text:
How to determine the concentration of a base through indirect titration?

$5.267~\mathrm{g}$ of $\ce{Na2CO3}$ was dissolved in $250.00\ \mathrm{mL}$ of water. $10.00\ \mathrm{mL}$ of this solution was titrated with $\ce{HCl}$. The end-point occurred at $21.30\ \mathrm{mL}$. This acid was titrated with $25.00\ \mathrm{mL}$ of $\ce{Ba(OH)2}$. The end-point occurred at $27.10\ \mathrm{mL}$. Calculate the concentration of the $\ce{Ba(OH)2}$.
  [Answer: $0.1013~\mathrm{M}$]

Using the mass they gave me ($5.267~\mathrm{g}$) I worked out the moles of $\ce{Na2CO3}$ which was $0.0497\ \mathrm{mol}$.
I then worked out the concentration of $\ce{Na2CO3}$ using the moles I just worked out and the volume they gave me ($0.25~\mathrm{L}$) and I got $0.1988~\mathrm{M}$.
The question says that only $0.01~\mathrm{L}$ was used, so I worked out the number of moles this would be which was $0.001988\ \mathrm{mol}$.
Using stoichiometry, I multiplied this value by two to get the moles of $\ce{HCl}$ that would have reacted with this, which was $0.003976\ \mathrm{mol}$.
It says that the end-point occurred at $21.30~\mathrm{mL}$, which if you subtract the volume of $\ce{HCl}$ which was already there, I worked out that $11.3~\mathrm{mL}$ or $0.0113~\mathrm{L}$ of $\ce{HCl}$ must have been used.
From here, I worked out the concentration of $\ce{HCl}$ (using $0.0113~\mathrm{L}$ and $0.003976\ \mathrm{mol}$) and got $0.35186~\mathrm{M}$.
It says that for the next reaction, $25~\mathrm{mL}$ of $\ce{Ba(OH)2}$ was used and the end-point occurred at $27.10~\mathrm{mL}$. From this information, I worked out that $0.0021~\mathrm{L}$ of $\ce{HCl}$ must have been used in this reaction.
By knowing the volume of $\ce{HCl}$ used, I worked out the amount (moles) of $\ce{HCl}$ used in the reaction, using the previously found concentration. I got $0.0007389\ \mathrm{mol}$.
Using stoichiometry, I worked out the amount of $\ce{Ba(OH)2}$ which must have reacted with this many moles of $\ce{HCl}$. I divided the $\ce{HCl}$ moles by $2$ and got $0.00036945\ \mathrm{mol}$.
By using the volume of $\ce{Ba(OH)2}$ used which they give me ($0.025~\mathrm{L}$) and the moles I just found ($0.00036945\ \mathrm{mol}$) I finally worked out the concentration of $\ce{Ba(OH)2}$, which was $0.0148~\mathrm{M}$.

My answer, $0.0148~\mathrm{M}$ is different to the answer the question gives ($0.1013~\mathrm{M}$), and I'm not sure why.

It says that the end-point occurred at $21.30\ \mathrm{mL}$, which if you subtract the volume of $\ce{HCl}$ which was already there, I worked out that $11.3\ \mathrm{mL}$ or $0.0113\ \mathrm{L}$ of $\ce{HCl}$ must have been used.

I think, this is where you went wrong. If a question says something along the lines of

The end-point occurs at $\dots~\mathrm{ml}$

then you should generally assume that that volume of whatever you are titrating with was used. In this case, that would mean a full $21.3~\mathrm{ml}\ \ce{HCl}$. The same thing of course happened at your point 7, where you write:

It says that for the next reaction, $25\ \mathrm{mL}$ of $\ce{Ba(OH)2}$ was used and the end-point occurred at $27.10\ \mathrm{mL}$. From this information, I worked out that $0.0021\ \mathrm{L}$ of $\ce{HCl}$ must have been used in this reaction.

Working through this bit by bit, and starting at your point 5:
$$c(\ce{HCl}) = \frac{n(\ce{HCl})}{V(\ce{HCl})} = \frac{0.003976~\mathrm{mol}}{0.0213~\mathrm{l}} = 0.1867~\mathrm{M}$$
For the titration of $\ce{Ba(OH)2}$:
$$n(\ce{HCl}) = c(\ce{HCl}) \times V(\ce{HCl}) = 0.1867~\mathrm{M} \times 27.10~\mathrm{ml} = 5.059~\mathrm{mmol} = 0.005059~\mathrm{mol}\\
n(\ce{Ba(OH)2}) = \frac{n(\ce{HCl})}{2} = \frac{5.059~\mathrm{mmol}}{2} = 2.529~\mathrm{mmol} = 0.002529~\mathrm{mol}\\
c(\ce{Ba(OH)2}) = \frac{n(\ce{Ba(OH)2})}{V(\ce{Ba(OH)2})} = \frac{2.529~\mathrm{mmol}}{25~\mathrm{ml}} = \frac{0.002529~\mathrm{mol}}{0.025~\mathrm{l}} = 0.1012~\mathrm{M}$$
The difference still there can be attributed to a rounding error somewhere along the way.

This is how I work out the answer which is 0.1012 M.

It is better to use MathJax to have an actual rendered text like I did in my answer rather than upload an image …

Echoing @Jan, please typeset.


text:
Can silver metal be used to electrochemically reduce copper?
I have a solution of copper(II) ions and several cations.  I am trying to reduce the copper ions to copper metal with electricity.  My Ideas was to use silver as the anode and a copper wire as the cathode.  I have seen people say to do this using platinum as the anode, however this is expensive.  I was wondering if silver could be used as an alternative anode because it has a higher reduction potential than copper.  If not, could any metals besides gold and platinum work.  I have attached a picture of my proposed cell.  I have tried the cell, but the silver began to corrode and form a silvery deposit, probably a silver salt.

No as it is less reactive than copper, moreover if you look at these reactions you will see that Ag2SO4+Cu=CuSO4+Ag

In principle, electrically depositing copper(II) ions onto a silver electrode should be possible. As it should also be possible to deposit those ions on just about every other cathode in existence as long as there are no more noble cations in solution.
However, depending on what you want to do with the copper, you may run into problems. Do you want copper-plated silver? That should be fine. Do you want to redissolve the copper to get a pure solution by acidifying? In that case, you cannot use the typical $\ce{HNO3}$ method because that would oxidise the silver, too, and you would end up with a mixture of copper(II) and silver(I) nitrates in solution.

No, I am trying to get the copper as a powder by reducing it with a copper cathode and silver anode.  I started with copper ore, dissolved it in acid, and I am now trying to bring it to the metallic state.  I wanted to make sure that the silver would not oxidize or corrode.

@popgalop You won’t get copper as a powder, you will get a metallic plating on your cathode.

Yes, but I have heard that it mostly flakes off.  I was going to have the copper in a plastic tube with only the end exposed and scrape the copper off of the end of the wire.  The idea is to obtain the pure copper from the ore.

@popgalop Can’t comment on that but my lab course experiment (with platinum cathodes) had a fine sheet of copper on the cathode that you needed to get off with $\ce{HNO3}$.

Ok, I just wanted to make sure the silver wouldn't corrode in the strongly oxidizing conditions.

@popgalop You have reducing conditions if you want to reduce copper …

I attached a sketch of my proposed cell.  I plan on reducing the copper ions at a copper cathode, and using the silver metal as an inert anode.

@popgalop Ah! In that case that should be no problem and oxygen should be generated at the anode given the correct conditions. Sorry for misunderstanding!

Or maybe $\ce{Ag}$ would dissolve to some extent, and that would be a problem. Normally people use copper anode for this purpose, so as not to bother about impurities. Copper in, copper out.

Otherwise, use carbon (graphite). It will oxidize, but who cares. It won't reduce anyway.

I don't want to use carbon, because it will contaminate the product.  And I don't want to use copper because I am trying to quantitatively determine the amount of copper in the solution.

Then still use carbon. It won't contaminate anything (that is, except atmosphere).

Carbon will flake off as a powder and contaminate the copper from watching videos

Then put the carbon electrode in another vessel and join them by a salt bridge.

I tried the silver, but it began to corrode and flake off into the solution.

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Why are full and half filled orbitals the most stable?
Why are degenerate orbitals (restricted to a single spin) less stable when neither fully filled nor completely empty? Why, in most molecules, are half-filled and fully-filled shells more stable than partially filled shells? I understand that transition metals often promote electrons to their d orbital to make it half filled because this makes them "more stable." Why would this be the case? Relatedly, why are orbitals filled with single electrons in one spin before doubling up to form pairs and complete orbitals? Why must degenerate orbitals follow hund's rule? It seems like it would be more stable for electrons to pair up as they are added to an orbital rather than to become additional radicals in the orbital. Is hund's rule a result of maximizing stability, or is there some other principle at work?

