
Where :
$z$ is the difference lengh between the top of the cup and the top of the water, $\rho$ is the density of water at $T_0$, $M_{H_2O}$ is the molar weight of water, $\mathcal{D_{w/a}}$ is the diffusion coefficient of water in the air and $y$ is the molar fraction of water in the air.
If we consider for example that until room temperature only $0.5 \text{ cm}$ of water evaporates then $t=5426025 \text{ s}$ I just need know the real value of how much centimeters really escaped.

Sorry for the lengh of the question and thank you in advance for your answer !

OK, maybe you are over complexifying the problem.  If the cup does not allow heat transfer (completely insulated) then would the heat lost by the remaining amount of  liquid water have to equal the heat gained by the amount of water that vaporizes?

@JosephHirsch you might haven't understood something. During the time the water will cool from 100°C to room temperature, maybe only one millimiter will escape maybe two. My question is how much millimeters of water will escape during this time. When I say the cup have no effect on heat transfert I mean for the heat transfer you can consider there is no cup between the water and the air (but not for the mass transfer). :)

Got it.    So it is really a rate problem.


text:
When can I use concentrations instead of activities in equilibrium calculations?
To calculate an equilibrium constant, it is said that you should actually use activities $a_i$ instead of concentrations $c_i$. But it is also said that within a certain range, you can use concentrations instead. This is also what is usually done because it's easier.
So what is this range? When can I safely use concentrations and when should I definitely switch to using activities?

To calculate any equilibrum constant we need to be sure of the fugacity of each species. Note that fugacities are defined for every common state of matter (gas, liquid and solid), although they are most commonly used for gases. Look at the end of this answer for more details.

The activity of the coumpound $i$ written $a_i$ is defined as:
$$a_i=\exp\left(\frac{\mu_i -\mu_i^{\circ}}{RT}\right)$$
where $\mu_i$ is the chemical potential, which depends on the fugacity.
However this expression can be hard to use if you have not a strong idea of what you are doing because the are a lot of different ways to calculate things, depending on the formulas and laws or theories used.

What you must remember all the time
Activity can also be defined as $$a_i=\gamma_i \cdot \frac{X_i}{X_i^\circ}$$
where $X_i$ can be a concentration, a partial pressure, a molar fraction, a molality, and so on, and where $\gamma_i$ is the activity coefficient. Be careful; this coefficient depends on what $X_i$ is.
The role of $\gamma_i$ is to correct for the interactions between each constituent of the mixture.
If you are in the ideal regime, which means that the amount of your solute is almost negligible with respect to the entire mixture, then $\gamma_i \approx 1$. Only then can you use $X_i$ in place of $a_i$.
In general, the determination of this coefficient is not quite simple, especially for ions: see the example here for hydrogen ion.

Ionic solutions
Electrolyte solutions (e.g. a solution of $\ce{NaCl}$) can be modelled by Debye–Hückel theory. There are different versions of this model depending on the precision you want. One of the most used is given in the paper about hydrogen ion:
$$\log(\gamma_i)=-\frac{A_m z_i^2\sqrt{I_m}}{1+\sqrt{I_m}}$$
where $I_m$ is the molar ionic strength, given by:
$$I_m=\frac{1}{2}\sum_i z_i^2m_i$$
where $m_i$ is the molality of $i$.

To answer the other part of your question, whether your system is behaving "ideally" or not ultimately depends on the precision you want.
For example if you have a reaction constant expressed as:
$$K=\frac{a_\ce{A}^2 a_\ce{B}^3}{a_\ce{C}}=\frac{\gamma_\ce{A}^2 \gamma_\ce{B}^3}{\gamma_\ce{C}} \cdot \frac{(c_\ce{A}/c^\circ)^2 (c_\ce{B}/c^\circ)^3}{c_\ce{C}/c^\circ}=\frac{\gamma_\ce{A}^2 \gamma_\ce{B}^3}{\gamma_\ce{C}} K^\circ$$
which corresponds to the reaction $\ce{C -> 2A + 3B}$, you will have a large error if you assume you are under ideal conditions when you aren't.
For example if you assume ideality in calculating the $\mathrm{pH}$ of $\pu{5 mol/L}$ of $\ce{HCl}$ solution, like I did here, the answer will not be the same as if you were to consider the activity of the ion, like this.
So at least for strong acid, if the concentration is much than $\pu{1 mol/L}$, you must use activity coefficients. Depending on your problem you'll need to think by yourself and do calculations with and without assuming ideality. If you are unsure as to whether your solution can be considered as being ideal, it's good to do a check (and feel free to use a computer in hard cases).

I would add, as an example, if one of your compounds is 10% of the solution in mass, the solution must no longer be considered as ideal. But, as said earlier, the limit of what is ideal is fixed by you, depending on the precision you want.

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sort: votes


text:
Why doesn't curie temperature follow a periodic trend in ferromagnetic materials
I was writing up an answer to a question which was asking why the Curie Temperature's of $\ce{Fe, Co, Ni}$ don't follow a periodic trend. The OP voluntarily deleted his question before I could finish and post my answer.
The OP has reasoned as follows: Since Iron, Cobal and Nickel have electron configurations $\ce{[Ar] 3d^6 4s^2, [Ar] 3d^7 4s^2\ \text{and}\ [Ar] 3d^8 4s^2}$, giving them 4 , 3, and 2 unpaired electrons respectively. This according to the OP should suggest a periodicity in Curie Temperature's which isn't observed.
Clearly this isn't the case. For your reference, the Curie temperatures for the three metals are (iron, cobalt, nickel):    $1043\ \mathrm{K},  1400\ \mathrm{K},  627\ \mathrm{K}$
Since this did indicate some ignorance about the origins of magnetism, I decided to to write up a short answer, describing the salient features and at the same time correcting OP's misconception. The answer was kept relatively simple, and was nearly done but OP chose to delete his question.
Not wanting to let a answer I put some effort in go to waste, I am posting it here. Additionally, since the goal of this site is to build a Q&A, it would be great if others could chime in with their views, derivations etcetera.

Man, I've got a duplicate.
https://physics.stackexchange.com/questions/301652/why-arent-the-curie-temperature-of-iron-cobalt-and-nickel-periodic

TL;DR : There's no reason for it to show any periodicity. It doesn't depend in a simple way on the outer electron configuration of atoms. The specifics of crystal structure are important in ferrmoganetic (also in antiferromagnetic, and ferrimagnetic) materials.

This by no means is a complete description of the phenomenon. If you're interested, consult textbooks, the internet etcetera. and come back with more specific questions. The person who originally asked the question seemed like he was a high school student (or lower) so I tried to tailor an explanation that he would understand. I welcome more complete desscriptions
First of all, ferromagnetism is a property that goes beyond the chemistry of a material, and is highly dependant on crystalline structure and microstructure.
Existence of several ferromagnetic metal alloys whose constituents are not themselves ferromagnetic (Heusler alloys), and conversely, the existence non-magnetic alloys, composed almost exclusively of ferromagnetic metals (for example some types of steel) are evidence of this fact.
The OP originally spoke of unpaired electrons, so I am guessing he knew something about electron spin.
Spin, just like electron charge is an intrinsic property, and its behaviour is described using quantum mechanics. Spin and charge on the electron give rise to a magnetic dipole moment (people like to talk about classical charges spinning and generating a magnetic dipole, however, I would like to discourage this analogy).
Due to its quantum nature, the spin of the electron can be in one of only two states; with the magnetic field either pointing "up" or "down" (what I call up and down is somewhat arbitrary; what's important here is that there are two arbitrary, mutually exclusive eigenstates).
The spin of the electrons in atoms is the main source of ferromagnetism, although there is also a contribution from the orbital angular momentum of the electron about the nucleus.
Materials made of atoms with filled electron shells have a total dipole moment of zero, because the electrons all exist in pairs with opposite spin, every electron's magnetic moment is cancelled by the opposite moment of the second electron in the pair.
So of course, a ferromagnetic material needs to have spin unpaired electrons, but that alone isn't sufficient. Hund's rules, posit that the first few electrons in a shell tend to have the same spin, thereby increasing the total dipole moment.
These unpaired dipoles (often called simply "spins" even though they also generally include angular momentum) tend to align in parallel to an external magnetic field, an effect called paramagnetism. Such materials are hence called paramagnetic. Molecules with spin paired electrons are called diamagnetic.
Ferromagnetism involves something extra. In a few substances the dipoles tend to align spontaneously, giving rise to a spontaneous magnetisation, even when there is no applied field. This is ferromagnetism.
Now let's develop some specifics. For instance, how and why do spins align? Additionally, why can we find iron in both magnetised and unmagnetised states, shouldn't it always be magnetised?
When two nearby atoms have unpaired electrons, whether the electron spins are parallel or antiparallel affects whether the electrons can share the same orbit as a result of the quantum mechanical effect called the exchange interaction. The exchange interaction is related to the Pauli exclusion principle, which says that two electrons with the same spin cannot also have the same "position".
When the orbitals of the unpaired outer valence electrons from adjacent atoms overlap, electrostatic repulsions are reduced when spins are parallel to each other. In simple terms, the electrons, which repel one another, can move "further apart" by aligning their spins, so the spins of these electrons tend to line up. This difference in energy is called the exchange energy. In ferromagnetic materials, the exchange interaction is much stronger than the competing dipole-dipole interaction. For example, in iron (Fe) the exchange force is about 1000 times stronger than the dipole interaction.
Therefore, below the Curie temperature virtually all of the dipoles in a ferromagnetic material will be aligned. (we will get to Curie temperature in a minute)
So why are iron and other ferromagnets ever found in an "unmagnetised" state ?
The reason for this is that a bulk piece of ferromagnetic material is divided into tiny regions called magnetic domains. Ferromagnetic materials spontaneously divide into magnetic domains because the exchange interaction is a short-range force, and over long distances of many atoms the tendency of the magnetic dipoles to reduce their energy by orienting in opposite directions wins.
If all the dipoles in a piece of ferromagnetic material are aligned parallel, the space surrounding it contains a lot of magnetostatic energy, which the material wishes to  reduce, and it does so energy by splitting into many domains pointing in different directions, so the magnetic field is confined to small local fields in the material, reducing the volume of the field.
This means that within each domain, the spins are aligned, but (if the bulk material is in its lowest energy configuration, i.e. unmagnetised), the spins of separate domains point in different directions and their magnetic fields cancel out, so the object has no net large scale magnetic field.
The domains are separated by thin domain walls a number of molecules thick, in which the direction of magnetisation of the dipoles rotates smoothly from one domain's direction to the other
Although the exchange interaction keeps spins aligned, it does not align them in a particular direction. Without magnetic anisotropy, the spins in a magnet randomly change direction in response to thermal fluctuations and the magnet is super-paramagnetic. There are several kinds of magnetic anisotropy, however, let's not go into that here
If one now places such a material in a strong enough external magnetic field, the domains will reorient so more of the dipoles are aligned with the external field. Moreover, the domains will remain aligned when the external field is removed, creating a magnetic field of their own extending into the space around the material, thus creating a "permanent" magnet.
The domains do not spontaneously return to the original minimum energy configuration when the field is removed because the domain walls tend to become fixed in place by the crystal structure, and defects in the bulk solid. Thus the magnetisation, and the resulting magnetic field, is baked into to the crystal structure of the material, making it very difficult to demagnetise (it is metastable in this sense).
As the temperature increases, thermal motion, competes with the ferromagnetic tendency for dipoles to align, and when the temperature rises beyond a certain point, (the Curie temperature), the system can no longer maintain a spontaneous magnetisation, as reorientation takes place.
This is an example of a second order phase transition
Thus the material looses its ferromagnetism; unpaired electrons are still present, and thus it is still paramagnetic.
Now clearly, you can see that ferromagnetic properties don't depend in a simple way on unpaired spins, and there is no reason for the curie temperature to follow a periodic trend the way you expect it to.

