Structural representation of substituted benzene Recently, I was watching a video about separating o-chlorotoluene and p-chlorotoluene from its solution. Following was a reaction demonstrated in the video:

I was curious about the structure of chlorotoluene. In this, the chlorine substituent was directly inserted into the benzene ring rather that attached to carbon atom. What does the bonding signifies? Is this structural representation correct? Does IUPAC recommend it? I couldn't find such type of image on internet not in any literature and thus I am skeptical about this representation.

Yes, this graphical representation of a substituted benzene is in accordance with Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008).

  GR-9.4 Variable attachment location

  In addition to allowing the type of attachment to vary, it may also be convenient to indicate that the attachment’s location is variable as well. This type of notation should be restricted to substituents that are known to be bonded to a specific ring, but at an unspecified or unknown atom of that ring. The substituent will always replace a hydrogen atom on one of the ring atoms, and cannot be bound to any atom that lacks an attached hydrogen atom. Unless explicitly specified otherwise, such diagrams imply that the substituent may be bonded to any ring atom that has an attached hydrogen atom. (…)


Strong and weak acid I know sulfuric acid is a strong acid and acetic acid is a weak acid. However when I write them in their chemical symbol.

Sulfuric acid: $\ce{H2SO4}$

Acetic acid: $\ce{CH3COOH}$

I realise that acetic acid has 4 atoms of hydrogen whereas sulfuric acid has only 2 atoms of hydrogen. So, shouldn't acetic acid be stronger than sulfuric acid?

Looking at the structures of those two acids might help. In both cases, the acidic hydrogens are part of hydroxyl groups. Because of polarity and the inductive effect, the O-H bond is weaker than C-H bonds which are very strong. As a result, it's nearly impossible for Carbon to lose a hydrogen without some kind of elimination reaction, whereas the O-H bond can autoprotolyze with other O-H bonds (even water). The weakness of the O-H bond is the basis for the acidity of those hydrogens. C-H bonds are essentially non-acidic, having pKa values around 45-50.


sp, sp2 and sp3 hybrid orbitals are actual solutions of Schrodinger's equation or only a heuristic? According to Wikipedia's article about orbital hybridisation:

[...] today it is considered an effective heuristic for rationalising the
structures of organic compounds. [...] Hybrid orbitals are assumed to be mixtures of atomic orbitals, superimposed on each other in various proportions.

So are $\mathrm{sp}$, $\mathrm{sp^2}$, $\mathrm{sp^3}$ hybrid orbitals 'real'? Are they solutions of Schrodinger equation? Are they a superposition of atomic orbitals?

As soon as you move to systems with more than one electron them all orbitals are approximations (i.e. not real). Hybridized orbitals are no more "heuristic" than atomic orbitals.

The transformation between hybrid and atomic orbitals can be mathematically, though not uniquely, defined (e.g. as the minimization of inter-orbital electrostatic repulsion) and the transformation does not affect the total electron density.

See for example http://molecularmodelingbasics.blogspot.dk/2010/03/canonical-and-localized-molecular.html and http://molecularmodelingbasics.blogspot.dk/2010/04/double-bonds-banana-and-otherwise.html


How can I filter dissolved PVA from water I'm working with a 3D printer that uses PVA filament as a support material.  After a 3D printed part is made, it is put in water for some time to dissolve away the PVA supports.

I'm trying to speed up the process by using a heated magnetic stirring plate with a pump connected to a filter.  The filter is for removing the already dissolved PVA from the water solution so that the solution doesn't get saturated.

I plan on using a 2L container of water and heating it to about 60-80 degrees C.  There would be about 10-50 grams of PVA being dissolved at a time.

I have two main questions:

What kind of filter(s) could I use for PVA in a water solution?
Do I really need a filter?  I.e. would 50 g of PVA have issues dissolving in 2L of 80C water?

Rig up a column that you can fill with granular activated carbon (GAC). Throw batting on either end to prevent carbon fines from entering your dissolution chamber. Replace the carbon when you reckon it is no longer working. If you can afford the heat, throw a cooling coil in front of the filter.

Another method is indicated in this patent: US4078129. They add 0.25g sodium borate and 0.6g sodium sulfate per liter to precipitate between 0.03 and 3% PVA.
This may not be a good solution if you are prone to overdosing, as it would say, goop your soup, when you go to do the next batch.

A third method seems to be already popular at the large scale: ultrafiltration. There are many papers discussing its usage for waste water discharge applications. Use the smallest pore UF you can find (50kDa is what I've seen in the literature). The pressures required to operate it are pretty low. Be sure the temperature of the solution is <50 degrees C prior to entering the UF (almost all membrane applications fail >50 degrees C). Run it outside-in. Don't dead-head it. You will generate a small concentrate stream. Perhaps try method 2 on that to reduce waste generation.


Why do we use J(J+1) instead of J² in the rigid body rotor energy equation? I looked for the reason, but it seems to be taken for granted that it must be $J(J+1)$ instead of $J$.

I'm making use of the analogy of the particle in a box energy where

$$E=\frac{n^2h^2}{8mL^2}$$

This arises from the quantum mechanical properties of the angular momentum operator (vide infra). Since the operators for angular momentum and translational energy are different, there is no reason to expect a similarity between the form of the eigenvalues.

Spectra of the angular momentum operators

First we note the classical relationship between the angular momentum $L$ and the rotational kinetic energy

$$E_\mathrm{rot} = \frac{1}{2}I\omega^2 = \frac{L^2}{2I} \tag{1}$$

since $L = I\omega$, with $I$ being the moment of inertia and $\omega$ being the angular velocity. By analogy, the quantum mechanical Hamiltonian for rotational motion is

$$\hat{H} = \frac{\hat{L}^2}{2I} \tag{2}$$

where $\hat{L}^2$ is the operator for (the square modulus of) orbital angular momentum. Because $L$ is a vector, we have

$$\hat{L}^2 = \hat{L}_x^2 + \hat{L}_y^2 + \hat{L}_z^2 \tag{3}$$

where $L_x$ is the operator for the projection of orbital angular momentum onto the $x$-axis, and so on. The classical angular momentum is given by $\vec{L} = \vec{r} \times \vec{p}$; by expanding the cross product we can obtain the equations $L_x = yp_z - zp_y$, $L_y = zp_x - xp_z$, and $L_z = xp_y - yp_x$. Using these relations as well as the classical commutation relationship $[x,p_x] = [y,p_y] = [z,p_z] = \mathrm{i}\hbar$, we can derive the commutators of all four operators in equation $(3)$:

$$\begin{align}
[L_x,L_y] &= \mathrm{i}\hbar L_z; & [L_y,L_z] &= \mathrm{i}\hbar L_x; & [L_z,L_x] &= \mathrm{i}\hbar L_y; \tag{4} \\
[L^2,L_x] &= 0; & [L^2,L_y] &= 0; & [L^2,L_z] &= 0 \tag{5}
\end{align}$$

Conventionally, angular momentum eigenstates are chosen to be simultaneous eigenstates of $L^2$ and $L_z$: since $[L^2,L_z] = 0$, it is guaranteed that there is a complete set of states which are eigenstates of both operators. We denote these states by $|\lambda,\mu\rangle$ where $\lambda$ and $\mu$ are related to the eigenvalues of these states:

$$\begin{align}
L^2|\lambda,\mu\rangle &= \lambda\hbar^2 \tag{6} \\
L_z|\lambda,\mu\rangle &= \mu\hbar \tag{7}
\end{align}$$

In this paradigm, we then define the raising and lowering operators $L_+$ and $L_-$ (the meaning will become clear in due time)

$$L_\pm \equiv L_x \pm \mathrm{i}L_y \tag{8} $$

Now consider the effect of $L_\pm$ on any arbitrary state $|l,m\rangle$ which is simultaneously an eigenstate of $L^2$ and $L_z$. Since $L^2$ commutes with both $L_x$ and $L_y$, we have

$$\begin{align}
L^2L_\pm|\lambda,\mu\rangle &= L_\pm L^2|\lambda,\mu\rangle \\
&= L_\pm \lambda\hbar^2|\lambda,\mu\rangle \\
&= \lambda\hbar^2 L_\pm|\lambda,\mu\rangle \tag{9}
\end{align}$$

or in other words, $L_\pm|\lambda,\mu\rangle$ is also an eigenstate of $L^2$ with the same eigenvalue $\lambda\hbar^2$ as $|\lambda,\mu\rangle$ itself. Also, using the commutators in equation $(4)$,

$$\begin{align}
L_z L_\pm |\lambda,\mu\rangle &= (L_zL_x \pm \mathrm{i}L_zL_y)|\lambda,\mu\rangle \\
&= [(\mathrm{i}\hbar L_y + L_xL_z) \pm \mathrm{i}(L_yL_z - \mathrm{i}\hbar L_x)]|\lambda,\mu\rangle \\
&= [\hbar(\mathrm{i}L_y \pm L_x) + (L_x \pm \mathrm{i}L_y)L_z]|\lambda,\mu\rangle \\
&= [\pm\hbar(L_x \pm \mathrm{i}L_y) + (L_x \pm \mathrm{i}L_y)L_z]|\lambda,\mu\rangle \\
&= L_\pm(L_z \pm \hbar)|\lambda,\mu\rangle \\
&= L_\pm(\mu \pm 1)\hbar|\lambda,\mu\rangle \\
&= (\mu \pm 1)\hbar L_\pm |\lambda,\mu\rangle \tag{11}
\end{align}$$

which tells us that $L_\pm|\lambda,\mu\rangle$ is also an eigenstate of $L_z$, but with a different eigenvalue: the new eigenvalue is $(\mu \pm 1)\hbar$. The raising operator (for example) therefore has the effect of conserving the total angular momentum $L^2$, but increasing the $z$-projection of this angular momentum $L_z$.

However, for this to make physical sense, there has to be a limit to how far this can go: or else we will have no maximum value of $L_z$, even for a given value of $L^2$. Note that since $L_x^2$ and $L_y^2$ are both real and non-negative (which follows from the Hermiticity of $L_x$ and $L_y$), equation $(3)$ sets a maximum limit on $L_z$:

$$L_z^2 \leq L^2 \tag{12}$$

Mathematically, the only way this can be fulfilled is if there is a "highest" state $|\lambda,\mu_\mathrm{max}\rangle$ which the raising operator turns into zero:

$$L_+|\lambda,\mu_\mathrm{max}\rangle = 0 \tag{13}$$

Now consider

$$\begin{align}
L_-L_+|\lambda,\mu_\mathrm{max}\rangle &= (L_x - \mathrm{i}L_y)(L_x + \mathrm{i}L_y)|\lambda,\mu_\mathrm{max}\rangle \\
&= [L_x^2 + L_y^2 + \mathrm{i}(L_xL_y - L_yL_x)]|\lambda,\mu_\mathrm{max}\rangle \\
&= [L^2 - L_z^2 + \mathrm{i}(\mathrm{i}\hbar L_z)]|\lambda,\mu_\mathrm{max}\rangle \\
&= (L^2 - L_z^2 - \hbar L_z)|\lambda,\mu_\mathrm{max}\rangle \\
&= (\lambda\hbar^2 - \mu_\mathrm{max}^2\hbar^2 - \mu_\mathrm{max}\hbar^2)|\lambda,\mu_\mathrm{max}\rangle \tag{14}
\end{align}$$

However, we already know that $L_+|\lambda,\mu_\mathrm{max}\rangle = 0$. When $L_-$ acts on zero, it simply returns zero, and so the "eigenvalue" in equation $(14)$ must be equal to zero:

$$\begin{align}
\lambda\hbar^2 - \mu_\mathrm{max}^2\hbar^2 - \mu_\mathrm{max}\hbar^2 &= 0 \\
\lambda &= \mu_\mathrm{max}^2 + \mu_\mathrm{max} \\
&= \mu_\mathrm{max}(\mu_\mathrm{max}+1) \tag{15}
\end{align}$$

Fundamentally, that is where the term comes from. There are several steps now to close some loopholes. I will not go through them in detail because it is far too long, but I will outline what is needed.

