text:
Born-Oppenheimer Approximation reloaded
I'm an undergraduate Chemistry student and I'm currently writing my Final Dissertation.  I've written a form of the Born-Oppenheimer approximation with the only initial assumption of parametric dependence on nuclear coordinates. Let me know  what do you think about it. Thanks in advance for any help you are able to provide.
For a given molecular system, within the non-relativistic approximation, the molecular Schrodinger equation can be written as $$ \hat{\mathcal{H}}_{\mathrm{tot}}\Psi\left(\mathbf{R},\,\mathbf{r}\right)=\mathcal{E}_{\mathrm{tot}}\Psi\left(\mathbf{R},\,\mathbf{r}\right)\quad\left(1\right) $$ where $$\hat{\mathcal{H}}_{\mathrm{tot}}=\hat{\mathcal{T}}_{\mathrm{e}}\left(\mathbf{r}\right)+\hat{\mathcal{T}}_{\mathrm{N}}\left(\mathbf{R}\right)+\hat{\mathcal{V}}_{\mathrm{ee}}\left(\mathbf{r}\right)+\hat{\mathcal{V}}_{\mathrm{NN}}\left(\mathbf{R}\right)+\hat{\mathcal{V}}_{\mathrm{eN}}\left(\mathbf{R},\mathbf{r}\right)=\hat{\mathcal{T}}_{\mathrm{N}}+\hat{\mathcal{H}}_{\mathrm{e}}$$

Without loss of generality, if there exists a complete basis $ \left\{ \Psi_{k}\left(\mathbf{R},\,\mathbf{r}\right)\right\}
$ , the molecular wave function $ \Psi\left(\mathbf{R},\,\mathbf{r}\right) $ can be expanded as $$ \Psi\left(\mathbf{R},\,\mathbf{r}\right)={\displaystyle \sum_{k}}c_{k}\Psi_{k}\left(\mathbf{R},\,\mathbf{r}\right) $$
The nuclei in a molecule are heavy enough that, to a good approximation, they behave as classical particles. Thus, it is completely meaningful to talk about nuclear configurations, or in other words to assume a parametric dependence from nuclear coordinates.
$$
\hat{\mathcal{H}}_{\mathrm{tot}}\Psi\left(\mathrm{R},\,\mathbf{r}\right)=\mathcal{E}_{\mathrm{tot}}\left(\mathrm{R}\right)\Psi\left(\mathrm{R},\,\mathbf{r}\right)
$$
For a given complete basis set $ \left\{\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)\right\} $, the molecular wavefunction is determined only by the coefficients $ c_{k} $. Thus, the coefficients $ c_{k} $
depends on nuclear configuration:
$$
\Psi\left(\mathrm{R},\,\mathbf{r}\right)={\displaystyle \sum_{k}}c_{k}\left(\mathrm{R}\right)\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)\quad\left(2\right)
$$
Now by substituting eq.(1) into eq.(2)

$$
\begin{array}{c}
{\displaystyle \sum_{k}}\left(\hat{\mathcal{T}}_{\mathrm{N}}+\hat{\mathcal{H}}_{\mathrm{e}}\right)c_{k}\left(\mathrm{R}\right)\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)=\mathcal{E}_{\mathrm{tot}}{\displaystyle \sum_{k}}c_{k}\left(\mathrm{R}\right)\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)\\
{\displaystyle \sum_{k}}\hat{\mathcal{T}}_{\mathrm{N}}c_{k}\left(\mathrm{R}\right)\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)+{\displaystyle \sum_{k}}c_{k}\left(\mathrm{R}\right)\hat{\mathcal{H}}_{\mathrm{e}}\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)=\mathcal{E}_{\mathrm{tot}}{\displaystyle \sum_{k}}c_{k}\left(\mathrm{R}\right)\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)
\end{array} $$

First term of eq.(3)
$$
\begin{array}{c}
{\displaystyle \sum_{k}}\hat{\mathcal{T}}_{\mathrm{N}}c_{k}\left(\mathrm{R}\right)\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)=-{\displaystyle \sum_{a,k}}\dfrac{\hbar^{2}}{2M_{a}}\left\{ \nabla_{\mathrm{N}_{a}}\cdot\nabla_{\mathrm{N}_{a}}\left[c_{k}\left(\mathrm{R}\right)\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)\right]\right\} =\\
=-{\displaystyle \sum_{a,k}}\dfrac{\hbar^{2}}{2M_{a}}\left\{ \nabla_{\mathrm{N}_{a}}\cdot\left[c_{k}\left(\mathrm{R}\right)\nabla_{\mathrm{N}_{a}}\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)+\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)\nabla_{\mathrm{N}_{a}}c_{k}\left(\mathrm{R}\right)\right]\right\} =\\
=-{\displaystyle \sum_{a,k}}\dfrac{\hbar^{2}}{2M_{a}}\left\{ c_{k}\left(\mathrm{R}\right)\nabla_{\mathrm{N}_{a}}^{2}\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)+\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)\nabla_{\mathrm{N}_{a}}^{2}c_{k}\left(\mathrm{R}\right)+2\left[\nabla_{\mathrm{N}_{a}}c_{k}\left(\mathrm{R}\right)\nabla_{\mathrm{N}_{a}}\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)\right]\right\}
\end{array} $$
Multiplying from the left by $ \Psi_{j}^{\star}\left(\mathrm{R},\,\mathbf{r}\right) $ and integrating over electron coordinates, remembering that $ \left\langle \Psi_{j}\left(\mathrm{R},\,\mathbf{r}\right)|\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)\right\rangle =\delta_{jk} $
$$
\begin{array}{c}
-{\displaystyle \sum_{a}}\dfrac{\hbar^{2}}{2M_{a}}\nabla_{\mathrm{N}_{a}}^{2}c_{k}\left(\mathrm{R}\right)-{\displaystyle \sum_{a,k}}\dfrac{\hbar^{2}}{2M_{a}}\left\{ c_{k}\left(\mathrm{R}\right)\left\langle \Psi_{j}\left(\mathrm{R},\,\mathbf{r}\right)\left|\nabla_{\mathrm{N}_{a}}^{2}\right|\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)\right\rangle +2\nabla_{\mathrm{N}_{a}}c_{k}\left(\mathrm{R}\right)\left\langle \Psi_{j}\left(\mathrm{R},\,\mathbf{r}\right)\left|\nabla_{\mathrm{N}_{a}}\right|\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)\right\rangle \right\} =\\
=\hat{\mathcal{T}}_{\mathrm{N}}c_{k}\left(\mathrm{R}\right)-{\displaystyle \sum_{a,k}}\dfrac{\hbar^{2}}{2M_{a}}\left\{ c_{k}\left(\mathrm{R}\right)\left\langle \Psi_{j}\left(\mathrm{R},\,\mathbf{r}\right)\left|\nabla_{\mathrm{N}_{a}}^{2}\right|\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)\right\rangle +2\nabla_{\mathrm{N}_{a}}c_{k}\left(\mathrm{R}\right)\left\langle \Psi_{j}\left(\mathrm{R},\,\mathbf{r}\right)\left|\nabla_{\mathrm{N}_{a}}\right|\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)\right\rangle \right\}
\end{array}
$$
Second term of eq.(3)  Multiplying by $\Psi_{j}\left(\mathrm{R},\,\mathbf{r}\right)$ and integrating over electron coordinates
$$
{\displaystyle \sum_{k}}c_{k}\left(\mathrm{R}\right)\left\langle \Psi_{j}\left(\mathrm{R},\,\mathbf{r}\right)\left|\hat{\mathcal{H}}_{\mathrm{e}}\right|\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)\right\rangle =\mathcal{E}_{k}\left(\mathrm{R}\right)c_{k}\left(\mathrm{R}\right)
$$
Final form of the molecular electronic function
$$
\left(\hat{\mathcal{T}}_{\mathrm{N}}+\mathcal{E}_{k}\left(\mathrm{R}\right)\right)c_{k}\left(\mathrm{R}\right)-{\displaystyle \sum_{a,k}}\dfrac{\hbar^{2}}{2M_{a}}\left\{ C_{2a,jk}c_{k}\left(\mathrm{R}\right)+2C_{1a,jk}\nabla_{\mathrm{N}_{a}}c_{k}\left(\mathrm{R}\right)\right\} =\mathcal{E}_{\mathrm{tot}}c_{k}\left(\mathrm{R}\right)
$$  where
$$
\begin{array}{c}
C_{1a,jk}=\left\langle \Psi_{j}\left(\mathrm{R},\,\mathbf{r}\right)\left|\nabla_{\mathrm{N}_{a}}^{2}\right|\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)\right\rangle \\
C_{2a,jk}=\left\langle \Psi_{j}\left(\mathrm{R},\,\mathbf{r}\right)\left|\nabla_{\mathrm{N}_{a}}\right|\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)\right\rangle
\end{array}
$$
By neglecting the coupling terms
$$
\left(\hat{\mathcal{T}}_{\mathrm{N}}+\mathcal{E}_{k}\left(\mathrm{R}\right)\right)c_{k}\left(\mathrm{R}\right)=\mathcal{E}_{\mathrm{tot}}c_{k}\left(\mathrm{R}\right)
$$

Final conclusion
Within this picture, the $c_{k}\left(\mathrm{R}\right)$
  behave, in a natural way, as a nuclear eigenfunction.
$$
\begin{array}{c}
\left(\hat{\mathcal{T}}_{\mathrm{N}}+\mathcal{E}_{k}\left(\mathrm{R}\right)\right)\Phi_{k}\left(\mathrm{R}\right)=\mathcal{E}_{\mathrm{tot}}\Phi_{k}\left(\mathrm{R}\right)\\
\hat{\mathcal{H}}_{\mathrm{e}}\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)=\mathcal{E}_{k}\left(\mathrm{R}\right)\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)
\end{array}
$$
Questions

While computing the Second term of eq.(3), I assumed $ \Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)$
as an eigenfunction of the electronic Hamiltonian. How can this assumption be demonstrated?
Do you think that the Assumption 3 is correct?

Hi Im just wondering how you got your final two equations? I understand your proof up to the point where you neglect the coupling terms but then how do you generate two final results? Didn't you already use the electronic Hamiltonian when you took its eigenvalue $\epsilon _k$ into the nuclear expression you quote?