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text:
Number of two electron integrals
I was trying to write a C++ code following the instruction given in this website. The problem is in the two electron integral file, only 228 entries for 7 molecular orbitals are provided. But my code says it should be 406 (please see below)
int NFI=0; //NFI=number of four electron integrals
for (int i=0; i<NAO; i++){
 for (int j=0; j<=i; j++){
  for (int k=0; k<=i; k++){
   for (int l=0; l<=k; l++){
    int ij=i*(i+1)*0.5+j;
    int kl=k*(k+1)*0.5+l;
     if (kl<=ij){NFI += 1;}
   }
  }
 }
}
printf("%d\n",NFI);

I'm using 4 criteria as can be seen in this code snippet. Can anyone explain me why the number of unique integral is 228, not 406?
As per suggestion, I compared my index with the index provided, here they are
1 1 1 1                   1     1     1     1
2 1 1 1                   2     1     1     1
2 1 2 1                   2     2     1     1
2 2 1 1                   2     1     2     1
2 2 2 1                   2     2     2     1
2 2 2 2                   2     2     2     2
3 1 1 1                   3     3     1     1
3 1 2 1                   4     4     1     1
3 1 2 2                   5     5     1     1

Left 4 numbers are generated from my code and right 4 numbers are given in the input file. (3,1,1,1) is missing in the data file provided (I checked the whole file), but it fulfills the criteria of $ij \geq kl$.

You can see all the integrals here http://sirius.chem.vt.edu/~crawdad/programming/project3/h2o_sto3g/eri.dat  Print out your indices and compare

I’m not sure whether there’s a need to delete the question.

OK, I'm deleting the first part. Still I couldn't figure it out. Mathematically also it should be 406 (as 78/2=28 and 2829/2=406)

Have you considered the symmetry of H$_2$O or the NAOs? <sp||ss> = 0 if on the same center. Same goes ofr <gg||gu> (g even under symmetry operator, u odd under symmetry operator, e.g., $C_2$).

You have a good point, I considered that possibility. As I am a chemical engineering graduate student, I'm not sure how to take symmetry into account. I didn't want to put substantial efforts into that because I'm doing it in my leisure time. Fortunately, I got one complete file that shows lots of zeros.

Update:: I think that 228 integral was a mistake, I found another project file from Dr. Crawford which listed 406 unique integrals for the same system ($H_2O$ with 7 molecular orbitals). In case if anyone wants to work on that project, they can use this data file.
-----------------------------------------------------------------------
      PSI4: An Open-Source Ab Initio Electronic Structure Package
                          PSI 4.0.0-alpha Driver
T. D. Crawford, C. D. Sherrill, E. F. Valeev, J. T. Fermann, R. A. King,
M. L. Leininger, S. T. Brown, C. L. Janssen, E. T. Seidl, J. P. Kenny,
and W. D. Allen, J. Comput. Chem. 28, 1610-1616 (2007)

                     Additional Contributions by
Francesco Evangelista, Andrew Simmonett, Justin Turney, Jeremiah Wilke
-----------------------------------------------------------------------

Using LocalCommunicator (Number of processes = 1)

Memory level set to 256.000 MB

./plugin_aointegrals.so loaded.
Calling plugin plugin_aointegrals.so.
    Molecular point group: c2v
Geometry (in Angstrom), charge = 0, multiplicity = 1:

   Center              X                  Y                   Z
------------   -----------------  -----------------  -----------------
       O          0.000000000000     0.000000000000    -0.071151380605
       H          0.000000000000     0.757939245855     0.564612021746
       H          0.000000000000    -0.757939245855     0.564612021746

Number of unique atoms: 2

Atoms equivalency:
   unique atom 0: 0
   unique atom 1: 1 2

Nuclear repulsion energy:       8.90770810

## Overlap ##
             1            2            3            4            5            6            7

1    1.0000000    0.2367039    0.0000000    0.0000000   -0.0000000    0.0500137    0.0500137
2    0.2367039    1.0000000    0.0000000    0.0000000   -0.0000000    0.4539953    0.4539953
3    0.0000000    0.0000000    1.0000000    0.0000000    0.0000000    0.0000000    0.0000000
4    0.0000000    0.0000000    0.0000000    1.0000000    0.0000000    0.2927386   -0.2927386
5   -0.0000000   -0.0000000    0.0000000    0.0000000    1.0000000    0.2455507    0.2455507
6    0.0500137    0.4539953    0.0000000    0.2927386    0.2455507    1.0000000    0.2510021
7    0.0500137    0.4539953    0.0000000   -0.2927386    0.2455507    0.2510021    1.0000000

## Kinetic ##
             1            2            3            4            5            6            7

1   29.0031999   -0.1680109    0.0000000    0.0000000   -0.0000000   -0.0045400   -0.0045400
2   -0.1680109    0.8081280    0.0000000    0.0000000   -0.0000000    0.1137475    0.1137475
3    0.0000000    0.0000000    2.5287312    0.0000000    0.0000000    0.0000000    0.0000000
4    0.0000000    0.0000000    0.0000000    2.5287312    0.0000000    0.1993355   -0.1993355
5   -0.0000000   -0.0000000    0.0000000    0.0000000    2.5287312    0.1672036    0.1672036
6   -0.0045400    0.1137475    0.0000000    0.1993355    0.1672036    0.7600319    0.0083249
7   -0.0045400    0.1137475    0.0000000   -0.1993355    0.1672036    0.0083249    0.7600319

## Potential ##
             1            2            3            4            5            6            7

1  -61.6882822   -7.4363118    0.0000000    0.0000000   -0.0186797   -1.6150635   -1.6150635
2   -7.4363118  -10.1101908    0.0000000    0.0000000   -0.2221598   -3.6569582   -3.6569582
3    0.0000000    0.0000000   -9.9595668    0.0000000    0.0000000    0.0000000    0.0000000
4    0.0000000    0.0000000    0.0000000  -10.0957534    0.0000000   -2.0901916    2.0901916
5   -0.0186797   -0.2221598    0.0000000    0.0000000  -10.0553869   -1.8259929   -1.8259929
6   -1.6150635   -3.6569582    0.0000000   -2.0901916   -1.8259929   -5.7165220   -1.5685885
7   -1.6150635   -3.6569582    0.0000000    2.0901916   -1.8259929   -1.5685885   -5.7165220

## One Electron Ints ##
             1            2            3            4            5            6            7

1  -32.6850823   -7.6043227    0.0000000    0.0000000   -0.0186797   -1.6196036   -1.6196036
2   -7.6043227   -9.3020628    0.0000000    0.0000000   -0.2221598   -3.5432106   -3.5432106
3    0.0000000    0.0000000   -7.4308356    0.0000000    0.0000000    0.0000000    0.0000000
4    0.0000000    0.0000000    0.0000000   -7.5670222    0.0000000   -1.8908561    1.8908561
5   -0.0186797   -0.2221598    0.0000000    0.0000000   -7.5266557   -1.6587893   -1.6587893
6   -1.6196036   -3.5432106    0.0000000   -1.8908561   -1.6587893   -4.9564901   -1.5602636
7   -1.6196036   -3.5432106    0.0000000    1.8908561   -1.6587893   -1.5602636   -4.9564901

