text:
How to calculate wavenumbers of normal modes from the eigenvalues of the Cartesian force constant matrix?
I would like to get the wavenumbers (Frequencies -- in Gaussian) from the eigenvalues of the force constant matrix (Force constants in Cartesian coordinates: in Gaussian).
Could someone please explain how to perform this calculation? The Cartesian force constants are given in Hartrees per Bohr squared, while the wavenumbers are in $\pu{cm^-1}$.

In short: http://sirius.chem.vt.edu/wiki/doku.php?id=crawdad:programming:project2

@Wildcat That's a good start, but it leaves out a lot of details -- why mass-weight? How does the units conversion work? Answer pending as soon as I have the time to write it.

Am I being just too naive to suggest that for a harmonic oscillator $E=k\Delta x^2/2$ and so k in units of Hartree/Bohr^2  are to be expected  as the same as $\pu{J/m^2} $ (with conversion constants)  and that $2\pi c\nu = \sqrt{k/\mu}$  has $\nu$ in $\pu{cm^{-1}}$  ?

@porphyrin No, you're right, the analogy is fairly direct. The mechanics of the calculation and the specifics of the units transformation are highly non-obvious, though. As noted, answer is pending. Possibly not until the weekend -- things are kind of hectic ATM.

Short answer: (1) mass weight each force constant (see below). (2) Diagonalize the matrix. (3) Convert the eigenvalues ($\{ k_1^{\prime\prime} \}$) to wavenumbers $\tilde{\nu}=\sqrt{k_i^{\prime\prime}}/(2\pi \times 137 \times 5.29\times 10^{-9})$.  137 is the speed of light in atomic units and $5.29\times 10^{-9}$ converts bohr to cm.
More detail
Harmonic vibrational frequencies come from solving $F=ma$ assuming $E(x)=\frac{1}{2}k(x-x_0)^2 = \frac{1}{2}k^\prime l^2$ where $l=\sqrt{m}(x-x_0)$ and $k^\prime = k/m$. $F=-(\partial E/\partial x)$ so $F=ma$ becomes
$$ \frac{\partial^2 l}{\partial t^2}+k^\prime l = 0 $$
The solution to this differential equation is
$$ l = K \cos(\sqrt{k^\prime}t) +\varepsilon$$
where $K$ and $\varepsilon$ are arbitrary constants.
The solution tells us that the displacement coordinate will increase and decrease in a repeating fashion every $\sqrt{k^\prime}/2\pi$ times, which is the definition of frequency ($\nu$). The units of $k^\prime$ are energy/(mass length$^2)$ or 1/time$^2$, because energy = (mass)(length/time)$^2$.  So the units of $\nu$ is 1/time.  This can be converted to a wavenumber ($\tilde{\nu}$, 1/length) by dividing $\nu$ by the speed of light.
Diatomic molecule
You have 6 Cartesian coordinates $\{q_i\}$ and 6x6 Cartesian force constants (some have the same value),
$$ k_{ij} = \left( \frac{\partial^2 E}{\partial q_i \partial q_j} \right) $$
and it's not clear what to do with the cross terms.
So you diagonalize the matrix to get rid of the cross terms, but you need to mass-weight first for the reasons described above.
$$k^\prime_{ij} = \frac{1}{\sqrt{m_i m_j}}k_{ij}$$
After diagonalization of $\mathbf{k^\prime}$ you have 3N-5(6) non-zero force constants $\{ k_i^{\prime \prime}\}$ , which you can convert to wavenumbers by $\sqrt{k_i^{\prime \prime}}/2\pi c$

Thank you! Could you please help me, how to determine the normal mode which converts the minimum into the transition state numerically? (This normal mode corresponds to the one of the imaginary frequency.)

This calculation is extraordinarily common in computational chemistry packages, but despite that I had an extraordinarily hard time finding a clear exposition of the method. I ended up having to piece it together myself, drawing from a couple of papers by Vincenzo Barone$^{1,2}$, an arXiv submission of George Phillies, the classic Wilson, Decius & Cross, and the method as described on the Gaussian.com website. The below exposition is written for the analysis of a nonlinear molecule; suitable adjustments to the formulae must be made for linear systems.

The first thing to do is to convert the Cartesian Hessian  (force constant matrix) $\require{begingroup}\begingroup\def\mbf#1{\mathbf{#1}}\mbf{F}$ to vibrational normal coordinates; the first step of this process is to mass-weight it. This is necessary in order to establish the proper physical relationship between the frequencies of motion and the force constants in the normal-mode system of displacements. A useful analogy can be constructed to a one-dimensional mechanical system with two equal masses and three springs of equal force constant; for this system, the defining equation is:
$$
\left(\omega^2m - 2k \right)^2-k^2 = 0, \tag{1}\label{det-eqn}
$$
where $\omega$ is the vibrational frequency, each object is of mass $m$, and each spring has force constant $k$. As can be seen in Eq. $\eqref{det-eqn}$, the units of $\omega^2$ are related to those of $k$ by a factor of $m^{-1}$.
For vibrational analysis, this adjustment is accomplished through the use of the expanded atomic mass matrix, $\mbf M$.  As there are $3N$ degrees of freedom to a molecular system with $N$ atoms, the dimension of $\mbf F$ is $\left(3N\times3N\right)$.  Accordingly, the matrix $\mbf M$ used for mass-weighting must have the same dimensions.  The elements of $\mbf F$ are usually clustered such that:
$$
\def\pardsq#1#2{{\partial^2 #1 \over \partial #2 ^2}}
\def\pardxy#1#2#3{{\partial^2 #1 \over \partial #2 \partial #3}}
\mbf F = \left[
\begin{array}{cccc}
\pardsq{E}{x_1} & \pardxy{E}{x_1}{y_1} & \cdots & \pardxy{E}{x_1}{z_N} \\
\pardxy{E}{y_1}{x_1} & \pardsq{E}{y_1} & \cdots & \pardxy{E}{y_1}{z_N} \\
\vdots & \vdots & \ddots & \vdots \\
\pardxy{E}{z_N}{x_1} & \pardxy{E}{z_N}{y_1} & \cdots & \pardsq{E}{z_N}
\end{array}
\right], \tag{2}\label{f-mtx}
$$
where $x_i$ is the $x$-coordinate of atom $i$ in the system. Therefore, $\mbf M$ is usually constructed from the atomic masses $m_i$ as the diagonal matrix:
$$
\mbf M =
\left[
\begin{array}{ccccccc}
m_1 & 0 & 0 & 0 & \cdots & 0  & 0\\
0 & m_1 & 0 & 0 & \cdots & 0  & 0\\
0 & 0 & m_1 & 0 & \cdots & 0  & 0 \\
0 & 0 & 0 & m_2 & \cdots & 0  & 0\\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & 0 & 0 & \cdots & m_N & 0 \\
0 & 0 & 0 & 0 & \cdots & 0 & m_N \\
\end{array}
\right] \tag{3}\label{m-mtx}
$$
The mass-weighted Hessian $\mbf f$ is then calculated as the following:
$$
\mbf f = \mbf M^{-1/2} \mbf F \mbf M^{-1/2} \tag{4}\label{weighted-f-mtx}
$$
This results in each element $f_{ij}$ being divided by the product $\sqrt{m_am_b}$, where $m_a$ and $m_b$ are the atomic masses of the atoms whose geometric coordinates are involved in defining that element $f_{ij}$. In particular, it's important to note here that each factor of $m_a^{1/2}$ is weighting a matching length unit associated with atom $a$.  To see this, note that each $F_{ij}$ is a second derivative of the energy with respect to the displacements of two atoms $a$ and $b$, where $a=b$ is possible, with units of Hartrees per Bohr squared $\left(\pu{E_h}/\pu{B2}\right)$. The weighting approach effected by Eq. $\eqref{weighted-f-mtx}$ means that $f_{ij} = \pardxy{E}{w_im_a^{1/2}}{w_jm_b^{1/2}}$, where atoms $a$ and $b$ are associated with displacements $w_i$ and $w_j$, respectively, and $\overline w=\left[x_1, y_1, z_1, x_2, \ldots \right]$. Thus, it is actually the displacements that are being mass-weighted, each by the square root of the mass of its associated atom.