@RedPen: The second equation follow from the fact that, as pointed out by Wildcat, we choose to expand, without any approximation, the molecular wave function as
$$ {\displaystyle \sum_{ij}}c_{ij}\Psi_{\mathrm{N},i}\left(\mathbf{R}\right)\Psi_{\mathrm{el},j}\left(\mathbf{r}\right) $$
where the electronic wavefunction is by definition an eigenfunction of the electronic Hamiltonian

Coincidentally, I'm also currently writing my dissertation and I do also have section discussing Born-Oppenheimer approximation. :) My description is quite similar to yours, so I do not see any critical issues.
With respect to your question, we do not assume that $\Psi_k(R,r)$ is an an eigenfunction of the electronic Hamiltonian, we know that for sure. If you look at equation (2) then essentially what is going on here is the following. Any well-behaved function of two independent variables $\Psi(R,r)$ can always be expanded over the complete set of functions $\{ g_{i}(R) \}$ and $\{ h_{j}(r) \}$ in the following way
$$
\Psi(R,r) = \sum\limits_{i} \sum\limits_{j} c_{ij} g_{i}(R) h_{j}(r) \, ,
$$
or by defining
$$
c_{j}(R) = \sum\limits_{i} c_{ij} g_{i}(R) \, ,
$$
as
$$
\Psi(R,r) = \sum\limits_{j} c_{j}(R) h_{j}(r) \, .
$$
Now as a complete set $\{ h_{j}(r) \}$ of functions of electronic coordinates we can choose the set of electronic wave functions $\Psi_k(R,r)$ which are by definition eigenfunctions of electronic Hamiltonian. The fact that these function carry parametric dependence on nuclear coordinates does not matter here.
And yes, this expansion is correct. For more on that story check this question at Physics.SE I asked about a year ago.

Thank you so much for your answer! Your question on Physics.SE was enlightening. Another question: According to what you have said, it is not strictly necessary to assume that electronic WF depends on the nuclear coordinates. Am I wrong?

@DavideLaVardera The parametric dependence of electronic wave function on nuclear coordinates is not an assumption but rather a consequence of an assumption that we could treat the motion of a relatively fast moving subsystem (electrons) as if it depends only on the static configuration of a relatively slow moving one (neclei). Once we have made this assumption we obtain the electronic Schrödinger equation and we know that the electronic Hamiltonian is different for different nuclear configurations due to $V_{nn}$ and $V_{en}$ terms.

@DavideLaVardera Consequently we know a priori that solutions of the electronic Schrödinger equation will be different for different nuclear configuration. We express this ides in other words by saying that the electronic Schrödinger equation, the electronic Hamiltonian, the electronic wave functions and the electronic energies are parametrized by nuclear coordinates.

@DavideLaVardera What we have to assume is that this parametric dependence of the electronic wave function on nuclear coordinates is continuous and differentiable, and that both first and second derivatives of it with respect to nuclear coordinates are in general non-zero.

Now I fully understand! Again, thank you so much!


text:
How can Iodine bond with 5 Fluorines in Iodine Pentafluoride?

According to this lewis diagram, Iodine has 12 valence electrons? How is the possible? Only thing I can think of is that I'm mistaken in counting the covalent bonds as two valence electrons each. Do they only count as one in this scenario?

This may help http://chemistry.stackexchange.com/questions/13949/how-can-sulphur-and-phosphorus-expand-their-octet

Explanations of bonding and hybridization in non-metals that involve d-orbitals have generally fallen out of favor for two reasons:

Usually d-orbitals are considerably higher in energy than the s- and p-orbitals they would interact with leading to poor mixing, and
d-orbitals are very diffuse leading to poor overlap with other orbitals and any resulting bonds would be very weak

A reasonable hybridization alternative involves what is termed hypercoordinated bonding; where 3 center, 4 electron bonds are involved.  Such bonding involves p-orbitals with overlap geometry and molecular orbitals as pictured in the following diagram.

In the case of $\ce{IF5}$, the molecule has the following geometry.

Using the hypercoordinated bonding approach, one would say that the 4 identical "equatorial" fluorines (they're not quite equatorial, we'll return to that later) and 2 p-orbitals on iodine would participate in this type of bonding.  Each of the two iodine p-orbitals contains two electrons from iodine and bonds to two fluorines which each contribute one electron.  We wind up with two linear hypercoordinate bonds at right angles to each other that lie in a plane.  Each of these bonds is comprised of 3 atoms ($\ce{F-I-F}$) and contains 4 electrons as described in the MO diagram above.  The only orbitals remaining on iodine are an s- and a p-orbital.  These two orbitals mix in traditional fashion (like the carbon in acetylene) to produce two $\ce{sp}$ orbitals.  these two orbitals are arranged in a linear fashion (again, like acetylene) and are perpendicular to the plane containing the two 3 atom - 4 electron hypercoordinate bonds.  One of these $\ce{sp}$ orbitals contains an iodine lone pair of electrons and the other forms a bond to fluorine using one iodine electron and one fluorine electron.
Because of the asymmetry introduced by having a lone pair on one side of the molecule and a fluorine on the other side, our 4 "equatorial" $\ce{I-F}$ bonds distort from planarity tilting away from the electrostatically repulsive lone pair as illustrated in our figure (above) showing the geometry of $\ce{IF5}$.
In this bonding scenario the central iodine is sharing 4 electrons in each of the two hypercoordiante bonds (total=8) and two electrons in the apical $\ce{I-F}$ bond; plus there is a lone pair on the iodine.  Therefor the iodine is neutral (7-[(8+2)/2]-2 = 0), but is it surrounded by an octet of electrons?  The answer is, yes.  Look again at the MO diagram for the hypercoordinate 3 center - 4 electron bond.  Notice that the nonbonding orbital ($\Psi_2$) contains 2 electrons, but these electrons are centered entirely on the fluorine atoms, not the central iodine.  Therefore, around iodine we have 2 (lone pair) plus 2 (apical $\ce{C-F}$ $\ce{sp}$ hybridized bond) plus 2 times the two $\Psi_1$ electrons in our hypercoordinate bonds, yielding a total of 8 electrons around the iodine.

Ah, iodine pentafluoride. This is a lovely example of a molecule that doesn't follow simple Lewis theory. You will find that this is often the case for heavier elements, where electrons from multiple subshells can contribute to the bonding.
In the case of iodine, the 5p and 5s subshells and the 4d subshells are actually quite close in energy to each other, and are ALL valence subshells. So, iodine can use all of these electrons in chemical bonding: 2 in the 5s, 5 in the 5p and up to 10 in the 4d, although it only needs to use 4 of its 4d electrons to do the bonding in this molecule. Hence it can have more than 8 electrons involved in its bonding.
Technically, the molecule has sp$^3$d$^2$ hybridized bonding, but in simpler terms, iodine (roughly!) forms one normal covalent bond to a F atom, four dative covalent bonds to 4 F atoms, leaving one lone pair remaining.
As you might therefore expect from valence shell electron repulsion theory (VSEPR), there are 4 equivalent F atoms lying in a plane, with one vertically above the plane with a slightly shorter bond length.

How does iodine form dative covalent bonds to fluorine atoms when fluorine atoms already have 7 electrons to begin with? Clearly, by your description, the fluorine atoms around I in this molecule would end up with 9 electrons in its valence shell...

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How will liquid nitrogen affect a steel instrument's ability to function?
I want to retrieve cryotubes from the bottom of a liquid nitrogen freezer without taking all the freezer racks out and draining the liquid nitrogen. I think I can accomplish this using a tool like this but I am not sure how well the tool will function at temperatures as low as -160 Celsius. Also, this particular tool is probably only about half as long as it needs to be.
Assuming I can find a longer version of such a tool, how can I determine whether or not I need to be concerned about using to retrieve cryotubes from a liquid nitrogen freezer? My main worry is that parts of the tool may become brittle and break or that the claw's ability grab may somehow become compromised. I am not concerned about the metal becoming excessively cold because I will be wearing gloves intended for such work.
Is there a chart or other resource that I can use to get a sense of metal's shear strength at cold temperatures?

Metals become brittle as low temperature (and that's why you cool your Parr bomb in a dry ice bath, never in liquid nitrogen when you condense gases in).  But the worst thing that can happen is that a claw might break off, which shouldn't cause any damage to equipment.  Why not try in a small dewar first.  Liquid nitrogen is cheap.
A good alternative to racks of storage boxes are storage canes, such as these.

Thats a good idea to experiment in a small dewar first.
I assume steel is a good metal for this kind of tool to be made of because its durable but I have no real understanding of how much variety there is to how brittle metals become in cold temperature or how quickly metals become brittle.

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text:
Why does in the hydrolysis of acetylsalicylic acid with NaOH the solution has to be simmered gently?
Why does in the hydrolysis of acetylsalicylic acid with NaOH the solution has to be simmered gently?
What could happen if I would strongly boil it.
Would the OH- react into the carboxyl groups to form a diol?

What could happen if I would strongly boil it

Mechanistically, nothing different will occur.  The same reaction that occurs at lower temperature will still occur at higher temperature.  The only difference will be the rate of the reaction; the reaction will go faster as you increase the reaction temperature.  In a classroom setting the procedure might involve simmering rather than boiling because

The reaction is usually followed by making uv-vis absorbance measurements every 15, or so, minutes.  Depending upon the reaction temperature and the concentration of base, the reaction might proceed too fast at boiling to allow for convenient measurements (e.g. the reaction might be complete in 10 minutes).
Safety reasons, if something goes wrong (the flask cracks) a warm basic solution won't burn the student as bad as a boiling basic solution would.


text:
How is the energy of reactants in a chemical reaction found ?
A pretty rudimentary question, but I'm reviewing the concept of energy diagrams and activation energy and I'm curious to know how chemists find the energy of reactants and of a transition state in a laboratory.
What kinds of instruments are used ? As well, aren't transition states only something we can imagine ? So how is it that chemistry are able to find their energy in a lab ?

Energy itself is usually not measured, but rather change in energy ($\Delta E$) is the quantity of practical importance. One may be interested in change in free energy as well, $\Delta G$, changes in enthaly $\Delta H$, and even change in entropy $\Delta S$.
A common way to evaluate the increased energy of transition state ($\Delta E^{\ddagger}$) above the reactant is via the Arhennius Plot.  Here, one evaluates the rates of reaction ($k$) at a set of different temperatures ($T$ in Kelvin).  It is fairly easy to show that a plot of the $\ln(k)$ vs. $1/T$ gives a line whose slope is related to $\Delta H^{\ddagger}$, and where the intercept is related to $\Delta S^{\ddagger}$. (The ideal gas constant is in here too, but it's easy to accomodate.)
(There is an analogous plot for energy change for a reaction, which is called the van't Hoff plot, but it generally requires both reactant and product to be in equilibrium such that the concentrations of both are observable.)
Especially the enthalpy of reaction can be found by methods such as bomb calorimetry.  This measures the heat that is liberated upon reaction.


text:
What are the dangers of handling 0.12N sulfuric acid?
I'm handling 0.12N of sulfuric acid for a science experiment that I am carrying out for a school project. I hear about all these terrible effects if it's ingested, inhaled, or gets in contact in skin. But how dangerous is it?
My safety equipment that I'm using currently are a pair of polyvinyl gloves, goggles, long sleeve shirt, apron, jeans, and closed toe shoes. However, I do not have a respirator and wondering if I should get one.
Would it be safe to handle this without a respirator? I will not be spraying the sulfuric acid directly in the air. However, I will be adding approximately 0.5mL of the 0.12N sulfuric acid into 500mL of water, putting it in a spray bottle, and be watering some plants with it. Will it still be dangerous even spraying a low concentration in the air?
Would the 0.12N concentration still be enough to cause chemical burns on my skin? To what extent and severity? Also, just to confirm, it is safe to add drops of the acid to water, but not vice versa? I would really appreciate the advice that you can provide me.