Two-electron Integrals
( 0  0 |  0  0) =    4.785065404705502
( 1  0 |  0  0) =    0.741380351973408
( 1  1 |  0  0) =    1.118946866342470
( 1  0 |  1  0) =    0.136873385354388
( 1  1 |  1  0) =    0.256633394730974
( 1  1 |  1  1) =    0.817206321526058
( 2  0 |  0  0) =    0.000000000000000
( 3  0 |  0  0) =    0.000000000000000
( 4  0 |  0  0) =   -0.000000000000000
( 2  1 |  0  0) =    0.000000000000000
( 3  1 |  0  0) =    0.000000000000000
( 4  1 |  0  0) =   -0.000000000000000
( 2  0 |  1  0) =    0.000000000000000
( 3  0 |  1  0) =    0.000000000000000
( 4  0 |  1  0) =   -0.000000000000000
( 2  1 |  1  0) =    0.000000000000000
( 3  1 |  1  0) =    0.000000000000000
( 4  1 |  1  0) =   -0.000000000000000
( 2  0 |  1  1) =    0.000000000000000
( 3  0 |  1  1) =    0.000000000000000
( 4  0 |  1  1) =   -0.000000000000000
( 2  1 |  1  1) =    0.000000000000000
( 3  1 |  1  1) =    0.000000000000000
( 4  1 |  1  1) =    0.000000000000000
( 2  2 |  0  0) =    1.115813812152427
( 3  2 |  0  0) =    0.000000000000000
( 3  3 |  0  0) =    1.115813812152427
( 4  2 |  0  0) =    0.000000000000000
( 4  3 |  0  0) =    0.000000000000000
( 4  4 |  0  0) =    1.115813812152427
( 2  0 |  2  0) =    0.024477412258099
( 3  0 |  2  0) =    0.000000000000000
( 3  0 |  3  0) =    0.024477412258099
( 4  0 |  2  0) =    0.000000000000000
( 4  0 |  3  0) =    0.000000000000000
( 4  0 |  4  0) =    0.024477412258099
( 2  2 |  1  0) =    0.256683985810103
( 3  2 |  1  0) =    0.000000000000000
( 3  3 |  1  0) =    0.256683985810103
( 4  2 |  1  0) =    0.000000000000000
( 4  3 |  1  0) =    0.000000000000000
( 4  4 |  1  0) =    0.256683985810103
( 2  1 |  2  0) =    0.037808607416361
( 3  0 |  2  1) =    0.000000000000000
( 4  0 |  2  1) =    0.000000000000000
( 3  1 |  2  0) =    0.000000000000000
( 3  1 |  3  0) =    0.037808607416361
( 4  0 |  3  1) =    0.000000000000000
( 4  1 |  2  0) =    0.000000000000000
( 4  1 |  3  0) =    0.000000000000000
( 4  1 |  4  0) =    0.037808607416361
( 2  2 |  1  1) =    0.817022605320914
( 3  2 |  1  1) =    0.000000000000000
( 3  3 |  1  1) =    0.817022605320914
( 4  2 |  1  1) =    0.000000000000000
( 4  3 |  1  1) =    0.000000000000000
( 4  4 |  1  1) =    0.817022605320914
( 2  1 |  2  1) =    0.180518392104632
( 3  1 |  2  1) =    0.000000000000000
( 3  1 |  3  1) =    0.180518392104632
( 4  1 |  2  1) =    0.000000000000000
( 4  1 |  3  1) =    0.000000000000000
( 4  1 |  4  1) =    0.180518392104632
( 2  2 |  2  0) =    0.000000000000000
( 3  2 |  2  0) =    0.000000000000000
( 3  3 |  2  0) =    0.000000000000000
( 4  2 |  2  0) =    0.000000000000000
( 4  3 |  2  0) =    0.000000000000000
( 4  4 |  2  0) =    0.000000000000000
( 3  0 |  2  2) =    0.000000000000000
( 3  2 |  3  0) =    0.000000000000000
( 3  3 |  3  0) =    0.000000000000000
( 4  2 |  3  0) =    0.000000000000000
( 4  3 |  3  0) =    0.000000000000000
( 4  4 |  3  0) =    0.000000000000000
( 4  0 |  2  2) =   -0.000000000000000
( 4  0 |  3  2) =    0.000000000000000
( 4  0 |  3  3) =   -0.000000000000000
( 4  2 |  4  0) =    0.000000000000000
( 4  3 |  4  0) =    0.000000000000000
( 4  4 |  4  0) =   -0.000000000000000
( 2  2 |  2  1) =    0.000000000000000
( 3  2 |  2  1) =    0.000000000000000
( 3  3 |  2  1) =    0.000000000000000
( 4  2 |  2  1) =    0.000000000000000
( 4  3 |  2  1) =    0.000000000000000
( 4  4 |  2  1) =    0.000000000000000
( 3  1 |  2  2) =    0.000000000000000
( 3  2 |  3  1) =    0.000000000000000
( 3  3 |  3  1) =    0.000000000000000
( 4  2 |  3  1) =    0.000000000000000
( 4  3 |  3  1) =    0.000000000000000
( 4  4 |  3  1) =    0.000000000000000
( 4  1 |  2  2) =    0.000000000000000
( 4  1 |  3  2) =    0.000000000000000
( 4  1 |  3  3) =    0.000000000000000
( 4  2 |  4  1) =    0.000000000000000
( 4  3 |  4  1) =    0.000000000000000
( 4  4 |  4  1) =    0.000000000000000
( 2  2 |  2  2) =    0.880159093375045
( 3  2 |  2  2) =    0.000000000000000
( 3  2 |  3  2) =    0.047444445118384
( 3  3 |  2  2) =    0.785270203138277
( 3  3 |  3  2) =    0.000000000000000
( 3  3 |  3  3) =    0.880159093375045
( 4  2 |  2  2) =    0.000000000000000
( 4  2 |  3  2) =    0.000000000000000
( 4  2 |  3  3) =    0.000000000000000
( 4  2 |  4  2) =    0.047444445118384
( 4  3 |  2  2) =    0.000000000000000
( 4  3 |  3  2) =    0.000000000000000
( 4  3 |  3  3) =    0.000000000000000
( 4  3 |  4  2) =    0.000000000000000
( 4  3 |  4  3) =    0.047444445118384
( 4  4 |  2  2) =    0.785270203138277
( 4  4 |  3  2) =    0.000000000000000
( 4  4 |  3  3) =    0.785270203138277
( 4  4 |  4  2) =    0.000000000000000
( 4  4 |  4  3) =    0.000000000000000
( 4  4 |  4  4) =    0.880159093375045
( 5  0 |  0  0) =    0.159028953813045
( 5  1 |  0  0) =    0.381788700431517
( 5  0 |  1  0) =    0.029107255214247
( 5  1 |  1  0) =    0.088639046604491
( 5  0 |  1  1) =    0.054208207239172
( 5  1 |  1  1) =    0.311911628346485
( 5  2 |  0  0) =    0.000000000000000
( 5  3 |  0  0) =    0.208975170450133
( 5  4 |  0  0) =    0.175289465612119
( 5  0 |  2  0) =    0.000000000000000
( 5  0 |  3  0) =    0.001028596856434
( 5  0 |  4  0) =    0.000862792421254
( 5  2 |  1  0) =    0.000000000000000
( 5  3 |  1  0) =    0.049151047350638
( 5  4 |  1  0) =    0.041228155506758
( 5  1 |  2  0) =    0.000000000000000
( 5  1 |  3  0) =    0.005091750233326
( 5  1 |  4  0) =    0.004270986718219
( 5  0 |  2  1) =    0.000000000000000
( 5  0 |  3  1) =    0.001923027401919
( 5  0 |  4  1) =    0.001613045439387
( 5  2 |  1  1) =    0.000000000000000
( 5  3 |  1  1) =    0.184701361441018
( 5  4 |  1  1) =    0.154928467698285
( 5  1 |  2  1) =    0.000000000000000
( 5  1 |  3  1) =    0.038683534029509
( 5  1 |  4  1) =    0.032447950603007
( 5  2 |  2  0) =    0.009336058122476
( 5  2 |  3  0) =    0.000000000000000
( 5  2 |  4  0) =    0.000000000000000
( 5  3 |  2  0) =    0.000000000000000
( 5  3 |  3  0) =    0.011813922734513
( 5  3 |  4  0) =    0.002078445792233
( 5  4 |  2  0) =    0.000000000000000
( 5  4 |  3  0) =    0.002078445792233
( 5  4 |  4  0) =    0.011079469320272
( 5  0 |  2  2) =    0.054104219459936
( 5  0 |  3  2) =    0.000000000000000
( 5  0 |  3  3) =    0.054291571211162
( 5  0 |  4  2) =    0.000000000000000
( 5  0 |  4  3) =    0.000157151628507
( 5  0 |  4  4) =    0.054236039067604
( 5  2 |  2  1) =    0.054125156914812
( 5  2 |  3  1) =    0.000000000000000
( 5  2 |  4  1) =    0.000000000000000
( 5  3 |  2  1) =    0.000000000000000
( 5  3 |  3  1) =    0.075855681518665
( 5  3 |  4  1) =    0.018227677656995
( 5  4 |  2  1) =    0.000000000000000
( 5  4 |  3  1) =    0.018227677656995
( 5  4 |  4  1) =    0.069414628237947
( 5  1 |  2  2) =    0.306655599347507
( 5  1 |  3  2) =    0.000000000000000
( 5  1 |  3  3) =    0.315738293535417
( 5  1 |  4  2) =    0.000000000000000
( 5  1 |  4  3) =    0.007618611374202
( 5  1 |  4  4) =    0.313046130402530
( 5  2 |  2  2) =    0.000000000000000
( 5  2 |  3  2) =    0.008064485483452
( 5  2 |  3  3) =    0.000000000000000
( 5  2 |  4  2) =    0.006764532589140
( 5  2 |  4  3) =    0.000000000000000
( 5  2 |  4  4) =    0.000000000000000
( 5  3 |  2  2) =    0.176295759644369
( 5  3 |  3  2) =    0.000000000000000
( 5  3 |  3  3) =    0.197638924560966
( 5  3 |  4  2) =    0.000000000000000
( 5  3 |  4  3) =    0.011138225736852
( 5  3 |  4  4) =    0.179964436324773
( 5  4 |  2  2) =    0.147877804962089
( 5  4 |  3  2) =    0.000000000000000
( 5  4 |  3  3) =    0.152251498109802
( 5  4 |  4  2) =    0.000000000000000
( 5  4 |  4  3) =    0.011733162163856
( 5  4 |  4  4) =    0.164484175321776
( 5  5 |  0  0) =    0.517125360977304
( 5  0 |  5  0) =    0.006270522616106
( 5  5 |  1  0) =    0.122013556886528
( 5  1 |  5  0) =    0.019208705902818
( 5  5 |  1  1) =    0.491873027324534
( 5  1 |  5  1) =    0.146642827474200
( 5  5 |  2  0) =    0.000000000000000
( 5  5 |  3  0) =    0.009188483135799
( 5  5 |  4  0) =    0.007707347696814
( 5  2 |  5  0) =    0.000000000000000
( 5  3 |  5  0) =    0.011128525245716
( 5  4 |  5  0) =    0.009334665162233
( 5  5 |  2  1) =    0.000000000000000
( 5  5 |  3  1) =    0.095810960370799
( 5  5 |  4  1) =    0.080366734512074
( 5  2 |  5  1) =    0.000000000000000
( 5  3 |  5  1) =    0.105474819021288
( 5  4 |  5  1) =    0.088472829676139
( 5  5 |  2  2) =    0.471913115835107
( 5  5 |  3  2) =    0.000000000000000
( 5  5 |  3  3) =    0.506704669305872
( 5  5 |  4  2) =    0.000000000000000
( 5  5 |  4  3) =    0.029183336960891
( 5  5 |  4  4) =    0.496392251464638
( 5  2 |  5  2) =    0.021048752318672
( 5  3 |  5  2) =    0.000000000000000
( 5  3 |  5  3) =    0.096048971720199
( 5  4 |  5  2) =    0.000000000000000
( 5  4 |  5  3) =    0.062910576176908
( 5  4 |  5  4) =    0.073818472547383
( 5  5 |  5  0) =    0.026686622886860
( 5  5 |  5  1) =    0.289333860384054
( 5  5 |  5  2) =    0.000000000000000
( 5  5 |  5  3) =    0.237637271677700
( 5  5 |  5  4) =    0.199331385983227
( 5  5 |  5  5) =    0.774605943919898
( 6  0 |  0  0) =    0.159028953813045
( 6  1 |  0  0) =    0.381788700431517
( 6  0 |  1  0) =    0.029107255214247
( 6  1 |  1  0) =    0.088639046604491
( 6  0 |  1  1) =    0.054208207239172
( 6  1 |  1  1) =    0.311911628346485
( 6  2 |  0  0) =    0.000000000000000
( 6  3 |  0  0) =   -0.208975170450133
( 6  4 |  0  0) =    0.175289465612119
( 6  0 |  2  0) =    0.000000000000000
( 6  0 |  3  0) =   -0.001028596856434
( 6  0 |  4  0) =    0.000862792421254
( 6  2 |  1  0) =    0.000000000000000
( 6  3 |  1  0) =   -0.049151047350638
( 6  4 |  1  0) =    0.041228155506758
( 6  1 |  2  0) =    0.000000000000000
( 6  1 |  3  0) =   -0.005091750233326
( 6  1 |  4  0) =    0.004270986718219
( 6  0 |  2  1) =    0.000000000000000
( 6  0 |  3  1) =   -0.001923027401919
( 6  0 |  4  1) =    0.001613045439387
( 6  2 |  1  1) =    0.000000000000000
( 6  3 |  1  1) =   -0.184701361441018
( 6  4 |  1  1) =    0.154928467698285
( 6  1 |  2  1) =    0.000000000000000
( 6  1 |  3  1) =   -0.038683534029509
( 6  1 |  4  1) =    0.032447950603007
( 6  2 |  2  0) =    0.009336058122476
( 6  2 |  3  0) =    0.000000000000000
( 6  2 |  4  0) =    0.000000000000000
( 6  3 |  2  0) =    0.000000000000000
( 6  3 |  3  0) =    0.011813922734513
( 6  3 |  4  0) =   -0.002078445792233
( 6  4 |  2  0) =    0.000000000000000
( 6  4 |  3  0) =   -0.002078445792233
( 6  4 |  4  0) =    0.011079469320272
( 6  0 |  2  2) =    0.054104219459936
( 6  0 |  3  2) =    0.000000000000000
( 6  0 |  3  3) =    0.054291571211162
( 6  0 |  4  2) =    0.000000000000000
( 6  0 |  4  3) =   -0.000157151628507
( 6  0 |  4  4) =    0.054236039067604
( 6  2 |  2  1) =    0.054125156914812
( 6  2 |  3  1) =    0.000000000000000
( 6  2 |  4  1) =    0.000000000000000
( 6  3 |  2  1) =    0.000000000000000
( 6  3 |  3  1) =    0.075855681518665
( 6  3 |  4  1) =   -0.018227677656995
( 6  4 |  2  1) =    0.000000000000000
( 6  4 |  3  1) =   -0.018227677656995
( 6  4 |  4  1) =    0.069414628237947
( 6  1 |  2  2) =    0.306655599347507
( 6  1 |  3  2) =    0.000000000000000
( 6  1 |  3  3) =    0.315738293535417
( 6  1 |  4  2) =    0.000000000000000
( 6  1 |  4  3) =   -0.007618611374202
( 6  1 |  4  4) =    0.313046130402530
( 6  2 |  2  2) =    0.000000000000000
( 6  2 |  3  2) =   -0.008064485483452
( 6  2 |  3  3) =    0.000000000000000
( 6  2 |  4  2) =    0.006764532589140
( 6  2 |  4  3) =    0.000000000000000
( 6  2 |  4  4) =    0.000000000000000
( 6  3 |  2  2) =   -0.176295759644369
( 6  3 |  3  2) =    0.000000000000000
( 6  3 |  3  3) =   -0.197638924560966
( 6  3 |  4  2) =    0.000000000000000
( 6  3 |  4  3) =    0.011138225736852
( 6  3 |  4  4) =   -0.179964436324773
( 6  4 |  2  2) =    0.147877804962089
( 6  4 |  3  2) =    0.000000000000000
( 6  4 |  3  3) =    0.152251498109802
( 6  4 |  4  2) =    0.000000000000000
( 6  4 |  4  3) =   -0.011733162163856
( 6  4 |  4  4) =    0.164484175321776
( 6  5 |  0  0) =    0.154983826095369
( 6  0 |  5  0) =    0.006178562861131
( 6  5 |  1  0) =    0.036324481156051
( 6  1 |  5  0) =    0.018637075964466
( 6  0 |  5  1) =    0.018637075964466
( 6  5 |  1  1) =    0.138899129503383
( 6  1 |  5  1) =    0.120424410513661
( 6  5 |  2  0) =    0.000000000000000
( 6  5 |  3  0) =    0.000000000000000
( 6  5 |  4  0) =    0.002735070642527
( 6  2 |  5  0) =    0.000000000000000
( 6  3 |  5  0) =   -0.009851358272803
( 6  4 |  5  0) =    0.009063088333113
( 6  0 |  5  2) =    0.000000000000000
( 6  0 |  5  3) =    0.009851358272803
( 6  0 |  5  4) =    0.009063088333113
( 6  5 |  2  1) =    0.000000000000000
( 6  5 |  3  1) =    0.000000000000000
( 6  5 |  4  1) =    0.024748716021254
( 6  2 |  5  1) =    0.000000000000000
( 6  3 |  5  1) =   -0.059473510026833
( 6  4 |  5  1) =    0.070735331310826
( 6  1 |  5  2) =    0.000000000000000
( 6  1 |  5  3) =    0.059473510026833
( 6  1 |  5  4) =    0.070735331310826
( 6  5 |  2  2) =    0.135540657848790
( 6  5 |  3  2) =    0.000000000000000
( 6  5 |  3  3) =    0.138097677983239
( 6  5 |  4  2) =    0.000000000000000
( 6  5 |  4  3) =    0.000000000000000
( 6  5 |  4  4) =    0.142908608378693
( 6  2 |  5  2) =    0.016764912164332
( 6  2 |  5  3) =    0.000000000000000
( 6  2 |  5  4) =    0.000000000000000
( 6  3 |  5  2) =    0.000000000000000
( 6  3 |  5  3) =   -0.013036508898613
( 6  3 |  5  4) =   -0.034131389240033
( 6  4 |  5  2) =    0.000000000000000
( 6  4 |  5  3) =    0.034131389240033
( 6  4 |  5  4) =    0.057682230376614
( 6  5 |  5  0) =    0.007800931718220
( 6  0 |  5  5) =    0.025610632874342
( 6  5 |  5  1) =    0.065308066611221
( 6  1 |  5  5) =    0.196819899854417
( 6  5 |  5  2) =    0.000000000000000
( 6  5 |  5  3) =    0.040452941573134
( 6  5 |  5  4) =    0.044287849234219
( 6  2 |  5  5) =    0.000000000000000
( 6  3 |  5  5) =   -0.085489074048849
( 6  4 |  5  5) =    0.126882792323867
( 6  5 |  5  5) =    0.133977082851046
( 6  6 |  0  0) =    0.517125360977304
( 6  0 |  6  0) =    0.006270522616106
( 6  6 |  1  0) =    0.122013556886528
( 6  1 |  6  0) =    0.019208705902818
( 6  6 |  1  1) =    0.491873027324534
( 6  1 |  6  1) =    0.146642827474200
( 6  6 |  2  0) =    0.000000000000000
( 6  6 |  3  0) =   -0.009188483135799
( 6  6 |  4  0) =    0.007707347696814
( 6  2 |  6  0) =    0.000000000000000
( 6  3 |  6  0) =   -0.011128525245716
( 6  4 |  6  0) =    0.009334665162233
( 6  6 |  2  1) =    0.000000000000000
( 6  6 |  3  1) =   -0.095810960370799
( 6  6 |  4  1) =    0.080366734512074
( 6  2 |  6  1) =    0.000000000000000
( 6  3 |  6  1) =   -0.105474819021288
( 6  4 |  6  1) =    0.088472829676139
( 6  6 |  2  2) =    0.471913115835107
( 6  6 |  3  2) =    0.000000000000000
( 6  6 |  3  3) =    0.506704669305872
( 6  6 |  4  2) =    0.000000000000000
( 6  6 |  4  3) =   -0.029183336960891
( 6  6 |  4  4) =    0.496392251464638
( 6  2 |  6  2) =    0.021048752318672
( 6  3 |  6  2) =    0.000000000000000
( 6  3 |  6  3) =    0.096048971720199
( 6  4 |  6  2) =    0.000000000000000
( 6  4 |  6  3) =   -0.062910576176908
( 6  4 |  6  4) =    0.073818472547383
( 6  6 |  5  0) =    0.025610632874342
( 6  5 |  6  0) =    0.007800931718220
( 6  6 |  5  1) =    0.196819899854417
( 6  5 |  6  1) =    0.065308066611221
( 6  6 |  5  2) =    0.000000000000000
( 6  6 |  5  3) =    0.085489074048849
( 6  6 |  5  4) =    0.126882792323867
( 6  5 |  6  2) =    0.000000000000000
( 6  5 |  6  3) =   -0.040452941573134
( 6  5 |  6  4) =    0.044287849234219
( 6  6 |  5  5) =    0.342541413419711
( 6  5 |  6  5) =    0.035634211559360
( 6  6 |  6  0) =    0.026686622886860
( 6  6 |  6  1) =    0.289333860384054
( 6  6 |  6  2) =    0.000000000000000
( 6  6 |  6  3) =   -0.237637271677700
( 6  6 |  6  4) =    0.199331385983227
( 6  6 |  6  5) =    0.133977082851046
( 6  6 |  6  6) =    0.774605943919898