You need to show that for orbital angular momentum the value of $\mu_\mathrm{max}$ is restricted to non-negative integer values; we define this quantity to be $l$ and we can replace the quantum number $\lambda$ with $l$ because $\lambda$ is related to $\mu_\mathrm{max}$, which is in turn equivalent to $l$. The eigenvalues of the total angular momentum operator $L^2$ are $\lambda \hbar^2 = \mu_\mathrm{max}(\mu_\mathrm{max}+1)\hbar^2 \equiv l(l+1)\hbar^2$.

Note that for orbital angular momentum only integer values of $l$ are allowed; half-integer values are not allowed. This condition can only be derived by solving the differential equations corresponding to the eigenvalue equations of the angular momentum operators (see further reading). For example, you need to solve this:

$$\begin{align}
L_z f(x,y,z) &= (xp_y + yp_x)f(x,y,z) \\
&= \left(-x\mathrm{i}\hbar\frac{\partial}{\partial y} - y\mathrm{i}\hbar\frac{\partial}{\partial x}\right)f(x,y,z) = \mu\hbar f(x,y,z) \tag{16}
\end{align}$$

Half-integer values for spin are allowed because the exact form of the spin angular momentum operators are not known. However, they obey the same commutation relations as the orbital angular momentum operators, so the rules governing them are the same except for the constraint on integer values.
You need to show that the allowed values of $\mu$ are integers between $-l$ and $l$ respectively. We have already shown that $\mu$ increases in integer steps, and that $\mu \leq l$, so what remains is to show that $\mu \geq -l$: the proof is exactly analogous to that in equations $(13)$ through $(15)$, except that you need to consider $L_+L_-|\lambda, \mu_\mathrm{min}\rangle$.

Conventionally the quantum number $\mu$ is renamed to $m_l$ and the eigenstates are labelled with $|l,m_l\rangle$. So now we have shown that $l$ is the maximal value of $m_l$, and that $m_l$ ranges from $-l$ to $l$ in integer steps, as we expected
The total angular momentum $J$ is the sum of the orbital and spin angular momenta, i.e. $J = L + S$. For the rigid rotor system where you ignore spin, $J = L$ and hence we can simply replace $L$, $l$, and $m_l$ with $J$, $j$, and $m_j$: we therefore have $J^2|j,m_j\rangle = j(j+1)\hbar^2|j,m_j\rangle$.
Finally, the energies are then given by

$$\begin{align}
E &= \langle j, m_j|H|j,m_j \rangle \\
&= \left< j,m_j \middle| \frac{J^2}{2I} \middle| j,m_j \right> \\
&= j(j+1)\frac{\hbar^2}{2I} \tag{17}
\end{align}$$

where $j$ is a non-negative integer (i.e. $j = 0,1,2, \ldots$).

Further reading

Chapter 4 of Molecular Quantum Mechanics, 5th ed., Atkins.

Chapter 7 of The Physics of Quantum Mechanics, Binney & Skinner for an alternative derivation.


Thermodynamic properties software I am looking for a software (can be freeware or purchased) to look up water and steam thermodynamic properties (e.g. density, enthalpy, entropy) preferably a macro in excel. What do you use for this application?

I have access to several programs or similar solutions for the work with so-called steam tables. I usually prefer to use the NIST Reference Fluid Thermodynamic and Transport Properties Database (REFPROP) – NIST Standard Reference Database 23, Version 9. It is a program that can calculate the thermodynamic and transport properties of water and steam as well as some other important fluids and their mixtures. It was developed by the National Institute of Standards and Technology (NIST) and is commercially available here.
Although I usually don’t do this myself, REFPROP can be used with Microsoft Excel. This is described in the corresponding Help file as follows:

Excel Spreadsheets
The REFPROP software includes dynamic link library ('.dll') support that allows a broad variety of applications to take advantage of REFPROP's capabilities.  REFPROP is distributed with front-end software that links Microsoft Excel applications to fluid property functions.
The REFPROP Excel front-end is supplied as a Microsoft Visual Basic module.  Excel automatically incorporates Visual Basic as a macrofunction execution, scripting and interface utility.  When the REFPROP front-end module is imported into an Excel spreadsheet, REFPROP functions become accessible to the user.  The proceedure for doing this is described below.
The information given below demonstrates the front end usage; however, the user is not bound to this format; rather, the preferred method would be to write one's own visual basic code that calls the DLL directly.  A dedicated module will increase the speed of the application and give the ability to use the REFPROP functions to their fullest.
The following supplementary instructions are intended for users who wish to create and modify their own spreadsheets that utilize the REFPROP functions based on the front-end module.  The Excel application included with the software is called REFPROP.xls and is located in the directory where the REFPROP program was installed.  Sample FORTRAN and Visual Basic files are also included.
Properties are calculated by calling various functions.  For example, the following line would be placed in a cell to calculate the density of argon at 300 K and 10 MPa:
=Density("argon","TP","SI",300,10)
Other properties can be calculated by using any of the following functions:
(…)


Why doesn't an sp3s hybridisation exist? Trying to prove that nitrogen doesn't split its lone pair to form 5 bonds, I thought of a situation that I couldn't rule out; the paired electron being excited to the 3s orbital, so that five bonds could be formed- in an "$\ce{sp3s}$" hybridisation.
I know hybridisation is just an observation-based explanation, but I'd like to know how to rule wacky situations like this out.
Can this - and all other nonstandard hybridisations you can come up with - be ruled out by reasoning that the electron excitations involved take up too  much energy?

The counter question that might help you rule this out is: "Which 3s orbital?" Note that atomic Nitrogen doesn't have a 3s orbital as N itself belongs to the 2nd period(n=2).

Also note that in hybridization, Atomic orbitals of same or almost similar energies intermix to form same number of new Hybrid orbitals of same shap and equivalent energy.
That being said, you cant expect 2s or 2p orbitals to intermix with a 3s orbital
(if it was a case. Since this case isn't possible with N)


Why is a third body needed in the recombination of two hydrogen atoms? In the article Discuss. Faraday Soc. 1962, 33, 205, the authors say that in order to form molecular
hydrogen from collision to atom we need a third body to remove the excess energy. That is we have the reaction $\ce{H + H + M -> H2 + M}$.
I suppose the reason is because if we have the excess this energy will break the bond. If it this the case why the energy can not be transferred to the electron and so would have $\ce{H + H -> H2^*}$, where $\ce{H2^*}$ denotes an excited state of $\ce{H2}$?

Yes there is validity to the finding that two atoms of hydrogen form molecular hydrogen at very slow rates.  Review Gould, R.J., and Salpeter, Edwin E., Astrophysical Journal, Vol 138, pg. 393.  It is publicly available at: http://articles.adsabs.harvard.edu/pdf/1963ApJ...138..393G
The exact answer to your question is found on page 138 of that work, under the heading “II. Mechanisms for the Formation of Molecular Hydrogen.”
In space, molecular hydrogen is destroyed (by cosmic radiation) at rates that exceed its formation without some type of surface catalyst.  This conundrum motivated significant research to find the actual mechanisms of molecular hydrogen formation in natural settings.  More recent work can be found by Gianfranco Vidali, et.al., Journal of Physics: Conference Series:6, 2005:
https://iopscience.iop.org/article/10.1088/1742-6596/6/1/003/pdf
 This more recent work cites the Gould and Salpeter work. It appears to be still accepted as the explanation for the slow rate of formation of molecular hydrogen from atomic hydrogen.  According to this explanation, after ground state hydrogen atoms collide, the new hydrogen molecule will be in a triplet state, and it cannot decay into a singlet state because this is a forbidden transition.  A cursory search on the PES's of triplet states of diatomic hydrogen show that indeed these are shallow curves which would lead to rapid dissociation.


Has a carbon analogue of poly(dimethylsiloxane) been synthesized? Consider poly(dimethylsiloxane) (PDMS), a member of the family of silicone polymers:

Now, imagine its carbon analogue:

I've not been able to find any evidence that this carbon analogue has been experimentally prepared. Searching for "poly(dimethylalkoxane)" got me nowhere.  I eventually searched for "dimethyl polyoxymethylene," which turned up the unsubstituted variant of the above, polyoxymethylene (POM):

as well as methyl-capped POM oligomers, polyoxymethylene dimethyl ethers:

But, no "poly(dimethyloxymethylene)" was forthcoming.

Has the carbon analogue of poly(dimethylsiloxane) been synthesized? If so, what name has it been given?

The comments to the original question do well to demonstrate that acetone polyacetal (or polyacetone; names in the literature vary, but these are the most common) is extremely difficult to produce by typical polymerization methods; this is mainly due to thermodynamic reasons, though acetal dimers and trimers of acetone have been produced at very low temperatures.1

Polyoxymethylene is the polymeric form of paraformaldehyde (formaldehyde oligomers), which forms spontaneously in aqueous solutions of formaldehyde. The methyl-capped paraformaldehyde and polyoxymethylene would form if methanol is added to the solution; indeed, this is actually typically done to limit the length of paraformaldehyde oligomers. Polyoxymethylene, as a homopolymer, is quite stable and machinable, provided end-capping is performed; this is typically done with acetic anhydride.2

Burnop, V. C. E. Polymer 1965, 6 (8), 411-417. DOI: 10.1016/0032-3861(65)90025-X
Boyer, Lyndon D.; Sorenson, Wayne R.. Polyoxymethylene process. U.S. Patent 3355428, issued November 28, 1967.


Evaporation velocity of ethanol: adding ethanol vapour to a gas mixture I need to add a known amount of ethanol vapour to a gas mixture (He/Ar). To do that I let the mixture bubble in liquid ethanol kept at a known temperature (14 C).

The system is made of a large steel cylinder (d=20 cm; l=138cm) lying in horizontal position and half filled with liquid ethanol. the gas is bubbled trough the liquid on one side of the cylinder and extracted on the other side.

The gas-flow trough the system is usually pretty low (~100ml/min) but I would like to increase it to ~10l/min and I need to know if this leaves enough time for the ethanol vapour to approach its equilibrium vapour pressure.

Please note that I have never done this.

But are you sure that you want to rely on approximation/assumptions?

I realize that it needs additional changes to the experimental setup, but is measuring the ethanol concentration out of question? Gas sensors for ethanol are commercially available and a setup, in which they steer mass flow controllers is conceivable.

As a side note, you might be interested to see that ethanol sensors can be build in the lab:

A Novel Flexible Room Temperature Ethanol Gas Sensor Based on SnO2 Doped Poly-Diallyldimethylammonium Chloride.


Side product obtained in reduction of cuprous oxide with carbon So my book sates that in the extraction of cuprous oxide, the following reaction takes place.
$$\ce{Cu2O + C -> 2 Cu + CO}$$
This makes sense to me, if it is seen as the following two reactions coupled.
$$
\begin{align}
\ce{Cu2O &-> Cu + 1/2 O2} \\
\ce{C + 1/2 O2 &-> CO}
\end{align}
$$
But as per the Boudouard reaction, carbon monoxide is the dominant form of oxide of carbon at higher temperatures, however using carbon for reduction is also possible at lower temperatures, which can be seen from Ellingham diagram.
So, are both $\ce{CO}$ and $\ce{CO2}$ possible as side products depending on temperature at which the extraction of $\ce{Cu}$ is done?

@ Paras Khosla. Your first equation is correct, but not your second one : $\ce{Cu_2O}$ reacts with $\ce{C}$ without getting decomposed previously into $\ce{Cu + O_2}$. $\ce{Cu_2O}$ is more stable than $\ce{CuO}$ above $1000$°C
Furthermore, $\ce{Cu_2O}$ does not react easily with Carbon in the solid state. The contact between the reacting solids is not good. The contact is better if $\ce{Cu_2O}$ is liquid. But $\ce{Cu_2O}$ melts at $1026$°C. So, the reaction starts at this temperature, forming $\ce{CO}$ which is more stable than $\ce{CO_2}$ at such a high temperature.


Did my teacher specialist about sulfur compounds made an error in part of Ramberg-Bäcklund rearragement mechanism? I have a lot of troubles with what I circled in red with the big question mark near. I am not speaking about stereochemistry but about conservation of number of hydrogen atoms, and also there is actually no reason for the electrons on the carbon near $R_1$ to attack the other carbon. In my lesson, the teacher didn't show how the extra hydrogen was removed.