You don't need a respirator since sulfuric acid you are using is 300 times less concentrated than the concentrated sulfuric acid which is much more hazardous. You shouldn't be harmed if a little got on your skin if you wash it off quickly. You may want to have a solution of baking soda (5g/100mL) to put on any spins that get on your cloths. The solution that you will be making (pH of 4, this is the acidity of pickles) is even less hazardous and small amounts could probably be inhaled without any problems (but of course, I wouldn't want to do so).

This is not a question about personal medical advice but lab safety practice and as such I consider it perfectly on topic for this site.

This is a good example of the principle of dose response: If something is harmful, then a bigger dose is more harmful and a smaller dose is less harmful. The alternative view (not borne out by the science) is that if something is harmful at all, it must be harmful in any quantity.

0.12 Normal = 0.06 Molar.
For comparision, concentrated sulphuric acid is 18 Molar.
So the 0.12 N sulphuric acid is over 99% water.

Would it be safe to handle this without a respirator?

Yes, the vapor pressure of sulphuric acid is low, and it has already been diluted with considerable water, so there is no need for a respirator.

My safety equipment that I'm using currently are a pair of polyvinyl gloves, goggles, long sleeve shirt, apron, jeans, and closed toe shoes.

Goggles are the most important safety equipment.  An eye wash station and shower are also standard lab safety equipment.
The gloves are good because liquids will not pass through them.  Ordinary clothing isn't really helpful because if you did spill something on that part of your body, the substance would absorb into the cloth and you would just be delayed in washing that part of your body by having to take off the clothes.

Would the 0.12N concentration still be enough to cause chemical burns on my skin? To what extent and severity?

Pronged contact may cause some damage, but nothing comparable to the concentrated acid.  If you spill some on yourself, stay calm, and wash with soap and water. If you spill on your clothes, take the wet clothes off first, then wash with soap and water.

Also, just to confirm, it is safe to add drops of the acid to water, but not vice versa?

This is always a good rule to keep in mind.  0.12N is dilute enough that little heat will be released from 0.5mL, but just follow the rule.

However, I will be adding approximately 0.5mL of the 0.12N sulfuric acid into 500mL of water, putting it in a spray bottle, and be watering some plants with it. Will it still be dangerous even spraying a low concentration in the air?

Now you are diluting another 1000 fold.  The solution will be around pH 4, similar to some rain water.  I would still want you to wear goggles and not spray in the direction of anyone.

That dilute of $\ce{H2SO4}$ won't remotely burn your skin. You can wear goggles just in case you spray it in your face for some reason.

Q1) For the first question, we can calculate the molarity of the sulphuric acid.(Molarity of acid × volume of acid = molarity of the dilute acid × the volume of water)
0.12N × 0.5ml = molarity of dilute acid × 500ml
Molarity of dilute acid = 0.12N × 0.5ml / 500ml
                        = 0.00012N
Thus, you can see that the molarity of the acid is quite low, meaning the ph value is nearly at 4 or 5. I can say that is not necessary to wear the respirator but remember to prepare something base or alkaline for cases like the acid start to burn the gloves or your skin.
Q2) Yes. You must add the drop of acid into the water and not vice versa. It is because if you instantly add water to acid, the solution will form an exothermic reaction and release a lot of heat, causing the acid to boil and splatter itself everywhere.
However, if you add drops of acid to water. The heat produce is not enough to cause the solution to boil. Hence, it is must more safer.
I hope that it will help you on experiment.

In the question, not the molarity was given, but the "normality", i.e. equivalence concentration


text:
Can protons be swapped in a chemical reaction?
Can the protons of two elements be swapped in a chemical reaction?

Do you mean "two elements" or "two compounds?"

Roughly speaking, no.   This would amount to nuclear fission/fusion, and is generally not considered "chemistry" in any colloquial sense of the term, but rather physics.
In general, chemical reactions consider nuclei to be tightly bound by the strong force, and immutable over the course of the reaction.
Disclaimer: there is an ill-defined border between physics and chemistry, and so one could argue that I am wrong about this.  Indeed, alchemists had/have been trying to e.g. convert lead to gold for centuries, and this is related to the reaction that you propose.

umm... proton = hydrogen cation. You don't need fusion to exchange protons.

@JanDvorak Yes, and do you normally see hydrogen cations diving into other nuclei?

The question isn't speaking about nuclei

As I've mentioned a few times, there is actually "nuclear chemistry," but I agree that outside of bombarding nuclei with protons or alpha particles, this isn't a typical reaction.

@GeoffHutchison Thanks for pointing this out.  This was the origin of my rather guarded language.  long time no talk.  hope all is well :-)


text:
Boundary lines in phase diagrams and the lever rule
My first question is: What exactly happens on a point situated on a line? Or on a point such as E in this figure?

Does point E mean that the $\delta , \gamma + \delta , \delta + \epsilon , \gamma + \epsilon $ phases all coexist here? If so, then why would it be considered a eutectoid invariant point where three phases coexist if in reality there are more than three phases coexisting?
My second question is: How can I apply the tie line and lever rules on a point on a point situated on top of a line, if I want to find Liquid percentage Solid Percentage, and maybe concentrations of one component in the liquid phase and solid phase? Is it undefined on top of the line?

Also, regarding the nature of the lever rule, is it some sort of empirical rule without any real theoretical foundation? At first glance, I can't believe it's actually accurate. Why is it true?
Thanks and sorry if my questions are unclear...

I have tried to make the title a bit more specific, but please correct it if I've misinterpreted anything.

"γ+ϵ" is not a phase, it's a field consisting of two phases: "γ" and "ϵ". That said, the question that immediately follows is not clear. You might want to comment or add your question so we will be able to answer it properly.

Lots of questions wrapped in one, but they do all come down to understanding what is represented on a phase diagram.  Up front, I'll suggest Porter and Easterling's classic text on 'Phase Transformation in Metals and Alloys' - it has lots of good info for lovers of thermodynamics and phase transitions.
In a system like Cu-Zn that you have pictured above, there are a number of different phases observed.  You could imagine that at a given temperature you could construct a free energy curve for each phase as a function of composition. At each composition, only one will have the lowest free energy, and one might think you could identify the phase diagram that way. However, a straight line across the G vs. composition plot is special - it represents where the chemical potentials of both elements are constant.  So a line that is tangent to two free energy curves represents a range of compositions where the elements in the two phases have the same chemical potential.  This then is how you construct the phase diagram - find the minimum free energy across composition, then find the common tangents (if any). An example is shown here:
The top plot is the free energy curves for all the phases, the bottom plot is the phase boundaries determined by the common tangents.  The dashed vertical lines up top are where the common tangent points are, coming down to their position on the phase diagram.
Now the $\delta$, $\gamma$, $\epsilon$ are specific phases.  In regions with just one of them listed, that is the only phase stable at the temperature and concentration.  Regions such as $\delta + \gamma$ are regions where both are stable. There relative amount has to follow the lever rule, which is just a linear variation across the two-phase region from, say, all $\gamma$ right on the boundary of $\gamma$ with $\gamma + \delta$, to all $\delta$ on the other side.
The horizontal lines are there mostly to emphasize when a new phase appears or disappears.
So, at point E on you diagram above, there is one specific temperature and composition at which $\delta$, $\gamma$, and $\epsilon$ are in equilibrium.  Go up in temperature, and it wants to be all $\delta$; come down in temperature and you enter the two-phase $\gamma$-$\delta$ region with the relative abundance from the lever rule. At P, the horizontal line is to point out that $\epsilon$ disappears into the $\delta$-liquid region.
So, at the eutectoid E, there is one point where the three phases coexist (Gibbs phase rule). Moving away from that point moves into single or two-phase regions. Everywhere along the horizontal line through E is $\gamma$+$\epsilon$, except at exactly point E.
As for the lever rule, that follow directly from coming along the common tangent - the ratios of the two phases have to vary linearly or else you are deviating from the common tangent.

You might have answered my question, but I still need some time to digest it since this seems beyond me at the moment. Thanks though :)


text:
Mixing liquids when applying a voltage
I'm thinking in the following experiment:
A) Two (non mixable (like water and oil)) liquids are going to lie in a container, in the extremes of the container there are two electrodes, those electrodes will have a potential difference $\Phi$. Just as the figure $1.a)$.
I'm looking for  some interesting liquids that mix, change of color when a voltage difference is present.

B) The next one is really similar, now there are some solid particle present, analogously I look for a mixture that has different properties when a voltage difference is present.
I'm a physics student that want to study different properties of mixtures when applying different voltage differences, and maybe study the diffusion of the solid particles when they mix, if the diffusion coefficient is grater, etc.
Any help with the liquids that has this property is really  appreciated.
Cross posted Physics stack Exchange

I was just looking up some ancient papers on electrophoresis (1950s) before they had gels or even paper electrophoresis. They had to watch the fluid with special optics to catch when the refractive index changed as the material moved through the apparatus. When I'm back in lab I'll try to post the links, might be interesting for you.

Here is a paper about KCl diffusion in water using a device to image the change in refractive index. This paper discusses electrophoresis of proteins in solution.