In short, the so-called Fermi correlation which is a consequence of the Pauli principle and applies only to electrons of the like spin keeps such electrons further apart comparing to the case of electrons of unlike spin. Naively one might think that a larger average distance between electrons of like spin reduces electron-electron repulsion energy which is the reason for the stability of high spin states. And while it is true that for each and every pair of electrons electron-electron repulsion energy is decreased as the average distance increases, it was shown that in the whole electron-electron repulsion energy for the molecular system in a high spin state is actually greater that in a lower spin spate.
So, the modern explanation of the lower energy of high spin states is that a larger average distance between electrons also effectively leads to less screening of the nucleus, i.e. each electron experiences a larger nuclear effective charge. As a result the unpaired electrons can approach the nucleus more closely and the electron-nuclear attraction is increased. And this increase in the electron-nuclear attraction energy (which is negative) overcomes the above mentioned increase in the electron-electron repulsion energy (which is positive) leading to a greater stability of high spin states.
For details, see Boyd, R. J. Nature 1984, 310, 480-481, or any modern textbook on quantum chemistry.

I have a quantum exam on monday, and this is one of the concepts being tested. This is an attempt to justify my procrastination, and an attempt at convincing myself that this counts as review. I am writing this to complement @Wildcat's already excellent answer.
The point that I am trying to prove here is that in general, for a given electron configuration a state in which the spins are unpaired has lower total energy than one in which the spins are paired.
To make this point, I consider the case of an excited helium atom with the electron configuration $1\mathrm{s}^1 2\mathrm{s}^1$
The possible spin pairings:
$$\begin{array}{cc}
\hline
1\mathrm{s} & 2\mathrm{s} \\ \hline
 \uparrow & \uparrow\\
\uparrow & \downarrow \\
\downarrow & \uparrow \\
\downarrow & \downarrow\\
 \hline
\end{array}$$
This gives rise to $^1S$ and $^3S$ terms (ignoring spin-orbit interaction).
For the singlet case and triple case (assuming the orbital approximation, wherein the exact many electron wavefunction is approximated as a product of one electron wavefunctions. In that, sense $1s$ , $2s$ etc. given below are similar to hydrogenic atomic orbitals but aren't exactly the same. On second thought, I think the notation can be misleading.),
$$\psi_{\text{singlet}} \approx \frac{1}{\sqrt2}[1s(1)2s(2) + 2s(1)1s(2)]\frac{1}{\sqrt2}[\alpha(1)\beta(2)-\beta(1)\alpha(1)]$$
$$\psi_{\text{triplet}} \approx \frac{1}{\sqrt2}[1s(1)2s(2)- 2s(1)1s(2)]\times
  \left\{
\begin{array}{ll}
      \alpha(1)\alpha(2) \text{or} \\
      \beta(1)\beta(2) \text{or} \\
      \frac{1}{\sqrt2}[\alpha(1)\beta(2) + \beta(1)\alpha(2)] \\
\end{array}
\right.$$
You can check that these wavefunctions have been construted to obey Pauli's Exclusion Principle.
Notation: $1s(.), 2s(.)$ etc. are the spatial orbitals and $\alpha(.), \beta(.)$ are the spin functions. The number in parantheses is a label for the electron. For example, $1s(1)$ means electron one is in the $1s$ orbital, and $\alpha(1)$ means electron one is spin up.
The Hamiltonian is $$ \hat{H} = \hat{H}_1+ \hat{H}_2 + \frac{e^2}{4 \pi \epsilon_0 r_{12}}$$
where, $\hat{H}_i$ is the hamiltonian for the single elctron $i$ neglection repulions. The third term accounts for inter electron repulsions in this 2 electron case.
$$E = \langle \psi|\hat{H}|\psi \rangle $$
and using the approximate expressions we calculated above to find the energy for the singlet and triplet case. I have dropped the spin functions because the hamiltonian doesn't depend on spin here.
$E_{\text{singlet}} = \frac{1}{2}\int \int ([1s(1)2s(2) + 2s(1)1s(2)])^* \left ( \hat{H}_1+ \hat{H}_2 + \frac{e^2}{4 \pi \epsilon_0 r_{12}} \right)([1s(1)2s(2) + 2s(1)1s(2)]) \mathrm{d} \tau_1 \mathrm{d} \tau_2$
$E_{\text{singlet}} = E_{1s} + E_{2s} + \frac{1}{2}\int \int \mathrm{d} \tau_1 \mathrm{d} \tau_2 ([1s(1)2s(2) + 2s(1)1s(2)])^* \left( (\frac{e^2}{4 \pi \epsilon_0 r_{12}}) \right )([1s(1)2s(2) + 2s(1)1s(2)]) $
$E_{is}$ is simply shorthand for $\langle \phi_{is}|\hat{H}_i|\phi_{is} \rangle$
Now, the last integral breaks up into two integrals
$$J_{12} = \frac{e^2}{8 \pi \epsilon_0} \int \int \mathrm{d} \tau_1 \mathrm{d} \tau_2 [1s(1)]^2 \left(\frac{1}{r_{12}} [2s(2)]^2\right)$$
This is called the Coulomb integral. The second is called the exchange integral and is given below.
$$K_{12} = \frac{e^2}{8 \pi \epsilon_0} \int \int \mathrm{d} \tau_1 \mathrm{d} \tau_2 [1s(1)2s(2)] \left(\frac{1}{r_{12}} [1s(2)2s(1)]\right)$$
Both these integrals are positive quantities. You can convince yourself of this fact if you examine the integrands closely.
Based on this $$E_{\text{singlet}} = E_{1s} + E_{2s}+J_{12}+K_{12}$$
A similar calculation for the triplet state yields (I am omiting the tedious algebra),
$$E_{\text{triplet}} = E_{1s} + E_{2s}+J_{12}-K_{12}$$
the $E_{is}$'s are negative terms (they are just orbital energies). The two positive terms $J$ and $K$ come from introducing coulombic interaction and Pauli's exclusion principle in our system.
You can clearly see that the triplet state with has a lower energy. You can generalise this result to conclude that for a given configuration spins unpaired is always lower in energy than spins paired.
The reason is something @Wildcat has already adresssed. Unpaired elctrons lead to poorer screening, and the increased electron-nucleus interaction is what leads to lowering of energy (despite an increased electron-electron repulsion in this case).

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Ionic charge factors
New to chemistry so forgive the fundamental question. What exactly determines the element's charge? Oxygen has 2- and hydrogen has 1+. Oxygen wants to receive 2 electrons to complete a shell when hydrogen wants to donate 1 electron. For example why wouldn't hydrogen want to gain 1 to complete its shell? What exactly is the methodology to  this of any given element (non transitional obviously). I understand that the groups of the periodic can be used and memorized but whats the actual scientific theory as to why a given element donates rather than accepts and vise versa.

There's no such thing as "element's charge" and saying that hydrogen wants to get rid of electrons is nonsense.

@Mithoron then when we say oxygen is 2- what is that referring to? Its ionic form? Can u better explain?

You can say for example that Na2O consists of sodium cations and oxygen anions, but it's simplification.

Since you are new to chemistry, I assume the concept you are searching for is electronegativity.
Each element has an electronegativity, which in short is its tendency to attract electrons. Hence, since oxygen has electronegativity 3.4 and carbon has electronegativity 2.6, the $\ce{CO2}$ molecule can be seen as one $\ce{C^{4+}}$ and two $\ce{O^{2-}}$.
Of course this is a gross oversimplification and not actually how the $\ce{CO2}$ molecule works, so don't get attached to this view as you learn more chemistry.
To go forward, I suggest you check the different kinds of chemical bonds.


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Clarification on absorption spectra and crystal field theory
My textbook, Chemistry: The Central Science, gives this figure to explain the purple color of the hexaaquatitanium(III) ion:

Such absorption spectra are common to chemistry and biochemistry. However, as the crystal field theory explains, the peak wavelength absorbed is equal to the splitting energy $\Delta$ between the two split energy levels of the d subshell. This is illustrated in the figure below.

So the splitting of the d subshell explains the peak of the absorption spectrum. However, I’m not comprehending why the curve slowly descends on either side of that peak—if only 495 nm light is absorbed, then shouldn’t the absorption spectrum show absorption only at 495 nm, with no absorption anywhere else? Why does absorption slowly decrease as you move in either direction away from 495 nm? What mechanism allows these other wavelengths to be absorbed as well?
Moreover, normally, when things absorb light, they reemit the energy in the form of infrared. That’s why that wavelength is taken out of the wavelengths that reflect back to your eyes; if the same wavelength were reemitted, you would perceive it too. I assume infrared is radiated because electrons can take alternate pathways to move back to ground state, jumping down in small steps that release less energy each time. However, with d subshell splitting, there is only one possible way for the electron to climb down to ground state. Shouldn’t that mean that the wavelength it absorbed will be reradiated out? Shouldn’t we see that wavelength, then?