There are 406 unique integrals

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Does UV water purification cause cancer?
I found out that a relative is using a machine that performs UV water purification at home (for drinking).
There seems to be a distinct difference from the water that is not purified by it.
Is it possible that this use of UV radiation may lead to cancer in humans?

No, it does not

I believe that UV air purifiers make some ozone which might be linked to cancer.  Water seems different because by the time you drink it I don't think that it would have radicals.  UV is used to sanitize medical equipment and sunlight is putting uv into water all the time.

Thank you joseph-hirsch for your answer. geafix, In general, UV rays are known to cause skin cancer, so that's why I was confused.

That's right, but it's not like your water now contains "dissolved" UV rays or something. They are gone.

I'm glad I got the answer to this question. But I'm surprised as to why it is off topic and "personal". I only found out about this due to a relative using it. But I sure believe this applies to the entire human kind (or for that matter all living creatures). So I would like this to remain for the benefit of humans. After all, the tag 'water-treatment' is 100% relevant.

Is it possible that this use of UV radiation may lead to cancer in humans?

No!
The lamps used for water treatment are usually low pressure mercury vapour lamps. These lamps emit in the UV-C range with a maximum at $\lambda = 254~\mathrm{nm}$.

No mercury can escape from an intact lamp.
In principle, UV-C radiation may cause severe sun burns and harm your eyes. However, the lamps are housed and the consumer is protected from the direct exposure to radiation.
There is no change to the water due to UV irradiation.
UV-C radiation is efficient in killing microorganisms in water.

Obviously, the UV is not doing harm to the water and the person who drinks it. Just make sure, that no person gets in contact with the hard UV radiation. Still, when contaminations are present in the water, the UV might alter them to make them even more toxic. For example Br- ions might react to Bromate BrO3- which would indeed be related to cancer. I think in California, some drinking water reservoirs were covered with shade balls, to make sure no UV radiation reaches the water, to eliminate the risk of toxic products being formed.


text:
Is trans-1,2-dimethylcyclobutane chiral?
If we look at its mirror image about a mirror which is perpendicular to plane of the molecule, the two $\ce{-CH3}$ groups which are pointing upwards and downwards (implied by the wedge and the dash) will interchange their positions:

But also the mirror image would be perfectly superimposable on 1,2-dimethylcyclobutane. Hence, it should be achiral, but the answer key says it is chiral. Is this compound an exception?

Related: Does trans 1,3-dichlorocyclobutane have zero dipole moment? and Are these two structures of 1,3-dichlorocyclobutane enantiomers?

One thing you should always remember is that you should always consider the actual three-dimensional structure when determining chirality. Sometimes there are elements of chirality that are not visible on paper but clear if the full structure is considered. For cyclobutane, this means you should consider the puckered structure of the core structure. However, this often rapidly equilibrates between different practically identical configurations and does not add a layer of complexity to this molecule. Thus, we are able to analyse with the paper plane only.
If you put a plane of symmetry through the centre of the $\ce{CMe-CMe}$ bond, at first glance it may look like it’s a perfect fit: all four carbon atoms of the ring are put on top of another and the wedge and hash simply switch positions.
However, note that wedge and hash mean something entirely different. In one case, the methyl group is above the paper plane (wedge), in the other, it is below it (hash). The corresponding opposite position is occupied by a hydrogen atom that completes the four valences of the cyclobutane’s carbon. Thus, this plane of symmetry you propose transforms hydrogen onto methyl and vice-versa. It is not a symmetric transformation.
You can maybe visualise this by taking a sheet of paper, marking the bottom left corner with a filled dot (the wedge-methyl group) and the bottom right with a circle (the hash-methyl group). Turn it around and put a circle behind the dot and a dot behind the circle, so that you may identify it from both sides. You may already notice that in both cases the dot is bottom-left while the circle is bottom-right — this already indicates that we have a $C_2$ axis, i.e. we can rotate around an axis that is perpendicular to the $\ce{CMe-CMe}$ bond and in the ring plane.
Prepare a second piece of paper but with the reverse orientation: this is the molecule you created by mirroring along the plane. Now attempt to orient the two pieces in such a way that dot aligns with dot and circle aligns with circle. Spoilers: you will fail. anti-1,2-dimethylcyclobutane is chiral.
You can retry the experiment with the syn-1,2-dimethylcyclobutane, i.e. two dots at the bottom of the sheet of paper (and two circles on the back side). But you should immediately see that another piece of paper with identical dots/circles can immediately be aligned perfectly with the original. Hence, syn-1,2-dimethylcyclobutane is achiral.
This can be explained well with point groups. The anti isomer has the point group $C_2$ which does not feature a plane or centre of symmery. The syn isomer has the point group $C_\mathrm{s}$ whose only element of symmetry is a plane, hence it is achiral. You can read more about point groups in this question.

The given compounds is chiral as there is no element of symmetry. If you take a plane going between the two methyl groups it can be evidently seen that the compound is not symmetrical about the plane.
For more details you can go through these videos:(watch in order)
1.https://www.youtube.com/watch?v=n1n-J5v_I7I
2.https://www.youtube.com/watch?v=6WytiwAcpZ8

Thanks for the answer.But why is the mirror image not superimposable. I mean that just the $\ce{CH3}$ which was pointing down the plane now becomes above the plane and vice versa in mirror image.All other characteristics are still same.what if instead of the other $\ce{CH3}$ being adjacent was diagonally opposite having the same direction as earlier?It must be achiral then then why it is not in this case?

Draw the compund and its mirror image first. Try rotating the mirror image anyway it wont superpose with the original image. Trying flipping the mirror image upside down it would still not superpose. When you turn the mirror image upside down and try to superpose you would see the the corresponding methyl groups are facing opposite direction, ie, one would pointing into the plane of the paper and the other would be pointing out of the plane of the paper.Thus it is clear the the mirror image is not superposable on the original compound!