I won't comment on B for the moment, but liquid-liquid interface electrochemistry is quite a vibrant field. There's a review of it here, if you're interested. The simplest interaction involves ions: if one applies a potential across the interface of sufficient strength and polarity, the energy barrier of transferring a given ion from one phase to the other can be overcome. If you do this with something like a solvatochromic dye, it will change colour as it moves from one phase to the other. (It would have to be something that doesn't partition into the second phase easily on its own, of course) One could also have some reagent in one phase that reacts with the ion being pushed across the interface that produces some coloured product. (iodide-starch, perhaps?) There are also voltage-sensitive dyes, though I don't have any experience with them.
One thing to note is that neither water nor oils and other immiscible solvents are generally very conductive. To keep the cell resistance down to a reasonable level, it will be necessary to dissolve electrolytes in both. Things like Tetrabutylammonium hexafluorophosphate are commonly used as electrolyte in non-aqueous electrochemistry.


text:
Does the amine used to generate the enolate affect the selectivity of an Evans aldol reaction?
In an Evans aldol reaction, the enolate is typically generated by addition of a boron Lewis acid (e.g. $\ce{(n-Bu)2BOTf}$) and an amine, which acts as a weak base. I'd like to know whether there's any difference between

Triisobutylamine, $\ce{(i-Bu)3N}$
N,N-Diisopropylethylamine (Hünig's base), $\ce{(i-Pr)2NEt}$
Triethylamine, $\ce{Et3N}$

in the selectivity of the enolate formation and subsequent aldol reaction. I'm thinking about the different $\mathrm pK_\mathrm a$ values (1>2>3) and the different steric demand (1>2>3). Probably the base with the highest $\mathrm pK_\mathrm a$ will be kinetically fastest. Also, the cis-enolate should be favoured as it avoids a destabilising syn-pentane interaction between the terminal methyl group and the auxiliary. However, do all of the bases lead to the same ratio of cis/trans-enolate formation (and hence diastereoselectivity in the subsequent aldol reaction)?

I'll have to dig up my notes on this, but I believe amides are always stereoselective for (Z)-enolate formation. It's likely due to an A1,3 strain. If you have access, check JACS 1982, 104, 1737.

Not 100% sure, but first thing that occurs to me is that the less bulkier the amine the more proportion of cis-enolate formed? If we are talking about a thermodynamic product, the one that should yield more Z proportion, is likely to be the one that gives rise to an enolate of the less bulky ammonium salt. For the same reason I also think that lithium amides will yield high Z isomer proportion, since OLi group is small and will cause lower repulsion to the methyl group than any amine.

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


text:
Why does potassium permanganate turn colourless when reacted with sulfur dioxide?
I learnt at school that when sulfur dioxide gas, $\ce{SO2}$, is combined with potassium permanganate, $\ce{KMnO4}$, the purple solution will become colourless.
What happens in the solution to cause this to happen?

This is a redox reaction. The permanganate ion (purple) is reduced to manganese(II) ion which is a very pale pink, or almost colourless.
$$\ce{MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O}$$
while sulfur dioxide is oxidized to the sulfate ion, which is colourless.
$$\ce{SO2 + 2H2O -> SO4^2- + 4H+ + 2e-}$$
So, the solution is decolourised. The overall equation is
$$\ce{2MnO4- + 5SO2 + 2H2O -> 2Mn^2+ + 5SO4^2- + 4 H+}$$

But what happen to the potassium ion ? Does it combine with the sulfate ion or still remains as an ion?

Potassium ion is an spectator ion. It doesn't participate in the redox equation. Its role is to keep the solution electrically neutral.


text:
How to calculate the rate constant at different temperature for the decomposition of dinitrogen pentoxide?

For the reaction
  $$ \ce{2N2O5(g) -> 4NO2 + O2(g)} $$
  the rate law is:
  $$ \frac{\mathrm{d}[\ce{O2}]}{\mathrm{d}t} = k[\ce{N2O5}] $$
  At $\pu{300 K}$, the half-life is $\pu{2.50E4 s}$ and the activation energy is $\pu{103.3 kJ/mol}$.
  What is the rate constant at $\pu{350 K}$?

I know there is something fishy about the rate law, but I can't make sense of it.
\begin{align}
\frac{ln2}{k} &= \pu{2.50E4}\\
k &= \pu{2.773E-5}\\
\frac{\pu{2.773E-5}}{k_2} &= \frac{A\cdot\exp\left\{\frac{-103300}{8.314\times300}\right\} }{ A\cdot\exp\left\{\frac{-103300}{8.314\times350}\right\}}\\
\end{align}
Finding $k_2$ from this gives me a weird value: $k_2 = 0.0103$.
The answer for this question is $\pu{7.47E-8 s^-1}$.
Where have I gone wrong?

I get the same answer you do. Reality check: the hotter the reaction, the faster it'll go. Therefore, your k value should be growing as your temperature increases. The given answer is 3 orders of magnitude smaller at 350 K than at 300 K. I think your book/solution guide is wrong.

If looks like your book divided k_1 by the ratio of k_2/k_1 instead of multiplying it (see my response below). As @tralston says, k should increase with temperature.

I arrived at the same $k_{\pu{300 K}}$ as you did. I find it a little weird that there's a 3 order of magnitude decrease in rate constant for a $\pu{50 K}$ increase in temperature.
Per the note below, the original equation I put had a mathematical error. The below should be correct:
\begin{align}
\frac{k_{\pu{350 K}}}{k_{\pu{300 K}}} &= \exp\left\{\frac{E_\mathrm{a}}{R\left(\frac{1}{T_{\pu{300 K}}} - \frac{1}{T_{\pu{350 K}}}\right)}\right\}\\
\frac{k_{\pu{350 K}}}{k_{\pu{300 K}}} &= \exp\left\{\frac{103000}{8.314\left(\frac{1}{300} - \frac{1}{350}\right)}\right\}=371.14
\end{align}
If you take $k_{\pu{300 K}} = \pu{2.773E-5}$ and multiply by that factor above (as you should) you get your answer, if you take $k_{\pu{300 K}}$ and divide by the factor above, you get the book's answer, which is where I believe their mistake is.

When using Arrhenius equation: you have to multiply the activation energy by $1000$, because it must be in $\pu{J}$ and not $\pu{kJ}$.
you must also divide by $RT$ and not by $T$ as you did.


text:
Hydrocarbons with only 4 carbon atoms
How many different organic structures (from the pure theoretical viewpoint) can be drawed with only 4 (exact) carbon atoms and with/without hydrogen? Polycyclic compounds and bridged compounds are also allowed. Please, name them all as well! :)
Remark: I have drawed the chemical graphs in a piece of paper (the skeletons) and I get 37 graphs (including linear with simple, double and triple bonds, and planar and nonplanar compounds), without geometric isomerism (cis/trans). Is that OK?

I top out at 23. I'd be curious to see your list.

Are you including non-existent compounds like cyclo-C4, cyclobutadiyne, for example?

exactly 4 carbons, or 1 to 4 carbons?  Does this include ions? Radicals?

Exact 4 carbons, including non-existent (non "made" compounds, i.e., theoretical, as I stated)...I got 36...Like my current age, and I want to know if I am right...

I think pentavalent carbon is more realistic than some of those: http://www.scs.illinois.edu/denmark/presentations/2007/gm-2007-04-17.pdf

There is also cis/trans 2-butene in real world. In theoretical world you could draw cis/trans isomers for some cyclobutenes too

@DavePhD  Dave, how would pentavalent carbon play into this when there are only 4 carbons?

@Ron Because carbon can bond hydrogens, not just other carbons.

What about the same question with 5 carbons? I guess the number of chemical graphs become larger...

https://oeis.org/A134818 Here's the latest list, stereoisomers ignored.

Final update, all earlier edits incorporated.
Groundrules: Considering compounds with:

only carbon and hydrogen
only 4 bonds to carbon

There are 37 isomers without considering trans isomers; 49 when trans isomers are included.  Also, many of these compounds seem extremely unstable and therefore unlikely to exist.
Note to self: check back in 20 years and see how many of the unlikely ones have been detected.

Names by row:

butane, isobutane
but-1-ene, but-2-ene, 2-methylpropene
buta-1,3-diene, buta-1,2-diene, buta-1,2,3-triene
but-1-yne, but-2-yne, but-1-ene-3-yne, buta-1,3-diyne
methylcyclopropane, 2-methylcyclopropene, 1-methylcyclopropene, methylenecyclopropane, methylenecyclopropene, methlycyclopropadiene
methylcyclopropyne, methlenecyclopropyne
cyclobutane, cyclobutene, cyclobuta-1,2-diene, cyclobuta-1,3-diene, cyclobutatriene, cyclobutatetraene, cyclobutyne, cyclobutenyne, cyclobutadiyne
bicyclobutane, bicyclobut-1(3)-ene, bicyclobut-1(2)-ene, bicyclobuta-1,3-diene, bicyclobuta-1,2-diene
tetrahedrane, tetrahedrene, tetrahedradiene

Comparing with my skeletons, you lack a 1methyl-cyclopropene, and the cyclobutatriene with 3 double bonds (irrespectively it existed or not, you have not drawed it). Also, you have not drawed the C4 (the square with 4 double bonds). Thus, I get 36 (or 37, if I include the cis/trans versions of 2-butene, that I had not considered diferent so far).

@ron You might want to include a note for posterity that some of these "compounds" are incredibly strained and likely unstable. (This was why I was including ?? marks in my answer.)

@GeoffHutchison  A couple of things 1) I'd suggest leaving your answer in place, it is nice work, 2) regarding 1,3- butadiene are you talking about s-cis and s-trans about the 2-3 bond?

@GeoffHutchison There is a higher than normal barrier to rotation about the 2,3-bond in 1,3-butadiene due to a contribution from the resonance structure with a 2-3 double bond. However, the barrier is still low enough that the s-cis and s-trans forms are generally considered rotamers.

Now I get 37 (without cis-trans): you are left a bicyclo[1.1.0]but-1,1-diene, I think...Indeed, I also forgot it in my first skeletons!

cyclobuta-1,2-diene could have enantomers http://en.wikipedia.org/wiki/Allene

Is it legitimate to refer to cyclobutatetraene and cyclobutadiyne as different compounds, or are they just different resonance structures?  Both are just a ring of 4 carbon atoms and no hydrogen.

@DavePhD I pictured cyclobutadiyne as having 2 sets of non-equivalent bonds, whereas the bond lengths in cyclobutatetraene were all equivalent.  So they're not resonance structures, but probably not much of a barrier separating them.

@ron like Fig. 3 of http://pubs.acs.org/doi/pdf/10.1021/ja0123002

@DavePhD I can't get to fig. 3, but do you understand my argument about the 2 molecules having different bond lengths, therefore they can't be resonance structures?

@ron, yes I understand what you are saying, that cyclic C4 with all bond lengths equal would be one compound and C4 with alternating short and long would be a different compound (but Fig. 3 of article refers to these as resonance structures in another cyclic pure-carbon compound).  Another article makes calculations for linear C4 and cyclic C4 and says the calculated electronic structure of cyclic C4 is alternating short and long bonds.  http://www.sciencedirect.com/science/article/pii/000926149500801A#

@DavePhD  Do they say how much higher in energy the cumulenic isomer is and what barrier separates the 2 isomers?