You would expect a sharp peak for a well-defined structure which has bond lengths and angles that do not significantly change. In a crystal structure at low temperatures you may be able to observe sharp peaks rather than broad bands.
In solution, however, there is a number of effects that broaden the peaks, making the entire thing less sharp:

aqua ligands can dissociate and reassociate; if the photon hits the complex while the structure is different, the wavelength will be, too
typically, the ligands of a complex vibrate both symmetrically and antisymmetrically. Both vibrations lead to structure changes, therefore electronic changes and thus changes in the absorption wavelength.

Different solvents will give different peak broadening (I saw a graph once, but I can’t remember where it was).
The titanium(III) spectrum is also more complex, by the way. If you look closely, you can see a shoulder on the low-energy side of the main peak. This is because the excited state is Jahn-Teller distorted, and there is not one but two transitions possible that slightly differ in energy. (The ground state is very, very weakly Jahn-Teller distorted, because there is not much energy to be gained.)

After a molecule has absorbed the energy of a photon and went into an excited state, there is not only the simple rerelaxation that would emit a photon of the same wavelength to return to the ground state. Rather, by a set of complicated mechanisms, the energy can be transformed into vibration (more and more vivid than before) and thus slowly released in form of heat. That is why a solution that absorbs light will warm slightly.


text:
Calculation correct binding energy for a system?
I am doing DFT calculations (VASP) to calculate correct binding energy for an amorphous silica surface and silver adatoms. From MD simulation, we found that a binding energy of 0.2 eV gives results which can be matched to experimental ones.
Since this is an amorphous surface, how do I go along to measure the correct binding energy. I have tried 18 runs where I have relaxed the adatom at different positions and at different heights but how many runs to do this as each result has a different value?

Maybe I misunderstand your question, but why you don't stick to MD simulations?

What, precisely, do you want to know? For an amorphous surface one would expect a range of binding energies because each surface 'site' is a little different from the others. So, you are looking at a spread of binding energies and you would need to characterize the shape of that spread. Now, you will also have to consider the possible errors and variation in DFT itself, which can be considerable depending on the system.

Yes, the binding energy value is an average over the surface. I want to do DFT so as to see whether I get a similar value of binding energy that I did in MD. Also to know the bond length between the atoms but what I have observed is that the spread is too large and the values are quite high. I am confused on how to accurately determine these two.


text:
Solubility specification in a standard
I'm reading a standard where under the specification for a substance it says:

Solubility in Alcohol: 1 ml in 3 ml 50% alcohol remains clear to 8
  ml.

Just wanted to confirm my interpretation: I make a 50-50 Ethanol + Water solution (w/w) and then add 1 ml of test-sample to 3 ml of this solution. This should remain clear.
Then I continue adding more of the alcohol solution till total quantity is 8 ml & throughout the mixture must stay clear?
Is this right? Any comments?
Edit:
Here is a supplier specification from Sigma Aldrich which has a variation of the same wording:

I'm not sure what you are doing but that doesn't sound right.  Are you preparing a standard (of what) from a concentrate of the substance?  What is the substance?  If you can be specific I'm pretty sure we can come up with the answers you need.

@airhuff Thanks. No, not preparing a standard. Just reading a published standard for specifications of a chemical and this is one of the tests mentioned. I am just trying to interpret the test procedure correctly. From the wording it wasn't clear to me what the procedure should be.  I will post an image of the actual standard if I can.

That would be very helpful!

@airhuff I can't find my copy of the original standard but Sigma Aldrich is a regular supplier of the chemical and their version of the specifications also lists the same (but slight modification in quantities) test. Does this help? All I need to know is what the wording of the test means. i.e. What is to be done and how are the results to be interpreted. I have given my interpretation of the test procedure. Does that look right to you?

What you have said seems correct.  I don't know why you are doing that though, but so long as you do all is good I guess.  Is this a turbidity standard, or indicator?  Sigma's "Solubility (Turbidity)" looks a bit unfamiliar to me.  Also, they say the solubility is 1 ml in 2 ml, so where you say w/w I think should be v/v.  I'm still very curious what the chemical is.  I'm not sure I've helped you much here.

@airhuff Thank you! The why is simple: I have a sample of this material from a Manufacturer that wants to make sure its product meets the spec.

Ohhh, interesting, thanks for clarifying!  Glad to be of some help anyway and good luck.


text:
Can we use prefixes like iso, neo, etc in IUPAC nomenclature of organic compounds?
Shouldn't the name of the compound in the picture be 5-(1-ethyl-2-methylpropyl) nonane instead of the given name?

Related: Are common names of substituents accepted by IUPAC?

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) various prefixes are retained for use in general nomenclature. However, many prefixes are no longer recommended.

Trivial, common, and traditional prefixes have always been an integral part of organic nomenclature. However, as systematic nomenclature develops and becomes widely used, many of these prefixes fall by the wayside. Accordingly, each set of IUPAC recommendations contains fewer of these traditional prefixes.

For example, the prefixes “isobutyl”, “isopentyl”, “neopentyl” were contained in the 1993 recommendations but are no longer recommended as approved prefix. The prefix “isopropyl”, is still retained for use in general nomenclature; however, for the preferred IUPAC name (PIN), the preferred prefix is “propan-2-yl”. (The prefix “1-methylethyl” may be used in general nomenclature).
Anyway, the name “5-(1-ethyl-1-isopropyl)nonane”, which is given in the picture in the question, is clearly wrong since no parent structure is given for the substituents 1-ethyl and 1-isopropyl:
“5-(1-ethyl-1-isopropyl what?)nonane”
If you really want to use such a name including the substituents 1-ethyl and 1-isopropyl, the corresponding systematic name would be “5-(1-ethyl-1-isopropylmethyl)nonane”. However, such a name would be highly unusual and not in accordance with the basic principles of IUPAC nomenclature.
The name “5-(1-ethyl-2-methylpropyl)nonane”, which is proposed in the question, describes the correct compound and may be used in general nomenclature. However, the PIN for this compound is “5-(2-methylpentan-3-yl)nonane” since prefixes of the “alkanyl type” are preferred if the free valence is not in position 1 of the substituent group (see this similar question).


text:
Why can toluene be chlorinated at either the benzylic position or the ring depending on the conditions?
My book mentions a reaction for the preparation of benzaldehyde with toluene in which side chain chlorination of toluene gives (dichloromethyl)benzene, which upon hydrolysis gives benzaldehyde:

Why does the chlorination occur at the methyl group and not the aromatic ring in this case? Also, why chlorination with $\ce{FeCl3/Cl2}$ lead to 2-chlorotoluene (and 4-chlorotoluene), without any chlorination of the methyl substituent?

The halogenation of toluene is possible by two mechanisms. In German, they are referred to by the shorthands KKK and SSS, where KKK stands for Katalysator, Kälte, Kern (catalyst, cold and core) while SSS is Sonne, Siedehitze, Seitenkette (sun, sizzling heat and side chain).
At low temperatures and in the presence of an activating catalyst such as $\ce{FeCl3}$, halogenation of toluene will typically occur as phenyl ring halogenation following the electrophilic aromatic substitution mechanism.

Scheme 1: Catalytic chlorination of toluene with $\ce{Fe^3+}$ as a catalyst. ortho- and para-products expected; (para not shown) mesomeric structures of the intermediate not shown.
Und higher temperatures and irradiation, halogenation takes place according to a radical chain mechanism.

Scheme 2: Radical chlorination of toluene under irradiation. Multiple chlorination possible. Mesomeric structures of the benzyl radical not shown.

The electrophilic aromatic substitution is, of course, not as easily possible on the benzylic carbon; we would somehow need a free electron pair to appear there — not likely.
On the other hand, if a chlorine radical attempted to remove a hydrogen of the phenyl ring, that would result in a radical that cannot be delocalised (the $\ce{C-H}$ bond is orthogonal to the π-system) which is therefore very unstable. Another way of looking at it is to consider the $\ce{C–H}$ bond strengths: weaker bonds are more easily cleaved homolytically, as is the case in free-radical halogenation. Aryl $\ce{C–H}$ bonds are roughly $\pu{100 kJ/mol}$ stronger than benzylic $\ce{C–H}$ bonds (Wikipedia), so radical abstraction preferentially occurs at the benzylic position.