At this point, in an ideal world, we could just diagonalize $\mbf f$ to obtain the eigenvalues we need (analogous to $\omega^2$ in the above example) to calculate the vibrational frequencies.  This diagonalization yields the diagonal matrix $\mbf \Lambda$ containing the eigenvalues, with the accompanying eigenvectors matrix $\mbf L$:
$$\mbf \Lambda=
\left[
\begin{array}{cccc}
\lambda_1 & 0 & \cdots & 0 \\
0 & \lambda_2 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & \lambda_{3N}
\end{array}
\right]
~;~~
\mbf L=
\left[
\begin{array}{cccc}
| & | & & | \\
\overline \ell_1 & \overline \ell_2 & \cdots & \overline \ell_{3N} \\
| & | & & |
\end{array}
\right] \tag{5}\label{eigen-noproj}
$$
Since molecules are isolated systems, in theory the energy should be independent of both translational and rotational motion. In this ideal circumstance, for a frequency analysis of a species at its equilibrium geometry, the first six eigenvalues $\lambda_1$ through $\lambda_6$ would be identically zero, corresponding to the absence of an effect on the second derivative of the energy from translational and rotational motions.
However, in reality, energy, gradient and Hessian calculations can never be perfectly converged and so small amounts of "contaminating" translational and rotational dependence will be present in the original Hessian $\mbf F$.  This leads to nonzero, though usually rather small, values for $\lambda_1$ to $\lambda_6$. In order to "purify" the Hessian of these non-physical numerical artifacts, a projection operation is typically applied.  A projection matrix $\mbf D$ is constructed for this purpose, composed of defined "normal modes" for translation and rotation, and trial normal mode vectors for the vibrational motion.
The "normal modes" of translational motion are straightforward to define.  If a molecule is translating, then all of the atoms are moving in the same direction. The simplest way to express this in 'normal modes' form is by using translation along the coordinate axes $e_x, e_y, e_z$:
$$
\def\vbuf#1{\left(\overset{~}{\underset{~}{#1}}\right)}
\def\vecex{\left(\begin{array}{c}1 \\ 0 \\ 0 \end{array}\right)}
\overline d_1=
\left[\begin{array}{c}\vbuf{\vec e_x}_1 \\ \vbuf{\vec e_x}_2 \\ \vdots \\ \vbuf{\vec e_x}_N \end{array}\right] =
\left[\begin{array}{c}\vecex_1 \\ \vdots \\ \vecex_N \end{array}\right]
~;~~
\overline d_2=
\left[\begin{array}{c}\vbuf{\vec e_y}_1 \\ \vbuf{\vec e_y}_2 \\ \vdots \\ \vbuf{\vec e_y}_N \end{array}\right]
~;~~
\overline d_3=
\left[\begin{array}{c}\vbuf{\vec e_z}_1 \\ \vbuf{\vec e_z}_2 \\ \vdots \\ \vbuf{\vec e_z}_N \end{array}\right]
\tag{6}\label{trans-d-vecs}
$$
Any set of three orthonormal vectors $e_1, e_2, e_3$ would work, however. For reasons I won't go into in detail here, the "normal mode" vectors for rotational motion can be constructed in a roughly similar fashion, using the cross products of the three orthonormal vectors (representing the three independent axes of rotation) with the displacement vectors $\vec r_a$ of each atom $a$ from the center of mass of the system:
$$
\def\vbuf#1{\left(\overset{~}{\underset{~}{#1}}\right)}
\def\vecex{\left(\begin{array}{c}1 \\ 0 \\ 0 \end{array}\right)}
\overline d_4=
\left[\begin{array}{c}\vbuf{\vec e_x \times \vec r_1} \\
     \vbuf{\vec e_x \times \vec r_2} \\ \vdots \\
     \vbuf{\vec e_x \times \vec r_N} \end{array}\right]
~;~~
\overline d_5=
\left[\begin{array}{c}\vbuf{\vec e_y \times \vec r_1} \\
     \vbuf{\vec e_y \times \vec r_2} \\ \vdots \\
     \vbuf{\vec e_y \times \vec r_N} \end{array}\right]
~;~~
\overline d_6=
\left[\begin{array}{c}\vbuf{\vec e_z \times \vec r_1} \\
     \vbuf{\vec e_z \times \vec r_2} \\ \vdots \\
     \vbuf{\vec e_z \times \vec r_N} \end{array}\right]
\tag{7}\label{rot-d-vecs}
$$
The corresponding columns of $\mbf D$ are obtained by normalizing each mass-weighted $\overline d_i$:
$$
\overline D_i = {\mbf M^{1/2}\overline d_i \over
     \left| \mbf M^{1/2}\overline d_i \right|} \tag{8}\label{D-mtx-assy}
$$
Note that the mass-weighting has a power of positive one-half here, since each element of a $\overline D_i$ indicates the contribution of the associated atomic displacement to the changes in the inertial configuration of the system when it undergoes deformation along mode i. To note, the normalization of Eq. $\eqref{D-mtx-assy}$ eliminates the dimensionality of this subset of $\mbf D$, converting its elements into unitless quantities.
The remainder of $\mbf D$ is filled with $3N-6$ arbitrary 'trial' normal vibrational mode vectors, which must be mutually orthonormal as well as orthogonal to the translational and rotational "modes" $\overline D_1$ through $\overline D_6$. This can be accomplished by seeding the remainder of the matrix with random values, checking to ensure linear independence, and performing any suitable orthonormalization procedure such as Gram-Schmidt.
By construction, $\mbf D$ is an orthogonal matrix, and thus can be used as a coordinate transform (a 'projection') from the Cartesian force constant matrix $\mbf f$ to a normal-mode force constant matrix $\mbf{\hat f}$:
$$
\mbf{\hat f} = \mbf D^\mathrm T \mbf f \mbf D \tag{9}
$$
At this point, all elements in rows 1-6 and columns 1-6 of $\mbf{\hat f}$ should be small, reflecting the translational and rotational artifacts present in the original Hessian.  Only the elements $\hat f_{i,j}$ where both $i>6$ and $j>6$ should have appreciable magnitudes.  We can now coerce the spurious translational and rotational content to zero and extract the lower-right sub-matrix of $\mbf{\hat f}$ that contains only the vibrational information of interest:
$$
\mbf{\hat f}\Rightarrow
\left[
\begin{array}{cc}
\mbf{0} & \mbf{0} \\
\mbf{0} & \mbf{\hat f'}
\end{array}
\right]~;\quad
\mbf{\hat f'}=
\left[
\begin{array}{cccc}
\hat f_{7,7} & \hat f_{7,8} & \cdots & \hat f_{7,3N} \\
\hat f_{8,7} & \hat f_{8,8} & \cdots & \hat f_{8,3N} \\
\vdots & \vdots & \ddots & \vdots \\
\hat f_{3N,7} & \hat f_{3N,8} & \cdots & \hat f_{3N,3N} \\
\end{array}
\right] \tag{10}
$$
Now we can perform the same diagonalization on $\mbf{\hat f'}$ as was used to generate Eq. $\eqref{eigen-noproj}$, obtaining the translation- and rotation-purified eigenvalues and eigenvectors for the vibrational modes:
$$
\mbf \Lambda'=
\left[
\begin{array}{cccc}
\lambda_7' & 0 & \cdots & 0 \\
0 & \lambda_8' & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & \lambda_{3N}'
\end{array}
\right]
~;~~
\mbf L'=
\left[
\begin{array}{cccc}
| & | & & | \\
\overline \ell'_7 & \overline \ell'_8 & \cdots & \overline \ell'_{3N} \\
| & | & & |
\end{array}
\right] \tag{11}\label{eigen-proj}
$$

With eigenvalues $\mbf \Lambda$ and/or $\mbf \Lambda'$ in hand, we now can turn to the actual calculation of the wavenumbers of the vibrational frequencies. Consider a representative eigenvalue $\def\hala{\hat\lambda}\hala$, which carries units of:
$$
\def\uniteq{\enspace\left[=\right]\enspace}
\hala \uniteq {\pu{E_h}\over\pu{amu B^2}}.\tag{12}\label{eigen-units}
$$
The Hartree is the atomic unit of energy. Just as we can decompose the Joule into its composite fundamental units, we can do the same with the Hartree $(\pu{m_e}$ is the electron mass, and $\pu{T_a}$ is the atomic time unit, approximately $\pu{2.4E-17 s})$:
$$
\pu J = {\pu{kg m2}\over\pu{s2}}~;~~\pu{E_h} = {\pu{m_e B2}\over\pu{T_a^2}} \tag{13}\label{energy-units}
$$
One of the most subtle aspects of figuring out how to perform this wavenumber calculation correctly was to note, as shown above, that the atomic unit of mass is the electron mass, and not the atomic mass unit, despite the near identity of the terminology.  Thus, the units of $\hat\lambda$ are actually:
$$
\require{cancel}
\hat\lambda \uniteq {\pu{E_h}\over\pu{amu B2}} = {\pu{m_e} \cancel{\mathrm{B^2}} \over \pu{T_a^2}} \cdot {1\over \pu{amu} \mathrm{\cancel{B^2}}} =
{\pu{m_e}\over\pu{amu}} \cdot {1\over\pu{T_a^2}}
\tag{14}\label{lambda-units}
$$
For example, say one of the eigenvalues exhibited by a system has a value of $\hala = \pu{0.1000 E_h/amu B2}$. The value of $\hala$ in fundamental atomic units is thus:
$$
\begin{align}\hala = 0.1000\, {\pu{E_h}\over\pu{amu B2}} &= 0.1000 \cdot {\pu{m_e}\over\pu{amu}} \cdot {1\over\pu{T_a^2}} \\ &=  0.1000 \cdot \pu{5.4858E-4} \cdot {1\over\pu{T_a^2}} = \pu{5.4858E-5 T_a^-2}\tag{15}\label{hala-Ta-val}\end{align}
$$
The factor of $\pu{5.4858E-4}$ is the ratio $\pu{m_e / amu}$ per, e.g., NIST reported values.
The angular frequency of oscillation of this hypothetical vibrational mode is then just $\def\hanu{\hat\nu}\def\haom{\hat\omega}\def\brenu{\breve\nu}
\haom = \sqrt{\hala}$, or $\haom = \pu{0.007407 T_a^-1}$. The cyclic oscillation frequency can be obtained via dividing by the usual factor of $2\pi$:
$$
\hanu = {\haom\over2\pi} =
0.001179\,{\pu{cyc}\over\pu{T_a}}, \tag{16}\label{hanu-ta-val}
$$
where $\pu{cyc}$ ("cycles") is included in the units as a reminder that $\hanu$ is a cyclic frequency.
Per standard physical chemistry derivations (e.g., at Wikipedia), the spacing between the energy levels of a quantum harmonic oscillator is equal simply to $\hbar\haom$ (or, equivalently, $h\hanu$). This represents the energy of photon required to excite the system along the associated vibrational mode $\hat\ell$. Per the normal expression for photons of $E=h\brenu$, the straightforward relationship of $\hanu=\brenu$ is obtained.
Thus, all that remains is to convert $\brenu$ to the desired wavenumber units of $\pu{cyc/cm}$.  As noted by Jan Jensen, the speed of light expressed in atomic units has the value $c = \pu{137 B/T_a}$; combining this value for $c$ with a conversion from $\pu B$ to $\pu{cm}$ yields:
$$
\begin{align}
\brenu &= 0.001179\,{\pu{cyc}\over\pu{T_a}} \\
{\brenu\over c} &= \pu{8.604E-6}\,{\pu{cyc}\over\pu{B}} =
   1626\,{\pu{cyc}\over\pu{cm}}
\end{align}\tag{17}\label{final-result}
$$
Thus, in the end, the conversion is as simple as that described by Jan. The above provides deeper insight into why that calculation gives the correct answer.

EPILOGUE: If I understand correctly, Gaussian prints the wavenumber values for the vibrational frequencies both before $(\lambda_i)$ and after $(\lambda_i')$ the rotational and translational noise is projected out of $\mbf F$. This is why two similar but non-identical sets of frequencies may appear in the output.

$^1$ Barone, V. J Chem Phys 122(1): 014108 (2005), doi:10.1063/1.1824881
$^2$ Barone et al. Int J Quantum Chem 112(9): 2185 (2012), doi:10.1002/qua.23224
$\endgroup$

Thanks for the insights. However, when I read the book by Frank Jensen, he describes the approach by mentioning that we need to solve the nuclear schrodinger equation. He did mention the diagonalisation of the mass-weighted Hessian. But because it is an SE of the molecule, it contains the KE operator as well and a the end, the approach involves solving 3N 1-dimensional SEs and then obtains the eigenvalues, which are related to the vibrational frequencies by v = 1/2pi * root(eignenvalue).

But in your approach, there is no nuclear schrodinger equation mentioned anywhere. Is it because his approach is based on QM while yours is classical?

@TanYongBoon The nuclear vibrational SE is that stated by Jan Jensen in his answer, and both his answer and mine are based on it. Thus, both Jan's answer and mine do provide quantum solutions.

You are incorrect: this nuclear vibrational SE does not contain a kinetic energy term, as it is usually assumed that the nuclear $\frac{1}{2}mv^2$ quantities are small enough to be neglected and thus only the Hooke's Law potential energy term is present.

I'm not sure how to convey this to you. But perhaps, if you have the book by Frank Jensen, you can look at p. 312, under Ch. 13, Change of Coordinate System.

Let's continue in chat.

@hBy2Py Do you have a reference for why $\bar{d}_4$,$\bar{d}_5$, and $\bar{d}_6$ can be written with the help of the standard basis vectors not only with the eigenvectors of the inertia tensor?

@hBy2Py Furthermore, if the molecule is linear, then one of $\bar{d}_4$, $\bar{d}_5$, and $\bar{d}_6$ should be a zero vector (at least numerically)?

@TobiR Re standard basis versus inertia tensor eigenvectors, I do not. All I know is that, in the handful of cases that I tested, using the standard basis vectors successfully reproduced ORCA's translationally-rotationally-purified Hessian.

@TobiR If you were to use the inertia tensor eigenvectors as the rotational axes, then yes: I would expect to get a zero vector for the $\overline d_i$ falling parallel to the molecule. Using the standard basis vectors, I'd only expect a zero vector if the molecule were aligned with one of the coordinate axes.  I haven't looked at this in a while, though, so I may be mistaken.


text:
Iodine value calculation
A $\pu{1.1g}$ sample of an oil was treated with Wij’s solution and excess potassium iodide solution. The liberated iodine reacted with $\pu{7.3 cm3}$ of $\pu{0.1 mol/dm3}$ sodium thiosulfate solution. The blank titration required $\pu{43.5cm3}$ of sodium thiosulfate. What is the iodine value?
I've been asked this question but there are no examples of calculating iodine value in my textbook. I'm aware that the iodine value is the mass of iodine to react with $\pu{100g}$ fat/oil and the question gives a $\pu{1.1g}$ sample of an oil. Does this mean $1.1$ should be divided by $\pu{100g}$? I'm just a bit confused how to even start this question.