Can you explain what my teacher didn't show or forgot to include?

Thank you in advance!

As discussed above, the pathway shown on the left is incorrect because it essentially shows insertion into a $\ce{C-H}$ bond by an anion with loss of $\ce{X^-}$, which is not the easiest path for this reaction to take by a long shot.

The actual mechanism for this reaction is as shown below.

First, the non-halogenated alpha-carbon is deprotonated.
The carbanion formed arranges itself such that the it can attack the halogenated carbon and in such a way that the two R-groups are cis to each other.
Attack occurs, with loss of $\ce{X^-}$. Product is cis.
In water, due to lack of solubility, decomposition is faster than redeprotonation and the (Z)-alkene forms.
In organic solvent, base deprotonates an alpha-carbon, which can then scramble.
The main reprotonation product is the trans product.
The trans product decomposes, and the (E)-alkene is the major product.

A few notes:

The red path makes almost completely (Z) product. The blue path makes a mix of the two with the (E) product as the main product.
Purple star: the reason the anion attacks to form the cis ring is unknown, but there are several theories, apparently.
Green stars: cheletropic ring decomposition is symmetry forbidden, so this is theorized to be a two-step decomposition through either diradical bond fissions or a dipolar mechanism.


Superoxide dismutase of Mn (SOD) I have a little problem with my paper about Mn superoxide dismutase. On this picture we have "B" behind the 3 OH ions, I really don't understand what is this. Where is this "B" from? Please, help me, my lecture will not accept my work today.

That ain't MnSOD, it's arginase.

B = boron

The molecule in red is 2(S)-amino-6-boronohexanoic acid.

See figure 2 of Arginase−boronic acid complex highlights a physiological role in erectile function Nature Structural Biology 6, 1043 - 1047 (1999)


Why is 2-methyl-2,3-pentadiene achiral (allene)? 2-Methyl-2,3-pentadiene is achiral. Why?

  Centre of symmetry
  Plane of symmetry
  C2 axis of symmetry
  Centre and Plane of symmetry.

I've drawn the structure:

So far I've dealt with planes of symmetry in determining whether a compound is chiral or achiral and nothing more. So I'm relatively new to axes and centers of symmetry.  I can't visualise the rotations in the axis of symmetry.
What I don't understand is: How can you say a compound is chiral/achiral based on it's axis of symmetry?

I'll start by linking to Symmetry @ Otterbein, as it's a very useful tool for looking at various structures and seeing planes an axes of symmetry.  If you're having issues visualising rotations and such, then this website should certainly help.

I suppose the first step would be to make sure that you're always, always, always visualising the structure in 3D. I'm not saying that you haven't been, but I just want to make sure.

A general explanation for testing for chirality would be to take your two structures and check to see if they are superposable. If they are, then they aren't chiral, but if they're not superposable, then chirality isn't ruled out.

If you were to take models of (R)- and (S)-bromochlorofluoromethane, you'd find that under no conditions are they superposable. They are indeed chiral.

Now consider your structure:

There is a mirror plane going right through the middle of the molecule (cutting carbons 2, 3, 4, and 5 in half, as well as one of the hydrogens on carbon 5).

Let's view it from above, with the mirror plane going through the centre of the image:

Now let's consider one of the core concepts of symmetry. If you were to only see what's on one half of the mirror plane, could you draw the entire molecule, knowing with complete confidence that you didn't draw any kind of isomer?

The answer should be yes, because that's what a mirror plane implies: it's the same on both sides of the plane, so if you know what's on one side, then you know what's on the other.

If you were to take the mirror image of this molecule through that same plane, you'd end up getting the identical molecule. If you mirror this molecule through a perpendicular axis, you'd end up getting the same thing, but with the methyl and hydrogen on carbon 5 switched around.  With the symmetry of the molecule, though, you could just rotate the molecule 180° to get the original one back.  That would make them superposable, and therefore not chiral.

When you recognise any symmetry, you'll know (or eventually intuitively see without thinking) what you need to do to generate the superposable conformation.  A reasonable analogy would be that if you write a message on white paper using black permanent marker and hold it up in the mirror, you'll see a mirrored image; BUT, realising that the paper itself becomes a mirror plane when the ink soaks through to the other side, you'll know that all you need to do is flip the paper over, showing the mirrored writing to the mirror, and see forward writing in the mirror.  This is very related to what you'd do with molecules.

I think that your best bet is to play with as many 3D models (physical or computational) as possible, doing rotations and mirror operations whenever possible.  At some point, it should click and be completely intuitive when just thinking about the molecule.  Try it on all sorts of things.  Try making a pizza with C6v symmetry.  Think about what your pizza would look like if it had D6h symmetry instead (note: very messy to make and to bake).


How does calcium chloride act as a sequestrant (food science)? This Wikipedia page lists common sequestrants (with respect to food science). As stated on the page:

  A sequestrant forms chelate complexes with polyvalent metal ions, especially copper, iron and nickel, which can prevent the oxidation of the fats in the food.

To me, this behavior makes sense for every example listed except calcium chloride. I can certainly see chloride complexing these metal ions, but even then, I typically don't expect the chloride ion to reduce catalytic capabilities much.

I assume the calcium cation has an effect on these processes somehow, too. Otherwise, I would expect regular table salt to have the same effect, but, while discouraging bacterial growth, I don't think sodium chloride reduces the rate of oxidative rancidification.

Could it be as simple as the desiccant effect of calcium chloride keeping the polyvalent metal ions from dissolving? It would seem like a massive amount of calcium chloride would be needed to have the desired effect if that was the preservation mechanism involved.

Food Scientist here. Calcium chloride is typically used in high concentrations in pickles to keep them firm (give that desired bite) and also add a slight salty flavor without increasing sodium content. It is also used in potato chips to reduce formation of acrylamide. Calcium chloride is highly soluble in water and will dissociate into Ca2+ and 2Cl-. It is used to sequester metal cations. The metal ion complexes with the free Cl-. It has also been found that calcium ions may reduce iron absorption, and is a divalent cation that can form complexes of its own. Calcium chloride can stabilize colors, flavors, antioxidants, and also improve oxidative stability. Fats and oils readily oxidize in the presence of metal ions, so a sequestrant such as calcium chloride is used to form a complex with the metal. Additionally CaCl2 is used to stabilize gels such as those formed with alginate. Another popular sequestrant, or chelating agent is EDTA.


Calculating ratio of conjugate base and acid when pH and pKa are given Buffer solutions are used by biological mammalian systems to maintain the $\mathrm{pH}$ of blood plasma within a narrow range. In these systems, the compound from which this solution is obtained is $\ce{CO2}$, produced in cell respiration, which is converted into $\ce{HCO3-}$ and $\ce{H2CO3}$ inside the red blood cells. Using the Henderson-Hasselbalch equation, answer: $\mathrm{p}K_\mathrm{a} = 6.35$. What is the ratio between $\ce{HCO3-}$ and $\ce{H2CO3}$ in a blood sample with $\mathrm{pH} = 7.4$?

The problem is I try to make the exercise in this way:

$$\mathrm{pH}= \mathrm{p}K_\mathrm{a} + \log \left(\frac{\ce{[HCO3-]}}{\ce{[H2CO3]}} \right)$$

BUT in the solution they make $$\mathrm{pH} =\mathrm{p}K_\mathrm{a} - \log \dfrac{[\ce{HCO3-}]}{[\ce{H2CO3}]}$$

They make the inverse of $$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \dfrac{[\ce{H2CO3}]}{[\ce{HCO3-}]}$$

And this don't make sense for me. I don't understand.

The solution given is in the white box on the left side.

Recall the fundamental properties of the logarithm
$$+\log \left(\frac{a}{b}\right)$$

is the same as

$$-\log \left(\frac{b}{a}\right)$$

so Henderson Hasselbach can have plus or minus signs. The correct equation is

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \left(\frac{\text{conjugate base}}{\text{acid}}\right)$$

or

$$\mathrm{pH}  = \mathrm{p}K_\mathrm{a} - \log \left(\frac{\text{acid}}{\text{conjugate base}}\right)$$

Now sort out whether the textbook solution is right or not.


Is cyclononatetraenylium antiaromatic? I was thinking that it might be non aromatic since the ring is very big and it might bend just like COT (cyclo-octatetraene) does to avoid destabilization due to antiaromaticity but I don't know if the given does that or not.

Cycloenneatetraenyl cation, using the Greek root for nine, is certainly not antiaromatic.  A ring that large will in general be nonplanar unless strong aromatic coupling is there to drive a planar shape, and if the planar form is antiaromatic so much the better for nonplanarity.  By classical rules such a nonplanar form would be nonaromatic.

Whether this nonplanar form is Mobius aromatic is open to interpretation, or at least it seems that way given the twisted nature of the ring.  In any event Mobius aromaticity is unlikely to give the strong stabilization we see in the Huckel version because the p orbitals do not overlap perfectly.  Most likely the structure and stability of the ion  are more determined by sigma bond interactions (minimizing steric strain) than pi bond interactions.


Hybridization of nitrogen in a ring Is there a well defined way to discern the hybridization of a nitrogen atom in ring, like pyrrole? How can you know whether the nitrogen's lone pair are in the conjugated system or not?

If an $\alpha$-$\beta$ unsaturated heteroatom isn't already participating in a $\pi$-bond and it has a lone pair of electrons, these electrons will be delocalized and in the $\mathrm{p}$-orbital.

Let's compare pyridine and pyrrole:

$\hspace{5cm}$

The nitrogen in pyridine is already participating in a double bond, so it is clearly $\mathrm{sp^2}$ hybridized. The nitrogen in pyrrole is $\alpha$-$\beta$ unsaturated, so its lone pair is delocalized. Both of these molecules are heterocyclic aromatic rings, but these are not the only cases in which a nitrogen atom's lone pair are delocalized. They are also delocalized in amides (which resemble the $\alpha$-$\beta$ unsaturation of pyrrole):

$\hspace{4.6cm}$

If you want to know more about criterion for aromaticity, read about Hückel's rule.


Does hydrogen bonding only occur between water molecules? Does hydrogen bonding only occur between water molecules?

Is the dipole-dipole intermolecular interaction know as hydrogen bonding because it is between water molecules, or is called that because of the hydrogen molecules present in the bond?

NO, Water is not only the example for hydrogen bonding. It's most common example for basics.

Hydrogen bonds occur in DNA, proteins, polymers, etc. But for introductory level it is easy to explain with water.

DNA: Hydrogen bonding between guanine and cytosine, one of two types of base pairs in DNA.

Polymer: Para-aramid structure

Images: Wikipedia


Why can't the reaction of grignard reagent with carboxylic acid give a ketone? I understand that a Grignard reagent can act both as a base and a nucleophile.
It acts as a base in the presence of a proton source (eg: alcohols, amine, water, etc). But when reacting with a ketone or an aldehyde, it acts as a nucleophile and uses a specific reaction mechanism. Why isn't that same reaction mechanism not viable for reaction with carboxylic acid?

Why is this reaction mechanism wrong? Is the formation of a diol not possible due to repulsions or is there some other reason?

Huy Ngo says it all.  A Grignard reageant will react first with just about any hydrogen that has even a modicum of proton donation capability, before "resorting" to the slower nucleophilic attack.
To get a ketone (in most cases), avoid the dissociable proton by using a nitrile instead, assisted by a copper salt. This forms an intermediate imine which is hydrolysed to a ketone after water/acid workup:
$\ce{CH_3MgI + CH_3COOH -> just a salt}$
$\ce{CH_3MgI + CH_3CN ->[Cu^+] (CH3)2C=N-M ->[H2O] (CH3)2C=O}$


Does hydrazine crystallize when freezing? When discussing problems of keeping hydrazine from freezing on spacecraft we stumbled upon a question: can pure hydrazine separate through fraction-freezing from the stabilizing substances like UDMH? That would largely depend on what form it takes. If it crystallizes, it's a risk - it's quite unstable when pure. If it turns amorphous, embedding the stabilizer within the structure, it's moot. Publicly available resources are quite skimpy on descriptions of freezing hydrazine (or I just suck at searching). So, -

Does hydrazine crystallize when freezing? Can it trigger fraction-freezing separation if it's the first freezing substance?