Upon excitation, homolytic cleavage of the $\ce{Cl-Cl}$ bond occurs:
$$\ce{Cl2 ->[$h\nu$] 2Cl*}$$
The resulting chlorine radical abstract a hydrogen atom (not a proton!) from the methyl group of toluene, yielding a benzyl radical.

$$\ce{Ph-CH3 + Cl* -> Ph-CH2* + HCl}$$
The benzyl radical may either

react with $\ce{Cl2}$:

$$\ce{Ph-CH2* + Cl2 -> Ph-CH2Cl + Cl*}$$
or undergo radical recombinations, such as
$$\ce{2 Ph-CH2* -> Ph-CH2-CH2-Ph}$$ or
$$\ce{Ph-CH2* + Cl* -> Ph-CH2Cl}$$
Taking into consideration that the radical concentrations are typically much lower than that of $\ce{Cl2}$, these recombination steps are much less likely than the reaction of the benzyl radical with a chlorine molecule.
Benzyl chloride, the primary reaction product of the photochlorination may again serve as a hydrogen atom donor in a further reaction with chlorine radicals:

$$\ce{Ph-CH2Cl + Cl* -> Ph-CHCl* + HCl}$$

After dichlorination on methyl group of toluene, why can't a third chlorine attach on the benzylic carbon?

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text:
Formation of Lactone from 5-amino-decanoic acid?
As expressed in the title, I am trying to figure out how to form a lactone from 5-amino-decanoic acid. The thing that comes to my mind as a start is the attachment of the amino group ($\ce{NH2}$) to the carboxyl carbon and then immediately break the double bond of this carbon with oxygen … But the process gets more likely to form a lactame, where my exercise is asking a Lactone. Can anyone help?

It would be easier if the Oxygen in COOH would attach to the Carbon bonded with NH2, but that carbon is also bonded with the other Carbons on the decane chain, so it is not likely to be attached as much as the Carbon of COOH, which has a big delta plus due to the Oxygens, and the NH2 has a couple of unpaired electrons to use.. But still this is more likely to give a Lactam rather than a Lactone

Yes that's what I think as well, but that's what the exercise is asking..

I doubt your book is asking you to convert this molecule into a lactone. This would be very hard to do. It could, however, be easily converted into a lactam. 5 and 6 membered lactams are formed by simple dehydration. For your γ-lactam, this can be achieved by heating the amino carboxylic acid, forming the thermodynamically favourable, stable 6-membered lactam. Similarly, lactones can be be interconverted by adjusting the pH.

IF you really wanted to convert the amine into a lactone, it would involve some pretty uncommon reagents. This blog post mentions that pyrylium salts can be used to make a primary amine into a leaving group, which would, presumably, allow you to substitute a hydroxyl group, thereby giving you access to the lactone.
http://curlyarrow.blogspot.hu/2009/09/how-to-turn-amine-into-leaving-group.html

I believe there was an error of printing as I perfectly agree with you. So I'll just take your answer and write a Lactam, which is the only normal product there..

Good point @Curt. F. I guess my fingers remember the word 'dehydrogenation' too well these days :P.

After my initial comment I reconsidered and thought it through a lot. You can technically by not-so-common reactions turn your 5-aminodecanoic acid into a lactone if:

You convert the amine into a leaving group — diazo jumps to mind (add $\ce{HNO2}$) but a triple alkylation would work, too.
You then make sure that the carboxylic acid is nucleophilic enough to attack the ex-amine carbon in an $\mathrm{S_N2}$ manner (i.e. add a base) while making sure that:

No $\mathrm{E2}$ elimination will occur
There are no other nucleophiles around that could attack.

If, unluckily, your amine eliminated, you will need to attack the double bond with a suitable leaving group. I don’t think a carboxylate is strong enough to attack double bonds to form a lactone.
You look the thing you just did up in SciFinder and if it doesn’t exist, you publish it just because wow and why. Don’t aim for Angewandte or JACS, though, unless you did something really cool. ;)

A second option would be diverting after step 1 adding a hydroxide nucleophile so that you generate 5-hydroxydecanoic acid, which you can then transform into the lactone by about a million methods (Yamaguchi, Mitsunobu, Steglich just to name three).

With the alkene you can do an iodolactonisation then remove the iodine with e.g. Bu3SnH/AIBN although I guess you would probably get significant amounts of 5-ring product, don't know of any way to get selectivity with this kind of acyclic compound. Even the elimination would probably give you a mixture of isomeric alkenes

what about the triphenylpyrylium salt method? You can use triphenylpyrylium trifluoroacetate to form the ROCOCF3 ester corresponding to the primary amine, and then you can hydrolyze that to yield the alcohol, followed by lactone cyclization. It says here that this reaction is more 'clean and efficient' than diazotization: http://reag.paperplane.io/00002962.htm

@gannex Looks cool, but I don’t personally trust ‘clean and efficient’ mechanisms any more after having done too much total synthesis ;)

I wouldn't blame you. I think it looks like a nifty reaction though (on paper that is).

Yes but it only gives the amminoacid, and it asks to build a lactone out of it, so I believe it meant lactam instead, although it was useful to see such reactions through this question. Thank you very much sir! Perhaps trying photons as a catalyst might do the work, but I ain't trying very soon haha


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Why is the specific heat of ethanol not representing experiments
I am in my last year of high school completing an assignment hot and cold packs, and particular the enthalpy behind them.
My major lack of understanding is why my calculations are working for heat of water but not for heat of ethanol.
In the past week, I have tested $0.1$–$0.4~\mathrm{M}$ of calcium chloride, magnesium sulphate and potassium chloride and ammonium chloride, in $100~\mathrm{ml}$ of distilled water.
Before doing these experiments I did some basic calculations Using  $\Delta H= mc\Delta T$
for each solute and its respective concentration, I assumed water temperature to be $25~\mathrm{^\circ C}$ to just get a rough understanding of the temperature I could expect and got some calculation which looked like this:

I got the heat of solution from a PDF I found.
I then did the experiments and got results which were very close to what these calculations showed.
However, this week I decided to test another solvent, ethanol. I found its specific heat to be $2.44\ \mathrm{J/g}$ and did the calculations again. I saw that almost all the estimated temperature increased/decreased drastically depending on if it was an exothermic reaction or endothermic reaction. However, after doing this experiment the results I got was very different. I decided to just test $0.2~\mathrm{M}$ of each solution and my calculations estimated these temperatures

But my calcium chloride temperature ended up only reaching $42\ \mathrm{ \circ C}$ whilst the potassium chloride and ammonium chloride barely changed at all, if anything they went up by $0.1~\mathrm{^\circ C}$.
So where is the problem with my thinking of what I have done incorrectly. I'm guessing ethanol can't be modelled the same as water but I cant find any research supporting that.

Ethanol can be modeled like water all right (the same formula linking $\Delta\rm H$ to $\Delta\rm T$, etc.), but ethanol is not water. The enthalpies of dissolution of salts (all of them!) will be very different from those in water, to the point that knowing the latter gives you absolutely no clue as to the former. To add more to the confusion, your ethanol is most likely a water/ethanol mixture, but that's another story.


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Rate determining step
The formation of the cation is the rate-determining step. we can look at this in two ways. Either  that a cation is an unstable species and so it will be formed slowly from a stable neutral organic molecule, or that the cation is a very reactive species and so all its reactions will be fast, regardless of the nucleophile. why Both arguments are correct.

What's wrong with that? Yes, a cation is an unstable species and hence very reactive. Both arguments are right, and both lead to the same conclusion.

@IvanNeretin but here a carbocation is stable due to hyperconjugation .

There are so many different grades to "stable". Yes, it is stable compared to other cations, much like a porcelain vase is stable compared to a house of cards. But it is nowhere near any real molecule. Those are like buildings of concrete and steel.

I don't know anything about this specific reaction, but maybe what it says is something like this:
1) cation forms slowly so it's rate determining - "and so it will be formed slowly from a stable neutral organic molecule"
2) The formation of the cation is the slowest step in the reaction because after it has already formed the rest of the steps are fast in comparison - "cation is a very reactive species and so all its reactions will be fast"

The reaction that you are looking at is known as an $S_N1$ reaction. In the first step of the reaction, the leaving group (in this case bromine) falls off the carbon chain, forming a carbocation. In the second step, some sort of nucleophile (in this case a hydroxide group) attacks the carbocation, effectively replacing the leaving group. The reaction is called an $S_N1$ reaction because the rate law (and thus the rate determining step) involves the concentration of a single compound - the substituted alkyl group. This justifies that the rate determining step must be the formation of the carbocation.
This conclusion should also be logical. In order to form the carbocation, you not only need to separate two opposite charges, but to break the full octet of the carbon that becomes a carbocation. These steps require a lot of energy, making this part of the reaction very slow. The next step of an $S_N1$ reaction remedies both of the issues created in the first step. Not only are a negative and a positive charge brought together, but the octet of the carbocation once again becomes full. While it is true in your diagram that the formed carbocation is tertiary, which makes it more stable as a result of hyper-conjugation, this does not change the fact that it is a non octet filled carbon, and thus an unstable intermediate. After the carbocation is formed, it will almost instantaneously react with a nucleophile, as this is such an exothermic step.