Just to be clear, there is a symmetry element, a ${C_2}$ axis.  However, a rotational axis by itself is not sufficient to make a molecule achiral.

Please note that saying ‘watch these videos’ does not constitute a good answer. A good answer has the resources here so that they may remain even if the video links get deleted.

This answer is just plain wrong and very misleading. There are symmetry elements and the existence of symmetry elements does not constitute a chiral molecule making the whole answer very week. There are no $S_n$ elements so the molecule is chiral. That's it. That's the only precise answer to the question.

@aniline If it helps, here is a MolyMod I made of the molecule in its two chiral forms: Mirror image
And here you can see that they are not superimposable upon one another.

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


text:
Why does chlorine become stable?
We have an equation in our book
$$\ce{2Cl- (aq) + 2H2O -> 2OH-(aq) + H2(g) + Cl2(g)}$$
Why does chlorine become stable if it is electronegative?
Chlorine has excess of electrons, but later loses electron to become stable.
Is chlorine more stable than chloride?

For this process we need energy, and in this process called the chlor-alkali process, the energy is given by a electrolysis of the choride solution.

For more info Read this https://en.wikipedia.org/wiki/Chloralkali_process

@PrachurjyaBiswas Thank you. So isfree energy positive? And will chlorine be more stable in an ionic form.

Yes the free energy of this rxn. is positive

Chemistry is not about finding the most stable state. In fact, a ‘most stable state’ has no meaning; it always depends on the conditions and other substances around.
Rather, chemistry is about following the pathway which leads to the greatest accessable reduction in Gibbs free energy $\Delta G$. If energy, in your case electricity, is supplied from the outside, the system’s Gibbs free energy may even be minimised by reacting towards substances with a lesser standard heat of formation.


text:
Help with determining the rate constant of a reaction
The question asked was,

The initial concentration of $\ce{N2O5}$ in the first order reaction $$\ce{N2O5 -> 2NO2 + 1/2O2}$$ is $1.24\cdot 10^{-2}~\mathrm{mol\,L^{-1}}$ at 318K.
The concentration of $\ce{N2O5}$ decreases to $0.20\cdot 10^{-2}~\mathrm{mol\,L^{-1}}$ after 1 hour.
Calculate the rate constant at this temperature.

How i attempted to solve this is as follows:-
The answer given, however, is $4.82 \times 10^{-4}\ \mathrm{s{-1}}$. What have I understood incorrectly?

You assumed incorrectly that the rate of the reaction does not change as the concentration of the reactant decreases.  You need to solve this problem as a differential equation.

This Khan Academy Video explains it very nicely.
So since this is a first order reaction
$$\ce{ -\Delta N2O5/\Delta t = k[N2O5]}$$
This gives you the rate of reaction at the very start. A second later however,  $\ce{[N2 O5]}$ decreases which will in turn change the rate of reaction. So lets apply calculus. We can write this as
$$\ce{ \frac{-d [N2O5]}{d t} = k[N2O5]}$$
Then we can integrate to find out the total change over a period of time.
$$\int_{[N2O5]_0}^{[N2O5]t} \frac{d[N2O5]}{[N2O5]}= \int_0^t-k dt$$
since $-k$ is constant, we can take it outside of the integral
$$\int_{[N2O5]_0}^{[N2O5]t} \frac{d[N2O5]}{[N2O5]}= -k\int_0^tdt$$
So then we get
$$ \ln([N2O5]_t) - \ln([N2O5]_0) =-kt$$
So putting in your values
$$\ln(0.20\cdot 10^{-2}) - \ln(1.24\cdot 10^{-2}) = -k\cdot 60\cdot 60$$
So I'm pretty sure the answer is $$5.068\cdot10^{-4}~\pu{s^{-1}}$$

$k=\frac{2.303}{t}*\log{\frac{a}{a-x}}$
For first order reaction . here $a,(a-x)$ are initial and final reactant concentration.
$k\to rate constant$
$t\to time$
Instantaneous rate is proportional to instantaneous concentration of the reactant . So, u can't take the value of conc. as ini conc. and $dx$ is small change in conc. so u can't take it as $C_{1}-C_{2}$


text:
how can i clear that secret Veneer

Hi
my friend gave me a mysterious 10 dollars and on the back of it there's a secret code hidden but theres something weird on it covering it what i don't know how to erase it. This is the picture i feel like its thicker than an actual dollar this is all i know.
Could anyone help me solve the problem ? How and with what can i erase it?

Please tell us some more information about what is covering it. Is it hard? Is it brittle? Is it smooth?

Can you read the code through the bill by shining a bright light through it?  Maybe its not meant to be removed.  You can, however, run a bill through the washing machine.

Use household tools!

Can you bend or fold the note? Is the layer brittle and does break off?
Can you scratch the layer off with a sharp blade?
Place the note between paper (kitchen roll) and use a hot flat iron. Does the layer melt?
Can you wash the layer off using a cotton bud and

water
methylated spirit
nail polish remover

tanX for help its soft paper and

tanX for help its soft paper and i try this solutions but dos't work

Try dissolving it in ethanol or some kind of alcohol. Alcohol is polar so it would dissolve the cover if the substance is polar. The dollar bill is non polar and wouldn't be dissolved.

special tnx but alcohol acid and nail polish remover dosnt work on it


text:
How can two same reactions have a different reduction potential
According to wikipedia the reduction of AgCl (s) to give Ag (s) and $Cl^-$ (aq) has reduction potential of 0.22 V . But it also says tha the reduction of $Ag^+$ (aq) to give Ag (s) has reduction potential of 0.80 V . In the former reaction the oxidation state of +1 and gets reduced to zero oxidation state and chloride acts as spectator ion whereas the same happens in the latter reaction just without the chloride ion. Also I believe silver chloride is a strong electrolyte and dissociates completely. So my question is If both the reactions are exactly identical then why isn't there reduction potential same ? I am super confused, any help will be highly appreciated.

AgBr not AgCl? Check again perhaps? 2) AgCl isn't a strong electrolyte at all 3) See Why the reduction potential of the reaction AgCl (aq) /Ag + Cl- is +0.22V

Sorry that was a typo it is agcl

Ok if agcl is insoluble in water then how is it electrolysed is it done in molten state?

AgCl will dissolve better in a NaCl (aq) or NH4OH solution.


text:
What is "dancing resonance"?
While searching the Internet for information about the tricyclopropylmethyl carbocation, I found a site (askiitians.com) where it said that its stability was due to "dancing (superfast) resonance".
What is "dancing resonance"? How does it take place, and what are the prerequisites for it?

Related: What is the reason for the exceptional stability of the cyclopropylmethyl carbocation?

Are you asking about the homoallyl cation? You might want to look at nonclassical carbocations.

Related: Stability Comparison between the Tropylium and Tricyclopropylcarbinyl Carbocation

Resonance is not ‘dancing’. Resonance is often confused by undergraduates (and sometimes even professors) as some kind of oscillation back and forth between multiple structures. That is not the case.
When explained properly, resonance is a consequence of our unability to draw adequate depictions of what the bonding situation in the molecule truly looks like. For example, take a benzene ring. We can only draw single and double bonds. The structures should be identical whether the double bonds are in 1,3,5 or 2,4,6 positions. Our depictions predict different structures because double bonds are shorter than single ones. The actual benzene structure, however, has equal bond lengths between all carbons which could be described as 1.5-bonds. It is incorrect to think of the double bonds "resonating" between 1,3,5 and 2,4,6; rather, both depictions should be considered as two extreme pictures that by themselves cannot capture the truth.
This may be relatively easily understood in simple cases such as benzene or the cycloheptatrienyl cation, where the resonance depictions are all very similar. It is a lot harder to understand it in the case of a cation such as the tricyclopropylmethyl cation. There are a number of answers here explaining exactly which resonance forms stabilise it; it can and should also be seen as a nonclassical cation of the pentavalent carbonium type. You can, for example, check this question.
However, that does not mean that the resonance in these non-classical cations has a different nature. It is not correct to say that the resonance here is "faster" than in benzene, because resonance is simply not a oscillation or vibration between different structures: there is no "speed" or "frequency" to speak of.

To add a personal comment: I have never seen any reputable source use the phrase "dancing resonance".

There's also those people who draw a phenyl moiety as a hexagon with a ring inside it. Come to think of it, that might actually be the better way to do it, too.


text:
Finding the nearest neighbour, from an atom within a crystal
Is there any software can give me the information of nearest neighbour, nest nearest neighbour,....(and their distance) from an atom within a crystal (described by a .cif file)? Perhaps some visualization software like VESTA?

I don't use VESTA and there are probably a few tools that can do this.
My suggestion would be to use the open source Python package ASE or "Atomic Simulation Environment." developed in part by the group of Karsten W. Jacobsen.
There are several algorithms available for generating neighbor lists and iterating through them:
https://wiki.fysik.dtu.dk/ase/ase/neighborlist.html
Here's an example from the documentation:
# get neighbors of atom 42
indices, offsets = nl.get_neighbors(42)
for i, offset in zip(indices, offsets):
    print(atoms.positions[i] + dot(offset, atoms.get_cell()))

I checked ASE, but that seems require LINUX. I have only WINDOWS system

ASE requires Python (e.g., ASE install instructions ) and certainly Python and ASE can be installed on Windows: https://www.python.org/downloads/windows/

I have got Pythod under windows....so how should I proceed to install ASE? get Cywin? I have also got it......

As indicated in my comment above, if you have Python installed, you can install ASE using pip install --upgrade --user ase

Where should I run this pip command? Under Python or under window command? or under Cywin? I have tried them all but none successed.

I don't use Windows, so I'm not a good person to ask. I'd find a Python-specific forum (e.g., Python on Windows) and ask there. On Mac or Linux, you can simply run pip from the command-line. Heck, you might want to ask the ASE forums.


text:
A drop of water in a tin of sugar: Which one's the solvent, the sugar or the water?
The other day, when we were dealing with the chapter Solutions, our teacher asked us this:

If I add a drop of water, to a tin full of sugar (without mixing it in), what's the solvent here? The water or the sugar?

Naturally we were taken aback by the 'unorthodox' nature of the question. Seeing that none of us were in a position to answer that any time soon, he gave us the "answer":

It's the sugar. If you remember the definition we learnt ( solute + solvent = solution; between the solute and the solvent, the solvent is one present in a larger quantity and is in the same phase/state of matter as the solution ) you ought to have realized that, since there's more sugar than water, and at the end of the process the tin's filled with a solid.

Now that may seem like a clever way of testing our command of terminologies and definitions, but I saw a hitch there.
A solution is defined as a homogeneous mixture.
Adding a drop of water to a tin of sugar (without mixing it in) does not result in a homogeneous mixture. So as interesting as the question is, I feel it's flawed on this account... because what we're dealing with isn't technically even a solution.
Now I told this to my teacher, but he (quite deftly) side-stepped my query...a subtle way of indicating that he's not comfortable discussing this.
So I guess my question here boils down to this:
Was my teacher's "answer" correct? Or was the question seriously flawed to begin with?

EDIT-
I wouldn't call this a duplicate. Sure, I mean, both my question and the other one that was linked with this was about identifying the solvent and solute but I feel my question is distinct because:
1) It isn't answered well enough in the other question (that answer was too generalized... I've given a specific instance here)
2) I also want to know if a drop of water added to a tin full of sugar (without mixing it) can be considered a solution.

Related: What is the solvent and what is the solute?

Nature hardly cares about how humans decide to name things.

Out of the nine definitions for solvent on TFD that all say ~ ["A substance (usually a liquid) capable of dissolving another substance."] only one of them ends with "or the component that is present in excess." If you want to be colloquial, then the answer's water. If you want an A, then it's w/e they say it is.