I assume the normality of sodium thiosulphate is 0.1 N since no explicit mass was given.

Assuming you are given normality of sodium thiosulphate as 0.1 N (or sometimes called standard), you could use this formula to find the iodine value:
iodine number=$$\frac{(mL 0.1N Na_2S_2O_3 blank  – mL 0.1 N Na_2S_2O_3 test) (12.7)(100)}{(1000) (1.1)}$$

Where “mL 0.1 N Na2S2O3 Blank” = the volume of sodium thiosulphate required to titrate
the blank solution,
Where “mL 0.1 N Na2S2O3 Test” = the volume of sodium thiosulphate required to titrate the
test solution that contained either solid fat or oil as the sample,
Where “12.7” is the amount of grams of iodine contained in one litre of 0.1 N iodine,
Where “1.1” is the amount of fat taken in grams  and,
Where “100” converts the sample to per 100 grams of fat.

is 12.7 just a standard to multiply by? Thanks for your help!

@Amy Its a gram gram equivalent of weight iodine (iodine has molar mass 127), so its already accounted for in this formula. (in 0.1 litre there is 12.7g of iodine which gives 0.1N iodine) so if "standard" solutions are used then this value becomes a constant

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sort: votes


text:
How to get the SMILES from structure without software?
Usually I use online programs and structure drawing tools to find the SMILES of a compound. How to find the SMILES without software? Where to find resources to learn it?

How could be software anything but logical...

@Mithoron My bad, I guess logical isn't the right word. How do you find SMILES without software.

@PrittBalagopal In the mirror, ;-) "cheese" .

@Buttonwood Molecules aren't gonna be smiling for my camera :).

https://chemistry.stackexchange.com/questions/47242/what-is-this-smiles-notation-csicc-11 https://chemistry.stackexchange.com/questions/20486/what-would-be-smiles-notation-for-a-compound-with-delocalized-bonding

You simply need to know how SMILES works.

@Mithoron Could you direct me to sources that explain how SMILES work?

@PrittBalagopal - If you see the links above, there are related questions. There's the Daylight SMILES tutorial and theory and OpenSMILES specification all from Google searches. The concept of SMILES was to be easy to write by a human, once the syntax is understood.

here a brief tutorial http://molecularmodelingbasics.blogspot.dk/2016/05/a-brief-introduction-to-smiles-and-inchi.html

In this website you can type SMARTS and it generates a visual representation:
https://smartsview.zbh.uni-hamburg.de

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text:
What is the significance of the 2 and 4 positions in 2,4-DNP?
(2,4-Dinitrophenyl)hydrazine, a popular laboratory reagent is used to identify carbonyl compounds by forming 2,4-dinitrophenylhydrazones, which are often red or yellow colored precipitates.

Why is the specific 2 and 4 positions that cause precipitate formation with carbonyls, and make it a useful reagent? Can (2,6-dinitrophenyl)hydrazine work this way?

Probably any DNP could work if you could get it.  But, because of the directing  character of substitutents in electrophilic substitution you inherently favor the 2,4 isomer:

the hydrazine group (or more accurately the chlorine atom which is converted to a hydrazine group) is ortho/para directing.
the nitro groups are deactivating and meta directing.

So you get the nitro groups meta to each other and ortho/para to the hydrazine group.  Therefore 2,4 DNP.

So in this way, a 2,4,6-TNP should also work right?

If you can get it, try it.  The nitro groups are deactivating and getting three of them on the ring is tough.  Might having that much nitro in the molecule present other hazards?  (TNT.)

Also, sterics probably come into play with 2,6-DNP.  With nitro groups on both sides, I think that hydrazine has a hard time getting to the electrophile with the right orientation.

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text:
Real boiling temperature
I've been thaught that water at 1 atm water boils at 100 °C (the so called “normal boiling point”).
However, I came across a more rigorous description of the boiling process saying that in order to fulfill thermodynamic equilibrium between liquid and vapour the pressure inside the bubble must be higher than that outside and accordingly the liquid results superheated (to be at the same temperature of the vapour).
I also understand the nucleation process, but here for sake of simplicity I assume that the bubble is already present.
So my question is why we say that water boils at 100 °C (at 1 atm)?

Related: https://chemistry.stackexchange.com/questions/23566/will-the-water-at-the-bottom-of-the-tube-boil-at-the-same-temperature-as-the-wat

Simply speaking: The rising bubbles in your pot shrink wherever it is too cold (< 100 °C), and grow whereever it is too hot (> 100 °C), equillibrating the temperature, by exchanging some heat of evaporation. If you keep your thermometer at a distance above the heat source, it will show (mostly, see below) exactly 100 °C, or whatever is the boiling point at the given outside pressure.
Very small bubbles have to have a measurably increased inside pressure, but while the water is boiling, the bubbles grow, and so this difference vanishes. You can calculate it with the surface tension of water (0.06 N/m at 100 °C) using the https://en.wikipedia.org/wiki/Kelvin_equation, and it is negligible for bubbles of millimeter size.
Of course with increasing distance to the water surface, the temperature rises  because of the pressure increase of ~ 0.1 bar/m. Slightly less, because water at 100 °C is less dense.

Thanks for your answer! I understand your considerations but my question refers to a slightly different point. I've added a Pv diagram in which are put in evidence the two states for liquid and vapor. If the boiling point temperature is taken at atmospheric pressure (i.e. at the one of the liquid, neglecting the dependence on the depth) it will result lower than that needed to sustain a (mean diameter) bubble of vapour. Maybe because are some degrees (1 to 10 K) this difference can be neglected and conventionally the two values are assumed equal?

For a gas bubble of a millimeter diameter, the additional pressure to overcome the surface tension is probably in the millipascals. It's a simple calculation.

The boiling point is not identical to the temperature at which a material boils. Unless, of course, you define it that way. I don't know what your technical background it, so I'll mention a couple of basics which you can easily find out more about using widely available on-line resources.
The first point I'll make is that significant figures matter. The correct technical interpretation of your claim about the BP of water at 1 atm is that you meant 1.0 ± 0.5 atm. This is, of course an enormous range of atmospheric pressures. So, while the technical interpretation is clear, you probably didn't intend it. (See Wikipedia for both a discussion of significant figures and for one on atmospheric pressure (range).) The statement that water (assumed to be substantially pure and free of trace contaminants) boils at 100 °C at 1 atm is correct. The statement that the boiling point of water at 1 atm is 100 °C is also correct. Hence, to the accuracy of the values provided, there is no discrepancy.
As an aside, I am impressed that you understand the nucleation of vapor bubbles; I don't. Also, I've experienced (unfortunately) the explosive boiling of a cup of water I placed in a microwave oven. I don't recall its purity. After the heating cycle was complete, I opened the door and apparently the vibrations from that was enough to allow enough of the superheated (liquid) water to vaporize to blow most of the water out of the cup. Not an experience I recommend.
The point here is that while the equilibrium vapor pressure = gas phase (overhead) pressure is the criteria for the definition of boiling point, it just isn't true that water always boils at that temperature (and pressure).
There are two effects that are generally ignored when discussing (thermodynamic) boiling points: surface tension and isotopic composition. Not all water has the same isotopic composition, this fact is used in the study of a variety of historical, geological, and biological data. Surface tension can also be significantly affected by small trace contamination which are often quite difficult to remove. Other important factors are the height of the water column (assuming heating is from below), mixing of the liquid (rarely will you see isothermal samples of water, on a macroscopic scale), and even the composition of the gas phase (it is the Free Energy which determines the equilibrium; N2, O2, Ar, and CO2 all have different energies of interaction with H2O.) Surface tension is important in bubble formation. Small bubbles are especially sensitive to its value. As surface curvature flattens (radius gets bigger), surface tension becomes much less of a factor in the growth of bubbles. Vienna Standard Mean Ocean Water is the 'standard' water typically used in discussions of water's BP. (It is a pure water (not salt water) with a specified isotopic composition.) Depending on which temperature scale is used, at 1.00 atm water's BP is 99.99 °C.
The question of when and where bubbles nucleate and how fast they grow depends on convection, heating rate, water column pressure, the liquid's vessel's interior surface (both roughness and composition), and the purity of the water, both inorganic and organic contaminants, as well as dissolved gasses and particulates present. You are correct that in order to form a small bubble (here, small means on the order of 10 to 100 microns) surface energy (surface tension) requires a temperature and pressure higher than the equilibrium over the (flat) surface. Once bubbles attain sizes of 100 microns or so, the decrease in surface energy as the bubbles continue to grow is negligible. (Growth scales with surface area which scales at r², which is why large bubbles form preferentially to the same volume of small bubbles.) You can look at this as work, in the thermodynamic sense, as long as you realize that thermodynamics was long established before the atomic theory of matter (and before quantum mechanics). The two views, looking at the thermodynamics (energy and entropy) and looking at the molecular forces involved, are separate and complimentary perspectives. It's simplest not to mix them. From the simple equilibrium thermodynamic point of view, the process of boiling violates the assumption of "near equilibrium". That is, we need to be very careful about using thermodynamics in the discussion of bubble formation. All this is background to my last statement on the subject. Water boils near enough to 100 °C at 1 atm, given the typical ranges of composition, air pressure, temperature calibration, heating, vessel, etc.etc., to make it a piece of useful information.

This answer is an unreadable text worm. How can you even write like this?

Thanks to your answer! There are some points not fully clear but I've tried to follow your reasoning which goes in very detailed considerations while I was looking for a more idealized situation(e.g. assuming that the value of the instruments is the "true" one, pure water and so on and so forth). I agree that bubbles are not in equilibrium since they can grow or collapse (homogeneous nucleation is metastable while heterogeneous in unstable).

However, thinking to maintain a bubble of a mean diameter (among all the possible which depends o the superheating of the liquid) is correct to say that the boiling temperature, if referred to the liquid pressure at 1 atm, is somewhat different from 100 °C ?


text:
Is SO3 an ion or a molecule?
Earlier I thought that $\ce{SO3^{2-}}$ (sulfite) is an ion with -2 charge but today I read on internet $\ce{SO3}$ (sulfur trioxide) is a molecule with no charge. How can this be possible? Please explain knowing that I am only a student of 10th standard (use normal chemistry concepts as far as possible).

Search in Google for their molecular structures. You'll understand the difference.

For one thing, $\ce{SO3^{2-}}$ has two extra electrons, so it's an ion. $\ce{SO3}$ is a neutral molecule. There are a different number of electrons, so they're just different.

The key is that they actually have different molecular formulae and are thus different compounds. Remember that charge counts ;)

It's pretty hard to explain this to a 10th grader. Just wait for a while, 11th grade teaches you all about different types of bonds. :)

I'll give you the structures. Then you'll understand everything:
Resonance structures of $\ce{SO3^{2-}}$ ion:

Structure of Sulfur Trioxide:

The absence of a double bond between two oxygen atoms in sulfite ion shows why it is bivalent.

I don't think that is the structure of $\ce{SO3}$. The real structure has one double bond and 2 dative bonds.


text:
pKa and pH graphically
I'm quite a newbie in chemistry and I'm trying to learn how logarithmic diagrams work. I have been told that you can calculate pKa and pH graphically:

pKa, when [HA] = [A-]
pH, when [H+] = [A-]

Why does this happen?

$\mathrm{p}K_{\mathrm{a}} = \mathrm{pH}$ when $\ce{[HA]} = \ce{[A-]}$. This is easy to see from the form of the Henderson-Hasselbalch equation. So, we the concentration of hydronium is equal to concentration of conjugate base, you can read off the $\mathrm{p}K_{\mathrm{a}}$ from whatever the $\mathrm{pH}$ reads. It's not clear what you're second statement means...