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How to measure the water content of opaque liquids?
Water content of (semi) transparent liquids could be approximated via absorption spectra. What method(s) can be used to measure the water content of an opaque liquid such as milk for example?

I would add Karl Fischer titration.
Of course for your milk example, you would have to dilute your sample quite a lot in a dry (or known-dryness) solvent before titration. Also, the strong redox solution might interact in complicated ways with other materials in your sample (proteins, etc?).

You could put your liquid in a sealed container to let water from your sample evaporate and saturate the air inside the container with water vapor.  Then you could measure the amount of water vapor in the air.  This will tell you the vapor pressure of the water in your sample and is a measure of water activity, which is slightly distinct from water content.
I suspect that for milk, it is OK to approximate the solution as ideal, which mens the water activity will be proportional to the water content.

Many different techniques would work; here are three:

Weigh, evaporate water, and weigh again.
Measure the dielectric constant. This requires calibration, perhaps by method 1, but can even be used on solids such as wood.
Put sample in a closed container with a hygrometer... also needs calibration.

(1) Not only water evaporates. (2) How is permittivity useful here? Assuming linearity? (3) How selective is a hygrometer to water?

Correct, if there are other volatiles, then water would need to be absorbed selectively and weighed, e.g. in NaOH.   2. See https://www.ces.ncsu.edu/nreos/wood/wpn/methods_moisture.htm for measurement of wood moisture.  3. Good question... that's why calibration is needed.


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Vapor pressure, volume and boling point
The saturated vapor pressure is designated as the gas pressure in a closed container when the evaporation process has reached an equilibrium (when there are as many molecules returning to the liquid as there are escaping it). Doesn't that mean that vapor pressure is dependant on the volume? Why not?
And how exactly is vapor pressure connected to the boiling point? I know that a liquid boils when its vapor pressure becomes equal to the external pressure but why is that? Is it because the pressure in the bubbles, that are formed by heating the liquid, is the same vapor pressure?
Thanks in advance.

Related: http://chemistry.stackexchange.com/questions/38653/a-question-about-vapor-pressure-and-boiling-point/38654#38654

So I was right? The pressure inside a bubble, which was formed by heating, is the vapor pressure? And what about my first question?

Yes, that's right. As for your first question: no, it doesn't. Gas with some liquid in it behaves quite differently from gas alone. You double the volume, but the pressure would not drop by a factor of 2. It would not change at all, for some liquid would evaporate and maintain the same pressure that was before.

authors: Αντώνιος Κελεσίδης


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Baking soda & magnesium chloride reactions

Woke to this interesting reaction.
Would love any explanation of this occurrence. If it is perfectly common then I apologize for my ignorance. The mag was trace in the clear glass suspended over the white glass jar that had a strong mixture of baking soda and water. They are not touching and are as I found them. Look carefully and you can see the "gel puddle" that formed on my dish drainer. There are slight bubbles in the gel.
Or is it ectoplasm from the ghost?  (Lol)

duplicate:-http://chemistry.stackexchange.com/questions/4046/what-is-the-reaction-between-magnesium-chloride-and-sodium-bicarbonate

It is difficult to reconstruct what actually happened based on a single picture.
Nevertheless, the essential effects seem to be the hygroscopy and deliquescence of magnesium chloride. Magnesium chloride has a strong affinity for moisture. When exposed to the atmosphere, the salt will absorb relatively large amounts of water from the air. Magnesium chloride can form several crystalline hydrates: $\ce{MgCl2.2H2O}$, $\ce{MgCl2.4H2O}$, $\ce{MgCl2.6H2O}$, $\ce{MgCl2.8H2O}$, and $\ce{MgCl2.12H2O}$. Furthermore, the salt absorbs moisture from the atmosphere until it dissolves in the absorbed water and forms a concentrated liquid solution.
Apparently, the concentrated liquid solution of magnesium chloride ran out of the glass vessel and formed a puddle on the dish drainer.


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swift API to convert smiles strings to 2D chemical structures
Is there a method or free api written in swift for converting smiles strings to the corresponding 2D structures?
I have used babel or ibabel with python, but how do i use it with swift?

Your answer is very specific and I dont know whether this is the right SE for this answer. Any way look into http://softwarerecs.stackexchange.com/

There's no Swift-based API for cheminformatics at the moment. (And I'm not sure if it's worth doing.)
My suggestion, since you mention using Babel and Python previously, is to use obabel on the command-line from Swift like this.
The appropriate shell command would be:
 obabel [file.smi] -osvg [...]

Thanks. Do you know what methods some apps such as 'Elemental' use to convert smiles into the 2D chemical structures on the mobile device.

@James AFAIK, Dotmatics, the developer of 'Elemental' has their own cheminformatics toolkit. There are certainly commercial toolkits suitable for iOS.

@James I'm not sure about other apps.


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Zeolytes as Catalyst in hydrocracking
What is the working principle that allows zeolites to be catalysts in hydrocracking?
Is it only their acidic ability to hold hydrogen and release to crack the oil?

most probably its  because of the huge surface area provided by zeolytes


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How do ion–dipole bonds manage to break apart an ionic lattice?
I was reading this answer and its reasoning behind the dissolution of NaCl seems to lie in entropic favorability. However, as it states, many, if not most, dissolution reactions are favored by enthalpy (i.e. exothermic). Reasoning with energy alone, without entropy, it seems that ion–dipole forces are greater than the Coulombic forces between the ions themselves. However, how can this be? Aren’t ionic bonds second only to covalent bonds? How can water molecules form ion–dipole bonds with ions that are stronger than the ionic bonds within the lattice?

could you plz simplify the question.


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Is the behavior of a dilute strong acid the same as the behavior of a weak acid?
My assumption is yes, as the $\mathrm{pH}$ levels of each will be similar and thus they have the same acidity. Is this correct or am I missing something?

The pH is the same, yes, but the weak acid has undissociated molecules "in reserve", so to speak, and thus can neutralize a lot more of any base.
Indeed, this concept of weak acids and their conjugate bases having molecules "in reserve", reacting only when called upon, is the principle behind acid-base buffers.

There are two main differences that can be observed, that Oscar's answer did not mention.
Firstly, when you add a strong base to the acid solution, the solution pH changes far more drastically when you had used a strong acid, simply because the amount of acid is very little to begin with.
Secondly, there is no buffer region for a strong acid solution. With a weak acid solution, at the rise in pH will slow down until a buffer region that lasts pretty much until nearly all the acid has been neutralized.
For instance a weak acid HA with disassociation constant $10^{-4.75}$ at concentration $0.1 \text{ mol} \text{ dm}^{-3}$ will have a pH of about $2.88$. A semi-strong acid with disassociation constant $10^{-2}$ will need to be at concentration about $0.0015 \text{ mol} \text{ dm}^{-3}$ to have the same pH. The pH of the solution as these two solutions are titrated against a strong base BOH can be easily proven to be the logarithm of the root of some cubic equation, which will have the following familiar shapes:

(This graph is when the added base is at concentration $0.2 \text{ mol} \text{ dm}^{-3}$. The blue curve is for the weak acid and the orange curve is for the semi-strong acid.)

I didn't show the graph for a strong acid because it would be squashed too much to the left and there's really nothing to see. According to Wikipedia, what I called a semi-strong acid is still considered a weak acid, though on the strong side, but anyway it is strong enough for illustrative purposes.

I am guessing the red line represents your "semi-strong" acid and the blue line represents the weak acid? It's a good idea to add it in imo

@orthocresol: Done. I had forgotten about that!

The main difference between those two kinds of acid is the way that they dissociation occurs. A strong acid will rapidly dissociate in its ions with nearly a 100% of conversion p ex: HCl. Instead weak acids like acetic acid remain in equilibrium with its ions HAc <=> H+ + Ac- This means that in order to achieve a value of pH with a weak acid you need to consider that the acid will only partially increase the concentration of H+ ions so the final concentration will be given by the equilibrium constant (Ka) for this acid. Instead the pH given by a strong acid will be given simply by the concentration of the acid.
So if you have HCl with a concentration of 10^-3 mol/L then the pH of that solution will be three.
I hope this was helpful :)

Just fyi, acetic acid is HOAc, AcH actually represents acetaldehyde


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Removal of octadecanoic acid from paraffin wax
I have some very plain candles. I don't think they contain anything but paraffin wax and octadecanoic (stearic) acid, since they are odorless and very hard to bend, tending to crack instead (even at around 45 degrees Celsius). I want to make some moldable wax plugs for bottles without heating over about 50 degrees Celsius (it'll burn my hands, I have no other tools to use). How can I get rid of the acid? I don't particularly care if the acid is destroyed; but it would be nice if it wasn't so I could make soap or something.
Some possible ways I've thought of:

Distillation. I can't do this because it stinks and clogs up my glassware. Also, stearic acid decomposes around the same temperature as paraffin wax boils.
Dissolution of octadecanoic acid in some alcohol. Wax won't be dissolved because alcohols are polar and wax is not. However, I don't know how to actually do this as the wax may block access to most of the acid.
Neutralization of the acid using aqueous solutions of sodium hydrogen carbonate, sodium carbonate, or sodium hydroxide. This seems like the fastest way of removing acid; however the base may not come into contact with the octadecanoic acid due to separation of the layers.