You could imagine stirring the sugar enough for the water molecules to be uniformly distributed throughout - it would then be homogeneous.
However, even then, to refer to the mixture as a solution of water in sugar is unhelpful, not least because referring to a slightly damp solid as a solution will only confuse.  A definition needs to be useful, as well as being a set of criteria to be met.
Consider adding more water, and stirring till uniformly distributed.  At some point, the mixture will become a thick syrupy liquid.  Add yet more, and it will eventually resemble an ordinary solution.  At what arbitrary point do you say one is a solution of the other?
As a device to get students thinking about the interactions that take place when a solid dissolves in a liquid, or a liquid in another liquid, it's an excellent question, but not one that has either a right or a wrong answer, other than "neither", or indeed "both".
What is definitely wrong is to insist that there's only one right answer to the question.

And for the particular case of sugar and water, as anyone who has experimented with cocktails can tell you, it's easy to make a syrup with equal parts (by initial volume) sugar and water, and even a ratio of water to sugar of 1:3.

Both answers are right. In the IUPAC Gold Book it states

A liquid or solid phase containing more than one substance, when for
  convenience one (or more) substance, which is called the solvent, is
  treated differently from the other substances, which are called
  solutes.

So either way can be used, but it is usually more conventional to call the substance that is in larger amounts the solvent.

The question is flawed since we do not know whether the resulting mixture consists of a single phase or not (I guess that there are two phases).
Moreover, the definition

solute + solvent = solution; between the solute and the solvent, the solvent is one present in larger quantities and is in the same phase/state of matter as the solution

seems to be a little bit impractical to me. E.g. a saturated solution of sucrose in water will always contain more than 64% mass of sucrose, but hardly anybody would call it a solution of water in sucrose. So I would prefer the IUPAC definition cited by Clangorous Chimera.

To be clear: the definition the Original Post (Question) gave is NOT one which is acceptable to me, or I'd guess, to a majority of the chemists of the world. The IUPAC definition clearly indicates that the "solvent" is assigned arbitrarily – the material in the mixture which may be "treated differently" depends on the CHOICE of treatment and so will vary depending on that choice – in fact can change from one instant to the next.
The question also suffers from vagueness: how much sugar is in a "tin"? Sugar is also ill defined. Table sugar is generally accepted to be the disaccharide sucrose. Sucrose is composed of the two hexoses fructose and glucose bonded together (their respective melting points are 103 °C and ~147 °C (but it depends on their stereochemistry) and sucrose has no melting point (as far as I know, I wonder if at high pressure this would still be true, but IDK), it decomposes at ~186 °C.
So, picture a container with molten (melted) fructose in it. There's no trouble adding a drop of water (a drop of water is often taken to be 1/20th of a gram since that's about what you get out of a glass eye-dropper (you get much less out of a plastic one, generally (due to less surface tension))). You'd not be wrong to claim you had a solution of water in a sugar. Allow the solution to freeze – does it suddenly stop being a solution? No – at least not necessarily. Take the solid and break it up into a powder. Now has it changed? Well, you have to be careful here. Often times grinding changes the chemistry (of the particles' surface), but if you've avoided that, you now have a granular solid solution of water in a sugar.
The problem is, as I've already said, that the definition your teacher gave you is inadequate in several ways. For instance what does "larger" mean? I can (easily) create a molecule (called a polymer) which weighs 100 grams (or 100 kg for that matter). If I add 1 molecule to 1,000,000,000,000,000,000,000 molecules of water, which is the solvent and which the solute? If I add 1 gram of water to 2 grams of tungsten which is the solvent? The water has a volume of 1 ml, the tungsten's volume is 0.1 ml so which is the "larger" amount the water or the tungsten? (Tungsten wouldn't dissolve in water at room temperature, so you'd have to heat this mixture as well). BUT. As an exercise in logic, given the definition you are supposed to rely on, then it is the water that is the solute.
I don't understand why you claim you weren't in a position to answer it. As far as I can see, you were. I also don't understand why you think that a homogeneous mixture MUST be continuous. I think you have confused two different concepts. A mixture of red marbles and blue marbles might be segregated or might be mixed randomly or might be in a well ordered array. Red-Blue-Red-Blue… (or some other order). The fact is, as you hopefully know, the atomic nature of matter means that matter is NOT continuous, but discrete (on a microscopic basis).


text:
How do I calculate the number of molecules in an equilibrium depending on a change in the equilibrium constant?

Consider 1000 molecules of $\ce{A}$ and 1000 molecules of $\ce{B}$ in a eucaryotic cell. The concentration of both will be about $\pu{10^-9 M}$.
  If the equilibrium constant $K$ for $\ce{A + B <--> AB}$ is $\pu{10^10}$, then at equilibrium there will be
  \begin{array}{ccc}
  \bbox[1ex,lightgreen]{270}
    & \bbox[1ex,lightgreen]{270}
    & \bbox[1ex,yellow]{730} \\
  \ce{A} & \ce{B} & \ce{AB}\\
  \text{molecules}&\text{molecules}&\text{molecules}\\
\end{array}
  If the equilibrium constant is a little weaker at $\pu{10^8}$, which represents a loss of $\pu{2.8 kcal/mol}$ of binding energy from the example above, or 2-3 fewer hydrogen bonds, then there will be
  \begin{array}{ccc}
  \bbox[1ex,lightgreen]{915}
    & \bbox[1ex,lightgreen]{915}
    & \bbox[1ex,yellow]{\phantom{0}85} \\
  \ce{A} & \ce{B} & \ce{AB}\\
  \text{molecules}&\text{molecules}&\text{molecules}\\
\end{array}

Figure 3-44 Molecular Biology of the cell 5/e (© Garland Science 2008)

I don't understand how in the above figure, the number of molecules is calculated.

Welcome to Chemistry.se! This seems like a homework question. Please share your thoughts towards a solution otherwise it will be closed in accordance with our homework policy. A "homework question" is any question whose value lies in helping you understand the method by which the question can be solved, rather than getting the answer itself.

I hope that my answer helps the person asking understand the methods used to solve this problem and others like it.

Setting out the problem like this using an ICE table:
\begin{array}{cccc}
  \text{species} & \ce{A}    & \ce{B}    & \ce{AB} \\\hline
  \text{I}       & 10^{-9}   & 10^{-9}   & 0       \\
  \text{C}       & -x        & -x        & +x      \\
  \text{E}       & 10^{-9}-x & 10^{-9}-x & x       \\
\end{array}
Where $\text{I}$ is the initial concentration, $\text{E}$ is the final concentration (concentration at equilibrium) and $\text{C}$ is the change in concentration.
This ICE table (Initial, Change, Equilibrium), is a standard method for solving these types of problems.
The equilibrium constant is defined as
$$K = \frac{[\ce{AB}]}{[\ce{A}][\ce{B}]},$$
where these are the concentrations at equilibrium.
We have our equilibrium concentrations as set out by the problem. Both $\ce{A}$ and $\ce{B}$ start out at a concentration of $10^{-9}M$ with their final concentrations being $\pu{10^-9 M} - x$ where $x$ is some change in the concentration.
Note, it mentions that 1000 molecules corresponds to a concentration of $\pu{10^-9 M}$. Therefore one can see that this corresponds to $10^{12}~\frac{\text{molecules}}{\pu{M}}$.
($\pu{1000 molecules} = \beta \cdot \pu{10^-9 M}$ where one can see $\beta$ has units $\frac{\text{molecules}}{\pu{mol}}$)
We can substitute into the expression for $K$ and noting that $[\ce{A}]=[\ce{B}]$
$$10^{10} = \frac{x}{(10^{-9}-x)^2}$$
I could not be bothered to work this out by hand, so Mathematica gives two values for $x$ as this is a quadratic:
\begin{align}
  x_1 &= \pu{7.3e-10 M}\\
  x_2 &= \pu{1.4e-9 M}
\end{align}
One can immediately see that $x_2$ is not a valid solution as it corresponds to a concentration of $\ce{AB}$ higher than 100% conversion.
Taking $x_1$ as the solution: $$\pu{7.3e-10 M} \times 10^{12}~\frac{\text{molecules}}{\pu{M}} = \pu{730 molecules}\text{ of }\ce{AB}$$
The same can be repeated for the second case where the equilibrium constant is given as $K=10^8$.

could you explain why we have to do the calculation in mol and then convert to molecule number. shouldn't the proportion of reacted/unreacted molecues stay the same since they represent kinda same thing?

also, in the 2nd example K >> 1 still. then how come we have less final product than  reactants?


text:
Is there any substance that is liquid at room temperature and pressure that doesn't contain H as one it's constituents apart from mercury?
I want to know if there's any substance that is liquid at room temperature and atmospheric pressure that doesn't contain HYDROGEN atom(s) in its make up. Any substance dissolved in water or any substance that contains HYDROGEN is not required

Why did you ask this and immediately self-answer? What are you trying to accomplish? A complete list of substances? That is impossible to get as there are too many. Voting to close as too broad.

I would suggest doing some research before asking a question. Simply searching up elements at room temperature would have given you an answer.

You are not going to get a list of compounds which fit your description. It's a very general question. However, many compounds of the form $\ce{C_nX_{2n+2}}$, where X is a halogen for example, are liquids at room tempterature, and contain no hydrogens.
Examples:
Perfluorooctane:

Melting point: $-25^o C$, Boiling point $103^oC$
Perfluoromethylcyclohexane

Melting point: $-37^o C$, Boiling point $76^oC$
Perfluorodecalin

Melting point: depends on the isomer, Boiling point $142^oC$
You can read about fluorocarbons. Many of these are liquids.

Nice, and many chlorocarbons as well.

And, if I recall correctly, oxygen is soluble in perfluorodecalin.  Due to its chemical inertness, we can theoretically "breathe" an oxygenated solution of this liquid.

@ChemProfMatt Very soluble apparently. "100 mL of perfluorodecalin at 25 °C can dissolve 49 mL of oxygen at STP"

Yes. Bromine. Melting point: - 7.2 degree Celsius, boiling point: + 58.8 degree Celsius
https://en.wikipedia.org/wiki/Bromine

Carbon disulphide and bromine are both liquids at room temperature and atmospheric pressure. They do not contain hydrogen atoms. Some elements become liquids at or near room temperature they include gallium, caesium and francium .I'm sure they're still other substances that are liquids at room temperature and atmospheric pressure.

Define room temperature.

One could make the argument that gallium is a liquid at room temperature, if you're willing to fudge a few degrees and go up to 30 C.  Say "room temperature on an unseasonably warm day".


text:
Difference between valency and oxidation state
I know the abstract and sketchy obscure differences but I cannot find anywhere the specific comparison between the concepts of oxidation states and valency including how they are used to describe chemicals.

Valency is a term used to describe the number of bonds a given atom undergoes in a certain environment. For example, the carbon atom in all of the following substances is tetravalent:

methane
methanol
difluoromethane
formic acid
carbon dioxide

An example for a formally trivalent carbon would be carbon monoxide. Note how valency only counts how many bonds and how many lone pairs an atom has.
The oxidation state is used as a formal measurement on how formally negative an atom should be under standard conditions. For this, each bond is split heterogenously and both electrons are counted as belonging to the electronegative partner only. Then, the electrons on each atom are counted, the difference to a neutral atom of the same kind is established which constitutes the oxidation number. The compound above — while all being tetravalent — have different oxidation numbers:

methane $\mathrm{-IV}$
methanol $\mathrm{-II}$
difluoromethane $\mathrm{\pm 0}$
formic acid $\mathrm{+II}$
carbon dioxide $\mathrm{+IV}$


text:
Galvanic cell structure and function
I was thinking about galvanic cells the other afternoon and I was wondering the situation
if a chemical engineer designs a galvanic cell using copper and zinc as electrodes, what would the structure, function and output voltage of the cell be, and how might you possibly increase the voltage?