@ron, actually after looking at a 3rd article http://pubs.acs.org/doi/abs/10.1021/ja00081a041, it looks like C4 has two possible isomers: a linear isomer with bond lengths 1.32, 1.30, 1.32; and a rhombic (cyclic or bicyclic) isomer with 4 bond lengths of 1.46 and two 63 degree angles and two larger angles, with the linear isomer being lower energy.

@ron Shouldn't 4th molecule from the left on the 5th row be named methylidenecyclopropane?

@Mockingbird  I don't think so.  If you google "methylidenecyclopropane" all you get back is "methylenecyclopropane".  "Methylenecyclopropane" is also the name used by Wikipedia.

@ron So why "methlenecyclopropyne" when "methylenecylclopropane"?

"https://pubchem.ncbi.nlm.nih.gov/compound/Methylenecyclopropane" says methylidenecyclopropane also works.

Cyclobutatetraene?? Are you kidding me?

@PrittBalagopal It fits within the stated "ground rules".

How many different organic structures (from the pure theoretical
  viewpoint) can be drawed with only 4 (exact) carbon atoms and
  with/without hydrogen?

We could make strict rules like each carbon has exactly 4 bonds and get a specific answer, but this is not reality.  There can be lone pair electrons and unpaired electrons.  The octet rule is not stictly followed.
$\ce {C_4}$ actually has been observed and is linear.
:C=C=C=C:
$\ce {C_4}$ has been the subject of numerous theoretical and experimetal papers because of its possible occurrence in nebulae. It has been debated whether a singlet linear, triplet linear or rhombic (kite shape) state is the lowest energy state.  http://www.sciencedirect.com/science/article/pii/S0009261400005765
Neither the linear nor rhombic states follow the naive rules.
Linear $\ce{C_4H}$ has been observed both in the lab and outer space.
In fact according to the University of Kohn lists Molecules in Space, linear $\ce{C_4H}$ is one of only 61 molecules and molecular ions found in extragalactic space and of only 190 found in the interstellar medium or circumstellar shells as of 2016.
For $\ce{C_4H_2}$ linear butadiyne in known.  Cyclobutatriene, cyclobutenyne and tetrahedrene have been ruled out theoretically as not represtenting any actual potential energy local minimum, while similar to $\ce {C_4}$, structures having lone pair or unpaired electrons and not following the octet rule (such as carbenes) have been calculated to represent actual minima.  See the following references for theoretical cyclic $\ce{C_4H_2}$ structures:
http://onlinelibrary.wiley.com/doi/10.1002/jcc.540020211/pdf
http://pubs.acs.org/doi/pdf/10.1021/jo060698k
http://pubs.acs.org/doi/pdf/10.1021/jo000941u
In outer space, not only has the usual HCCCCH isomer been found but also $\ce{H2CCCC}$
See Observations of cumulene carbenes, H2CCCC and H2CCC, in TMC-1

Yes, I was wondering as well how many of these structures are in fact known from the experimental side...


text:
Binding forces in crystalline solids
I am referring to rocks in this instance. Individual crystals grow together in an igneous rock as it solidifies. I know that the $\ce{SiO4}$ tetrahedron is the basic building unit of silicate minerals, and metal cations attach to ensure electroneutrality.
What I am wondering, however, is what force keeps the crystals together (as in bound to one another) in the solid? For comparison, sediment grains are held in place by a silica or calcite cement in fragmental rocks, but what stops individual crystals from breaking free in crystalline solids?

So, more 'Binding forces in polycrystalline solids' then?  The energy to introduce a grain boundary into a single crystal depends on the relative orientation of the two resulting grains, but there is some atomic rearrangement (strains or reconstruction), and you will end up with a number of bonds across the grain boundary.  The boundaries are usually weaker than the bulk crystal, which is why they preferentially fracture along the grain boundary.

Converting to answer: While the energy to introduce a grain boundary into a single crystal depends on the relative orientation of the two resulting grains, there will be some atomic rearrangement (strains or reconstruction), and you will end up with a number of bonds across the grain boundary. The boundaries are usually weaker than the bulk crystal, which is why they preferentially fracture along the grain boundary.


text:
Symmetry representations
I'm Having a real struggle trying to understand symmetry representations matrices and character tables. Is there anyone who would be kind enough to help me out here? I understand it is probably a little broad but nevertheless, I will list some areas that if anyone could shed some light on I would be very appreciative.
1) What is an irreducible representation? My book just starts going into $2\times 2$ matrices like:
\begin{equation}
\left( \begin{array}{cc}
1 & 0\\
0& 1\end{array}
\right)
\end{equation}
Saying that we can represent the effect of symmetry operations as such! When the operation actually affects the molecule by changing an atom then the diagonal of the matrix swaps round seemingly !!
2) Leading on from the first question, what is a reducible representation? Why the distinction?
3) What is a basis? My book introduces them via vectors down each of the cartesian axes and then a "set" of them together (why) to give four basis's for a $C_2v$.
4) I can use the reduction formula to see how many irreducible representations are contained within a particular reducible representation, but what am I doing!

What book are you using?

Introduction to molecular symmetry. J.S Ogden Oxford primer series. It is a very good book so far but I'm struggling past chapter 3.

Molecular Symmetry and Group Theory by Vincent is amazing, going into just the right amount of detail for a chemist:
http://eu.wiley.com/WileyCDA/WileyTitle/productCd-0471489395.html

Algebraic group theory is very useful because it allows us to capture in very abstract terms what the molecule does in terms of symmetry. The transformations within the group fully reflect the transformations within the molecule that make it look like itself.

Let's start with the basics:
A representation is a set of matrices that fulfill the multiplication table for the point group.
So I might be able to use a set of 3x3 matrices, since of course I can define any of the symmetry operations over Cartesian (x,y,z) space. e.g.:
$$E = \left(
\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \end{array}
 \right) \;
i =
\left(
\begin{array}{ccc}
-1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & -1 \end{array}
 \right)
$$
But usually, I can fulfill the multiplication table with much simpler (smaller) matrices.

An irreducible representation is a set of matrices that are as small as possible.
A reducible representation is a set of matrices that can be decomposed into a linear combination of irreducible representations. This can occur by multiple techniques, including block-diagonalization.

The distinction exists, because for chemical point groups, there are usually a finite number of irreducible representations (given in character tables), but of course there are an infinite number of reducible representations.
The power of symmetry and group theory is that any function of a molecule, like orbitals, vibrations, etc. can be described using a representation. Thus, we usually label orbitals and vibrations by their corresponding irreducible representation.

I'll have to hunt down my copy of Ogden, but usually people use the Cartesian axes or $p_z$, $p_x$, and $p_y$ to illustrate the different irreducible representations. That is, a basis is some function (like the Cartesian axes or an atomic orbital) used to generate a representation. We look at how the basis is transformed under the different symmetry operations and write down the matrix or characters from those transformations.

(I try not to use that term because students find it confusing and different guides and texts use subtly different definitions for it.)

The reduction formula (or algorithm) is a way of decomposing the reducible representation (which is really a big matrix) into a combination of the different irreducible representations, but only needing to act on the characters of the representations.

That is, I don't need to actually use the reducible matrices themselves. Using the reduction formula, I can get the linear combination of irreducible representations only using multiplication of characters of the representations. It's much easier to do the reduction formula than to figure out how to transform a large reducible matrix.


text:
How do I (further) lower the pH of a pyrrolidine solution?
Pyrrolidine ($2~\mathrm{mol}$) is slowly added to $164~\mathrm{ml}$ conc. $\ce{HCl}$ in $200~\mathrm{g}$ ice water to adjust the $\ce{pH}$ to $3-4$. When it is mixed, the $\ce{pH}$ is $14$, how do I lower the $\ce{pH}$ to $3-4$? I cannot add more $\ce{HCl}$ because it only calls for $164~\mathrm{ml}$.

A pH of ~ 4 would correspond to a solution of pyrrolidinium chloride.  Use less pyrrolidine or more hydrochloric acid to arrive at that point.
The commercial concentrated hydrochloric acid is 37 % (w/w), this would be a concentration of ~ 12 M, and 164 ml of that is slightly less than 2 mol HCl.  It's very strange that you are seeing a basic pH.

I absolutely agree, and would have thought the same thing. I'd suspect, the conc. HCl solution is not concentrated any more. Maybe too old.


text:
Explaning Cis & trans isomers through VSEPR
Total rewrite as was a panic-stricken 2am question.
Unfortunately I wrote on an application that VSEPR theory is a reason of Cis-Trans isomerism (specifically in fatty acids however I doubt that would make any difference). Although in my head it seemed to work at the time, I appreciated about a week after that I was actually thinking of the wrong type of thing.
My question is,

How can VSEPR and/or the electron configurations in an unsaturated fatty acid explain how cis and trans isomerism arises?

As far as I'm aware, the Pi bond prevents (resists?) rotation around the C=C link, but I feel as if I'm missing something.
Also,

Does the geometry caused by the repulsion of H's and alkyl electrons on either side of the C=C bond affect the isomers differently depending on whether or not they are Cis or Trans?

One of the below posters also highlighted a good question, does anybody have any insight on this?:
"I think the main question is whether there's a preference for cis or trans isomers in a fatty acid based on steric/VSEPR considerations"
(Thank you for your response)
Thanks for your help
Bonus related question:

Is there any preference for a specific stereoisomer in the dehydrogenation of a fatty acid's alkyl chain? If so, what are they and why?

What difference did you want to rationalize?

While I think VSEPR theory is not suitable for many things (it is a little bit too simple and outdated), I think it is especially not suited for this problem. I also backup @Dissenter's comment as I am not quite sure what you want to know. Maybe you can [edit] your question, to make it a little clearer. Do you want to rationalize the energy difference between cis/trans isomers or the structural differences in the surrounding environment of the double bond?

I think the main question is whether there's a preference for cis or trans isomers in a fatty acid based on steric/VSEPR considerations. In practice, I think both isomers are found in biological materials.

Answered my own bonus question (I believe): In case anybody was interested,
Trans seems to be favoured due to the methyl group on either side of a Cis bond repelling eachother (due to to both the methyl groups wanting to retain their tetrahedral structure, hydrogens are repelling one another and hence distorting things), and hence the trans is more energetically favourable.