What seems like the best or most effective method of removing the acid?

I honestly cannot tell why you would want to remove the octadecanoic acid, could you perhaps describe what you want to do with the cadles more exactly? How do you intend to mold wax without heating it?; it would still be solid.

In https://en.wikipedia.org/wiki/Paraffin_wax#History it is stated that octadecanoic acid is added to make the candles harder and to remedy their low melting point. My objective here isn't to have a putty-like substance, but rather to minimize the heat required to make it more soft. In addition to this, the acid will probably react with something inside the bottles (sodium carbonate solutions, etc) and contaminate my chemicals.

They sell paraffin in the grocery store to seal jam jars. Would that work?

@MaxW Where I live, they do not. Even if they did, this would still be interesting as an experiment.

Ok, you want to experiment? Boil with 0.1 molar NaOH for 20 minutes while vigorously stirring. Cool and let wax harden. Crack through wax and pour off NaOH solution. Repeat NaOH wash three times.

@MaxW: Isn't that creating soap, in a sense?

If that is your goal, why don't you just add an unsaturated fat? This is inherently easier than removing a substance. Also, wax has a rather sharply defined melting point, I don't quite understand you.

It should be noted that typically, candles are about 10% stearic acid and 90% paraffin. Also, there are many different grades of paraffin waxes which are generally identified by their melting points (often given as ranges); candles are usually made from paraffin graded from 53.3 °C (128 °F) to 65.6 °C (150°F), while paraffin of 76.6 °C (170 °F) is more common for molding and sculpting.
So I'm pretty sure that you didn't know what grade of paraffin you are after, and I would guess that you are looking for something more like cheese wax. While cheese wax is significanly softer, it actually has a generally higher melting range of perhaps 72 - 82 °C (161 - 179 °F).
Cheese wax is a blend of paraffins:

<30%  paraffin wax fume (Cas# 8002-74-2)
<30%  paraffin waxes and hydrocarbon waxes, microcryst. (Cas#
63231-60-7)
<20%  petrolatum (Cas# 8009-03-8)
30% Slack wax, petroleum, hydrotreated (Cas# 92062-09-4)

However, there may be multiple grades of paraffin used for your candles or the grade used may be of a low enough melting point (MP) to be useful for making the product you desire (in conjunction with other materials to make it softer for sculpting or molding).
The "analysis" and the separation procedure would be based on the melting point. So, (I would) try to use the melting points to separate the stearic acid from the lower temperature fractions.
Stearic acid's MP is 69.3 °C (156.7 °F)
Paraffin's MP may be as low as 37 °C (98.6 °F), but finding that grade of paraffin in your candles is unlikely. With a bit of luck, a low temperature grade of paraffin (48.3 °C (119 °F)) might show up; but more likely would be grades having a melting point of 53.3 °C (128 °F) and/or 61.1 °C (142 °F), which are common for candle making.
Step 1. Melt the candles in a pot and remove the strings. If possible, you would want to use a hot plate with accurate, good-quality temperature control.
Step 2. Insert a (candy) thermometer and a spoon or ladle in the pot (maybe a screen-type of ladle, like what is used for noodles and dumplings). You need the ladle to be warm, so that the wax doesn't freeze and stick to it when you use it to scoop out the stearic acid.

Step 3. Reduce the heat until precipitates begin to form. This should begin to happen somewhere around 70 °C. With an open pot, the cooler surface will probably cause the top to skim over. The higher melting point fractions might be what solidify first on the top, like slag on molten steel. The first precipitates or slag should be higher in octadecanoic (stearic) acid concentration. Use the ladle to scoop/filter out the precipitates (put the stearic acid in a in another pot or a bowl). You should keep collecting precipitates until it reached about 55 °C or until it all becomes solid. Be sure to make note of the final temperature. If the temperature at which it all solidifies is at or above 60 °C, then the kind of paraffin that you have is probably not what you wanted; and it will be quite difficult to separate, and this procedure is not very useful for separating the stearic acid from the paraffin.
Step 4. Even if the paraffin has a low MP, this will not be a 100% perfect procedure (far from it); you will probably need to repeat the (enrichment) procedure a couple, or maybe several, times. Also, you could probably recover some (low MP) paraffin from the precipitated/enriched stearic acid filtride. It might be favorable to have a few bowls or pots which represent collections at different temperature ranges... like, for example, the first bowl would contain solids from 70 - 65 °C; the second bowl contains solids from 64 - 58 °C; and the third bowl contains solids from 57 - 52 °C.
Step 5. Measure the new MPs. A good anylysis of the cleanliness of this procedure will be to compare the MPs. A pure solution of stearic acid will melt/solidify at exactly 69.3 °C, but obtaining a pure solution in this manner, is a pretty lofty dream. So, after the stearic acid fraction has melted, let it cool and  mark the temperature at which it begins to solidify and the point at which it is completely solid. If the two fractions that have been separated have distinctly different melting points, that's pretty good. Specifically regarding the paraffin, if it doesn't begin to solidify until it's below 50 °C then, great job, it's pretty clean and you obtained the low temperature paraffin that you desired!
Step 6. Depending on the cleanliness (melting point) and amount of the stearic acid that you have, you may want to try to salvage some of the paraffin from the stearic acid (filtride). I believe that this will be more difficult, and I would advise using a third pot (or second bowl) to collect the salvaged paraffin. And then work on cleaning the stearic acid out of the salvage, checking the cleanliness as in step 5, before combining the salvaged paraffin with the first "batch".
Step 7. Further chemical reactions for cleaning up the stearic acid (or converting it to soap) could be performed with better results from the enriched fractions.


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Density of the methane
I would like to know the density of the Methane with a temperature of 348.15 Kelvin because i guess it is not the same that we can find on Wikipedia. Someone know where can i find that ?
Thank you

The density of a gas at any temperature can be approximated using the ideal gas equation and the molar mass of the gas. Methane is a good candidate for a close approximation since it is nonpolar and relatively spherical. It should not deviate from ideality exept under extreme circumstances.
Density is the ratio of mass to volume:
$$D=\dfrac{m}{V}$$
If we assume that we have one mole of methane, we can use its molar mass to determine the mass of that one mole.
We can use the ideal gas equation to determine the molar volume of our gas (or any gas) at a given temperature and pressure. I presume you want the density at 1 atm or 1 bar, or you would have mentioned the pressure. How can we rearrange the ideal gas equation to solve for volume?
$$PV=nRT$$
Now it is just a matter of plugging in values. I am not going to do that for you.


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Boric acid ant poison
I made an ant poison using boric acid and peanut butter for grease eating ants.  They eat the peanut butter (mixed with some olive oil to get it closer to a liquid), but I see a white residue (crystals?) left behind.  How can I get the boric acid dissolved in an oil or grease?

Add a small amount of a liquid soap?

Boric acid is soluble in water, not oil.  Pick an emulsifier and add it to your ant poison.
Boric acid needs to remain acidic to retain its effectiveness as an ant killer.  Ants deaths due to boric acid poisoning is a topic still up for debate; however, it has been suggested boric acid is a neuro-toxin and that the acid damages one of the organs in an ants digestive system.
Peanut butter is usually slightly acidic; however, your particular mixture may be rendering the boric acid inert.  Solutions of sugar and boric acid with a high pH have been known to pass right through ants doing little harm. Perhaps using an acidic additive will increase the bait's effectiveness.
Boric acid in a 1-2% concentration may be ideal if the entire ant colony is targeted. To target the entire colony, enough poison should be used so the ant does not die before it makes it back it its colony.

Try to saturate a little amount of water with boric acid, and then create an emulsion with this saturated solution and your oily concoction. You will need a lot of vigorous agitation, so you perhaps should consider using a blender to produce the emulsion.


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Wavelength of an electron removed from an atom of hydrogen
I have the following exercise:

Determine the wavelength of an electron removed from an atom of hydrogen, in gas state, by a photon of wavelength of 45 nm.

So what I immediately thought was to apply De Broglie equation having
$ \lambda_{electron} = \frac{h}{m_{electron} v_{electron}}=\frac{h}{\sqrt {2 m_{electron} K_{electron}}}$
Ok then I thought that I need to calculate the kinetics energy of the electron.
Well if this was the photoelectric experiment with a metal we would have:
$K = h\nu - W$
But this is an atom we don't have an extraction energy W. Should I use the 1st ionization energy or is there other formula for calculating the kinetics energy of the electron.
Thanks!!

Yes, use the 1st ionization energy.