I would take a different position on answering the question, and look at the water-hydrazine binary phase diagram. Looking at J. A. McMillan and S.C. Los, J. Chemical Physics 42(1) 160-161 (1965) one finds the experimentally determined phase diagram. With enough water in the mix, one proceeds smoothly from the pure hydrazine melting point down to a slight eutectic near the monohydride, but only 3K lower in temperature.

Further, the same authors have another paper, J. Chem Phys 42(3) 829-834 (1965)  that discusses the nonequilibrium phase transformations. The first sentence in the abstract says

  Although neither hydrazine nor water supercool to a large extent, their mixtures do. Aqueous solutions of
  hydrazine may be strongly supercooled and stabilized as rigid glasses.

In the introductory paragraph, one also finds

  Since the phase diagram of this system
  is known, it was tempting to look for a correlation
  between the irreversible processes of crystallization and
  the equilibrium phase transformations. The hydrazine-water
  system is particularly useful for this purpose,
  since crystallization can indeed be induced in the entire
  range of concentration. Samples of different hydrazine
  content were then supercooled until glass stabilization
  occurred and then warmed up at different rates. The
  processes of glass transformation and crystallization
  were studied, and values of the free-energy barriers
  could be obtained for the latter process.

The rest of this second paper is devoted to measuring and describing thermodynamically the crystallization curves at different water-hydrazine compostions.

So, yes, you can definitely get a glass, if cooled quickly enough, but it just might decide to phase separate on you if it warms up a bit.

For the case of (only) hydrazine-1,1-dimethylhydrazine, the phase diagram is covered in J. A. McMillan, Journal of Chemical and Engineering Data 12(1) 39-40 (1967). This shows a very slow decrease in the liquidus temperature as 1,1-dimethylhydrazine is added, with the eutectic occurring at 94% mole percent 1,1-dimethylhydrazine at 214K, only 61K below the pure hydrazine melting point. There are no compounds. Further, the article states

  Unlike the systems of the components with
  water, the mixture hydrazine-1,1-dimethylhydrazine
  does not supercool

which indicates that phase separation and crystallization of the two components will readily occur. On the other hand, the samples were sealed in glass under nitrogen, and if kept in the dark were stable over a 6 month period.


What piezoelectric materials can be made at home? I would like to know a few examples of transparent materials which exhibit piezoelectricity and can be purchased online or synthesized at home with ease.

The top of my list for exhibiting piezo activity and can be easily purchased and is also transparent..

Quartz

That's about it. It's used everywhere - as an oscillator for quartz crystal watches, quartz crystal microbalances, etc.

Most of the other piezoelectric options:

Lead Zirconium Titanate (PZT)
Barium Titanate ($\ce{BaTiO3}$)
Zinc Oxide ($\ce{ZnO}$)

Aren't transparent. They're white ceramics. You can easily purchase them online from most scientific supply companies, but they don't meet your transparency criteria.

Polyvinylidene difluoride (PVDF) is claimed to be transparent (e.g., here) but you would need to apply a high electric field to pole it, and if the formulation isn't crystalline enough, you may not get a high piezo response.

My group is designing some materials that would be good, but they're still in the research stage. Stay tuned.


Suggestions for reaction conditions to prepare methyl 2,4,6-trimethylbenzenesulfonate I'm a beginner in organic chemistry and I wish to produce methyl 2,4,6-trimethylbenzenesulfonate.

My reactants are 2,4,6-trimethylbenzenesulfonyl chloride (solid crystals) and methanol (solvent as well as reactant). The product of this reaction should be methyl 2,4,6-trimethylbenzenesulfonate and HCl.

Now I've tried these many conditions: distilling the methanol, nitrogen inert atmosphere, slight heating about 40 °C, room temperature, closed system.

After trying these conditions and their combinations I'm not getting the product in the desired amount.

Kindly help me out and suggest a few tweaks and optimizations if possible. As far as my knowledge goes the problem seems to be with methoxy group.

This is the entire synthesis process, please check it out. I am stuck on step two of this synthesis.

Are you following a literature procedure? It looks like you should be able to find a reliable method that you can reproduce. Some general hints for optimisation:

Use a base to mop up the $\ce{HCl}$ that is produced. Either an
organic base like pyridine (preferably) or triethylamine or you can use an
inorganic base like $\ce{NaOH}$ or $\ce{Na2CO3}$ under
Shotten-Bauman conditions  i.e. as a biphasic aqueous-organic system
Increase the reactivity of the nucleophile. Methoxide is a much better nucleophile. You can use commercially available $\ce{MeONa}$ or you can create it by something like $\mathrm{^tBuOK/MeOH}$

EDIT: I am assuming OP tries to make methyl 2,4,6-trimethyl toluenesulfonate since he states that MeOH is a reagent.


Determining whether a compound is ionic or molecular in nature $\ce{HCl}$ is a molecular compound because there is a covalent bond between $\ce{H+}$(a proton) and $\ce{Cl-}$ (a chloride ion)
$\ce{NH4NO3}$ is an ionic compound because there is an ionic bond between $\ce{NH4+}$ (an ammonium ion) and $\ce{NO3-}$ (a nitrate ion) even though all the atoms comprising the molecule are non-metal.
Then, which of the following is true?
$\ce{H3C-COOH}$ is an ionic compound because there is an ionic bond between $\ce{C2H3O-}$ (an acetate ion) and $\ce{H+}$ (a proton)? OR
$\ce{H3C-COOH}$ is a molecular compound because there is a covalent bond between $\ce{C2H3O2-}$ (an acetate ion) and $\ce{H+}$ (a proton)?

How do you tell if a bond between the two ions are molecular or ionic? (regardless of whether they are polyatomic or monoatomic ions)

In principle, when studying a structure like HCl, CH3COOH, or any other compound, the first thing to do is drawing the symbol of the atoms in a reasonable geometry on a sheet of paper. Then consider the outer electron layers : put enough points around each symbol to get a reasonable Lewis structure. For example, one point near $\ce{H}$ and six points around $\ce{O}$. Then try to join atoms with their neighbors to make covalencies (or doublets). When all atoms are joined to neighbors, count the number of electrons around each atom, taking into account that the two electrons included in a covalence between A and B  are at the periphery of both atoms A and B. Check all atoms. If an atom respects the octet law, don't modify it : its structure is covalent. If an atom A does not respect octet law, try to remove one electron from its outer layer, and send this electron to the B atom, which becomes negatively charged. The covalency disappears from the region between A and B, and the bond AB becomes ionic with a positive charge on the A atom, which becomes $\ce{A+}$, and a negative charge on $\ce{B}$ which becomes $\ce{B-}$.
Using this method, you will see that $\ce{HCl}$ and $\ce{CH3COOH}$ are both made of covalencies, because all their atoms are respecting the octet rule. You will also see that $\ce{NaCl}$ may first be drawn in a covalent structure. But then the sodium atom does not fit the octet rule. It will if an electron is removed from the sodium atom, to be given to the chlorine atom, producing then $\ce{Na+}$ and $\ce{Cl-}$ ions.
Hydroxides contains two sorts of bonds. Hydroxides like $\ce{NaOH}$ are made of covalent bonds between $\ce{O}$ and $\ce{H}$ atoms, and of ionic bonds between $\ce{Na^+}$ and $\ce{O-}$.
Oxygenated acids are a special case. They may get two sorts of bonds between $\ce{A}$ and $\ce{B}$. For example sulfuric acid $\ce{H2SO4}$ can be drawn with this method to give first a covalent structure with two bonds $\ce{H-O}$, two bonds $\ce{O-S}$, and two double bonds $\ce{S=O}$. This is OK for all atoms except the central Sulfur atom, which is surrounded by six doublets or six covalences. At this point, some chemists are thinking that it is a case of hyper valence. Some think that the octet rule may be obtained by removing the second bond from each double bond from the sulfur atom, and sending it to the outer Oxygen atom. This makes a central sulfur atom charged $2+$ (respecting the octet rule) and two oxygen atoms charged $1-$. Both theories can be discussed.
The only problem with this reasoning is the fact that it does not work properly  for atoms of the third column. For example, $\ce{BF3}$ should be ionic making $\ce{B^{3+}}$ and $\ce{F-}$ ions. Indeed $\ce{BF3}$ remains covalent, and it is hard to justify. Maybe, on may admit that Boron atom prefers to stay covalent rather than to loose three electrons. Three is probably too much.


Optimum pH range for formation of Schiff bases What is the optimum range of pH for the formation of Schiff Bases by the addition of hydroxylamine to an aldehyde or a ketone?

According to Peter Sykes' A Guidebook to Mechanisms in Organic Chemistry,

"The initial addition step or the dehydration step can be made rate limiting at will, depending on the pH of the solution. At neutral and alkaline pHs it is generally dehydration, that is the slow and rate limiting step while at acidic pHs it is the initial addition which is slow and rate limiting. This clearly has significance in preparatory terms, and formation of such derivatives of carbonyl compounds tends to exhibit pH optima."

My doubts are :

How does the pH change the rate limiting step? That is, why is dehydration the rate limiting step at neutral and alkaline pHs and why is addition the rate limiting step at acidic pHs?
What is the significance in preparative terms that Sykes is taking about?
How will this significance determine the pH and what is that pH (or the range of pHs)?

Dehydration requires protonation of the hydroxy intermediate, this clearly is slower at neutral or alkaline pH. Hydroxylamine is basic so at acidic pH it protonates making it non-nucleophilic hence lower pH slows the rate.
Time taken for the reaction to go to completion. If the rate is slowed then other competing reactions (e.g. self-condensation) can and do reduce your yield.
The pH will vary for each substrate and needs to be experimentally determined. This is where a literature search saves you a lot of grief.


What is the origin of the bactericidal properties of silver in water? I often hear that water gets purified by being in a silver vessel, which sounds plausible because of bactericidal feature of silver. What doesn't sound plausible, though, is the way it's explained: that silver releases ions into the water. Since silver is a noble metal, why would any "reaction" at all occur with something as neutral as water?
Is the above explanation nonsense? Does the "disinfection" of water happen only on contact with the metal?

Silver is not as inert as gold. Tarnish is the name we give to the phenomenon when silver metal is oxidized and becomes a salt. Surfaces made of silver tend to disinfect themselves pretty quickly. As for disinfecting water poured into a silver cup, I imagine that would take a little longer since you have to wait for silver to diffuse away from the surface and into the solution. But even very trace levels of silver can have strong antimicrobial effects.


Relative acidity of aqueous LiF, SnCl2, and YCl3 Which one of the following salts make the highest and lowest pH when dissolved in water?
$$\ce{LiF, SnCl2, YCl3}$$

My approach was :
considering the corresponding aqua acids, the order of acidity should be $\ce{Li(H2O)_n+ < Sn(H2O)_n^2+ < Y(H2O)_n^3+}$ and $\ce{F-}$ is less stable than $\ce{Cl-}$. So $\ce{LiF}$ is the weakest acid and $\ce{YCl3}$ the strongest.

But considering aqua acid looks weird (do they exist?) and I cannot be sure this of solution. I don't even know the right answer. Can anyone give some help?

Salt solutions are acidic because of hydrolysis.

LiF is obviously neutral, because it's a salt of very strong acid and very strong base.

Ironically, Wikipedia is wrong about Sn2+: it's much more acidic in water than Ca2+ or Zn2+, and more acidic than Y3+,
because Sn2+ forms very stable and non-soluble hydroxo compounds and hence readily undergoes hydrolysis.

The answer is: LiF < YCl3 < SnCl2

Table data:

YCl3 has pH 4.5-5.8 (source),
SnCl2 has pH 1-2 (source 1, 2).