I have a feeling that looking at the question from the perspective of evolutionary biochemistry is more important than looking at it from purely bonding model perspective.
Cis fatty acids tend to have lower melting and boiling points than saturated fatty acids.
The cis geometry puts a "kink" or bend in the chain which makes efficient packing difficult, reducing the strength of the intermolecular forces holding the molecules to each other.  The weaker forces result in lower melting points.
Most living things produce a mixture of different saturated and unsaturated fatty acids, and this mixture has been refined by evolution to suit the needs of the species.
In most species of plants, we tend to see 25% or less saturated fat, and the rest as a mixture of different unsaturated fats.  This is not the case for palm kernel and coconut oils, which can have as much as 80% saturated fat.  Reference
In contrast, land animals like pigs and cows have 40%-50% of their fats as unsaturated.  Ducks, in contrast have only about 30% of their fat as saturated.
Individuals that produce these lower-melting fats had a competitive advantage over those that didn't, so the genes for the production of a greater percentage of unsaturated (cis) fats spread through the population.


text:
Is this a valid method to find percentage of pure ethanol in an impure sample?
To generalize the problem, I will just say:

5 gram of impure ethanol (or any other type of combustible material) is placed inside a bomb calorimeter where volume is constant. If the combustion produces a certain amount of energy, how should I find the mass in gram of pure ethanol contained in that quantity of impure ethanol?

Here is what I will do:

find the standard molar enthalpy of ethanol.
write a combustion equation.
find the energy produced in the combustion of one mole of ethanol.
then divide that amount of ethanol by the amount of energy stated in problem.

To find how many mole of pure ethanol, convert the mole into gram.
But there are two things that make me think that this way of solving the problem is not correct:

the quantity of impure ethanol is a mixture of pure ethanol and another unknown material. What if this unknown material produce part of the energy stated in the problem?
the standard molar enthalpy is measure under standard conditions (1 atm, 25 °C). However, the condition inside the bomb calorimeter is unknown.

What if this unknown material produce part of the energy stated in the problem?

It would be stated in the problem that the amount of heat that the unknown material produced.

The standard molar enthalpy is measured under standard conditions (1 atm, 25°C). However, the condition inside the bomb calorimeter is unknown.

We are given some information, the final temperature of the system. But you are right in that we do not know the pressure, and the specific heat of the remaining items is difficult to calculate.

That is impossible. If your compound is not pure and you do not know the amount of impurity and what kind of impurity, you cannot possibly calculate the amount of ethanol present. Unless we assume that the impurity is small or incombustible.
I wouldn't use a bomb calorimeter in this case because you want to work with enthalpy, not internal energy. An adiabatic flame calorimeter would be more suitable. If you need to use a bomb calorimeter, you can convert between $\Delta U$ and $\Delta H$ by tracking the pressure in the bomb. You can of course control the conditions inside the bomb at the start of the experiment.
The standard ethalpy of combustion can still be used, provided that the end temperature of the products of the chemical of physical process are not much higher than the starting temperature of the reactants. A large enough reservoir of heat absorbent or higher $C_v$ (calorimeter constant) should ensure this, but not too large or too high, else you won't be able to accurately read of the temperature difference (you see what the advantage is of an accurate thermometer in this case).


text:
What happens when oxidizing Ni(OH)2 with a strong base?
In an aqueous solution, Nickel oxyhydroxide is oxidised with a 1:1 ratio of aqueous Sodium hydroxide. My goal is to produce Nickel oxyhydroxide ($\ce{NiOOH}$).
Strong Base: $\ce{NaOH + H2O \leftrightharpoons Na+ + OH-}$
Add that to some stirred $\ce{Ni(OH)2}$:
$$\ce{ Ni(OH)2 + OH- \leftrightharpoons NiOOH + H2O + e-}$$
This is where things get difficult; wouldn't the spare electron be used by the sodium ions to make $\ce{Na (s)}$?:
$$\ce{ Na+ + e- -> Na (s) }$$
But then, $$\ce{Na(s) + H2O \leftrightharpoons NaOH }$$
That just brings everything back to the beginning.
I guess the real question is: will the initial oxidation of the $\ce{Ni(OH)2}$  happen? If not, how can I force it? I initially thought of just grounding the solution, as odd as that sounds.
Thanks for your help!
P.S. I'm a 12th grade high school student.

@JohnSnow In $\ce{...}$ you can simply substitute \leftrightharpoons with <=> and it will render correctly.

This is where things get difficult; wouldn't the spare electron be used by the sodium ions to make
  $\ce{Na (s)}$ ?: $\ce{Na+ + e- -> Na (s)}$

This reaction is, you might be knowing, the reverse of one of "the best" known favourable reaction, as the data suggests:
$$\ce{Na->Na+ +e-}\tag{$E^\circ=2.71\;V$}\\
2.71\;V\equiv-261.475\;\text{kJ/mol}\equiv1.0938\times10^{50}(\text{rate})$$
Just imagine the reverse reaction.

But then,
  $\ce{Na(s) + H2O -> NaOH }$

This then would never actually then happen. Intersetingly see this


text:
How do I calculate the concentration of sulphuric acid by a titration experiment with sodium hydroxide?
In my latest chem lab the objective was to create a primary standard of $\ce{NaOH}$ and use it to determine the concentration of sulfuric acid.
The first part of the lab was determine the molarity of the $\ce{NaOH}$ solution through a series of titrations.

A sample of KHP (abbreviated form of $\ce{KHC8H4O4}$) was placed into a flask with approximately $\pu{25 ml}$ of water.
Phenolphthalein was added to the flask as the indicator. $\ce{NaOH}$ was then titrated into the flask with a burette. From multiple titrations of this sort I was able to calculate the molarity of $\ce{NaOH}$.

Below I have included part of my table and calculations (Note: $\pu{1 mol}$ of KHP is equal to $\pu{1 mol}$ of $\ce{NaOH}$ in this experiment. If I have made any mistakes please tell me).
Trial 1:
Mass of KHP in flask $= \pu{0.5108 g}$
Volume of $\ce{NaOH}$ added to flask $ = \pu{21.73 ml}$
Calculation of molarity of $\ce{NaOH}$ for trial 1:
Molar mass of KHP $ =  \pu{204.23 g/mol}$
$\pu{0.5108 g}/\pu{204.23 g/mol} = \pu{0.002501 mol}$ of KHP, which is equal to $\pu{0.002501 mol}$ of $\ce{NaOH}$.
Molarity of $\ce{NaOH} = \pu{0.002501 mol}/\pu{0.02173 L} = \pu{0.1151 M}$
I did 3 other trials like this (in total 4) and calculated the average molarity of $\ce{NaOH}$ to be $\pu{0.1159 M}$.

The second half of the lab is the part I had trouble with.
We were given a sample of $\ce{H2SO4}$ with an unknown concentration. I took $\pu{10 ml}$ of this $\ce{H2SO4}$ and mixed it with $\pu{100 ml}$ of distilled water. This new diluted solution of $\ce{H2SO4}$ (I will refer to it as solution 2 now) was the solution used in the trials to determine the molarity. So $\pu{25 ml}$ of solution 2 was added to a flask with a few drops of phenolphthalein. A titration using $\ce{NaOH}$ (the same $\ce{NaOH}$ as used in the previous section) was performed.
My task is to now figure out the concentration of the original $\ce{H2SO4}$ solution. I have tried 2 different methods. The first method I attempted seems so flawed I didn't bother to put it on (it didn't even make sense to me). Each method seems incorrect and have yielded drastically different results. Below I have provided a sample of my table and one of my attempts to solve for the molarity of $\ce{H2SO4}$.
The net ionic equation of this procedure is:
$$\ce{H2SO4 +2NaOH <=> Na2SO4 + 2H2O}$$
Trial 1:
Volume of diluted acid (solution 2) in flask: $\pu{25.00 ml}$
Volume of $\ce{NaOH}$ added to flask: $\pu{23.81 ml}$
Attempt 1 at finding molarity:
Moles of $\ce{NaOH}$ added to flask: $\pu{0.02381 L} \cdot \pu{0.1159 M} = \pu{0.0027596 mol}$ $\ce{NaOH}$
Amount of of $\ce{H2SO4}$: $0.0027596/2 = \pu{0.0013798 mol}$ $\ce{H2SO4}$ (The 2 came from the net ionic equation above)
Molarity of diluted $\ce{H2SO4}$ (solution 2): $\pu{0.0013798 mol}/ \pu{0.025 L}= \pu{0.054172 M}$
(I may be using the wrong volume, is it possible that I have to add the $\pu{25 ml}$ to the $\pu{23.81 ml}$ and divide by $\pu{0.04881 L}$?)
\begin{align}
C_1V_1 &= C_2V_2\\
C_1&=?\\
V_1&= \pu{0.01 L}\\
C_2&= \pu{0.054172 M}\\
V_2&=\pu{0.1 L}\\
\text{Therefore:}\\
C_1&=(\pu{0.054172 M})\cdot(\pu{0.1 L})/(\pu{0.01 L})\\
C_1&=\pu{0.54172 M}\\
\end{align}
Molarity of original/stock $\ce{H2SO4}$.

You were supposed to measure 10ml of the acid in a 100ml capacity volumetric flask and make it up to mark with ditilled water. That is 10ml acid with 90ml distilled water. This will give you a dilution factor of 10.

Molarity of diluted $\ce {H2SO4}$ (solution 2): $\pu{0.0013798 mol}/\pu{0.025 L} = \pu{0.054172 M}$
  (I may be using the wrong volume, is it possible that I have to add the $\pu{25 ml}$ to the $\pu{23.81 ml}$ and divide by $\pu{0.04881 L}$?)

No, you used the correct volume since you want to know the concentration of the $\pu{25 mL}$ that you added to the flask.

$\pu{0.54172 M}$ - Molarity of original solution of $\ce {H2SO4}$.

Correct.

"I took $\pu{10 ml}$ of this $\ce{H2SO4}$ and mixed it with $\pu{100 ml}$ of distilled water"
Is $V_2=\pu{0.11 L}$ or $ V_2=\pu{0.1 L}$

@AdnanAL-Amleh If you take 10 mL of one solution and add 100 mL of another solution, you will always have more than 100 mL.


text:
Structure of carbon fiber
What is the structure of carbon fiber? Wikipedia doesn't give much insight.

The atomic structure of carbon fiber is similar to that of graphite, consisting of sheets of carbon atoms (graphene sheets) arranged in a regular hexagonal pattern, the difference being in the way these sheets interlock. Graphite is a crystalline material in which the sheets are stacked parallel to one another in regular fashion. The intermolecular forces between the sheets are relatively weak Van der Waals forces, giving graphite its soft and brittle characteristics.

The Wikipedia description begs the question of how exactly are the sheets of graphene interlocked. Are there pi-stacking interactions?