You should state the charge balance or the proton balance.
[H3O+] = [OH-] + [CH3COO-]
We take the logarithm of this and get:
log [H3O+] = log([OH-] + [CH3COO-]).
Now, you can plot this in your nice diagram.
log [H3O+] will follow the straight line in your diagram, i.e. log [H+].
The line for log([OH-] + [CH3COO-]) will at very high pH follow the line for log([OH-], but quite soon it will bend off and only follow the line for log [CH3COO-]. This is because [OH+] << [CH3COO-] at lower pH.
When [OH+] << [CH3COO-], you get log [H3O+] = log [CH3COO-]). When the line for log [H3O+] crosses the line for log [CH3COO-], you can calculate the pH of the solution from the diagram.
The lines are crossing each other at right angel: pH = (1.00 + 4.65)/2 = 2.82(5).
Very good, I am encouraging you to use logarithmic diagrams for solving acid-base equilibria.


text:
How do I calculate weight/number average relative molecular mass for polymers?
I have to estimate the weight/number average relative molecular mass of following diagram:

I know the equations:

I just can't interpret the diagram. Do I use the chain length as mass of the individual polymer?
I would then get $\pu{M_w = 52.305}$
and  $\pu{M_n = 68.227}$
Is this alright? I really appreciate any help.

They perhaps want you to make the calculations relatively, which means that even if you assume by looking at the first graph that the mass of the molecular chain length of $1$ (taken relatively and not as per scale) is $x$ units and thus, the mass of $10$ (again, taken relatively and not as per scale) such chains add up to $10x$ units and finally calculate $M_n$ to be $\approx 68.2273x$ it isn't helping us to get a meaningful answer.
A thing that should be noted is that even if you calculate it by following my example, when you'll finally calculate $\displaystyle \frac{M_w}{M_n}$ which (perhaps) the questioner wants you to calculate it won't make difference because this result is independent of $x$.

It appears that the questioner hasn't explicitly stated the mass of a polymer chain or that of the monomer units. They even haven't stated that all the polymers are considered identical with respect; it is highly probable that they want you to assume these things so as not to make things overly complicated.


text:
Is zero-point vibrational energy an intensive or extensive property?
The zero-point vibrational energy (ZPVE) is the non-zero amount of non-electronic energy that molecules have even at zero kelvin and is purely quantum mechanical in nature. For a molecule with $c$ countable normal modes, it is defined as
$$
E_{\text{ZPVE}} = \frac{1}{2} \hbar \sum_{c} \omega_{c}
$$
where $\omega_{c}$ is the energy or frequency of each normal mode.
Intensive and extensive variables have always confused me. This reference gives the following concise bulleted definition:

Extensive:

Additive
State variables (only sometimes)

Intensive:

Not additive
Field variables
Point variables (i.e., same at each point in a system at equilibrium)

with further examples of each (also on Wikipedia):

Extensive: $Q$, $W$, $U$ (and other free energies), $N$, $V$, $M$, $C$, $n$
Intensive: $P$, $T$, $\mu$, $\xi$ (electric field), any field variable, ratio of extensive/extensive

To use an example, the total energy of a system is extensive. If I have one molecule of caffeine, calculate $U$, and then take two caffeine molecules (even at infinite separation), and calculate $U$, they'll be different.
Now consider the ZPVE. For a nonlinear molecular system, there are $3N-6$ normal modes, where $N$ is the number of atoms. Take caffeine again, which has 24 atoms. There are 66 normal modes for one molecule and 138 for two molecules. If the ZPVE is considered to be a bulk or macroscopic property, then clearly it increases as the total number of normal modes increases, independent of whether or not the resulting normal modes are localized to one molecule. If the ZPVE is considered to be a single-molecule property (I hesitate to call it microscopic), then a single molecule is a single molecule, and it would make sense to "isolate" it in a way that doesn't make sense for other properties; that is, it doesn't make sense to only consider the temperature of a single molecule assumed to be in isolation that would in fact be part of some bulk phase.
So, perhaps a better wording of the question is: is ZPVE a microscopic (ugh) or macroscopic quantity? A property like polarizability also exists in a weird gray area for me as well.

Mathematically, an extensive property $f$ is one for which $f(\lambda x,\lambda y,z) = \lambda f(x,y,z)$ for extensive variables $x$, $y$, and intensive variable $z$. For example, the Helmholtz free energy $A(T,\lambda V,\lambda N) = \lambda A(T,V,N)$ is extensive.
Following this definition, the zero-point energy $$E_\text{ZPVE}(\{\omega_c\},\lambda n) = \lambda E_\text{ZPVE}(\{\omega_c\},n),$$ where $n$ is the number of normal modes, is extensive in the thermodynamic limit $N \to \infty$.
The polarizability should be able to be treated similarly.

It bears noting that a lot of macroscopic extensive properties are not extensive in the microscopic regime. For example, if we consider two identical systems, each with an energy $U$, then the combined system has energy $2U$ plus an additional energetic term $U_\text{sys-sys}$, accounting for superficial energetic interactions between the two subsystems. The surface, and hence the superficial energy term, scales as one dimension removed of the volume, and likewise vanishes in the thermodynamic limit.
Furthermore, it's perfectly valid to have properties that are neither extensive nor intensive, like the number of microstates that a system possesses.

...could I have read all this in Ken Dill's book?

@pentavalentcarbon, maybe, I don't have that book. My standard reference is Chandler.

Is the zpe not a term in U just like other vibrational and rotational energy levels? The word 'internal' is confusing in molecular terms as it refers to a thermodynamic 'system' not molecules (which do not need to exist as far as thermodynamics goes).  The internal energy is defined by First Law and so for an ideal/perfect gas of point like atoms it can have no other energy than from translational motion. After it was discovered that molecule have vibrational/rotational energy then it is possible that these energies can be changed by affecting the heat absorbed by and work done on the 'system' and so are part of the system's internal energy.
The point about intensive variables is that their value does not depend on how much 'stuff' there is, e.g. temperature and pressure.


text:
How to find a chemical by its formula?
There are (obviously) a lot of chemical compounds. Wikipedia is able to translate names to molecular formulas for a lot of them.
But is there a way to enter a molecular formula and find the chemical's name?
Coming from the computer science background, I realize that there should be some canonical textual representation of the molecular formulas for such search to be possible. So is there a language that can represent a full formula, including the structure? I could only find some latex-based languages.

recent related https://chemistry.stackexchange.com/questions/74646/how-to-get-the-smiles-from-structure-without-software

So, SMILES or even better SMARTS

https://chemistry.stackexchange.com/questions/34563/pubchem-inchi-smiles-and-uniqueness

chemspider.com has both a molecular formula search and a structural formula search.

I like chemspider.com. I wish such search was also integrated into the wikipedia. Unfortunately, wikipedia doesn't use chemical formulas in SMILES format for anything.

This answer is meant to address your second question about computerized representations of molecule structure since comments have already addressed your first question.
To truly represent the structure, mol and sdf formats are common: https://en.wikipedia.org/wiki/Chemical_table_file
Common string representations include SMILES and InChI. An issue with smiles is that there are multiple valid ways to represent a given molecule, which means a simple string match is not a good test for sameness. Canonical SMILES and InChI attempt to address this issue via additional rules. InChI additionally provides a standard way to convert the sometimes very long InChI strings into 27 character hashed version called InChIKey, which are better for searching.
Example Acetoin
SMILES string: CC(C(=O)C)O
InChI string: InChI=1S/C4H8O2/c1-3(5)4(2)6/h3,5H,1-2H3
InChIKey string: ROWKJAVDOGWPAT-UHFFFAOYSA-N


text:
What exactly is Q in the Nerst equation for a ¨lemon battery¨ of Cu and Zn in varying concentrations of acid?
I want to vary $\ce{[H+]}$ and see its affect on the voltage produced by a Zn-Cu in acid "lemon battery".  My question is about the expression for $Q$ in the Nernst equation for this cell.  I think $Q$ will have three species but I'm unclear about their values.

$\ce{[H+]}$ in mol/L, which is easy to quantify as its my independent variable.
$\ce{[Zn2+]}$ in mol/L which would be initially $0$, making Q undefined. Is this right?
The partial pressure of $\ce{H2}$, is this right, and in what units?
Can Q have partial pressures and molarities?

The 'Q' in Nerst equation is the same Q which you might've used while dealing with chemical equilibria, otherwise known as the Reaction Quotient. The expression is same as the equilibrium constant (K) and yes, they both include partial pressures and concentrations.
The units of partial pressure is the same as the unit of pressure (atm or equivalent), which can pinned down by knowing what partial pressure is in the first place. If I give you a container to begin with, filled with hydrogen and neon, and ask you about the partial pressure of hydrogen, I'd expect that you'd say:

"Had the container been filled with only  hydrogen and then the pressure exerted by hydrogen would've been measured, then I'd say that partial pressure of hydrogen (in the first case) would have been equal to this measured amount."

Now, to write down Q we need to have the equilibrium reactions with us:

At the cathode (copper in this case), hydrogen ions are forming molecular oxygen due to their reduction, so the situation appears to be like:
$$ \ce{2H+ \text{(aq.)} + 2e- -> H2 (g)}$$
At the anode, zinc is getting oxidized:
$$ \ce{Zn (s) -> Zn^2+ \text{(aq.)} + 2e-}$$

Recall that solid species are omitted from the equilibrium expression, so Q of the reaction would be written as:
$$\displaystyle\pu{\frac{P_{H_2} \cdot [Zn^{2+}]}{[H^+]^2}}$$


text:
Abiogenesis: Formation of Lipids from Montmorillonite
I have been really interested in scientific hypotheses regarding abiogenesis, particularly the lipid-world hypothesis. However, I could not find information on the chemical pathway that leads to the synthesis of amphiphilic molecules such as fatty-acids. All I've been able to find is that it involves the mineral montmorillonite which allows the step-wise formation of fatty acids using carbon monoxide (CO) and hydrogen (H). However, I want to know the exact chemical process. Thanks! Here's a link to a video explaining Abiogenesis. Note: I simplified the Montmorillonite structure and only included the parts that were involved (mainly the very negative oxygen atoms)
My hypothesis is the following: If alkanes were present in the earth's oceans, polymerization of fatty acids could have occurred via chain radical addition. However I'm not sure where the initial radical would come from on the montmorillonite clay particle. Just an idea.

I'm unsure as to how the termination step would work.


text:
Why boron trihydride is weaker lewis acid than boron tribromide and boron tri-iodide?
There is no back-bonding in case of boron trihydride so it is a stronger lewis acid than boron trichloride ( even though chlorine atom is more electronegative than hydrogen atom).
Then  why boron trihydride is weaker lewis acid than boron tribromide and boron tri-iodide?

Backbonding is ineffective in case of tribromides and triiodides since their 4p and 5p orbitals dont really overlap well with the 2p of boron. Morever they are electronegative and can make the boron more positive, and increase the tendency to accept a pair of electrons.

Predicting the acidic strengths of $\ce{BX_3}$ using back-bonding is one plausible theory. This article1 hints us to use bond-strength data to predict the same:

The weaker Lewis acid strength of $\ce{BF3}$ compared to $\ce{BCl3}$ with respect to bases such as $\ce{NMe3}$ is explained in terms of the ligand close-packing (LCP) model. The halogen ligands remain close-packed during the formation of the complex leading to an increase in the $\pu{B−Hal}$ bond length. Because a $\ce{B−F}$ bond is stronger than a $\ce{B−Cl}$ bond, more energy is required to stretch a $\ce{B−F}$ bond than a $\ce{B−Cl}$ bond. Hence $\ce{BF3}$ is a weaker acid than $\ce{BCl3}$.