How is Gibbs free energy related to stability? I have been learning thermodynamics for quite some time and I fail to relate Gibbs free energy and stability.
For example, how do I compare stability of metal oxides using the following data?
Also, do stable compounds decompose easily or hard?
$$
\begin{array}{clrr}
\hline
\text{#}  & \text{Compound} & \Delta_\mathrm{f}G^\circ/\pu{kJ mol^-1} & \Delta_\mathrm{f}H^\circ/\pu{kJ mol^-1} \\
\hline
1 & \ce{ZnO(s)} & -318.4 & -348.3 \\
2 & \ce{Cu2O(s)} & -146.0 & -168.8 \\
3 & \ce{HgO(s)} & -58.5 & -90.8 \\
4 & \ce{PbO(s)} & -187.9 & -217.3 \\
\hline
\end{array}
$$

The thermodynamic data refers to formation of the given compounds from the elements in their standard states. For instance for $ \ce{ ZnO(s)}$  it refers to the reaction
$ \ce{ Zn(s) +1/2O2(g) ->ZnO(s)} \tag{1}$
with each element or compound present at a pressure of 1 bar and (unless noted otherwise) a temperature of 298.15 K.
A more negative value of $\Delta G^\circ$ means greater product stability. This follows from the relationship
$$\Delta G^\circ = -RT\log K \tag{2}$$
where K is the equilibrium constant for the reaction at that temperature. An increased value of K implies an increased amount of products relative to reagents. By the relation in equation (2) a larger value of K also means a more negative value of $\Delta G^\circ$.
By definition a more stable compound decomposes less readily. Decomposition here refers to the reverse of reaction (1).
When comparing the tabulated oxides, a more stable oxide has a more negative free energy of formation. Therefore $ \ce{ZnO(s)}$ is the most stable of the oxides relative to oxygen and the reduced metal. Of the metals the mercury(II) oxide would require the least energy to reduce.
Note that for all of the reactions here
$$K = p^\circ/p_\ce{O2}$$
where $p^\circ=\pu{1 bar}$ (assuming the pressure is low). When starting from metal oxides, allowed to decompose until at equilibrium, a less stable oxide is at equilibrium with a higher partial pressure of oxygen (smaller K) due to more of the oxide decomposing. Therefore imagine you have an evacuated box containing the oxide of one metal and an amount of another metal in pure (reduced) form. If the equilibrium partial pressure $p_\ce{O2}(\text{eq})$ for the oxide is higher than $p_\ce{O2}(\text{eq})$ for the oxide of the second metal, then (if you wait long enough) all of the oxide will be reduced and the second metal oxidized until either the original oxide or pure metal is exhausted.
The enthalpies of formation refer to how much heat evolves when  1 mole of the oxide is formed from the reagents at constant T and p, with all compounds in their standard states. The more negative the value the more heat is released (the reaction is more exothermic). The oxidation of zinc is the most exothermic reaction (per mole of product formed) of those shown in the table.


Conversion between anisotropic temperature factors and geometric ellipsoids I am using the ANISOU of the PDB standard to represent ellipsoid shaped colloids.  I am writing a Monte Carlo simulation to transform the position and orientation of the ellipsoids and want to use a PDB file to represent the result.

I am not sure how to convert between a geometric representation of an ellipsoid to the anisotropic temperature factor tensor, $\mathbf{U}$.  I am able to define an ellipsoid aligned to the coordinate axes using the diagonal elements of $\mathbf{U}$: $\mathbf{U}(1,1)$, $\mathbf{U}(2,2)$, $\mathbf{U}(3,3)$, but I am not sure how to represent a rotated ellipsoid, i.e., I am not sure how to set $\mathbf{U}(1,2)$, $\mathbf{U}(1,3)$, and $\mathbf{U}(2,3)$.  I came across this resource which introduces the topic.  This second link mentions the similarity between tensor elements and geometric definitions and says the conversion is "rather complicated" but does not provide a link for further study.

Any guidance or direction to a good resource would be appreciated.

I found the publication titled Atomic Displacement Parameter Nomenclature which seems to have the details.


How to liberate propylthiouracil Two substances undergo a reaction that gives off propylthiouracil and another compound as products.

Is there an additional compound I could throw into that sludgy product to strain off the $\ce{C9H17O6}$ such as that used in the Miller Process, but instead would leave a clear propylthiouracil for my use? The chemical reaction is given below:

$\ce{2 C11H19NO6 + C3H6OS -> C7H10N2O6 + 2 C9H17O6}$

According to https://pubchem.ncbi.nlm.nih.gov, propylthiouracil has the following solubility properties:

  1 part dissolves in about 900 parts water at 20 deg C, in 100 parts
  boiling water, in 60 parts ethyl alc, in 60 parts acetone; practically
  insol in ether, chloroform, benzene (emphasis mine); freely sol in aq soln of ammonia
  and alkali hydroxides.

I would start with a small amount of your "sludgy" product. Stir it with ether as a first choice, chloroform as a second, for a while and see first of all whether you really get multiple phases or if just forms a less viscous sludge.  In a perfect world, after letting the solution set you would see the white propylthiouracil solid settle out at the bottom, with a sludge-discolored solvent layer on top.  If you see anything resembling this, then I would pursue this technique with more thorough stirring/shaking, maybe centrifuging to get better settling of your product, then further rinsing your product on filter paper with a vacuum flask, etc.  Try the other solvents if available if ether doesn't work.  The success/failure of this approach depends on the solubility of the sludgy stuff in the solvent.  It may fail, it may not give the purity you need, it may work wonderfully, but it is a cheap and simple way to test a method with a reasonable probability of success.


Experimental determination of pKa I have $20~\mathrm{cm^3}$ of a $0.1\ \mathrm{M}$ ethanoic acid solution and have to find out the $\mathrm{p}K_\mathrm{a}$.

In experiment this solution had a $\mathrm{pH}$ of $2.91$.

Do I use the formula $\mathrm{p}K_\mathrm{a}= 2 \mathrm{pH} + \log(c (\mathrm{acid}))$ ?

What I do not understand is that my $\mathrm{p}K_\mathrm{a}$ value varies. I measured many solutions with 18:2, 15:5, 10:10 sodium ethanoate/ethanoic acid composition. Which I calculated with the formula $\mathrm{p}K_\mathrm{a} = \mathrm{pH} - \log \frac{c(\mathrm{salt})}{c(\mathrm{acid})}$

I get close values but not the equal.  So the one above would be $\mathrm{p}K_\mathrm{a}=4.82$   and another value I obtained is $4.62$ (for the 15:5 solution with $\mathrm{pH}\ 5.11$).

Is this deviation purely due to experimental inaccuracy or does the composition matter.   Would it make more sense to use the literature $\mathrm{p}K_\mathrm{a}$ value for further calculation rather than those experimental ones?

Do I use the formula $\mathrm{p}K_\mathrm{a}= 2 \mathrm{pH} + \log(c (\mathrm{acid}))$ ?

No, for the pure acide use:

$$K_\mathrm{a} = \frac{([\ce{H+}])^2}{[\ce{HA}]_\mathrm{initial}-[\ce{H+}]}$$

Then, if you are still having problems, see this question and its answer:

Difference in calculated pH and the real pH of a phosphate buffer?


Why don't equivalent hydrogens cause splitting in NMR? When doing NMR spectroscopy, it is an observed fact that equivalent hydrogens do not split one another. Why don't equivalent hydrogens split each other's signals? For example, why is the NMR spectrum for ethane a singlet instead of a quartet or even a dodecuplet (due to the hydrogens on the same carbon)? What is so special about hydrogens being equivalent to one another that causes no splitting to be observed?

I will provide a full quantum mechanical explanation here.[1] Warning: rather MathJax heavy. Hopefully, this lends some insight into how the diagrams that long and porphyrin posted come about.

Finding the states between which transitions occur

The Hamiltonian for two coupled spins is

$$\hat{H} = \omega_1\hat{I}_{\!1z} + \omega_2\hat{I}_{\!2z} + \frac{2\pi J_{12}}{\hbar}(\hat{\vec{I}_{\!1}}\cdot\hat{\vec{I}_{\!2}}) \tag{1}$$

where $\omega_1$ and $\omega_2$ are the Larmor frequencies[2] of the two nuclei and $J_{12}$ is the coupling constant (in Hz) between the two nuclei. (The factor of $2\pi/\hbar$ is simply there to bring it into energy units.) $\hat{I}_{\!i}$ is the operator for the spin angular momentum of nucleus $i$ and $\hat{I}_{\!iz}$ is the operator for its projection along the $z$-axis.

However, in NMR, it is customary to work in frequency units instead of energy units. Since $E = h\nu$, we simply need to divide through by $h$. Bearing in mind that $\omega = 2\pi\nu$ and $h = 2\pi\hbar$, we get:

$$\hat{H}_\text{freq} = \frac{\nu_1}{\hbar}\hat{I}_{\!1z} + \frac{\nu_2}{\hbar}\hat{I}_{\!2z} + \frac{J_{12}}{\hbar^2}(\hat{\vec{I}_{\!1}}\cdot\hat{\vec{I}_{\!2}}) \tag{2}$$

On top of that, to make the maths easier, it is also quite common to set $\hbar = 1$. Therefore, we have

$$\hat{H}_\text{freq} = \nu_1\hat{I}_{\!1z} + \nu_2\hat{I}_{\!2z} + J_{12}(\hat{\vec{I}_{\!1}}\cdot\hat{\vec{I}_{\!2}}) \tag{3}$$

We will deal with two spin-$1/2$ nuclei here, where $|\alpha\rangle$ and $|\beta\rangle$ represent the "up" and "down" spins (recall we set $\hbar = 1$ so it doesn't appear in the eigenvalues):

$$\begin{align}
\hat{I}_{\!iz}|\alpha_i\rangle &= \frac{1}{2}|\alpha_i\rangle & \hat{I}_{\!iz}|\beta_i\rangle &= -\frac{1}{2}|\beta_i\rangle & (i = 1,2) \tag{4}
\end{align}$$

Furthermore since we are dealing with equivalent nuclei we can simply set $\nu_1 = \nu_2 = \nu$ and drop the subscript in $J_{12}$ just to make it a bit cleaner:

$$\hat{H} = \nu(\hat{I}_{\!1z} + \hat{I}_{\!2z}) + J(\hat{\vec{I}_{\!1}}\cdot\hat{\vec{I}_{\!2}}) \tag{5}$$

Now, we need to find the eigenstates and eigenvalues of $\hat{H}$. To do so, we will adopt the basis set of product functions $(|\alpha_1\alpha_2\rangle, |\alpha_1\beta_2\rangle, |\beta_1\alpha_2\rangle, |\beta_1\beta_2\rangle)$. From equation $(4)$ we have

$$\begin{align}
\hat{I}_{\!1z}|\alpha_1\alpha_2\rangle &= \frac{1}{2}|\alpha_1\alpha_2\rangle & \hat{I}_{\!2z}|\alpha_1\alpha_2\rangle &= \frac{1}{2}|\alpha_1\alpha_2\rangle \tag{6} \\
\hat{I}_{\!1z}|\alpha_1\beta_2\rangle &= \frac{1}{2}|\alpha_1\beta_2\rangle & \hat{I}_{\!2z}|\alpha_1\beta_2\rangle &= -\frac{1}{2}|\alpha_1\beta_2\rangle \tag{7} \\
\hat{I}_{\!1z}|\beta_1\alpha_2\rangle &= -\frac{1}{2}|\beta_1\alpha_2\rangle & \hat{I}_{\!2z}|\beta_1\alpha_2\rangle &= \frac{1}{2}|\beta_1\alpha_2\rangle \tag{8} \\
\hat{I}_{\!1z}|\beta_1\beta_2\rangle &= -\frac{1}{2}|\beta_1\beta_2\rangle & \hat{I}_{\!2z}|\beta_1\beta_2\rangle &= -\frac{1}{2}|\beta_1\beta_2\rangle \tag{9}
\end{align}$$

The action of the scalar product $\hat{\vec{I}_{\!1}}\cdot\hat{\vec{I}_{\!2}}$ is more complicated. We need to introduce the shift operators (or ladder operators).

$$\begin{align}
\hat{I}_{\!i+} &= \hat{I}_{\!ix} + \mathrm{i}\hat{I}_{\!iy} & \hat{I}_{\!i-} &= \hat{I}_{\!ix} - \mathrm{i}\hat{I}_{\!iy} \tag{10}
\end{align}$$

from which we can obtain

$$\begin{align}
\hat{I}_{\!ix} &= \frac{\hat{I}_{\!i+} + \hat{I}_{\!i-}}{2} & \hat{I}_{\!iy} &= \frac{\hat{I}_{\!i+} - \hat{I}_{\!i-}}{2\mathrm{i}} \tag{11}
\end{align}$$