Carbon fiber is also often a composite. That is, there is some binder or polymer keeping things together. It's also a very broad category of materials, sometimes protected by industrial trade secrets.

Products of carbonization may vary very strongly from source to source. Moreover, purging all heteroatoms from carbonaceous material is virtually impossible. Given that, it is safe to assume that despite short order existing for most carbon atoms, fragments of graphite-like structure are bent, interlocked and linked in various unpredictable patterns.

The key difference between graphite and carbon fibers is the degree (or range) of ordering.  In graphite there are large, extensive planes of carbon stacked one atop another resulting in long-range order. In the fibers, there are small ribbons of graphite-like (it is not all carbon, due to the way the fibers are prepared there is still some nitrogen present in the molecular array) material oriented along the fiber axis and there is only short range order.  As a consequence graphite with its extensive 3 dimensional planar array of ordered carbon atoms is crystalline.  Whereas the fibers, being ordered over a much smaller range,  are amorphous.  The amorphous nature of the fiber allows the smaller ribbons to bend and fold, and mechanically interlock with one another.  This mechanical interlocking is what serves to increase the fiber's strength.  See here for a very nice and concise description of how the fibers are prepared.  The pictures in the link make clear how the short range ordering in the fibers results in a highly folded and interlocked structure.  Further heating of the fiber will convert it to graphite.


text:
How to determine an unknown gas in a tube with a massless piston at equilibrium conditions?

There is a massless piston that can move freely in a horizontal tube that is closed at both ends. To the left, there is a sample of helium gas at $300~\mathrm{K}$ and $4~\mathrm{atm}$. To the right, there is a sample of an unknown gas $\ce{X}$ at $500~\mathrm{K}$ and $2~\mathrm{atm}$. At equilibrium, the piston is located at 1/9 the length of the entire piston from the right side. The molecules of $\ce{X}$ are homonuclear. What is most likely the identity of $\ce{X}$?

(From Chemistry Olympiad)
I'm confused by the equilibrium part. I tried $PV = n\mathcal{R}T$ to find the ratio of moles. But without molar mass, how do you solve for it? How would you solve it?

To start, I would have to assume that the piston started in the middle; unless you left that out. Then I would determine the ratio of the number of moles for each gas. Then use the molar mass of helium to find the other gas.

@LDC3 The question doesn't state whether the piston starts in the middle. Thus, we cant assume that.

Well, then I think there are too many unknowns to be able to identify X.

Could you cite the source for the question, that might clarify some things.


text:
Why are metal oxides basic and non-metals' acidic or neutral
Recently I came across the fact: Metal oxides basic and non-metals' acidic or neutral.
Searching for it, partly my query is solved, as I found:
Metal oxides or $\ce{O2-}$ forms $\ce{OH-}$ after hydrolysis in the following reaction so metal oxides are basic in aqeous solutions.
$$\ce{O2- + 2H2O->4OH-}$$
I think this might happen as metal form ionic compunds and non-metals form covalent and some of them may not even react tus neutral.
I've got only a little $5\%$ of the exact reason, therefore I wish if someone could explain the whole process/fact. I considered Arhenius concept.
Note My question is partly answered here, but is not satisfactory.

but is not satisfactory. || what the answer you linked lacks?

So how is this story goes with water?

related: http://chemistry.stackexchange.com/questions/28132/how-is-charge-density-a-useful-measure-of-polarising-power-and-polarisability, http://chemistry.stackexchange.com/questions/19763/why-is-pbcl%e2%82%84-covalent, http://chemistry.stackexchange.com/questions/10273/hardness-and-softness-are-terms-used-to-describe-which-aspect-of-acidic-behaviou

We have learnt that metallic oxides are basic nature. This is because the oxides of metals like $\ce{MgO}$ or $\ce{CaO}$ or $\ce{Na2O}$ form hydroxides in their aqueous solutions.
For example:
$$\ce{K2O + H2O -> 2KOH}$$
$$\ce{MgO + H2O -> Mg(OH)2}$$
But when non-metallic oxides dissociate into their constituent ions in water, they give $\ce{H+}$ ions in their aqueous solutions forming acids
For example:
$$\ce{CO2 + H2O -> H2CO3}$$
$$\ce{SO3 + H2O -> H2SO4}$$

This just restates information in a previous answer and the equations are also unbalanced. Perhaps you could clean this up a little and add some more detail.

I have improved the formatting of your post using $\LaTeX$. For more information on how to do this yourself please see here and here. Additionally, you can visit this chatroom for more assistance.

so they are acidic because they form acids? Sorry I dont know much about Arrehinius definition...only know the Brownsted Lowry one

Metal oxides like $\ce{CaO}$ and $\ce{Na2O}$ react with water and give hydroxides:
$\ce{CaO + H2O <=> Ca(OH)2}$
$\ce{Na2O + H2O <=> 2 NaOH}$
While non-metal oxides like $\ce{CO2}$ and $\ce{SO2}$ react with water and give acids:
$\ce{CO2 + H2O <=> H2CO3 <=> H+ + HCO3-}$
$\ce{SO2 + H2O <=> H2SO3 <=> H+ + HSO3-}$

can you use technical terms, thermodynamics, kinetics, bonding theories or other theoretical explanations?

It still didn't explain why.


text:
Does water have a smell?
I was washing my cup with hot water (without soap) and upon it nearing my nose, there was some sort of 'smell' (I lack a better word)-
However the 'smell' was different compared to when I was drinking just plain cold water.
So my question is, does water have a smell?

Water can't possibly have a smell as any olfactory receptor is constantly exposed to it. You're smelling molecules that were in the water.

@canadianer what does that mean?

That water as chemical (pure H2O) does not have any smell. What runs from your tap (or bottle) is not pure H2O and therefore this solution can have smell.

@ssavec oh i see- so what i presumably thought was the 'smell' of water are just the impurities in it?

@Nick: right. But chemist could call it impurities, physician could call it minerals. That means, it is not bad to drink "impure" water, on contrary, chemically pure (distilled) water is not really good for drinking. Hope I did not confused the stuff too much.

Does tap water then have a distinct smell? Or it varies from tap and country? @ssavec

Was it a plastic cup and very hot water? It's not unusual to notice a faint "plasticky" smell in such situations.

Water in its pure form, i.e. $\ce{H2O}$ does not have a smell, or at least no smell that we can distinguish because the receptors in our nose (and mouth) are continuously exposed to it.
What you smell are dissolved gases and other volatile impurities. The nature of these chemicals will vary mainly by location/source of the water, and might also be a bit influenced by your specific local plumbing.
For example, in Iceland near Myvatn there is a lot of volcanic activity and the water from the tap (in particular the warm water because it can dissolve more) smells like rotten eggs due to the sulfur. Another example is that in many countries the water will smell a bit like chlorine, because chlorine is used to keep the water safe to drink (kills germs).

It could be smell from substances on the cup (remains of tea, coffee etc).

Also possible, good point. I guess I forgot to mention that because the OP asked whether water itself had a smell.

I only wanted to add that in our community the water has a very high Iron content and carries with it an odor and distaste if not filtered. Northern Michigan. Commonly known as 'well water'.

there was some sort of 'smell' (I lack a better word)

So there was a sensation in your nose that you partly (but not confidently) identified as smell. It might not have been olfactory, though.
I'm no expert, but it seems likely to me that in addition to any impurities in the water that you really can smell, you're also detecting a sudden change in humidity (especially with the hot water) and/or temperature. Strictly speaking you can't "smell" water, but that's by definition of smell. You can nevertheless in the right circumstances detect water via the mucous membranes of your nose. As you inhale, the air passing over the membranes dries them (or doesn't) by a differing amount. Aside from the fact that you might falsely perceive this sensation as a smell, how dry the membranes are does affect your perception of anything you can smell in your environment. So a change in humidity can result in a change to what you smell.
Even the acoustics of the cup could be detectable (nose, ears, close together, attached to the rigid/resonating boney structure of your skull, you're putting your nose at the entrance to an enclosed space), and could be interpreted as "like a smell" if a smell is what you're expecting from your nose. Although I don't think that accounts for a difference between hot and cold.
So this could well be a biology/psychology question as well as a chemistry one ;-)

Well thought out of the box.

Haha perhaps I should ask them in the relevant stackexchanges mentioned

As answered above water in pure form is tasteless and odourless.
Due to various reasons there is change in taste and odour(smell) of water. Some of the different smell or taste are as following:

Chlorine taste or smell:

this occurs when the water is treated at the water treatment plant to
 disinfect it.

Metallic taste or smell:

If water is distributed using metal pipes it might get metal smell. Though nowadays plastic mostly plastic pipes are used.

Rotten egg smell:

It seems unusual but due to decaying of organic compounds water sometimes smells like rotten egg.

musty smell:

This type of unpleasant smell can be caused due to mixing of pesticide with water.

turpentine taste or smell:

This smell can be a result of MTBE contamination in your water.
P.S. As taste and smell works together it is not possible to distinguish between two.

Source: extoxnet

As stated by others, you're smelling whatever substances are in solution or suspension in the water.
One reason you may detect a difference between cold and hot water is that some homes have a water softener that replaces dissolved mineral salts with sodium or potassium salts. It's usually only used on the warm water line, because the main advantage is how it feels when bathing. "Soft" water may be less pleasant to drink than "hard" water. I never did like the taste of it.