Pulling up the bond strength data, I get:
\begin{array}{c|c}
\mathbf{Bond} & \mathbf{Strength (KJ/mole) } \\\hline
\text{B-H} & \text{389}\\
\text{B-F} & \text{613} \\
\text{B-Cl} & \text{456}\\
\text{B-Br} & \text{377}\\
\text{B-I} & \text{?}\end{array}
Building up on these two pieces of information, it can be said that since $\ce{B-H}$ bond is stronger than $\ce{B-Br}$ and $\ce{B-I}$ (see below), $\ce{BH3}$ is a weaker acid than $\ce{BBr3}$ and $\ce{BI3}$.
I'm afraid I couldn't find the data for $\ce{B-I}$ bond, but since we can compare the bond-strengths of boron trifluorides on the basis of back-bonding theory, we can be sure that the strength of $\ce{B-I}$ bond will be $\lt  \pu{377 kJ mol^-1}$.
References

Ligand Close-Packing and the Lewis Acidity of BF3 and BCl3
Bryan D. Rowsell, Ronald J. Gillespie, and George L. Heard
Inorganic Chemistry 1999 38 (21), 4659-4662
DOI: 10.1021/ic990713m

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


text:
Is dipole moment a vector?
Isn't it a misconception to assume that dipole moments are similar to vectors in their behavior? My reasons are as follows:
Let us take an example: Methane ($\ce{CH4}$)

Clearly, the hydrogen atoms pushing their electrons into the more electronegative carbon only enforce the effects of each other. But, we assume for no particular reason that dipole moments must behave like vectors. We then go on to add the vectors and finally decide that the net dipole moment is $0$. Why?

They are vector quantities.

Dipole moment is a concept introduced by man ryt? So why are they vector quantities? Or better: How did man know that they are vector quantities? How was that deduced?

We don't assume or deduce anything here. We define dipole moments as vectors, hence they are vectors, as good as vectors can be.

Whats wrong with them being defined by man? Basically every physical quantity is accurately defined by man, then according to you, everything is scalar?

@Ivan Neretin Why do we define them as vectors, whats the motivation?

@Pritt Balagopal Thats not what i meant. If a quantity is defined by man, it cannot be by-default vector. The properties are also assigned by man. So there must be some motivation for declaring them as vectors. Your first comment made it seem as if dipole moments were declared to be vectors by divine intervention or smtg

@Themysteryoflife seems more interested in philosophy than physics. He may be interested to know that while methane does not have a net dipole moment, it might have a net quadrupole moment. https://en.m.wikipedia.org/wiki/Quadrupole?wprov=sfla1

@electronpusher Great, thnx so much. That's exactly the kind of thing I wanted. But unfortunately the math is way over my head. But thnx anyways

You can probably still gain a geometric understanding of quadrupole moment from reading around. More importantly, please note that like dipole moment, quadrupole moment is an example of a human thought-invention, a mathematical model that we keep around because it is apparently useful in describing the natural universe.

yeah felt dipole moment was useless enough

Imagine applying every argument you've made here, @ThemysteryOflife, to the concept of distance as a vector quantity.

meaning? Distance as a vector quantity is meaningful enough. When I move somewhere, I move in a particular direction, there is no problem with that I think

We treat charge displacements as vectors because the charge is not located every where in space but associated with each atom and thus is localised. If experiments did not confirm that dipoles acted like vectors then their vector  properties would soon be abandoned.

Dipoles have a magnitude and direction, therefore they are vector quantities.

@ThemysteryOflife If you don't explicitly tag me in a comment, I won't know about it. My point exactly. One is intuitive and one is less so. That doesn't change the fact that they are by the way they are defined. If you can't reject distance as a vector quantity, then you can't reject dipole moment as a vector quantity.

we assume for no particular reason that dipole moments must be behaving like vectors

Ah, but there is a reason. Consider the interaction of a molecule with the scalar potential
$$
E_{\text{int}} = \int \rho(\mathbf{r})\phi(\mathbf{r}) \, \mathrm{d}^3 \mathbf{r}
$$
where the integral over all space is turned into an expansion:
$$
\int \rho(\mathbf{r})\phi(\mathbf{r}) \, \mathrm{d}^3 \mathbf{r} = Q^{[0]}\phi^{[0]} - \sum_{n=1}^{\infty} \frac{1}{n!} \mathbf{Q}_{j_{1}j_{2}...j_{n-1}}^{[n]}\cdot\mathbf{E}_{j_{1}...j_{n-1}}^{[n-1]}.
$$
Now to identify some terms. $\phi^{[0]} = 1$, so $Q^{[0]}\phi^{[0]} = E_0$, the interaction-free or field-free energy. $\mathbf{E}^{[n]}$ is the internal electric field or electrostatic potential generated by the atoms to the $n$th total order, and the part that would be the derivative term in a Taylor expansion is the set of electric multipole moments:
$$
\mathbf{Q}_{j_{1}...j_{n-1}}^{[n]} = \int r_{j_1} r_{j_2} ... r_{j_n} \rho(\mathbf{r}) \, \mathrm{d}^3 \mathbf{r}.
$$
The first-order term is the (electric) dipole moment:
$$
\mathbf{Q}_{i}^{[1]} = \int r_{i} \rho(\mathbf{r}) \, \mathrm{d}^3 \mathbf{r} = \mu_i.
$$
The second-order term is the (electric) quadrupole moment:
$$
\mathbf{Q}_{ij}^{[2]} = \int r_{i} r_{j} \rho(\mathbf{r}) \, \mathrm{d}^3 \mathbf{r}.
$$
Convert the continuous electric dipole integral back into a finite sum by discretizing all space into atomic positions and representing the interaction of the electrostatic potential at zero strength with the atomic density as the atomic charge:
$$
\mu_i = \sum_{a}^{N_{\text{atoms}}} \sum_{i\in\{x,y,z\}} r_{ia}q_{a}.
$$
The set of atomic coordinates is nicely represented by each atom as a vector pointing from the Cartesian origin (0,0,0) to the atom $a$'s position $\mathbf{r} = \vec{r} = (r_x,r_y,r_z) = (a_x,a_y,a_z)$ or however you want to notate it, leading to
$$
\vec{\mu} = \sum_{a}^{N_{\text{atoms}}} \vec{r}_{a}q_{a},
$$
which can be seen as a vector of length 3 from the reduction $r_{ia} \rightarrow \vec{r}_{a}$. The quadrupole moment can be seen as a matrix describing the non-uniformity of the internal electric field, or its gradient.

Dipole moment is a concept introduced by man ryt? So why are they vector quantities? Or better: How did man know that they are vector quantities? How was that deduced?

The dipole moment is a concept "introduced" by us in only two ways that come to mind. The first is that the electric field interaction is truncated to first order in the summation, which is mathematically inexact, even if higher-order moments happen to be zero. Nature includes this interaction to infinite order. Related to the first point, take a molecular sample and apply a single, uniform, finite electric field to it. The above answer deals with a system's permanent or internal moments, but one can also consider the dependence of the energy on an external electric field
$$
E(\mathbf{F}) = \sum_{n=0}^{\infty} \frac{1}{n!}E^{(n)}(\mathbf{a})\cdot(\mathbf{F}-\mathbf{a})^{n},
$$
where the electric field is $\mathbf{F} = \vec{F} = (F_x,F_y,F_z)$ and $\mathbf{a}$ is the expansion point. In practice, we always expand around $\mathbf{a} = \mathbf{0}$, so it is a Maclaurin series:
$$
E(\mathbf{F}) = \sum_{n=0}^{\infty} \frac{1}{n!}\mathbf{E}^{(n)}(\mathbf{0})\cdot\mathbf{F}^{n}.
$$
Assume that $E(\mathbf{F})$ is well-approximated by the first 3 terms of this expansion ($0 \leq n \leq 2$) and truncate it:
$$
E(\mathbf{F}) \approx E^{(0)}(\mathbf{0}) + \mathbf{E}^{(1)}(\mathbf{0})\cdot\mathbf{F} + \frac{1}{2}\mathbf{E}^{(2)}(\mathbf{0})\cdot\mathbf{F}^{2}
$$
I've been rather sloppy, but this is another way of writing the first derivation, except we are using external applied fields $\{\mathbf{F}\}$ rather than the internal fields $\{\mathbf{E}\}$. For a single applied electric field, only up to the first term $\mathbf{E}^{(1)}(\mathbf{0})$ survives, which is the induced electric dipole moment. I bring this up because assuming the applied field is strong enough that the induced dipole is stronger than the permanent dipole, the total electron distribution will most likely move uniformly or linearly with the field direction, so it makes physical sense to consider just the dipole (up to a point). (Unrelated, $-\mathbf{E}^{(2)}$ is the molecular polarizability $\alpha$.)
I would also say the above shows that we don't define the dipole moment as a vector, it appears naturally that way from mathematics. I also wouldn't call these "thought inventions".
So yes, the dipole moment is represented as a vector, the quadrupole moment is represented as a matrix, the octupole moment is represented as a 3rd-order tensor, etc.

Can you tell me what all i would need to learn in order to understand your answer and all that maths ??


text:
What does penicillin react with?
Just wondering what penicillin reacts with and if so, can you please write a word equation.
I've tried to do some research on it but had no luck. would mean a lot if someone could help me out

This is too broad, as penicillin undoubtedly reacts with a large number of things. While it's possible for an answer to list some of these reactions, it's not complete and Stack Exchange isn't suited for list-type questions.

To begin with, first pull up a structure of Penicillin (Image is in public domain):

Next, try to identify some of the functional groups, as an example I've marked two of them: one in yellow and one in blue.

The one which is marked in yellow is carboxylic acid ($\ce{R-COOH}$) and the blue one is an (secondary) amide ($\ce{R-NH-CO-R'}$). Do you remember any of the substances that react with these two groups? Some of the reactions you can think of are (I'm listing these for carboxylic acid, try thinking about amide yourself):

Sodium metal test
Acid base reactions (neutralization)
Decarboxylation reactions
Schmidt reaction ($\ce{R-COOH + HN3 -> R-NH2}$)
Reduction of acid to alcohol


text:
Are all monomers in a polymer covalently bonded?
I'm approaching chemistry and I read on Wikipedia that

The modern concept of polymers as covalently bonded macromolecular structures was proposed in 1920 by Hermann Staudinger"

But then the article goes on and it says:

Historically, products arising from the linkage of repeating units by covalent chemical bonds have been the primary focus of polymer science; emerging important areas of the science now focus on non-covalent links.

So now I'm wondering, are there polymers whose monomers are bonded in other ways?

Look up Titan and Hydro; also nanopolymers are an interesting read: http://physicsworld.com/cws/article/news/2007/jan/19/nanopolymers-make-their-debut

A very interesting possibility is to have a large amount of cyclic molecules interlocking with two (or more) neighbours each, like links in a chain. In this case the monomers (each individual ring) would not directly rely on any kind of bonding whatsoever between the rings to stay together, but would be held in such a way due to the "impenetrability" of matter. I think the main effect keeping the polymer together would be the Pauli exclusion principle, which is a rather unique situation. Though it would be marvellous to study such a polymer, it is a distant synthetic dream.

Circular DNA has the properties @Nicolau_Saker_Neto is talking about. The two strands of double-stranded DNA are twisted around each other, and in circular DNA, you can't separated them without breaking covalent bonds, but they are not connected by covalent bonds.