So, finally, we can write

$$\begin{align}
\hat{\vec{I}_{\!1}}\cdot\hat{\vec{I}_{\!2}} &= \hat{I}_{\!1x}\hat{I}_{\!2x} + \hat{I}_{\!1y}\hat{I}_{\!2y} + \hat{I}_{\!1z}\hat{I}_{\!2z} \tag{12} \\
&= \frac{\hat{I}_{\!1+}\hat{I}_{\!2-} + \hat{I}_{\!1-}\hat{I}_{\!2+}}{2} + \hat{I}_{\!1z}\hat{I}_{\!2z} \tag{13}
\end{align}$$

where in going from $(12)$ to $(13)$ one simply substitutes in $(11)$ and does a fair bit of algebraic manipulation. The action of the shift operators are

$$\begin{align}
\hat{I}_{\!i+}|\alpha_i\rangle &= 0 & \hat{I}_{\!i+}|\beta_i\rangle &= |\alpha_i\rangle \tag{14} \\
\hat{I}_{\!i-}|\alpha_i\rangle &= |\beta_i\rangle & \hat{I}_{\!i-}|\beta_i\rangle &= 0 \tag{14} \\
\end{align}$$

This allows you to work out the effect of $\hat{\vec{I}_{\!1}}\cdot\hat{\vec{I}_{\!2}}$ on our basis states. The actual maths is left to the reader and I will simply quote the results:

$$\begin{align}
\hat{H}|\alpha_1\alpha_2\rangle &= \left(\nu + \frac{J}{4}\right)|\alpha_1\alpha_2\rangle \tag{15} \\
\hat{H}|\alpha_1\beta_2\rangle &= -\frac{J}{4}|\alpha_1\beta_2\rangle + \frac{J}{2}|\beta_1\alpha_2\rangle \tag{16} \\
\hat{H}|\beta_1\alpha_2\rangle &= \frac{J}{2}|\alpha_1\beta_2\rangle - \frac{J}{4}|\beta_1\alpha_2\rangle \tag{17} \\
\hat{H}|\beta_1\beta_2\rangle &= \left(-\nu + \frac{J}{4}\right)|\beta_1\beta_2\rangle \tag{18} \\
\end{align}$$

Therefore in this basis the Hamiltonian matrix is

$$\mathbf{H} =
\begin{pmatrix}
\nu + J/4 & 0 & 0 & 0 \\
0 & -J/4 & J/2 & 0 \\
0 & J/2 & -J/4 & 0 \\
0 & 0 & 0 & -\nu + J/4
\end{pmatrix} \tag{19}$$

Finding the eigenvectors and eigenvalues of this matrix is again left to the reader (it is not a difficult task) and they are (eigenvalues denoted $E_i$)

$$\begin{align}
|1\rangle &= |\alpha_1\alpha_2\rangle & E_1 &= \nu + \frac{J}{4} \tag{20} \\
|2\rangle &= \frac{1}{\sqrt{2}}(|\alpha_1\beta_2\rangle + |\beta_1\alpha_2\rangle) & E_2 &= \frac{J}{4} \tag{21} \\
|3\rangle &= \frac{1}{\sqrt{2}}(|\alpha_1\beta_2\rangle - |\beta_1\alpha_2\rangle) & E_3 &= -\frac{3J}{4} \tag{22} \\
|4\rangle &= |\beta_1\beta_2\rangle & E_4 &= -\nu + \frac{J}{4} \tag{23} \\
\end{align}$$

The form of the eigenstates should be familiar: they are simply the triplet and singlet states of two spin-$1/2$ particles. These states arise from the coupling of two sources of angular momenta, $I_1$ and $I_2$, to form one overall angular momentum denoted $I$.

$$\vec{I} = \vec{I}_{\!1} + \vec{I}_{\!2} \tag{24}$$

The allowed values of $I$ are determined by the Clebsch-Gordan series:

$$I = I_1 + I_2, I_1 + I_2 - 1, \cdots, |I_1 - I_2| \tag{25}$$

Since $I_1 = I_2 = 1/2$, $I$ can take the values $1$ and $0$. The values of $M_I$, the projection of the total angular momentum along the $z$-axis, are as usual

$$M_I = I, I-1, \cdots, -I \tag{26}$$

so the states with $I = 1$ ("triplet") have $M_I = 1, 0, -1$ and the state with $I = 0$ ("singlet") has $M_I = 0$. One can use more quantum mechanics to work out which state is associated with which quantum numbers, but I will not do it here. They are:

$$\begin{array}{ccc} \hline
\text{State} & I & M_I \\ \hline
|1\rangle & 1 & 1 \\
|2\rangle & 1 & 0 \\
|3\rangle & 0 & 0 \\
|4\rangle & 1 & -1 \\ \hline
\end{array}$$

Selection rules

We have four different states, which leads to ${4\choose 2} = 6$ different possible transitions. However, not all of these transitions are allowed.

The intensity of the transition is proportional to the square of the matrix element $\langle \psi_\mathrm{f} | \hat{H'} | \psi_\mathrm{i} \rangle$ (the so-called "transition dipole moment"), where $\hat{H'}$ is the Hamiltonian for the process that induces the transition. In the case of NMR transitions, the transition arises due to a magnetic field aligned along the $x$-axis.[3] The corresponding Hamiltonian is therefore

$$\hat{H'} = \omega'(\hat{I}_{\!1x} + \hat{I}_{\!2x}) = \omega'\hat{I}_{\!x} \tag{27}$$

Exactly what $\omega'$ represents is not important here because we are only really concerned about whether the transition dipole moment is zero or not.[4] Making use of the relations established in equations $(10)$ and $(11)$,[5] one can find that the selection rules are

$$\Delta I = 0; \Delta M_I = \pm 1 \tag{28}$$

which means that the allowed transitions are $|4\rangle \leftrightarrow |2\rangle$ and $|2\rangle \leftrightarrow |1\rangle$. Transitions to and from the singlet state $|3\rangle$ are forbidden. The energies of the transitions are

$$\begin{align}
E_{4\leftrightarrow2} &= \frac{J}{4} - \left(-\nu + \frac{J}{4}\right) = \nu \tag{29} \\
E_{2\leftrightarrow1} &= \left(\nu + \frac{J}{4}\right) - \frac{J}{4} = \nu \tag{30} \\
\end{align}$$

i.e. the two transitions are degenerate and only one line in the spectrum at frequency $\nu$ is observed. This is exactly what is depicted in the diagrams posted in the other answers.

Notes and references

[1] I am assuming the reader has some knowledge of the quantum mechanical treatment of angular momentum, which is a topic that is treated thoroughly in most quantum mechanics textbooks. See, for example, chapter 4 of Atkins's Molecular Quantum Mechanics (5th ed.).

[2] The Larmor frequency is given by $\omega = -\gamma B_0$, where $\gamma$ is the magnetogyric ratio of the nucleus in question and $B_0$ is the strength of the external magnetic field. It represents the frequency with which a magnetic moment precesses about a magnetic field. See any textbook on magnetism for further details.

[3] I am glossing over some details here. The so-called magnetic field in the $x$-axis is a component of the radiofrequency pulse applied in the $xy$-plane. If you are interested please consult a textbook on the vector model of NMR. In particular I recommend Keeler's Understanding NMR Spectroscopy (2nd ed.).

[4] It is related to the strength of the magnetic field in the $x$-axis, $B_1$, by $w' = |\gamma|B_1$. The usual symbol is $\omega_1$, but I chose not to use this here to avoid potential confusion. Again, please consult a textbook on the vector model of NMR if you wish to find out more.

[5] A full proof can be found in J. Chem. Educ. 1982, 59 (10), 819. There is also some discussion of the selection rules in Gunther's NMR Spectroscopy (3rd ed.), p 156 onwards.


Calculating enthalpy change and energy required to convert lead sulphide to lead Consider the following:
  $$\begin{alignat}{2}
\ce{PbS(s) + 3/2O2(g) &-> PbO(s) + SO2(g)}\qquad&&\Delta H^\circ = -413.7~\mathrm{kJ} \\
\ce{PbO(s) + C(s) &-> Pb(s) + CO(g)}\qquad&&\Delta H^\circ = +106.8~\mathrm{kJ}
\end{alignat}$$
  a. Calculate the enthalpy of the following process:
  $$\ce{PbS(s) + 3/2O2(g) + C(s) -> Pb(s) + SO2(g) + CO(g)}$$

  b. What is the energy needed to make $454\ \mathrm g$ of $\ce{PbS}$ into $\ce{Pb}$?

My efforts:
a.

$\ce{PbS(s) + 3/2O2(g) -> PbO(s) + SO2(g)}$
$\ce{PbO(s) + C(s) -> Pb(s) + CO(g)}$
$\Delta H_\mathrm r^\circ = \Delta H_{\mathrm r(1)}^\circ + \Delta H_{\mathrm r(2)}^\circ$
$\Delta H_\mathrm r^\circ = -413.7~\mathrm{kJ~mol^{-1}} + 106.8~\mathrm{kJ~mol^{-1}} = -306.9~\mathrm{kJ~mol^{-1}}$

b. How do I find the energy needed?

Molecular mass of $\ce{PbS}$ is   $239.2650~\mathrm{g~mol^{-1}}$

so we have $\frac{454~\mathrm{g}}{239.2650~\mathrm{g~mol^{-1}}} = 1.897~\mathrm{mol}$ of $\ce{PbS}$

So energy required is $1.897~\mathrm{mol} \times 306.9~\mathrm{kJ~mol^{-1}} = 582.2~\mathrm{kJ}$

Part 1:

It's easier to imagine \begin{equation}\ce{A + B -> C + D} \tag{1}
\end{equation}
\begin{equation}\ce{D + E -> F + G} \tag{2}\end{equation}
\begin{equation}\end{equation}
Now if we wanted to write a 3rd equation where both the reactions 1 and 2 happen, we'd get
\begin{equation}\ce{A + B + E -> C + F + G} \tag{3}\end{equation}
Why?

D is both a reactant and a product. So, it's a reaction intermediate. It's like it's produced in 1 and again consumed in 2. So, if both reactions take place, it's neither on the reactants' side nor the products.
The conventional way of writing reactions doesn't show the mechanism of the reaction. You have to image the reaction happening, to completely understand why D isn't written.

So, as the third reaction is reaction 1 and 2 happening at the same place, $\Delta H_{r_3}$ is the algebraic sum of $\Delta H_{r_2}$ and $\Delta H_{r_1}$. Thus, your final result is correct: $$\mathbf{-413.7 ~kJ \times mol^{−1} +106.8 ~kJ \times mol^{−1}=−306.9 ~kJ \times mol^{−1}}$$

Part 2:

We have to take a more careful look at the unit of $\Delta H$. What does it mean? It means that for each mole of the reactant ($\ce{PbS}$), one mole of product ($\ce{Pb}$) is formed. (With regards to their stoichiometric coefficients in reaction 3) So, we've got to simply 'punch' the numbers in the formula $n= \frac{m}{M}$ (Where $n$ is the number of moles, $m$ is the given mass and $M$ is the molar mass) And then multiply that number by the $\Delta H_{r_3}$. Thus, your final answer is correct too.


What's the effect of temperature on buffering capacity? I did some research online but I still couldn't get any conclusion. Some said that the buffering capacity decreases when temperature increases because of the increase in ionisation of the weak acid molecules and therefore, some of the conjugate base ion will be used to neutralised the $\ce{H+}$ ion from the ionization of the weak acid molecules. Hence, only a small amount of strong acid is needed to complete remove the conjugate base ion and to break the buffering capacity. But, the problem is that the acid molecules ionise into both $\ce{H+}$ and the conjugate base so doesn't it mean that the number of conjugate base ion will remain the same since the $\ce{H+}$ ion is just reacting with the same conjugate base ion which they were initially bonded together ?