Hi! It can't be because I get to a negative kinetics energy

Then you are doing it wrong. How much is your $h\nu$, to begin with? How much is your $W$?

You're right, I got 1 calculation wrong. Thanks!

You should answer you own question so that it may be useful to future users.


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Partial pressure of gases in liquid
I was trying to understand what partial pressure of a gas in a liquid means and found the following sentence in a book I don't get it fully. What does it mean, can someone please explain? (What does it mean by a gas being in equilibrium with liquid?)

Partial pressure of a gas in solution refers to the pressure of the
   gas in the gas phase in equilibrium with the liquid.

Thank you in advance.

What illustration? Pressure out of equilibrium could have any value.

What do you mean by 'out of equilibrium'?

I have reworded my question.

Equilibrium of dissolution/vaporisation process...

@Mithoron I understand the purpose of your comment, but I think "What is your problem really?" was misunderstood to be insulting.

You have a gas species dissolved in a liquid.  The liquid is the solvent, and the gas species is the solute.  Suppose that the gas phase in contact with the liquid consists of mixture of several gases, and one of these gases is the gas species of interest.  In order for the concentration of the dissolved gas species in the liquid to remain constant, so that no more of the gas species dissolves nor does some of the gas species come out of solution, the partial pressure of the species in the gas phase must be at a certain equilibrium value.


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Melting and boiling point trend in Group II
The following picture shows the melting and boiling point trends down group II elements.

I have added question marks where the variability in data was rather disturbing (over two hundred degrees Celsius), or where the value is simply disputed. Nevertheless, this topic is not about the different results from measurements. Instead, it focuses on the "trend" in general.
For group I, we would say

the number of electron shells increases,
the increasing number of protons is unable to compensate for this fact,
therefore, atomic radius increases and
electronegativity, ionisation energies decrease,
hence the metallic bond is weaker $\Rightarrow$ melting and boiling points decrease
atomisation energy decreases.

Here are the atomic radii and ionisation energies of group II elements.

These (apart from radium) do seem to follow the same logic as in group I, however they do not correlate with the melting or boiling points. Some textbooks claim,

A good measure for the strength of a metallic bond is its atomisation energy.

Let us have a look.

Sure, this does agree better with the boiling and melting points. But note that it is actually not an explanation! Why do the atomisation energies behave in such a way?
This is where most resources either

ignore the problem altogether by not mentioning it,
give an explanation which is claimed by others to be faulty,
few acknowledge the issue and say no simple explanation is possible.

The faulty explanation
Beryllium and magnesium have crystal lattices that are hexagonal closest-packed (close-packed?). Therefore, they have their own trend. Then there are calcium and strontium which have face-centered cubic lattices. Finally, the pair barium and radium have body-centered cubic structures.
Others say this is not enough, because the differences should not be as significant, especially for magnesium but also for barium. Barium has a 8-coordinated structure but has a higher boiling point than 12-coordinated strontium!
It might also be stated that melting points themselves are not the best estimator of the strength of bonds. The bonds are not all broken, but rather rearranged and "made weaker" as a result.
The question
As of now, is there an explanation for melting and boiling point trends of group II elements? Is there consensus?

related http://chemistry.stackexchange.com/questions/32635/why-does-magnesium-have-an-exceptionally-low-melting-point

This paper seems to have a quantitative explanation https://journals.aps.org/prb/abstract/10.1103/PhysRevB.62.14818 but I don't really understand the paper from a brief skimming.

My 50 point bounty expired with no acceptable answer. Too bad, but it was worth the try.

@EdV I am inactive on Chem, but the bounty was appreciated, thank you. Indeed worth the try.

There can be no rigorous answer to such questions I think. We have no more than six data points, and each of them is just the way it is. The reason for the melting points is of course the crystal structure, and the bond properties therein. It can be nothing else, but of course that doesn´t explain much.

A clarifying note on the 'faulty' explanation - Be changes from hcp to bcc at 1527K, before melting at 1560K. Mg is the only one that goes from hcp->liquid.

Its goes down because the attraction between electrons in sea of electrons and cations in the metal lattice becomes weaker therefore less energy is required to interrupt the lattice and easier to melt .

What are you saying dude? Please read the question first

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


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Orbital radius: is 3d less than 3s?
So I was studying some graphics of the representation of orbitals in terms of radial dependence.
My graphic is similar to this one:
https://www.google.pt/search?q=radial+dependence+of+orbitals&espv=2&biw=1280&bih=728&source=lnms&tbm=isch&sa=X&ved=0ahUKEwi8kvOTuZvNAhWBshQKHYuNBw8Q_AUIBigB#imgrc=XTvxODPVjMkSHM%3A
Anyway, my book says that "the distance from the nucleus of the maximum value more distant from the nucleus is the radius of the orbital".
Well according to the graphic the maximum value of the orbital 3s corresponds to a greater value of the radius of the orbital than the orbital 3d. But that doesn't make sense, right? The orbital 3s should be closer to the nucleus, right?
This is making me confused I might be misreading the sentence? Can someone help me?

Maximum value means less than nothing. It is the average value that matters.

Sorry I didn't understand what you mean can you please elaborate more?

The answer depends on your definition of orbital radius or "closer".
Orbitals correspond to the probability distribution of finding an electron in a certain place around the nucleus. So you may define the radius as

the distance where the probability to find the electron has its
maximum, or
the average distance where the electron can be found, or
the radius of the sphere wherein the electron can be found with 50%,
or 90%, or 99% probability, or
some other fancy measure.

Which definition is best suited may depend on the type of comparison you make.

Ok I think I messed up the definitions of orbit and orbital. When I said closer I was thinking abou circular orbits. So based on your answer I would say that: if radius is the distance where the probability to find the electron has its maximum than the definition is correct, right? If radius is the average distance where the electron can be found, then the definition of the book, isn't right. Am I correct?


text:
How can you infer that density decreases when ice melts from the water phase diagram?
I know the phase diagram shows negative slope for water which is an exceptional case...  But how do you read that graph and get to the conclusion that density is increasing?
It's the one with pressure plotted against temperature
Thanks ..
sorry i couldn't attach the graph here...

https://en.wikipedia.org/wiki/Clausius%E2%80%93Clapeyron_relation

I know this equation..  But i thought maybe a different way to explain..  Thanks orthocresol

And water is not that exceptional at all. Amongst the elements at least Si, Ge, Sb, and Bi do as well (and at least one other that escapes me at the moment).

There is a nice phase diagram for water listed on physics.SE.

Draw an isotherm at a low enough temperature (< the "triple point temperature of 0.01°C), so that it passes through "all three" phases (ignoring the high-pressure ices that are not relevant to the question).
Note that as pressure goes up you go from solid to liquid.  Increasing pressure always favors a denser phase, so liquid is denser than solid for water.
Water is not quite unique in this respect. Silicon and germanium are two elements that do the same thing.


text:
Conversion of a value between g/gmol and lb/lbmol
I am doing some air emissions calculations and I really just have a bad feeling about my results.  By definition, molecular weights are given in $\pu{g/gmol}$.  For the substance I have, I was told by a technical contact at Citgo that this material has a molecular weight of $\pu{140 g/mol}$ (which should be $\pu{g/gmol}$).  Whenever I convert this $\pu{lb/lbmol}$ (which is required for this particular equation), I'm getting a value of $\pu{0.3086 lb/lbmol}$.  There are 453 grams in a pound.  Is it not correct to divide 140 grams by 453 grams to get the total number of pounds?
My final result for annual emissions is pretty low.

A gram-mole ($\pu{gmol}$),often just called a mole ($\pu{mol}$), is the amount of a material whose mass in grams equal to the numerical value of the atomic or formula mass. For example, a gram-mole of water has a mass of 18.017 grams, this number being the formula mass for water.
A pound-mole ($\pu{lbmol}$) is the amount of a substance whose mass in pounds-mass equals the atomic or formula mass. For water, one pound mole has a mass of $18.017$ lb-mass.
The pound-mole is bigger than the gram-mole by the same factor as a pound is bigger than a gram - so pounds per pound-mole equals grams per gram-mole.  You do not divide by anything.

If the amount of weight you are measuring it in is pounds then use pound moles if the amount your measuring in is grams use gram moles.
140g/g-mol = 140lb/lb-mol
that explanation is confusing to people the simplest way i have seen it explained is that
453.592gmol = 1 lbm-mol
so if you are given the weight of a substance say you have 1000lbm of hydrogen and you want to know the amount of moles.
simply do 1000lbm * lbmol/2lbm = (500lbmol)*453.592gmol/1lbm-mol = 226,796 gmol. If you want to convert the other way do the exact opposite.
btw lbm is lbm just means pounds weight not lbf pounds force.

https://www.calculand.com/unit-converter/?gruppe=Molar+Mass&einheit=Pound+per+pound-mole+%5Blb%2Flbmol%5D
1 lb / lb-mol is exactly 1 g / mol, because 1 lb-mol is 453,59 mol, which is the same ratio that there is between lb and g. The link above is a very useful converter.


text:
How does chromium(III) react to form chromium(VI) with hydrogen peroxide?
$$\ce{2Cr(OH)3 + 4NaOH + 3H2O2 -> 2Na2CrO4 + 8H2O}$$
How does this reaction occur? We suddenly get sodium chromate in the place of chromium hydroxide. I can't understand it. We mix two hydroxides and get a salt?
What role does the peroxide play?