The second quote is ambiguous. It could be talking about the importance of the (non-covalent) links between the polymer chains.

I think your second quote is referring to some polymers such as sodium polyacrylate, $\ce{[-CH2-CH(COONa)$-$]_n}$. This polymer is basically just the sodium salt of polyacrylic acid (or polymerized sodium acrylate). You can see below the ionic bond between the anionic carboxylate group and the sodium cation:

However, the quote could be referring to so-called 'chain-link' polymers, in which the monomers are intertwined and looped/knotted together, like links in a chain, instead of just consisting of a bunch of long strings of monomers, like a rope. However, chain-link polymers are highly uncommon and rather difficult to synthesize. Note that these are not the same as cross-linked polymers.


text:
Pressure increase due to the reaction of CaO and HNO3
Currently I'm facing difficulties in determining the pressure increase due to the reaction between $\ce{CaO}$ and $\ce{HNO3}$. To be more specific, we first fill a vessel with $\ce{H2O}$ and $\ce{CaO}$ which leads to the reaction between $\ce{CaO}$ and $\ce{H2O}$ forming $\ce{Ca(OH)2}$. Resulting in a very viscous solution which eventually stops reacting.
At a certain point diluted $\ce{HNO3}$ is added. (It is possible to fill the vessel with extraordinary amounts of $\ce{HNO3}$) This leads to reaction between $\ce{HNO3}$ with Ca(OH)2 which forms $\ce{Ca(NO3)2}$ and $\ce{H2O}$. Because the $\ce{HNO3}$ is diluted (38 mass%) it is likely that more $\ce{Ca(OH)2}$ will be formed with the $\ce{H2O}$ and the remaining $\ce{CaO}$.
We know that both reactions are exothermal and the most heat will be lost by heating up the mixture and evaporating the water. Since the $\ce{HNO3}$ is filled through the bottom of the vessel it is expected that the steam will be formed at the bottom of the vessel where the reaction takes place. This will result in the formation of a big bubble until either the viscous mixture cannot withstand the pressure and the bubble "explodes" upwards, which isn't a huge problem since it is an open vessel. Or the bottom of the vessel cannot hold the pressure and the bottom of the vessel will "explode".
Now is my question, does anyone know how to encouter such a problem theoretically? I want to calculate how much pressure will be build up if $\ce{HNO3}$ is added through the bottom. (because of the high viscosity I cannot assume a well-stirred mixture) Secondly, I want to know how much pressure it takes until the viscous mixture above the "bubble" will make it's place (through channels or slugging) for the steam to go through.
Addition to my question:
Thanks for the bernoulli equation, I completely forgot that one! Till this point I got an idea how to approach this:

Reaction mechanism determines the amounts of moles that will react/are produced.
Enthalpy of reaction determines the amount of energy produced.
Specific heat of water, $\ce{CaO}$ and produced $\ce{Ca(OH)2}$ and the vaporation enthalpy of water determines how much heat is leaving by heating up and evaporating water.
The viscosity then determines how "well-stirred" the vessel is, which is then a function of temperature and concentration. (I have no idea how to describe this in a formula)
The four things above are a model which has to be set up to determine the amount of moles that are used in the reaction. In my case the amount of water is limiting so almost all water is gone. (Not all the water, since the viscosity is causing the mixture to stop flowing which results in stopping the reaction)

When this is somehow set up it starts with the next problem, namely diluted $\ce{HNO3}$ is added from the bottom.

Reaction mechanism determines the amount of moles that will react/are produced for the reaction of $\ce{HNO3}$ with $\ce{Ca(OH)2}$. The amount of water released with this reaction is reacting further with the remaining $\ce{CaO}$ producing more $\ce{Ca(OH)2}$.
The water that comes with the $\ce{HNO3}$ causes the surrounding solids to dissolve, but then again $\ce{H2O}$ starts to react with the remaining $\ce{CaO}$ into $\ce{Ca(OH)2}$, this has to be in the model as well.
The enthalpies of both reactions determine the amount of energy produced.
The specific heat of water and all other components determines the temperature rise (probably up to 100 degree Celsius). Therefore the enthalpy of vaporation of water has to be used as well.
Then the viscosity is a function of temperature and concentration.
The produced steam cannot go through to viscous fluid. (this is expected by me) So the pressure will rise, causing the evaporation temperature of water to rise, causing the overall temperature to rise until the pressure exceeds either the critical pressure of the vessel or the critical pressure to create a bubble through the viscous mixture.
And then again, this has to be in a model which includes all functions mentioned above.

I don't know if this is even possible to model since it encouters so much variables which are a function of each other. But this pretty much sums up the problem I have. Including the fact that I don't know which formula I can use to find a correlation between the viscosity of this multicomponent mixture, the concentrations of these components and the temperature.

Looks like Bernoulli's equation can help. Perhaps, I'm simplifying your question.

What kind of vessel is this? I don't think you will stop steam from escaping the solution because you are constantly reducing the viscosity of the solution as more heat energy from the reaction goes into raising the solution temperature. If you want to model this, it is a complex problem that will require CFD, some thermodynamics modeling (i'm thinking some expression for the effect of surface tension), and some particular mass transfer equations. It might be easier to do experiments and build correlations than to go from first principles if a system behavior description is what you want.

Thank you for your comment. It's just a standard stainless steel vessel of approximately 35 m3. It is designed to be sealed, but due to practical reasons the lid isn't secured, causing the lid to open when the pressure exceeds 2 mbar. So it is considered as an open (atmospheric) vessel. If the steam indeed will escape than there is no need to model this, because the steam will leave the vessel through the lid. If it doesn't then I have to model this to know what's happening. So do you know where to find any proof about your argument which states that the steam will leave the solution? Thanks!


text:
Do ferrous sulfite and aluminum sulfite exist?
Source: Concise Chemistry-ICSE (Class X)SELINA PUBLISHERS Pg 207
In this book, it is mentioned that ferrous sulfite and aluminum sulfite does not exist.
I even tried searching these on Google but didn't get any satisfactory results.
Why do these compounds do not exist?

Atomistry gives a brief description about iron(II) sulfite:

Ferrous sulfite, $\ce{FeSO3.3H2O}$, is obtained by dissolving iron in
aqueous sulfurous acid in the absence of air. Upon concentration the
salt is deposited as colourless crystals. The reaction is somewhat
complicated, for no hydrogen gas is liberated during the solution of
the iron, the nascent hydrogen reducing some of the sulfurous acid
(or ferrous sulfite) to thiosulfuric acid (or ferrous thiosulfate).
Thus:
$$\ce{2Fe + 3H2SO3 -> FeSO3 + FeS2O3 + 3H2O}$$
The thiosulfate, being very soluble in water, remains in solution.
Ferrous sulfite also results when solutions of ferrous salts and
sodium sulfite interact, and when ferrous hydroxide is dissolved in
aqueous sulfurous acid. In these circumstances a red solution is
usually obtained, probably because of interaction with dissolved
oxygen. The colour quickly disappears, however, particularly on
warming. On concentration, the salt crystallises out.
On passing a current of sulfur dioxide into an aqueous suspension of
freshly precipitated ferrous sulfide, the latter passes into solution
and ferrous sulfite is gradually deposited:
$$\ce{FeS + SO2 + H2O -> FeSO3 + H2S}$$
Ferrous sulfite solution readily oxidises in air, yielding a red
solution. From its colourless solutions alkalies precipitate ferrous
hydroxide.

A chemist named Jacquemart prepared aluminum sulfite by passing a current of sulfur dioxide through a solution of sodium aluminate. The precipitate is collected and dissolved in excess of sulfurous acid. It also eliminates formation of basic aluminum sulfite as sodium bisulfite gets dissolved. Another method of preparing is by heating concentrated solution of sodium sulfide with aluminum sulfate which on cooling, sodium sulfate crystallizes out leaving behind aluminum sulfite.
A new method has also been discovered. When a solution of tetramethylammonium sulfite is added to a solution of aluminum chloride in liquid sulfur dioxide, voluminous amount of aluminum sulfite is formed.
$$\ce{2AlCl3 + 3(Me4N)2SO3 ->[liq. SO2] Al2(SO3)3 \downarrow + 6Me4NCl}$$
If excess of tetramethylammonium sulfite is added, the precipitate redissolves and a complex is formed. It can be fixed by adding thionyl chloride which reprecipitates back
$$\ce{Al2(SO3)3 + 3(Me4N)2SO3 -> 2(Me4N)3Al(SO3)3 }$$
$$\ce{2(Me4N)3Al(SO3)3 + 3SOCl2 -> Al2(SO3)3 \downarrow + 6Me4NCl +6SO2}$$
It is very soluble in water and when heated gives sulfur dioxide and when heated to extended period, a precipitate of $\ce{Al2O3.SO2.4H2O}$ is formed but at redness, it loses all sulfur atoms and gives out pure aluminum oxide.
Reference

The Manufacture of Alum and the Sulphates and Other Salts of Alumina and Iron, Lucien Geschwind Scott, Greenwood & Company, 1901
Journal of the American Chemical Society, Volume 5, American Chemical Society, 1883
Chemistry in Aqueous and Non-aqueous Solvents, Y. Mido, S. Taguchi
Discovery Publishing House, 2001

They don't exist because they are not energetically feasible.
I also think that what is represented at http://iron.atomistry.com/ferrous_sulphite.html is mostly wishful thinking.  Each "reaction" supposedly producing FeSO3 can be represented by an equilibrium which lies farther to the left.

FeSO3 is fine (if not straightforward). It is Fe2(SO3)3 that doesn't exist.


text:
How do phase transfer catalysts bring anions to organic phase?
Many anions are quite stable in the aqueous phase, but not so much in organic solvents where its necessary to perform some reaction (usually nucleophilic substitutions). I have heard that crown ethers are good phase transfer catalysts that bring ions to the organic phase. 18-crown-6 is used to coordinate to $\ce{K+}$ ions and bring them to organic phases.

While the cation can enter the organic phase, how does the anion enter it then? Does the anion simply "tag along" with the crown ether complex?
 The anion gains no solvation stability in organic phase, so the anion moving from aqueous to organic phase seems unlikely.
So my question is: How does the phase transfer catalyst ensure migration of anions to the organic phase?

I suspect the anion just tags along, forming a tight ion pair. In a sense, the anion does not need to be solvated by the solvent because it is largely "solvated" by the cation. Also, solutions do not necessarily need to be electroneutral, so in principle the organic layer could have a net positive charge from an excess of potassium ions, while the aqueous layer would have a net negative charge from an excess of anions, but this charge imbalance must be very small, otherwise an immense voltage difference would build up between the layers.

In addition, instead of the couple of crown ether + inorganic salt, you may use a couple of  organic cation and inorganic anion to get your anion into the organic phase.  Albeit not available as anhydrous salt, tetrabutylammonium fluoride (TBAF), used for desilylation in solvents like THF, comes to my mind as one example.

@Buttonwood while that's true, that doesn't seem to imply how inorganic anions enter organic phases.

There's always some solvation and almost always some solubility.

@Mithoron Pretty sure that slight solvation isn't greater that what water provides.