The other assumption is that temperature has no effect on buffering capacity since buffering capacity apparently only depends on two factors: concentration of the buffer solution and the ratio of pH to pKa. So even if the pKa changes due to the change in temperature, the pH also changes accordingly whereby pH increases when pkA increases and vice versa. So the ratio will remain the same no matter what.

So could anyone please give me a solution?

You question is too broad to answer since there is no "universal answer." In chemistry there is a "tends too" answer and then a list of exceptions.

By and large the root of the question centers around the change in the equilibrium constant as a function of temperature. The equilibrium constants do depend on temperature, but how much "tends" to depend on the particular equilibrium under consideration. However the Arrhenius equation can offer some insight.


Temperature as an independent variable in an endothermic reaction My experiment involves the reaction between sodium bicarbonate and acetic acid. I'll be changing the temperature between each trial and measuring the change in pressure by the gas produced. From here I hope to determine the rate expression, rate constant, activation energy then finally produce an Arrhenius plot.

My question is since this reaction is endothermic, how do I account for the change in temperature as the reaction proceeds? I also can't seem to find any published literature value for this reaction's activation energy or enthalpy change, and would love it if someone pointed out a database or something similar.

For this experiment to work, you need good control of your temperature profile.  In other words the temperature must be somewhat uniform.

If this is the case measure two temperatures.

$T_e$ is the isothermal temperature of the experiment.
$T_r$  is the temperature at the point of reaction.

If $T_e$ is 10°C, 20°C, 30°C and  $T_r$ is 9°C, 18°C, and 29°C  then the temperature change can be assumed to have a negligible effect. (You would want to use $T_r$ in the kinetics calculations.)

If $T_e$ is 10°C, 20°C, 30°C and  $T_r$ is 4°C, 6°C, and 8°C  then the effect is not negligible.  Accounting for the endothermic temperature change in the calculations is a poor option because it would add needless complication. A better approach is to keep things simple by revisiting the controls in the experimental design to see what can be done to minimize the endothermic effect.  One possibly would be using less reactants.


What is the unit of the Rydberg Constant? In the image, the Rydberg constant seems to have two units. How is that possible?

To convert between joules (units of energy) and inverse centimetres (conventional units of wavenumber) simply requires multiplication by a factor of $hc$. This is because

$$E = h\nu = \frac{hc}{\lambda} = hc\bar\nu$$

Of course, they are not literally equal, but they are physically equivalent.


Reason Behind Decrease In Potential Energy Atoms combine to gain stability. They gain stability when they lose energy. In my book, it's given that the potential energy of the system is reduced during chemical combination. But I am eager to know: how is the potential energy of the system reduced?

I want the theoretical reason.

The potential energy that changes during reactions is electrostatic energy (+/- attraction, +/+ or -/- repulsion) from the charged particles (electrons and protons) within the atoms. The theory is described by Coulomb's Law (involving the potential energy of charged particles):

http://guweb2.gonzaga.edu/faculty/cronk/CHEM101pub/energy.html

Generally, this comes from the attraction of electrons to nuclei and the repulsion between electrons and between nuclei. When atoms/molecules/the "system" becomes more stable, we will see either an increase in the attraction present (electrons are closer or less shielded from nuclei) or a decrease in repulsion (electrons or nuclei are further apart).

http://ch301.cm.utexas.edu/atomic/#bonding/bonding-all.php

More advanced theories will involve the changing of the allowed potential energy levels of the electrons either within atoms (atomic orbitals) or within molecules (molecular orbitals). Whichever theories are used, the net result is that there is more attraction and/or less repulsion in a more stable/lower potential energy state.

Interestingly, by the first law of thermodynamics, if the potential energy decreases, another type of energy must increase (so you don't end up with less energy than you started). It is normally thermal energy that increases and then is transferred as heat. This process ends up being one the most important ones in understanding why some processes happen and some do not.


Difference between Phase, heterogenous and state of matter and solution My question has already  been answered in parts at Difference between state  of matter and phase
But for deep and real understanding I have to ask for the difference between phase and a heterogenous mixture :
Say we have a bowl made of glass. The bowl is filled with a few different solid rocks on the ground, water in the middle and some oil on top.
I try to analyse this situation for understanding:

There is a phase of oil on the top
There is a phase of water in the middle
There is a phase of rocks on the ground
Since the rocks on the ground are not equally the same they are a bunch of single phases that do not mix up to each other
The rocks on the ground are also a heterogenous mixture
The air above the water is a phase of gas
The glass is a solid phase of silicon

Is that correct?

A deep and real understanding?!?

I don't think you'll find any definitions that you can't pick at.

So if in a large bottle you mix some water and benzene that don't fill the bottle then you'll end up with three phases. A gas phase that has air/water/benzene, a liquid phase that is mostly benzene with a tiny amount of water, and another liquid phase that is mostly water with a tiny amount of benzene.

So if I just take some random rocks (different kinds of rocks) and dump then into a bowl of water, then yes there is a solid phase. But as a chemist I wouldn't really think of the random rocks as a phase because there really isn't much of an equilibrium. Picking at this I'd have to admit that a minuscule amount of rock would dissolve in the water. Hence there would be an equilibrium of sorts between the various rocks too.

What we'd normally consider a glass is a mixture of silica and other minerals. However amorphous solids are also referred to as glasses.


Will compressed hydrogen float in air? Imagine a container of fixed volume is taken and hydrogen gas is filled in it, as a result the container floats in air. But if in the same container is filled with so much hydrogen that the hydrogen gas molecules had to compress to fit in the container, will the container still float in air ?

Yes. A simple example is an ordinary balloon, which places the gas inside it under pressure (because the tension of the rubber surface adds to the air pressure from outside the balloon), yet the balloon still floats readily.

What you're probably talking about is a theoretical container that can hold highly-compressed hydrogen, like from a supply cylinder, yet has negligible weight. At what point, then, would the hydrogen weigh more than air and the container would settle to the ground?

To answer that, we'll look at the ideal gas law: $PV = nRT$, or equivalently, $P = \dfrac{nRT}{V}$. For our purposes, we'll ignore the temperature increase of the compressed gas by saying this container, while having negligible weight, also has an efficient cooling system so the temperature delta is negligible.

At STP (273*K, 100kPa), one liter of volume filled with hydrogen gas contains:

$$100 = \dfrac{n(8.3145)(273)}{1}$$
$$100 = 2269.8482n$$
$$n = .0441\text{ mol}$$

With hydrogen having a molar weight of 1.00794 g/mol (and $\ce{H2}$ gas having double that), the density in g/L of hydrogen gas at STP is .08890 g/L.

Borrowing from ashu's answer, since his math and mine are within 1% error, the density of air at STP is 1.225g/L (1 m3 = 1000 L). So, in order to have the same mass of gas in the same volume, and thus the same density, you would need 13.78 times the amount of gas than you'd have in the same volume at STP (pretty much 1.225 mol). Because, given a unit volume, double the gas equals double the pressure, the resulting pressure of a container of negligible mass holding enough compressed hydrogen to not float in air is 1377.953kPa.

In SAE units, that's actually not much in relative terms; only 199.855 psi (call it 200 so it would actually sink in air and not be neutrally buoyant). That's the water pressure at a depth of about 460 feet (140.72m), which human freedivers have matched using dive sleds.


What makes Kleenex Cool Touch tissues cool to the touch, and is it safe? I love Kleenex Cool Touch tissues.  I'm just wondering what makes them actually feel cool.  For some reason it makes me think of refrigerant but I'm sure that's not in them.

Is it the combination of common ingredients that somehow gives it a moist feel, making it feel cooler than normal tissues?

Are the tissues safe to put to your face, if you are potentially inhaling the ingredient(s)?

Ingredients:  Stearyl heptanoate, stearyl alcohol, polyethylene, mineral oil, coconut oil, aloe

My first guess as to the nature of this behaviour is similar to water. Hypothermia is a realistic risk if you are drenched in water. This is because water can absorb large amounts of energy (it has a large specific heat capacity) to only rise in temperature slightly. Water itself evaporates into steam spontaneously, if some water is spilled then it will evaporate readily, regardless of the temperature. Likewise, the solids in the tissue are probably coming closer to melting temperature, which I believe is higher than your body temperature, but quickly absorbing a lot of energy from your body in the process.

Something similar to properties of water are probably characteristic to the tissue. Leave the tissue in the open in a cool room, along with a glass of water. After awhile, touching the glass of water and the tissue should invoke the same sense of coolness. This is because your body is at a higher temperature than the items in the room.

There is likely another property of matter besides specific heat capacity that is a measurement of how quickly heat is transferred, this is likely to play a role as well. I do not know what this property is called and it was never covered in my classes so far, but take metal for example. Metal has a very small specific heat capacity, it can not hold much energy. However, touching a metal object invokes a sense of coolness. I believe this is because heat is transferred very quickly through metal, although it warms very quickly due to its limited heat capacity. If you put a metal cup and a porcelain cup on the stove, the metal will burn you much quicker if you grab it than grabbing the porcelain/ceramic cup.

If you were very interested and inquistive you could test the specific heat of the tissue through experimentation. Also, try holding the napkin for a few seconds to try and find out how quickly the tissue warms to the touch. Chances are, it warms relatively fast and can absorb a lot of energy, giving you the impression that it is 'cooler' than most objects, to which marketers will be overwrought with the joy of hearing.


Corrosion of Aluminum by Baking soda An aqueous solution of sodium hydrogen carbonate corrodes aluminum foil quite readily. How do you explain this?

Sodium aluminate is produced. The reaction is slower than the usual $NaOH + Al$ and liberates $CO_2$.


Obtaining values for electrical mobility from ion transport numbers in a Hittorf cell I am following the experiments detailed in the links below in order to find the electrical mobility of a variety of ionic species:

Wikipedia article on ion transport numbers and Hittorf method
PHYWE series of publications - "Transport numbers" PDF
"Determination of Transference Number by Hittorf’s Method" - PDF

I know that the ion transport number of an ionic species is the ratio of the current carried by that particular species over the total current conducted through the Hittorf cell. The change in concentration of the electrolytes, the mass of the electrodes, the voltage and current of the cell can be used to calculate this number. But, using these results, how do I calculate the drift velocity of those ions and hence values for their electrical mobility?

I tried looking at the Einstein-Smoluchowski relation but the concepts are a bit over my head right now. Any other tips, notes and comments on this experiment would also be appreciated.

You can't directly derive electrical mobilities from ions from their transport numbers; transport numbers are a relative measurement of charge transfer that, precisely, reduces absolute mobilities for different ions to contributing fractions. Absolute mobilities are lost when we use transport numbers, so if all we know about an electrochemical system is transport numbers, we cannot recover ionic mobilities.

However, we can recover drift velocities if we have additional information on the system - specifically, total current $I_\mathrm{total}$ and distance between plates in the cell $d$. Alternatively, if we're interested in ion mobilities, we will also need to know voltage $V$.

Let's look at a simple system with only one electrolyte involved in the electrochemical reaction. Total current will be the sum of the contributions of the anion and the cation:

$$ I_\mathrm{total} = I_+ + I_- =  \cfrac{n_+ z_+ \nu_+ e}{d} +  \cfrac{n_- z_- \nu_- e}{d} $$

where $n_i$ is the number of ions $i$, $z_i$ is the charge of ion $i$, $\nu_i$ is the drift velocity of ion $i$, $e$ is the electron charge and $d$ is the distance between the plates. Per the requirement of electroneutrality,

$$ n_+ z_+ = n_- z_- $$

so

$$ I_\mathrm{total} = \cfrac{ n_+ z_+ e \left( \nu_+ + \nu_- \right) }{d} = \cfrac{ n_- z_- e \left( \nu_+ + \nu_- \right) }{d} \tag{1}$$

We can calculate transport numbers as

$$ t_+ = \cfrac{I_+}{I_\mathrm{total}} = \cfrac{n_+ z_+ \nu_+ e /d}{n_+ z_+ \left( \nu_+ + \nu_- \right) e / d} = \cfrac{\nu_+}{\nu_+ + \nu_-} $$

$$ t_- = \cfrac{I_-}{I_\mathrm{total}} = \cfrac{n_+ z_+ \nu_- e /d}{n_+ z_+ \left( \nu_+ + \nu_- \right) e / d} = \cfrac{\nu_-}{\nu_+ + \nu_-} $$

Note that we could use the definition of transport numbers to obtain a specific drift velocity, $\nu_i$, from the total drift velocity, $\nu_+ + \nu_-$:

$$ \nu_+ = t_+ \left( \nu_+ + \nu_- \right) $$

$$ \nu_- = t_- \left( \nu_+ + \nu_- \right) $$

Let's go back to $\mathrm{(1)}$ to derive an expression for total drift velocity, $\nu_+ + \nu_-$:

$$ \nu_+ + \nu_- = \cfrac{d}{n_+ z_+ e} I_\mathrm{total} = \cfrac{d}{n_- z_- e} I_\mathrm{total} $$

We can therefore derive drift velocity for an ion, $\nu_i$, from its transport number, $t_i$:

$$ \nu_+ = \cfrac{ t_+ d }{ n_+ z_+ e } I_\mathrm{total} $$

$$ \nu_- = \cfrac{ t_- d }{ n_- z_- e } I_\mathrm{total} $$

Note that it is necessary to know current $I_\mathrm{total}$ and cell size $d$ since precisely these values introduce the time and distance dependence necessary to recover drift velocity $\nu_i$ from transport number $t_i$.