If by voltage, you mean electric potential, then just analyze the nernst equation.
The standard reduction potential for both half-reactions is known, so add those to get the standard cell potential. Then realize that changing the concentrations of reactions and products (hence, the reaction quotient) can change the actual cell potential (the log term in the equation).

I think voltage is increase by increasing cocentration of zinc sulphate & Copper sulphate. Due to which we gain more sulphate ions and zinc and copper ions.


text:
Equivalent Wyckoff position in different groups
I am not a chemist, I am a physicist, so please bare with me if what I am asking is rather obvious.
If I have two symmetry groups ($P4/mmm$ and $P\!-\!1$), is there a way to find their respective Wyckoff positions for which they are equivalent? Meaning finding the Wyckoff position for each of those groups, so that those positions in specific share the same symmetries. Hopefully I made myself clear.
I really doubt it's as easy as to look at the Wyckoff groups symmetries but I don't know.


text:
What chemical reactions take place when mixing baking soda with citric acid and epsom salt and putting it into a microwave oven?
When I mixed in a bowl baking soda, citric acid and Epsom salt, then put it in the microwave between 30-60 seconds, the mixture will start to fizz and puff up, kind of like dough rising. What is happening here?
I've done test with all of the ingredients individually in the microwave and nothing happens. I've tried various combinations of any two of the three ingredients and nothing happens. Only when all three are mixed together does the mixture fizz and puff up. It seems like the catalyst is the Epsom salt. What is in the Epsom salt that is setting off the other two ingredients?
I know that baking soda plus citric acid together with water will also fizz, as $\ce{CO2}$ is released. I was wondering when I have the three ingredients in the microwave and it is puffing up, is $\ce{CO2}$ also being released?
Furthermore, the resulting mixture that had puffed up, if I put it in water it will still fizz and create $\ce{CO2}$, just as if had I never put the mixture in the microwave and just put it straight in the water. What I'm also wondering about if the microwave had hurt the ability of the mixture to create fizz in the water later, i.e., would there have been more fizz in the water otherwise?
After asking my question I found this YouTube video that said heating up Epsom salt releases $\ce{H2O}$ gas. Which would make sense why my mixture is puffing up. Still I'd love to hear anyone else's opinion on this.

The solid phase reaction of baking soda and citric acid to form monosodium citrate, water and $\ce{CO2}$ takes place very slowly, so you don't get the visible foaming and bubbling that you do when water is added, or if you were to use baking soda with an acidic juice or vinegar (~95% water).
The molecular formula for Epsom salt is $\ce{MgSO4\dot\ 7H2O}$, meaning it's actually more than 50% water by weight. At $\mathrm{150^oC}$ it starts giving up some of those waters of hydration. Although water is a gas at that temperature, enough of the water is able to wet the baking soda and citric acid to get that reaction going, which then releases more water:
$$\ce{NaHCO3 + C6H8O7 → NaC6H7O7 + CO2 ^ + H2O}$$
Once the above reaction is initiated by the release of enough water from the Epsom salt, the additional water the reaction releases makes it somewhat self-sustaining and the $\ce{CO2}$ provides the foam and bubble show. The fact that additional reaction was observed after adding more water just shows that there was unreacted baking soda and citric acid remaining in the solid "crust".


text:
Is 1,2-dihydro-1,2-diborete aromatic?
What kind of aromaticity does 1,2-dihydro-1,2-diborete have?

I read somewhere that 1,2-dihydro-1,2-diborete is an aromatic compound, but I believe that in order to be aromatic, it should have a continuous resonance. I also think that there should be a break at B that causes no π-electrons to be shared among the B—B atoms. So, I think this is non-aromatic. Is that correct?

http://pubs.acs.org/doi/pdf/10.1021/ja00295a033

FYI this paper^ says it's not aromatic, or even flat.

Annually the calculations are quite old and the level of theory quite crude, but most of it should hold. However, it should be easy to update these findings (if it has not been done already).

That molecule is indeed aromatic.
It satisfies Hückel's rule $(4n+2)$ with $n=0$. It's also planar.
Lastly, the boron atom is bonded to three other atoms, meaning it can exist with $\mathrm{sp^2}$ hybridization. The leftover p-orbital, whilst not actively forming π-bonds, can still complete a stable π-molecular orbital.

There is some delocalisation but it doesn't necessarily mean it's really aromatic. As for answer, there are orbitals not domains, and satisfying Huckel doesn't automagically mean the compound is truly aromatic.

Hückel's rules only work for the compounds that it is derived from: monocyclic hydrocarbons.

The compound you are asking for is Aromatic.
There are 3 conditions for aromaticity:

Compound must be cyclic and planar {Given compound is cyclic and planar since boron is $\pu{sp2}$ hybridised}
Compound must be conjugated {While checking for conjugation there is no need of π electrons. For conjugation the sole condition is there must be parallel p orbitals on adjacent atoms: whether they're participating in π bond formation, have lone pair or just empty p orbitals - (as in this case)}
There must (4n+2) π electrons in the largest conjugated periphery {There are 2 π electrons in this case which satisfies 4n+2 condition}.

PS:

Corrections would be highly appreciated.

MOT explanation for huckle rule: Huckle Rule & MOT

The same criticism as for the other answer applies here. Hence I have down voted here, too. Arguments towards an answer of this question should be based on observation of physical properties or rigorous theory. Hückel's rules are just a generalisation of an observation and a good guide for cyclic hydrocarbons.

@Martin-マーチン Hückel's rule can also be applied to molecules containing other atoms such as nitrogen or oxygen. As like in pyridine

It's not been derived from these cases but I'm aware of it's existence. It works coincidentally and probably accidentally for these cases, but it cannot be a substitute for a rigorous theory. Hückel's rules are a good guess estimate, but nothing more. Aromaticity is something that's still not fully understood and there is plenty of discussion still possible. The molecule here is simply to complicated to be treated with such a simple model.

I agree that aromaticity is yet something of a mystery in itself. But at an elementary theoretical level, huckle rules are maybe the best guess we have got. Yeah, experimentally we can do proton NMR spectrum to check the aromaticity, I'm not sure but i read in some research paper that given diborete resonates at about 7 ppm ,,,lil less aromatic then benzene ,,, but yeah, it is in aromatic category.


text:
Usage of ammonia as odorant in fuel gases
Usually, thiols such as ethanethiol are added to fuel gases to warn the user of a leak. Why is a few percent of ammonia not used instead?

Ammonia is flammable so it will burn to nitrogen and water in the flame.
Ammonia has a strong smell, and causes a stimulation effect which may wake sleeping people in the event of a gas leak.
Ammonia produces no sulfur oxides when it burns (although the usual amount produced with ethanethiol is insignificant).

Ammonia can be toxic also maybe corrosive.

In principle, you could use ammonia, but there are practical disadvantages.
The sulfur-based odourants are some of the smelliest compounds known and are detectable in very low concentrations. In practice, they are used in concentrations of perhaps 30ppm. Several percents of ammonia would be thousands of times more than this concentration implying greater expense. At this level, the concentration of ammonia would be problematic because of the corrosive nature of certain types of pipe and would probably create other safety concerns (it might even be a direct safety hazard at the concentrations where it is readily detectable by smell).
Typical sulfur-based odourants are detectable at such low concentrations that they are not a direct safety hazard. Nor is there much worry about the combustion products at such low concentrations.
Ammonia is also much more reactive than the typical sulfur-based odourants and would be much more likely to be stripped from the gas during transmission which defeats the purpose.

I think I should start a campaign to retain the proper (but deprecated) spelling of sulphur.


text:
What were some major breakthroughs in chemistry in 2016?
What breakthroughs have been made in the different fields of chemistry in the past 366 days? This might help me and others know advances made in different field.
I know this question is too broad, but however there are many  interesting advances in all sorts of branches that have been made this year, which I might have missed.
Inspired by "What were some major mathematical breakthroughs in 2016?", on Mathematics Stack Exchange

While this would be an interesting topic of discussion, I feel that it is too broad. Hence a cast a vote to close. Sorry.

http://yearinreview.cenmag.org


text:
what's the name of this org. compound?
What's the name of this compound? It is a semi product of Reimer-Tiemann reaction.
edit: I think its Sodium salicylaldehyde, but i'm not sure.

You might be interested in reading our guidelines on how to ask nomenclature questions.

That's supposed to be the sodium salt of salicylaldehyde, also known as sodium salicylaldehyde, sodium ortho-formylphenolate, 2-hydroxybenzaldehyde sodium salt or sodium 2-formylphenoxide. CAS number: 3116-83-4.
There's actually an error in the structure. The oxygen should have a negative charge if the sodium is written with a positive charge. The salt as a whole is neutral. What's written here is one lone pair, there should be 3 of those and a negative charge.

thanks, btw there is a negative charge on oxygen in the structure!

You mean the line above it? That's not a minus sign which it should be, it's too long and to make it clear it should be above and to the right of the atom symbol, just like the plus sign on sodium. What's drawn here is what is usually used for a lone pair, a longer line right above/on the side/below the atom.

i'm not sure, i used to think that the long line above (mostly oxygen) is a negative charge. this textbook is actually pretty confusing and many-a-times unclear. are you sure that it's used for a lone pair?

Well yes, that's what they mean but it's just a very, very bad (and in my eyes just plain wrong) way to do it, since you won't be able to differentiate between a negative charge and a lone pair. And yes, that's definitely a lone pair "symbol" which they just abuse as a negative charge for no good reason.

yeah i checked some other reactions and they put the charges on top of the symbols. imgur.com/a/5EbV8

The mentioned names (except “sodium salicylaldehyde”) are unambiguous, describe the correct compound and are in accordance with past or present nomenclature rules. However, the preferred IUPAC name according to current recommendations is missing: sodium 2-formylphenolate.


text:
Oxidation Number
What does Oxidation number actually signify?
If it means the number of electrons gained or accepted, why is it a fraction in some cases like in super oxides?

http://chemistry.stackexchange.com/questions/52010/are-non-integer-oxidation-states-possible

Quoting from Wikipedia:

The oxidation state, often called the oxidation number, is an
  indicator of the degree of oxidation (loss of electrons) of an atom in
  a chemical compound. Conceptually, the oxidation state, which may be
  positive, negative or zero, is the hypothetical charge that an atom
  would have if all bonds to atoms of different elements were 100%
  ionic, with no covalent component. This is never exactly true for real
  bonds.

Coming to fractional oxidation state:
It is not actually a fraction rather it is the average of various oxidation states of the element present in the compound.
For example:

$\ce{Fe3O4}$

Here the two Iron atoms that are bonded to three oxygen atoms have an oxidation state of +3 each, and the Iron atom that is bonded to two oxygen atom has an oxidation state of +1.
Thus taking the average (3+3+2)/3 = 8/3;
You would have got the same if you used the fact that $\ce{Fe3O4}$ is neutral.If you assume the oxidation state of Iron to be x, then 3x+4(-2)=0. Solving you would get x as 8/3(However this oxidation state would be the average oxidation state of Iron).

$\ce{KO2}$

In this case the oxygen on the left has an oxidation state of zero as it is only bonded to itself and the oxygen on the right hand side is in oxidation state of -1 (1 bond to another O and a charge of -1). In reality the -1 charge would be spread over the whole species - oxidation state is only a system of chemical " book-keeping". The oxidation state of -1/2 can be regarded as the mean of zero and -1.
There are many more such examples.
Image courtesy:ChemSpider

So the two oxygen atoms in $\ce{KO2}$ are different?