Actually, anions are not unstable in organic phases per se. There is no intrinsic physical property of organic phases that would make anions inherently unstable: the main problem is that most heteroatoms in organic solvents are more electronegative than carbon and therefore carry a slightly negative charge which attracts cations more easily than anions.
Most organic phases are best classified as ionophobe, however. They will not like cations or anions in their vicinity. Remember that organic phases are typically mainly hydrocarbons with the odd heteroatom, so they have a low polarisablility. Water, on the other hand, is a highly polarised dipole and able to stabilise both cations and anions easily with the appropriate side of the water molecule.
Now if we add a phase transfer catalyst to a biphasic mixture of water and an organic solvent — typical examples would be $\ce{Bu4N+ I-}$ or crown ethers — then these will readily transfer themselves into the organic phase. In the case of crown ethers, it is because they are basically already like the organic solvent: a hydrocarbon with a few heteroatoms. Crown ethers can, however, piggyback a coordinated cation at practically no extra cost: the cation is directly solvated by the crown ether’s oxygens. In the case of the tetrabutylammonium cation, the charge of the tetravalent nitrogen is somewhat shielded by the four butyl residues which interact much better with other organic residues than with water.
This means that we have a general positive charge in the organic phase due to the crown-ether solvated cations or tetrabutylammonium cations present in there. This in turn draws anions towards the cationic charge due to electrostatic interactions.
We can now also look at some basic principles explaining London interactions: generally, the more disperse the orbitals of a species are, the easier this species can be polarised and thus the better London interactions can be. Cations are species in which there is an overall deficiency of electrons so orbitals are drawn closer to the nuclei. Anions, on the other hand, have a surplus of electrons, meaning that these will repel each other more and orbitals will be more diffuse. Compare neutral atom radii to ionic radii to confirm this. The more diffuse anions will therefore have an easier time forming London interactions with the hydrocarbon solvent than the less diffuse cations.

Thus, in summary, one typically needs the ion pair to cross the phase boundary. An anion might transfer by itself but the cation needs help; thankfully, cations can be transferred easily with crown ethers or tetrabutylammonium compounds because these are more able to interact well with the solvent.

I still don't understand how the anion moves into the organic phase though. It's not coordinated by a crown ether, then what's making it to abandon it's cozy aqueous phase to enter the organic one?

Because the organic phase has a positive charge and it is not that uncozy.

Is it like a "I would leave my 10 close friends to follow my girlfriend into a party I didn't want to go" kind of way? That the anion prefers to stay with the cation as opposed to being solvated in the aqueous phase?

@PrittBalagopal You’re comparison is invalid, it’s rather leaving one or two friends to follow your girlfriend to another somewhat chill party. And it doesn’t really work on a 1:1 level imho; rather, the cations are so attracted to the organic phase that there is a charge imbalance that somewhat draws the anions.

So maybe it’s more like ‘Wow, there are tons of single girls at that party, let’s go there, the music’s not all too bad.’

The generally accepted mechanism by which phase transfer catalysts bring about the migration of anions is slightly different from that described by Jan. Phase transfer catalysts essentially help to solve the heterogeneity problem:

That is, if we have a mixture of a reactant that is only soluble in the organic phase and another reactant that is only water-soluble, how do we facilitate a reaction between the two?

Traditionally, this problem is solved with the use of a mutual solvent. We can possibly use a solvent in which both reactants have appreciable solubility in. This mutual solvent would then be the phase in which the reaction takes place in. An example of such a solvent is ethanol, which has both a hydrophilic hydroxyl group and a lipophilic ethyl group.
Phase transfer catalysis (PTC) offers another solution to this heterogeneity problem. Starks (1970) provides a good example to allow us to understand its working principle.

Consider a nucleophilic displacement reaction between 1-bromooctane and aqueous sodium cyanide. With a stirred heterogeneous mixture of the two reactants in separate phases, the only reaction is hydrolysis of $\ce {NaCN}$ to sodium formate after $\ce {2}$ weeks at $\ce {100}$ $^{\circ}$$\ce {C}$.

The use of a phase transfer catalyst helps to facilitate this nucleophilic displacement reaction. Reaction is brought about by the use of small quantities of an agent which transfers one reactant across the interface into the other phase so that reaction can proceed. The phase-transfer agent is not consumed but performs the transport function repeatedly. Thus, in the cyanide displacement on 1-bromooctane, a phase transfer catalyst may either transport the alkyl bromide into the aqueous phase for reaction, or transport cyanide ion into the organic phase. Organic-soluble quaternary ammonium or phosphonium cations, $\ce {Q+}$, were found to be excellent agents for the transport of anions from aqueous phase to an organic phase. Catalysis of the cyanide displacement on 1-bromooctane by quaternary salt is represented in eq 1. In this way a little organic soluble quaternary salt dramatically increases the displacement reaction rate.

The above image illustrates the general mechanism by which PTCs work. It primarily involves anion exchange and ion-pair migration.
Reference
Starks, C. M. Phase-transfer catalysis. I. Heterogeneous reactions involving anion transfer by quaternary ammonium and phosphonium salts. J. Am. Chem. Soc. 1971, 93, 195–199.


text:
How to determine which normal mode leads to the transition state (numerically)?
I would like to get the normal mode of vibration which leads to the transition state. Do you know any software for this problem? If no, could you offer an algorithm to generate it? By the way, is there any name for this vibration?


text:
Why does lead nitrate powder on heating starts turning yellow from the top rather than from the bottom?
Today while in school, I observed the thermal decomposition of lead nitrate. The peculiar thing I noticed was that the lead nitrate powder in the test tube starts turning yellow (due to the formation of lead oxide) from the top and then towards the bottom. This indicates that the certain decomposition temperature is first reached at the top and later at the bottom.
I tried to apply a logic that as soon as some heat was provided to the bottom of the powder in the test tube, it was quicky transferred to the powder present above it and so on till the top. At this point, since the heat had nowhere to go (since air is a bad conductor is heat) it got accumulated at the top and hence it reached a higher temperature.
However I'm unsure of my logic because for the top powder to reach the specified temperature, the lower powder must first reach either the same or a higher temperature (otherwise the powder present above it would not get heated).

Lead oxide is a peculiar type of salt. It's brown when it's hot and turns yellow when it's cool. The lower most part remaining white could be due to two possible reasons:

The top-powder becoming yellow indicates the region is more cooler than the salt with is present at the lower part of the test tube. The lower most part is in direct contact with the flame and the  upper part is receiving heat due to conduction and losing heat to the atmosphere; many factors could be at play here.
Another possible deduction could be the paucity of atmospheric oxygen at the bottom of the test tube than at the top and hence the rate of conversion to lead oxide to the top would be faster than at the bottom.

In case you haven't done the experiment and want to see what the OP is trying to convey, here's a youtube video that demonstrates the whole lab-based thermal decomposition of lead nitrate (it starts at 1.32  incase the timed setting of the link fails):
http://www.youtube.com/watch?v=csEnz9ZsAFM&t=1m32s

could it also be that the reaction is exothermic and it's also too hot directly at the flame?

@Fl.pf. By the second speculation, the reaction being exothermic adds up - the salt at the top cools down faster, and the salt at the bottom gains more heat because it's hot directly at the flame - so, yes.

@Fl.pf., what d'you mean by the reaction being exothermic? Isn't it supposed to be endothermic because it's a decomposition reaction?

@BerryHolmes, I think that the latter case is more dominant here. Also, even if the top-powder is cooler and hence yellow, it must have been at one point hotter than the powder at the bottom of the test-tube which is a contradiction to the logic which I stated in the question.

@AradhyeAgarwal Afaik the reaction is exothermic, lead nitrate is sometimes used in pyrotechnics, for instance fireworks.


text:
Double bond crossed lines representation
I have been studying organic chemistry for years, and I do not remember the spotting double bond represented with crossed lines, except maybe one ore two more times in addition to the following one

which is taken from the Aescin Wikipedia graphics here.
I guess it is a drawing or rendering mistake, or a representation of unspecific E/Z configuration. If the latter is true, where can I find more info on usage of this symbol?
UPDATE:
A comment by gilleain leads to GRAPHICAL REPRESENTATION OF
STEREOCHEMICAL CONFIGURATION (IUPAC Recommendations 2006). Relevant paragraph is:

...
  The most common convention for depicting double bonds with unspecified configuration in molecules of all sizes uses a wavy bond.
  ...
  A crossed double bond has sometimes been used to indicate unspecified double bond configuration or a mixture of configurations. This type of bond is not considered acceptable for general use, although it may still be required by some computer software.

There's still missing piece of information about which software and by what means it "requires" this representation.

Looks like it's the second one - E/Z https://en.wikipedia.org/w/index.php?title=Skeletal_formula#Stereochemistry

It is, for example, used by Marvin JS -- one of the editors in the GUI of Reaxys by Elsevier to query the database for structures, where you do not impose either (E)-, nor (Z)-configuration of the double bond.  Below an example if you were looking for both isomers of stilbene:

(demo version here).
Beside ChemDraw, ChemDoodle's sketcher (here) and PubChem's sketcher are other examples of "some software" equally allowing such a representation, too.  I speculate it has something to do how databases (like ISISBase, Beilstein/Reaxys, Scifinder, for example) traditionally store(d) chemical structures and reaction schemes in a machine-readable format.
One may, for example, draw (cis)-stilbene with ChemDoodle "in the standard way" and save it as *.mol format and set this file in comparsion with the result of an undefined double bond with the using the wavy bond towards one of the phenyl rings.  In the chemical table, there is no significant difference spot:

If the "standard (cis) *.mol" file is compared with the one where the double bond is crossed -- as in the OP figure -- there is a difference on line #26 (fourth column now reads a three, instead of a zero), while maintaining the other entries:

This picture, naturally, differs from the *.mol files of classically specified (cis) and (trans) configuration:

used structures:


text:
Heavy-oxygen water production
It is easy to find information about heavy-hydrogen water production, but not about heavy-oxygen water. How is heavy-oxygen water commercially produced (for both heavier isotopes)?

As far as I know, while fractional distillation would be a suitable technique to obtain O-18 water (yet, in very low yelds and in a way that is hard to control), the most used method nowadays is the membrane distillation technique: basically, a semipermeable and hydrophobic membrane is employed during the distillation (tipically made in PTFE), and the isotopes are separated by their ability to permeate through the membrane: lighter isotopes (O-16 water) tend to flow through the membrane, while heavier isotopes (O-18 water) are trapped in the apparatus and recovered.
You can find many detailed descriptions online, for instance:
http://www.ichtj.waw.pl/ichtj/library/Production%20of%20stable%20isotopes%20by%20membrane%20method.pdf
https://www.google.com/patents/US5057225

Worth noting that (at the lab scale, not industrial) it seems like fractional distillation has been used to enrich 17O https://pubs.acs.org/doi/abs/10.1021/ac1022887; if it works for the much rarer 17O, it presumably works for 18O as well?


text:
How exactly are ions in a Mass Spectrometer accelerated?
After a molecule is ionised, the ion is accelerated by an electric plate up to a certain kinetic energy by the attraction of the ion to the electric plate.
However, after the ion has passed the electric plate, won't the plate still exert an attractive force on the ion?  If so, by the attraction between the ion and the electric plate, the kinetic energy of the ion were attenuated, perhaps even to the point where there would not be a net gain in kinetic energy of the ion.

Related: https://chemistry.stackexchange.com/questions/4156/understanding-how-mass-spectroscopy-works?rq=1

The acceleration of the freshly generated ions is stepwise (similar to a linear particle accelerator), i.e. the beam of ions is guided an array of electrodes with a small opening in their centre, as schematically depicted below:

(image source)
Hence when ions approach the first accelerator electrode from the left hand side, their interaction with the subsequent accelerating electrode already phases in.  Ions passing the slit of the first accelerating electrode will experience an even stronger accelerating force by the subsequent accelerating electrode.  Eventually, their momentum will increase further; and on the same time, the ions preferentially will keep their direction of propagation towards the mass dispersive element (here: the magnet), regardless the attractive force between them and the last accelerating electrode.