Some people do have a water softener on the cold water line because they prefer the taste, it really depends on the individual and what's in the local water. They may have a different softener on the hot water system, both for the reason you say and to keep some of the crud from scaling up the water heater, so the point still stands.


text:
Balancing Ionic and Net Ionic equations
I am currently taking a chemistry course about acids and bases. One of the things I am a bit confused about, is the acid and base neutralization reaction, and how to balance it. For Example:
$$\ce{NaOH + HCl -> NaCl + H2O}$$ How do you write the ionic and net ionic equations for that?
Note: My main problem here is that my teacher requires us to first write the ionization of the $\ce{HCl}$ into $\ce{H3O+}$ and $\ce{Cl-}$. This is when I get confused, because I don't know how to balance the equations with the $\ce{H3O+}$
Also, I had another question in my exam, which I got wrong because of the balancing. Can you please tell me the correction of it:

Consider the neutralization reaction between nitric acid and calcium hydroxide. Write the balanced neutralization reaction and the ionic and net ionic reactions of it:

As $\ce{NaOH}$ and  $\ce{HCl}$ are strong base and acid respectively, they are completely dissociated in water. The salt $\ce{NaCl}$ is a  strong electrolyte and completely dissociated in water. So, the ionic equation is:
$\ce{(Na^+ +OH^- )}+\ce{(H^+ +Cl^- )}\rightarrow \ce{(Na^+ +Cl^- )}+\ce{H2O}  $
As the ions $\ce{Na^+}$ and$\ce{Cl^-}$ do not combine and remain in the solution as spectator ions. Their role is to keep the solution electrically neutral, the net ionic equation is:
$\ce{OH^- }+\ce{H^+ }\rightarrow \ce{H2O}  $.
On the other hand, we have to notice that the ion $\ce{H^+}$ is not stable in water, it combines with $\ce{H2O}  $ to form the ion hydronium $\ce{H_3O^+}$ . We usually write the complete dissociation of $\ce{HCl }$ as follows:
$\ce{HCl} +\ce{H_2O }\rightarrow$ $\ce{Cl^-} +\ce{H_3O^+}$
and the ionic equation is
$\ce{(Na^+ +OH^- )}+( \ce{H_3O^+}$$ +\ce{Cl^- })\rightarrow \ce{(Na^+ +Cl^- )}+2\ce{H2O}  $
And the net ionic equation is
$\ce{OH^- }+\ce{H_3O^+ }\rightarrow 2\ce{H2O}  $.
As for your exam questions:
$2\ce{HNO_3} + \ce{Ca(OH)_2 }\rightarrow \ce{Ca(NO_3)_2} + 2\ce{H_2O}$
 Now you can try to write the ionic and the net ionic equation to see if you understand what I have explained.

Consider the neutralization reaction between nitric acid and calcium hydroxide. Write the balanced neutralization reaction and the ionic and net ionic reactions of it

For simplicity consider this:
$$\ce{HNO3}\tag{nitric acid, ionic}$$$$\ce{Ca(OH)2}\tag{calcium hydroxide, ionic}$$
So, they'll break into their constituent ions:
$$\ce{HNO3<=>H+ +NO3-}\\\ce{Ca(OH)2<=>Ca^2+ +OH-}$$
Now $\small\ce{H+}$ exists as $\small\ce{H3O+}$ and after balancing by adding water to both sides and doing $\small\ce{H2O +H+\sim H3O+}$.Now multiply the first reaction by 2 after we notice the two hydroxide ions in the decomposition of calcium hydroxide, since we want in an integral multiple, the reaction $\small\ce{H+ +OH- ->H2O\equiv H3O+ + OH- ->2H2O}$
$$\begin{array}{}\ce{&2HNO3 +2H2O&<=>&2NO3-&+&2H3O+}\\\ce{&Ca(OH)2&<=>&Ca^2+ &+ &2OH-}\\\hline\ce{&2HNO3 +Ca(OH)2&->&\underbrace{Ca(NO3)2}_{Ca^2+ + NO3- <=> Ca(NO3)2}&+}&\underbrace{\ce{2H2O}}_{4-2=2}\end{array}$$

authors: Åmmårå Wåhéèd


text:
Extracting sodium from salt
At my school, my teacher tell us to extract sodium from normal salt ($\ce{NaCl}$) but without using electrolysis of molten sodium chloride. So is there a way to do that?

Most methods do not use sodium chloride, favouring alternative salts.

What did your teacher have in mind? I mean, what did they think was the correct answer?

I suppose there might be a chemical equilibrium reaction that could be achieved with free Fluorine but the process is unlikely to be safe.

I don't think it is safe or realistic to do that without extensive training and equipment, but yes, historically sodium metal has been made by chemical reduction of sodium salts at high temperature and condensing sodium vapor produced.  For example, sodium carbonate was heated with charcoal.  Even though equillibrium does not favor sodium metal, the fact that the small fraction of sodium produced is removed takes advantage of Le Chatelier's principle to keep making more sodium.

It has been done but it is not a trivial exercise.
NurdRage has documented one method on YouTube with a lot of extra ideas in the comments. Make Sodium Metal Without Electrolysis Using Domestic Chemicals
He has also succeeded in principle of isolating Sodium by the alcohol catalysed reduction with Magnesium (and sodium hydroxide).
Neither method uses sodium chloride though.


text:
Does silicon dioxide react with alkalis?
In my science class, I have been taught that silica (silicon dioxide/sand) has three properties, namely

It does not react with water.
It does not decompose with heat.
It does not react with acid.

However I think that if it does not react with acid, will it react with bases? I am just curious because if it does react, why does my textbook not say anything about that?

Silicon dioxide most certainly reacts with HF (hydrofluoric acid) - HF is the common liquid etchant for SiO2 in semiconductor processing. (HF is also nasty stuff, so be careful).

But I thought silicon dioxide is acidic. How it going to react with acid?

Silicon dioxide is not attacked by acids in general but it does react with HF.  The distinguishing feature here is F-.  So, yes, silica is acidic, and it reacts with a very special (Lewis) base.

You have to take all such statements with a grain of salt. Yes water doesn't dissolve an appreciable amount of silica in the lab. But heat and pressure will allow groundwater to dissolve much more. Quartz crystals in the earth are grown from such solutions cooling and depositing the supersaturated silica.

It really does react:
\begin{align}
\ce{4HF + SiO2 &-> SiF4 + 2 H2O}\\
\ce{SiF4 + 2 HF &-> H2[SiF6]}\\
\ce{SiO2 + 2 NaOH &-> Na2SiO3 + H2O}\\
\ce{SiO2 + 2 F2 &-> SiF4 + O2}\\
\ce{SiO2 + 2 C &-> Si + 2 CO}\\
\ce{(2,1)Na2O + (1,1)SiO2 &-> Na4SiO4/Na2SiO3}\\
\ce{SiO2 + Si &-> 2 SiO}
\end{align}

Will sio2 with 6hf directly give h2[sif6] or not?

$\ce{SiO2 + NaOH}$ where $\ce{NaOH}$ is taken in excess should give $\ce{Na4SiO4}$, am I correct?

$\ce{SiO2}$ definitely reacts with strong bases like $\ce{NaOH}$ or $\ce{KOH}$ which is why such solutions should not be stored in glass bottles.
Concentrated and hot aqueous solution of $\ce{NaOH}$ can damage glassware in the course of several days. $\ce{KOH}$ or $\ce{NaOH}$  in methanol is even more potent. Reaction is obviously faster with finely divided $\ce{SiO2}$
$\ce{4NaOH + SiO2 -> Na4SiO4 + 2H2O}$

Now, I realize that silicon dioxide is a very stable compound. However, there is an exception which is it will react react with a thing- Hydrofluoric acid ($\ce{HF}$)
The reaction is undergo like the equation below:
$$\ce{4HF(g) + SiO2(s) -> SiF4(g) + 2H2O(l)}$$
Only $\ce{HF}$ can react with sand ($\ce{SiO2}$), others chemical can't as sand is really that stable.


text:
Where can we buy chlorinated paraffin for lab application?
Any idea where to buy chlorinated paraffin for lab application? We need very viscous analog of $\ce{CHCl_3}$.
Should we think about synthesizing it?
I found a lot of places where I could buy 5 tons:) But we need only few liters for the rest of our life.

You ask any manufacturer/distributor if they will send you samples.

How about you tell your teacher to buy some for the whole class?

@Simon-Nail-It, that is nice. But we do not have teachers over here in the University. And basically my boss (teacher) asked me to buy clean one, but did not said where...

I'm voting to close this question as off-topic because it is not about chemistry.

Sigma-Aldrich is a global supplier of purified chemicals for research users.
They have a whole page of the haloalkane compounds they sell (in small quantities such as you desire).  You might be interested in compounds such as trichloropropane, pentachloroethane, etc.
A reason it might be difficult to get the higher-molecular weight compounds is that they are toxic and difficult to dispose of properly.  In the US, their use is subject to increasing regulatory requirements.


text:
Why methyl group is 2,4-directing?
Why methyl group is 2,4-directing? I am an A level student and this isn't required at this level. I am looking for a simple explanation which a high school student can understand.

Do you understand the rationale for other directing groups?

To solve this problem, draw the intermediates of reaction.  It will be obvious at that point-- or otherwise for the basis for a useful discussion.

Why methyl group is 2,4-directing?

When toluene undergoes electrophilic aromatic substitution the products are primarily the ortho and para isomers; usually only a small amount of the meta isomer is produced.  In order to explain this observation there are two effects to consider, inductive and resonance effects.
Inductively the methyl group releases electron density into the benzene ring.  This is because the methyl group, being $\ce{sp^3}$ hybridized, is less electronegative than the $\ce{sp^2}$ hybridized aromatic carbon.  The carbon with more s-character is more electronegative, electron density likes to flow from low to high s-character orbitals since the more s-character in an orbital, the lower its energy.  If inductive arguments were most important we would expect the ortho isomer to predominate followed by the meta and para isomers in that order.  This is not observed, the meta isomer is only a minor isomer.
Resonance arguments are based on the concept of hyperconjugation involving the $\ce{C-H}$ bonds in the methyl group.  The following figure shows some of the possible hyperconjugative resonance structures for toluene.

Additional resonance structures can be drawn using the other methyl $\ce{C-H}$ bonds as well.  As the resonance structures show, electron density is increased at the ortho and para positions, not the meta.  This hyperconjugative resonance effect makes the ortho and para carbons more susceptible to electrophilic attack.
Hyperconjugation explains how alkyl groups can be electron releasing in a resonance sense.  Hyperconjugation explains the mild activating effect of an alkyl group in electrophilic aromatic substitution and its ortho-para directing effect.

I find this argument is a lot more convincing when applied to the EAS intermediate, not the putative hyperconjugated-dissociated zwitterion.

@Lighthart Why should the argument be more or less convincing if we apply it (hyperconjugation) from the starting material or the intermediate side?  Hyperconjugation is applied to other neutralmolecule phenomenon like the rotational barrier in ethane and why more highly substituted alkenes are more stable.

Hyperconjugation is too quantum mechanicky for mere mortals to grasp easily.  I have found students grasp the energy of the resulting cation intermediate more quickly.  So the people I am convincing more easily are novices.  Your explanation is quite good, but very sophisticated.

@ron But I learnt that hyperconjugation occurs when there is partial overlap of C-H sigma bond and adjacent empty-p orbital of carbon. However your structures suggests to contradicts this as the C-H bond is fully broken, C=C bond is completely formed and a unit positive charge is involved in the resonance?

@ron Is it better to say that only partial electrons and partial positive charges are involved in the structures, assuming electron wave nature instead of particle nature?

@An_Elephant  What you say is correct, but what I've drawn is just a resonance structure.  It is an extreme form with charge separation, but it is meant to indicate that the $\ce{C-H}$ bond donates electron density into the ring (e.g. hyperconjugation).