No, they aren't and these pictures are quite bad...


text:
Reconcile the Beer Lambert Law with rate of absorption
The Beer-Lambert law gives the proportion of incident light intensity that will be absorbed. Intensity isn't strictly a rate but if the light is monochromatic and the path length is constant it sort of is.
$$I=I_0\exp(-\epsilon c l)$$
But the rate of absroption of photons is given by:
$$K=B\rho$$
Where B is the Einstein coefficient and $\rho$ is the spectral energy density which is proportional to the intensity of incident radiation.
Surely the second equation should mean that the intensity decays linearly? Would the absorbed intensity double if the concentration was doubled?

To answer your second question first, just use the Beer-lambert law. The first question needs us to go back a few steps and re-define the Einstein B coefficients in terms of the intensity which means that the 'new' B or $B^I=B^{\rho}/c$ and from now on for simplicity we just use B instead of $B^I$.
The intensity I is defined (per unit frequency interval) as flux across unit area per unit time thus has units  $\pu{W/m^2.s^{-1}}$ or mass time$^{-2}$ and thus is not a rate. The energy density $\rho$ means density of energy per unit volume and is related to I as $I=c\rho$ where c is the speed of light. (The rate of absorption is unchanged but written as $B^II$ rather than $B^{\rho}\rho$.)
The energy density $\rho$ has units $\pu{J.m^{-3}/s^{-1}}$ which is mass length$^{-1}$ time$^{-1}$. In these units the Einstein B coefficient has dimensions time mass$^{-1}$ or $\pu{s kg^{-1}}$ in SI units.
The Beer Lambert law derives from the differential form
$$-dI_{\nu} =I_{\nu}k_{\nu}dx$$ where for clarity  x is used to indicate length rather than l. The constant $k_{\nu} = \epsilon [c]$ at frequency $\nu$ and has dimensions of length$^{-1}$.
If the absorption band is wide we take a small element about frequency $\nu$, which is $\delta\nu$,  and calculate the rate of energy removal from the beam during absorption from state 1 to state 2 with populations $N_1$ and $N_2$ respectively, with Einstein coefficients for (stimulated) absorption $B_{12}$ and stimulated emission $B_{21}$
$$ -dI_{\nu}\delta\nu =  h\nu I_{\nu} [N_1B_{12}dx -N_2B_{21}dx  ]$$
rearranging gives
$$-\frac{1}{I_{\nu}}\frac{dI_{\nu}}{dx}\delta\nu=h\nu[N_1B_{12} -N_2B_{21}  ]=k_{\nu}\delta\nu $$
If the light intensity is low, the population of the upper level is only a minute fraction of that of the lower level and can be set to zero. In this case B can be obtained directly as $$B_{12}=\frac{k_{\nu}\delta\nu}{h\nu N_1}$$
Then integration over $\delta\nu$ gives the B value for the absorption band. The value of $N_1$ is effectively the total population.
This shows that the B values are related to the extinction coefficients as expected and that time does not come into the calculation. However, this is only part of the story.
A quantum calculation starting with the Schroedinger equation and using the dipole approximation and first order perturbation theory is necessary to explain the electric dipole interaction. The result of this calculation (see Atkins & Friedman Molecular Quantum Mechanics chapter 6 for gory details) is that the probability of absorption from state a to b is
$$P_{ab}= \frac{|V_{ba}|^2}{\hbar^2}\frac{\sin^2((\omega-\omega_{ab})t/2)}{(\omega-\omega_{ab})^2}$$
where $V_{ab}= <a|H|b>$ is the interaction energy, $\omega$ the radiation frequency and $\omega_{ab} = (E_a-E_b)\hbar$. This oscillating behaviour as a function of frequency is not ordinarily observed in thermal samples but has been observed in a cold molecular beam of HCN molecules. Time was made constant by passing the beam at constant speed through the fixed illuminated region (see Dyke et al. J. Chem. Phys. vol 57, 2277, 1972).
The rate of absorption is the time derivative of the probability which is
$$ R=\frac{|V_{ba}|^2}{2\hbar^2}\frac{\sin((\omega-\omega_{ab})t)}{\omega-\omega_{ab}}$$
and as the the absorption coefficient is directly proportional to the rate of absorption  $\gamma(\omega,t)=\beta R$ (where $\beta$ only contains constants) which indicates that near resonance absorption should increase in time. However, this is not what is usually observed for molecules at room temperature as a vapour or in solution. The reason for the generally observed behaviour is that the absorption is continually interrupted by first order processes, for example by collisions.
Suppose that the system is initially in state a and then the radiation field is turned on. The molecule is then in a superposition state involving a and b, for example $\Psi=c_a(t)\psi_a+c_b(t)\psi_b$ where the c are time dependent coefficients that move the system from a to b. A collision will destroy the superposition and return the molecule to state a or b each with a certain probability, then the radiation causes the superposition to grow again.
If we assume that the number of superposition states is n and that a first order process destroys them with lifetime $\tau$ , then $n(t)=n(0)\exp(-t/\tau)$. There is now a competition between the superposition state growing and being destroyed. The average absorption coefficient is obtained by averaging over the probability distribution of it being interrupted thus
$$\gamma(\omega)=\frac{\int_0^{\infty}\gamma(\omega,t)\exp(-t/\tau)dt}{\int_0^{\infty}\exp(-t/\tau)dt}$$
evaluating gives
$$\gamma(\omega) \propto  L(\omega_{ab}-\omega)$$
where L is the Lorenzian function
$$L(\omega_{ab}-\omega)=\frac{1}{\pi }\frac{\tau^{-1}}{\tau^{-2}+(\omega_{ab}-\omega)^2 }  $$
and examining the function the half-width  at half-height $\Delta \omega = 1/\tau$. The first order process thus explains why the absorption does not increase in time.  However, the experimentally line shapes on many molecules do not have a Lorenzian shape and the reasons for this are not that the idea of interrupting the superposition is wrong but that other effects dominate, doppler broadening for example, or in solution there is always inhomogeneous broadening. This is caused by interactions with solvent molecules which shift energy levels up and down and thus the solution effectively contains molecules with a mixture of energy levels. That this is so can be observed at very low temperatures (say 10K) where it is possible to burn a spectral hole in a sample and so remove a small part of the inhomogeneous population. The spectral hole produced is very narrow as expected from the calculation above.


text:
what's the name of this org. compound. RR'-C=NNH2
What's the name of this compound? It's a semi-product of Wolff-Kishner reduction.

The complete reaction is:

The (now deleted) answer by NotCorey has already explained that the compound that is shown in the picture is a hydrazone that is derived from acetone (propan-2-one).
According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the preferred IUPAC name (PIN) for a hydrazone is generated substitutively as ‘ylidene’ derivatives of hydrazine rather than by functional class nomenclature as in previous recommendations.

P-68.3.1.2.2 Hydrazones
Compounds having the general structure $\ce{RCH=N-NH2}$ or $\ce{RR'C=N-NH2}$ are called ‘hydrazones’ and are named in two ways:
(1) substitutively as derivatives of the parent hydride ‘hydrazine’, $\ce{H2N-NH2}$;
(2) by functional class nomenclature using the class name ‘hydrazone’.
Method (1) generates preferred IUPAC names.

Therefore, the PIN for the compound that is given in the question is (propan-2-ylidene)hydrazine.


text:
Mass chromatogram : Total Ion Chromatogram (TIC)
I am confused about what a Total Ion Chromatogram (TIC)  means. The definition I have says:

The total ion current (TIC) chromatogram represents the summed
  intensity across the entire range of masses being detected at every
  point in the analysis.

Does this mean it is a (mixture) sample injected into a MS without the preceding chromatographic separation (as is typical in, say, a GC-MS or an LC-MS)?  Are there MS instruments sold today that are not GC-MS's or LC-MS's etc. (i.e. mass spec. sans the preceding seperation?
What's the point in doing this? Wouldn't the resultant spectrum be very noisy since it's essentially the sum over all components of a mixture of unknown proportions.
Is the GC eliminated for speed of analysis?

Not a full answer but yes, there are many applications of MS that don't involve a GC system. The more obvious like HPLCMS and LCMS but also just MS systems alone.

@NotCorey LCMS / HPLCMS are analogous to a GCMS. You chromatographically separate peaks & then use the MS on essentially pure component peaks. Direct MS of a non-pure sample seems different.

Does this mean it is a (mixture) sample injected into a MS without the preceding chromatographic separation (as is typical in, say, a GC-MS or an LC-MS)?

No. A chromatogram is the measurement of something after separation by chromatography. So you got chromatography beforehand. In case of direct injection into MS the term is usually used too, if you record over time like you would do using chromatography.

What's the point in doing this?

Ok so let's imagine I'm running a sample of an unknown mixture and I'm recording m/z 100-1000. Now I don't know at all what to expect, so selecting randomly a specific mass for my chromatogram would, most likely, lead to nothing. What we can do is just integrate over the whole mass range and take the total number of all ions, so we see a peak this means there are more ions than for baseline noise. That should correspond to a substance reaching the detector.

Wouldn't the resultant spectrum be very noisy since it's essentially the sum over all components of a mixture of unknown proportions.

Depending on the method: yes that might be a problem. And if the noise is very high then even a strong signal of a single m/z wouldn't really contribute to the total number of ions, so we won't see a peak in the TIC.
A way to circumvent this problem is using a base peak chromatogram. Base peak is the highest peak in your spectrum, and you are plotting the intensity of the highest peak over time. Now for noise there might be a large amount of total ions, but the intensity of each m/z is usually small. If you got a specific signal from a substance you will get much higher signals there, so the base peak chromatogram will show a peak.

What's the alternative?

Collecting data in TIC mode is an alternative to collecting data in SIM mode. SIM stands for "single ion monitoring" or "selected ion monitoring". Let's say that you know your sample contains acetophenone, and the four largest peaks for acetophenone are at 120 (molecular ion), 105 (base), 77, and 51 m/z. Instead of integrating over all ions, you could monitor only those masses (or only one of them). You would know where the acetophenone eluted and could quantify it relative to a standard, but you would know little about the rest of the sample.
By analogy, consider a LC system with a UV/Vis photodiode array detector. The detector can monitor single wavelengths or it can capture a full absorbance spectrum at each time interval. A mass spectrometer collecting data in TIC mode is like a PDA detector collecting a full spectrum. A mass spectrometer collecting data in SIM mode is like a PDA monitoring individual wavelengths.

Great answer! Thanks. Can you elaborate more about the bit on "direct injection into MS if you record over time " When is this done? I thought the time dimension only comes from the GCMS or a TOF sensor. When would one record a directly injected MS over time?

Time doesn't stop, even when the injection port is directly connected (with no column) to the mass spectrometer.  "Flow injection analysis" or FIA is a fancy name that analytical chemists use when they inject a small volume of fluid from a sample into an analyzer.  The window of time in which the sample is actually going into the analyzer (e.g. the mass spec) is finite, and furthermore varies according to the flow rate and dead volume of the instrument's tubing.  The easiest solution: record spectral data over a long period of time, wider than when the sample goes in.

@curious_cat in case of MS systems used for liquid chromatography (HPLC most of the case), for example ESI-MS systems you can just inject an analyte solution by using a syringe. Since you need very little volume over time you usually use a syringe pump to do so. In this case the solution going into the MS doesn't change over time, like it would in case of chromatography, but you would still get a "chromatogram" with measurement over time.


text:
What's the name of the org. compound: Ar-C(O)CH3
What's the name of this compound?
*I know that Ar-C=O- is the Benzoyl group

It is acetophenone. The Ar-C=O is phenone group.https://en.m.wiktionary.org/wiki/phenone

I got the answer from the comment
It's Acetophenone

The name acetophenone is correct; however, according to current IUPAC recommendations, it is retained only for general nomenclature. The preferred IUPAC name is the systematically constructed substitutive name 1-phenylethan-1-one.