Your question since an ion is attracted to the second plate why doesn't this plate also attract after the ion has passed through it and bring it to a halt.
The ions are initially accelerated by a potential difference, which means there are different voltages between two metal plates separated by some distance. As a result the potential energy present due to the voltage between two plates is converted into kinetic energy of the ion. In energy terms if at the exit plate, leading into the spectrometer, the energy is $E_x=mv_x^2/2$ and initially the ion energy is $E_0=mv_0^2/2$ then $E_x = E_0 +W$ where $W$ is the work term, i.e. the kinetic energy added to the ion. This is $W=\pm qEd$ where q is the ion's charge, E field strength in volts/cm and d the separation of the two plates, i.e the distance over which the acceleration takes place. If W is positive the ion is accelerated, if not decelerated.
Thus you can see that if the potential were symmetrical, for example with three plates of potential +V  -V  +V through which the ion travels in sequence it would be brought to the same speed as initially at the final plate. But suppose that after the second plate the field ( volts/cm ) is arranged to be very small or even zero then the ion will travel at constant speed (ignoring the effect of gravity) into the rest of the spectrometer.


text:
Why does Li get darker in a glove box?
I'm having some issues with lithium getting darker in a glove box. Please see attached photos, Li gets dark overnight.
My sensors for $\ce{H2O}$ and $\ce{O2}$ are showing 0 ppm. I also checked oxygen and moisture contents using chemicals in case sensors were bad, and looks like there is no oxygen and moisture in the glove box. Just to make sure, I also replaced the nitrogen tank with argon (there are some reports in the literature saying Li reacts with nitrogen even at room temperature and gets darker in the presence of trace levels of moisture).
Does anyone had this problem before? Any recommendations would be of great help.

What chemicals did you use and are they really more sensitive than Li?

I do not use any chemicals in the glove box at the moment. Li is only exposed to the glove box atmosphere which is argon. Additional details, nothing is transferred from the ante-chamber during the past day.

How pure is the argon?  A little moisture or oxygen overnight goes a long way with lithium.  I suspect the lithium is catching something below your detection limits.

@Dervis I mean, those you've used to check oxygen and moisture contents. What are they?

Since argon is produced from air by cryogenic fractional distillation, it may still contain traces of nitrogen as also mentioned here.

Lithium can react with nitrogen giving a dark product.  https://en.wikipedia.org/wiki/Lithium

I suspect your sensors provide instantaneous readings, while the surface of the Li represents exposure integrated over time.

But your glove-box does use Argon and not nitrogen, right? Nonetheless, it's a pretty common even for new glove boxes. We seal all our products and important reagents in secondary tight containers in the glove box just to make sure. Whenever you add something new, at least some contamination will come in and you never know what the guy after you does and how careful he works.
I remember to have prepared a metal carbonyl hydride once using Schlenk-technique. It was bright yellow. We placed it into a glove-box where it turned orange. Although the sensors didn't show anything you could smell sulfur-like compounds. And even in our glove-box, and we only prepare dry, solid-state compounds, the air, that leaves when you open the box smells like selenium and sulfur. There is always some contamination in the air and once this happens your lithium is likely to react with it quite fast and you don't see much on the sensors.


text:
How do you neutralize pH of a solution by not knowing concentration
If I had a pH 10 in a 100ml solution of water and 93% sodium hydroxide how much mg/l of 93% sulfuric acid would we need to use to reach a pH of 4.
From my dated college chem books and a lot of google searching i know
1.) One mole of sulfuric acid will neutralize two moles of sodium hydroxide, as follows:
$$\ce{2NaOH + H2SO4 → Na2SO4 + 2H2O}$$  (the result is two water and one salt)
2.) 1 Gallon = 3785.41 mL or 378.541 L
3.) pH + POH = 14
4.) In one minutes time we will have 100 gallons or 378541ml (100 gallons * 3785.41 = 378541 mL)
5.) pH = 10 (of our water/sodium hydroxide solution.)
6.) The hydronium ion concentration can be found from the pH by the reverse of the mathematical operation employed to find the pH.
 10^(-pH) = hydronium ion concentration

 hydronium ion concentration = 10^-10 = 1e-10 M or 0.0000000001 mol/l

 378.541 L * 0.0000000001 mol/l = 3.78541e-8 moles

 we need a hydronium ion concentration of 10^-4 which is 0.0001 mol/l

How do i put this all together?

If you could use simple step by step with labeling each step would be amazing, I know I am asking a lot and I do not know if this is possible given my lack of variables. Any help is much appreciated.

After more research I have been told this is a titration curve problem and my concentration of sodium hydroxide needs to be known?

This is impossible to calculate unless you know the buffering capacity of the input. At best you could neutralize the added sulfuric acid, but that would only bring you back to around the input pH. (And you're also correct that it would be good to know the concentration of the sodium hydroxide solution.)

I'm voting to close this question as off-topic because it's asking about a situation which requires a licensed professional. We can not safely answer questions for your specific situation and you should consult someone with the appropriate certifications.

R.M, Thank you for your input, the boss ended up going though an engineering firm and its off of my plate. Should I delete this post or re-word as not knowing the answer is going to bother me.


text:
Can Water Oxidize Cu(I) to Cu(II)
Is DI water (also without $\ce{O2}$ gas dissolved in it after degasification) a sufficiently strong oxidizer to oxidize $\ce{Cu(I)}$ in $\ce{Cu2O}$ to $\ce{Cu(II)}$ forming $\ce{CuO}$?
Will this change at elevated temperatures (e.g., 200 °C)?

Why do you even ask this?

@Mithoron I'm wondering whether it is possible to obtain pure Cu2O films via vacuum deposition techniques that uses water vapor as the co-reactant, or wll it require post-deposition treatments to remove the Cu(II).

Then I guess you should add a lot of details to question. If you simply put Cu2O to water then almost nothing happens, but with superheated vapor, I dunno...

Might there be the opposite problem?  Water vapor oxidizes copper metal to +1 only with difficulty if at all.  There is an old heat treating test for air vs moisture leaks based on the fact that copper is inert to water vapor.  http://www.airproducts.com/industries/Metals/ask-the-experts-metals/heat-treating/oxidation-cooling-section-continuous-furnace.aspx


text:
How to determine an unknown metal based on the reaction products with oxygen and fluorine?
The following question is, as it will probably be obvious, a homework question, however I have absolutely no idea how to progress with it beyond what I have already done. This is the question:

$\pu{10 g}$ of an unknown metal reacts with oxygen to produce $\pu{16.58 g}$ of the metal oxide. $\pu{10 g}$ of the same metal reacts with fluorine to produce $\pu{25.63 g}$ of the metal fluoride. The valency of the metal is identical in the metal oxide and fluoride produced. Identify the metal after finding the mass of one mole.

My attempt:
\begin{align}
  m(\ce{O2}) &= \pu{ 6.58 g} & \implies&& n(\ce{O2}) &= \pu{0.206 mol}\\
  m(\ce{F2}) &= \pu{15.63 g} & \implies&& n(\ce{F2}) &= \pu{0.411 mol}
\end{align}
I also made the equations, where X was the unknown metal:
\begin{align}
  \ce{X + y/2 O2 &-> XO_y}\\
  \ce{X + z/2 F2 &-> XF_z}\\
\end{align}
Any help with where to go next would be massively appreciated.

Find the equivalent weight of the metal.

Your answer is Magnesium. How you get there requires you to first realize that the number of moles of Fluorine is double the number of moles of Oxygen. Since Oxygen will accept 2 electrons and Fluorine will accept one electron, you can deduce based off oxidation states and the number of moles that this metal has a 2+ charge when it is a cation. From this, you can write your chemical equations out X+1/2O2-> XO and X+F2->XF2 . Moving on, you can do some basic  stoichiometry to determine you have .412 moles of X present and use that fact that X's mass is 10g to determine the molar mass of 24.272 g/mols. Then look at your Periodic Table to find what corresponds to 24.272 g/mol and Magnesium does. Hope this helps


text:
Are bones generally harder to corrode than human skin?
I am making a fantasy universe and really I am also curious to know if bones generally are harder to corrode than human skin in our world and at what rate?
I was thinking to compare how a skeleton would react to corrosion compared to a regular human. Maybe some chemists had any insight to this. I see many fantasy universes incorporate "acid" as an element but neglect bases and corrosions as a whole.
Bonus question, is it possible to find a list on how different metals react to strong acids and bases. I have heard of the Reactivity series, but not much beyond that, or perhaps I just don't understand how to use it.

What kind of information are you looking for w.r.t. to your bonus question? Chemistry mechanism of how the metals react or if the reaction is violent or not? Even then, "violent" is a bit subjective.

Hi thanks for your comment, I realize I was a bit vague. Maybe I can clarify. Let's say someone where to splash some a strong base/acid on a substance made of bone vs. human skin tissue, then which of the two materials would have the best chance to not be eaten so to say. I know it is a bit of an open question.
So yes, I guess violent is a way of saying it.
I can also rephrase my metal question. I know this is poorly hypothetical, but if a person was to make an armor suit of any metal of his/her choice which would than be best at not being "eaten" by the acids/basses?

This question is definitely on-topic here, but definitely check out [Worldbuilding.SE], as they deal with a lot of these "theoretical"-type scenarios, too (I'm not an active member over there, but it is very popular).

Thank you, I will also post it over there:
https://worldbuilding.stackexchange.com/questions/81174/are-bones-generally-harder-to-corrode-than-human-skin
I hope either places can help me.
I am looking for a science-based answer nonetheless, so if anyone can contribute I will be grateful.

Just looking at your everyday skeletal remains of a dead person would imply that the skin definitely goes first. The bones are usually all that remain after some time, and that's in (somewhat) standard conditions (I'm going with a 6-feet under kind of thing here). You might get some traction looking into the effects of quicklime/hydrated lime on decomposition as in this abstract.

Cross-posting is usually discouraged (I should have mentioned that), but I think that this could get drastically different answers on each of the sites, so it may be okay.

I was not aware of the cross-posting, I am a bid new on this site. I hope it will be ok.
@ToddMinehardt
I was thinking something in the same line regarding the remains. However since the bones are on the inside and thus are not as much exposed as the outside skin, am I not then neglecting a factor?

I think this question has some interesting qualities, but it could also easily be considered too broad. You might want to grab one of the more basic chemistry texts, read up on acids and bases, and then tailor down the scope of this question. Even if you would only ask, how do acids and bases interact with skin, there are many kinds of different compounds out there, reacting with different mechanisms due to different properties, it is already quite broad.

what you need are oxidising agents, for example conc sulphuric & hydrogen peroxide, or piranha solution; dangerous if not use properly.

Metal reactivity with acids and bases is also a matter of oxidizing reagents along with hydrogen ion reactivity. Nitric acid can dissolve silver, but hydrochloric acid can't.

Acids and bases are considered corrosive in industrial chemical hygiene. While it is true on a technical level that harder tissue (like bone tissue) will be a bit harder to corrode compared to soft tissue (like skin), a corrosive substance that can cause damage to soft tissue will cause enough damage to harder tissue to be a medical emergency; this is certainly true when you have a highly concentrated substance. If in your world, things that are worn on the person are made from either soft or harder tissue, then the harder tissue would provide slightly better protection. But to be absolutely clear, both types of tissue will be eaten away, it's just a matter of how much.
Metal reactivity with acids and bases is nuanced because it isn't just about hydrogen ion reactivity. The alloy known as hastelloy (trademarked name) and tantalum alloys are metals used in industry for corrosive environments. Silver and gold are resistant to some common acids, but not all (e.g. nitric acid will dissolve silver).