Assuming the system can be approximated by parallel plates, i.e. assuming electric field is constant, ion mobility, $\mu_i$ can be derived from drift velocity through voltage, $V$, and distance between plates, $d$:

$$ \mu_i = \cfrac{\nu_i}{E} = \cfrac{\nu_i d}{V} $$

So the expressions for cation and anion mobility would be

$$ \mu_+ = \cfrac{t_+ d^2}{n_+ z_+ e V} I_\mathrm{total} $$

$$ \mu_- = \cfrac{t_- d^2}{n_- z_- e V} I_\mathrm{total} $$

For a more complex system with $N$ ions contributing to current,

$$ I_\mathrm{total} = \cfrac{e \sum\limits^{N} n_i z_i \nu_i }{d} $$

$$ t_i = \cfrac{ n_i z_i \nu_i }{\sum^{N} n_j z_j \nu_j } $$

and drift velocity and mobility can be computed as

$$ \nu_i = \cfrac{ t_i d }{ n_i z_i e} I_\mathrm{total} $$

$$ \mu_i = \cfrac{ t_i d^2 }{ n_i z_i e V} I_\mathrm{total} $$


Changes in UV-Vis spectra in host-guest complexes I'm doing some background reading before conducting an experiment, but am getting a little confused.  What serves as evidence for the formation of a host-guest complex when conducting analysis through UV-visible absorption?

From the sources that I have read and have personally observed in the lab, formation of host-guest complexes can push equilibrium to favour a specific species, like an acidic form or a basic form of the guest molecule due to stabilizing interactions within the host (think cyclodextrins + phenolphthalein).  This is reflected in absorption spectra as an increase/decrease in absorbance of the peaks to reflect the increasing/decreasing proportion of acidic/basic species of the guest molecule.

Another way of telling if the host-guest complex is formed is if the there is an increase in the molar absorption coefficient or a shift in the absorption peaks of the guest molecule.  Another conflicting source tells me that when a guest molecule is complexed, absorption of the guest molecule should decrease.  Can anybody clarify which of these changes serves as evidence for formation of a host-guest complex?

Can anybody clarify which serves as evidence for formation of a host-guest complex?

There's no single, general answer to this.

As you aptly described in the body of your post, there is a wide variety of changes to the UV-Vis absorption of a host-guest system on binding of the host and guest.  The changes to the recorded spectrum could fall into any of these categories, depending on the particular structural features of host and guest, and the exact manner that they bind to one another.

AFAIK, the only way to know how the spectrum changes when host and guest bind, is to perform careful experimentation and observe the changes that occur.


Software for predicting chemical reactions I wondered if it was possible to predict the product(s) of a reaction (or even the equilibrium) based on the reactants and the temperature/pressure.

I found an answer to a similar question here:

  No, there is no such thing as chemical reaction are not that easily
  predictable. You can get result right in some cases, but it is
  impossible to count all factors.

What (significant) factors also influence a reaction? Is it the arrangement or state of the chemicals? Is it too related to other physical properties to ignore them?

Please name the most important ones.

For predicting chemical reaction you should use classical mechanics to take in account the simple interactions between the molecules  and quantum mechanical for all the aspects concerning the molecule itself and complex interactions.
Knowing the principal parameter such as temperature and pressure that could describe the system with the classical mechanics (kinetic energy etc. etc.) is not enough for predicting the reaction with ab initio methods.
Adding the quantum mechanical contribute is mandatory and quite hard to do. A amateur chemist do not have neither the skills nor the computation power require to make such calculations. For the moment you can't do that with any software but the last Nobel was won by three theoretical chemists who are dealing with this topic. Have a look here for more information.

In your case where the reactions are well known you can use a chemical reactions database like the NIST one.
This is a simple query:

And this is the...


The pyridine/benzene stability 'paradox'? So, apparently pyridine is more stable than benzene. My lecture notes are saying that pyridine will undergo electrophilic aromatic substitution much more slowly than benzene will, which got me thinking - slower means the compound is more stable (higher activation energy). But isn't benzene more aromatic than pyridine?

I checked the heats of hydrogenation and it checks out - pyridine is indeed more stable than benzene. I'm a little bit lost though, because I don't get where this unexpected stability is coming from. Surely benzene is an ultra-stable compound because of the superb resonance it has (higher resonance energy than pyridine), electron density is (AFAIK) equally distributed around the ring, whereas this isn't the case in pyridine.

Also, the reason I call this a 'paradox' (in my head, at least) is because I've been learning about how benzene is a really stubborn compound when it comes to getting it to react - it doesn't undergo normal alkene addition reactions and all that - yet pyridine is quite a bit more stable than benzene and it has a million different weird reactions going on (acid/base, nucleophilic, EAS, N-oxide formation, metal complexes, etc. etc.)

Apparently pyridine is more stable, yet more reactive due to the lone pair electrons on the nitrogen (which allow lots of crazy stuff). In general, is it true that more stable = less reactive? That's the basic assumption I've been using until now - based on thermodynamics (?) - but I've seen a couple of things that make me think it's not as simple as that. Sorry for the rambling, just blurting my questions out!

When you think about stability, think of it as a relative value, meaning in comparison to atoms of other elements. Therefore, pyridine is more stable than the benzene but less stable than other elements, in general.

You are correct when you say stability is inversely proportional to reactivity. But because stability is relative, it is very difficult to comment on the relative reactivity.

EDIT

Heat of hydrogenation of an alkene in which the double bonds are conjugated is not additive because such an alkene is resonance stabilized, and, therefore, its Lewis structure, which the calculation would be based on, is a poor representation of the molecule.

I have found 2 aspects :

Pyridine is less reactive, than benzene toward electrophilic aromatic
substitution, because nitrogen is more electronegative, than carbon and acts like an
electron withdrawing group. As a result, the meta hydrogen is substituted.
Pyridine is more reactive than benzene because the presence of nitrogen
enables pyridine to react with nucleophiles. Nitrogen is an electron withdrawing
substituents enables the aromatic ring to participate in such reactions and the
substitution happens in the ortho and meta positions.


What does the term 'by mass' mean with respect to different contexts? I've seen the term 'by mass' being used in various sorts of contexts. Some, while explaining the law of definite proportions say, the 'same proportions by mass', some say 'constant chemical composition by mass'.
What does the term 'by mass' mean? Does it mean in a proportion/ratio, we have mass as both, the denominator and numerator?
When I say 'percentage by mass', what would that mean?

In $\ce{H2O}$, the proportion of $\ce{H}$ counted by atoms is $\ce{2/3 = 66.7}$ %. In the same $\ce{H2O}$, the mass of $\ce{2 H}$ is $2$ u. As the mass of $\ce{1 O}$ is $16$ u, the proportion of H by mass in $\ce{H2O}$ is $2/18 = 11.1$%. This percentage $11.1$% is rather different from $66.7$%


Products of Lead(II) nitrate decomposition It is known, that decomposition of Lead(II) nitrate is one of the ways of generating $\ce{NO2}$ for lab use.

I recently did this in order to acquire $\ce{NO2}$ in liquid form (ambient temperature was ~0 °C), and got quite strange results. First, liquid $\ce{NO2}$ had 2 visible layers, usual brown one at the top and transparent at the bottom:

Also, during condensation liquid initially had green color (I know that Graham condenser is not safe in this configuration - I wouldn't see this color change in conventional condenser setup).

Can anyone clarify why I got such variety of products instead of pure $\ce{NO2}$?

Check if the green condensate is $\ce{N2O3}$. It has been reported to be blue-green.
In the presence of water vapor and oxygen, the nitrogen dioxide that you are trying to obtain is in equilibrium with nitric acid and nitrogen trioxide, as explained here.
As reported in the Journal of Industrial and Engineering Chemistry, volume 12, number 6, page 531-538 (1920), condensation of gaseous nitrogen dioxide in a spiral condensor at -5 °C in the presence of oxygen and water vapor resulted in liquid with distinct upper and lower layers, the upper layer being a concentrated nitric acid layer (essentially no water) and the lower layer being a diluted nitric acid layer (27 %) water.
So the reason you are getting side products is likely that water vapor and oxygen are also present and causing equilibrium reactions that yield nitric acid, nitrous acid and nitrogen trioxide.
Reference

CCXIV.—The dissociation of gaseous nitrogen trioxide, Bernard Mouat Jones,  J. Chem. Soc., Trans., 1914, 105, 2310-2322, DOI: 10.1039/CT9140502310


How to name an unsaturated spiro compound with many substituents? I've been trying to name this compound, but I just can't, I've gotten a brilliant answer already, it's(8E,11Z,15Z)-4-amino-11-carbamoyl-3-chloro-2-(chlorocarbonyl)-9-cyano-5-formyl-4-hydroxy-13-methoxy-12-(methoxycarbonyl)-10-oxospiro[6.9]hexadeca-2,8,11,15-tetraene-8,14-dicarboxylic acid. I just want to know if there's any contrary opinion from the answer.

The parent structure is a spiro compound, which has a fixed numbering starting in the smaller ring.
The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-24.2.1 Monospiro alicyclic ring systems
Mono spiro parent hydrides consisting of two saturated cycloalkane rings are named by placing the nondetachable prefix ‘spiro’ before the name of the unbranched acyclic hydrocarbon with the same total number of skeletal atoms. The number of skeletal atoms linked to the spiro atom in each ring is indicated by arabic numbers separated by a full stop, cited in ascending order and enclosed in square brackets; this descriptor (called in these recommendations the ‘von Baeyer spiro descriptor’) is placed between the spiro prefix and the name of the acyclic alkane. Numbering starts in the smaller ring, if one is smaller, at a ring atom next to the spiro atom and proceeds first around that ring, then through the spiro atom and around the second ring.

In this case, the name of the parent spiro hydrocarbon ring structure is spiro[6.9]hexadecane.
The fixed numbering of the spiro hydrocarbon ring is not modified by the introduction of the double bonds and functional groups. However, low locants must be attributed if there is a choice.

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(a) fixed numbering in chains, rings, or ring systems, i.e., when the numbering of a system is fixed, (…), this numbering must be used, both in PINs and in general nomenclature;
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(e) saturation/unsaturation:
  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

Thus, low locants are first assigned to the principal characteristic group, which is carboxylic acid. This leads to the name spiro[6.9]hexadecane-8,14-dicarboxylic acid.
Next, locants are assigned to the double bonds according to Rule (e), which yields (8E,11Z,15Z)-spiro[6.9]hexadeca-2,8,11,15-tetraene-8,14-dicarboxylic acid.
Finally, locants are assigned to all other substituents. Therefore, the correct name for the complete compound is (8E,11Z,15Z)-4-amino-11-carbamoyl-3-chloro-2-(chlorocarbonyl)-9-cyano-5-formyl-4-hydroxy-13-methoxy-12-(methoxycarbonyl)-10-oxospiro[6.9]hexadeca-2,8,11,15-tetraene-8,14-dicarboxylic acid.