The peroxide oxidizes the Cr (III) to Cr (VI).

@f'' - why isn't it CrO3 then or Cr(OH)6?

Because there is NaOH around.

The hydrogen peroxide is a strong oxidiser. In basic conditions, this is partly due to $$\ce{H2O2->H2O + O\ \ }[*].$$ The monooxygen is very eager to gain electrons. Therefore, $$\ce{Cr^{III}->[\ce{[O]}]Cr^{VI} +3e-}.$$Chromium(III) hydroxide$^{[**]}$ is amphoteric, it reacts with bases, too.
$$\ce{Cr(OH)3 + 3NaOH <=>Na3[Cr(OH)6]}$$
This complex is then oxidised. For common chromium(VI) compounds, one would expect to see the equilibrium $$\ce{2CrO4^{2}^- <=>[\ \ \ce{H3O+}][\ \ \ce{OH-}]Cr2O7^{2}^{-}} \ \ \text{[***].}$$
Thus, in (strongly) basic conditions, the chromate ion $\ce{CrO4^{2}^-}$ would be favoured.
Extra
One thing that might be confusing are the equations themselves. In other words, they simply show

the most likely process (not always the case),
the end result (there are exceptions),
the formulae are not true to the actual structure (cannot be sure).

To make inorganic chemistry more tolerable, it is best to accept that there are multiple reactions occurring concurrently. Then it is easier to arrive at the "correct" end result by taking into account reaction conditions and stability.
[*] This is a simplified explanation taken only as a model for basic understanding. The actual decomposition mechanism is more complicated, depending on pH, other ions in solution, etc. For more information on specific situations, look at Kinetic Studies and Mechanism of Hydrogen Peroxide Catalytic Decomposition [---]. There you can find many other great references as well.
[**] Chromium(III) hydroxide might actually be $\ce{CrOOH*H2O}$, or $\ce{Cr2O3*}$ $n$$\ce{H2O}$. Even if we were to be dealing with the hydroxide, it actually comprises of $\ce{Cr(OH)6}$ octahedrons, and is claimed to be inactive (active form is the oxide hydrate). [N. S. Ahmetov's Inorganic chemistry]
I would make the claim that the hydroxide does not actually exist at all, and is, like for iron, mistaken for the oxidehydroxide hydrate. Furthermore, the reaction might not be direct, it could be a case of Le Châtelier – Braun principle.
[***] The overall transition is $$\ce{2CrO4^{2}^{-} +2H3O+ <=>Cr2O7^{2}^{-} +3H2O}$$
from where it is easy to see that adding hydroxide ions tilts the equilibrium to the left. Also, you might wish to look up different aqueous complexes of chromium and the structures of the ions above. That also helps to visualise the process.
Suggestion: Chromium complexes tend to polymerise ;).

Even if $\ce{CrO3}$ would form, as you suggested, it reacts rather well with water via
$$\ce{CrO3 + H2O \rightarrow H2CrO4, \\ 2CrO3 + H2O \rightarrow H2Cr2O7,}$$
and hydroxides:
$$\ce{CrO3 + 2NaOH \rightarrow Na2CrO4 + H2O}.$$


text:
Substances that pass visible light but absorb microwaves
Specifically, the substances should be water-miscible or water-soluble, but not water itself. Additionally, they should be absorptive but not opaque.

Water itself is such substance.

Ah right. I'm interested in substances other than water. I've edited my question.

Sure - ethanol and methanol (among others) are listed here as being "high absorbers" of microwave energy.


text:
Calculating Ratio of Stability Constants

Solution I contains a mixture of $\ce{FeCl2}$ and $\ce{FeCl3}$, and solution II contains a mixture of $\ce{K4Fe(CN)6}$ and $\ce{K3Fe(CN)6}$. The concentrations of iron-containing species satisfy the relations $[\ce{Fe^2+}] = [\ce{Fe(CN)_6^4-}]$ and $[\ce{Fe^3+}] = [\ce{Fe(CN)_6^3-}]$. The potential of platinum electrode immersed into the solution I is $0.652\ \mathrm{V}$, while the potential of platinum electrode immersed into solution II is $0.242\ \mathrm{V}$.
Standard redox potential for $\ce{Fe^3+ + e- -> Fe^2+}$ is $0.771\ \mathrm{V}$
Calculate the ratio of the stability constants $\beta [\ce{Fe(CN)_6^3-}] / \beta [\ce{Fe(CN)_6^4-}$]

My attempt
First I wrote out the Nernst Equation for both reactions:
$$0.652 = 0.771 - 0.059 \times\mathrm{log}\ce{ \frac{[Fe^2+]}{[Fe^3+]}}$$
$$0.242 = \mathrm{E_{cell}^\circ} - 0.059 \times\mathrm{log}\ce{ \frac{[Fe(CN)_6^{3-}]}{[Fe(CN)_6^{4-}]}}$$
Also I wrote out the expression for the ratio of stability constants"
$$\frac{\beta _1}{\beta _2} = \ce{ \frac{[Fe(CN)_6^{3-}][Fe^2+]}{[Fe(CN)_6^{4-}][Fe^3+]}}$$
Now I know that the ratio is just the sum of the 2 logs in the Nernst equations that I have wrote above. However I am not sure how to calculate the $\mathrm{E_{cell}^\circ}$ for the second reaction. Can we assume that $\mathrm{E_{cell}^\circ}$ for the second reaction also equals $0.771\ \mathrm{V}$?
The answer is $8.90 \times 10^6$
Any help is greatly appreciated.

You used all the information from the question except the line that would have helped you solve it

The concentrations of iron-containing species satisfy the relations $[\ce{Fe^2+}] = [\ce{Fe(CN)_6^4-}]$ and $[\ce{Fe^3+}] = [\ce{Fe(CN)_6^3-}]$

From your first cell you find that
$$\frac{[\ce{Fe^2+}]}{[\ce{Fe^3+}]} = 103.97$$
which means that in your second cell
$$\frac{[\ce{Fe(CN)6^4-}]}{[\ce{Fe(CN)6^3-}]} = 103.97$$
and $E^\circ$ for the cell with cyanide is $+0.361~\mathrm{V}$. (You have a typo in your equation, the 4- should be in the numerator and 3- should be in the denominator. Also your equation is missing units but I won't bother with that.) This is quite far off from the accepted literature value (around $+0.430~\mathrm{V}$) but let's move on.

Now I know that the ratio is just the sum of the 2 logs in the Nernst equations that I have wrote above

No, that is wrong. The ratio of the stability constants refers to that expression you have written, when all of the compounds are placed into the same solution. You cannot simply use the two equations earlier because those describe two separate solutions.
You have to use the difference in $E^\circ$ to work out the ratio of $\beta$.
The way to do this would be to realise that if you take the difference between the $E^\circ$ you get
$$\begin{align}
\ce{Fe^3+ + Fe(CN)6^4-} &\longrightarrow \ce{Fe^2+ + Fe(CN)6^3-} & E^\circ &= +0.410~\mathrm{V}
\end{align}$$
The equilibrium constant for this reaction is the quantity that you are looking for
$$K = \frac{[\ce{Fe^2+}][\ce{Fe(CN)6^3-}]}{[\ce{Fe^3+}][\ce{Fe(CN)6^4-}]}$$
and so
$$K = \exp\left(\frac{-\Delta_\mathrm{r} G^\circ}{RT}\right) = \exp\left(\frac{-nFE^\circ}{RT}\right) = 8.589 \times 10^6$$

Thanks for the great answer! You say that I can't use the two Nernst equations because those describe two separate solutions. I am curious to know if these two solutions were part of a galvanic cell, where one of them is the cathode and other is the anode. Would I able to use the  Nernst equations now since the two solutions are actually in equilibrium with each other.

It would have to be in equilibrium for that to happen (i.e. you have to let current flow until there is no more current). Galvanic cell systems reach equilibrium slowly because the current flows slowly, which is what allows the potential difference to be determined. Think about the Nernst equation - because the reaction proceeds very slowly, Q tends towards K very slowly, which makes E very stable. You would have to allow electricity to flow until Q = K before you could use the Nernst equation to determine K.


text:
Does boiling water really treat it to make it safer than filtered/chlorinated water?
It doesn't seem like it would, but some folks I know are adamant about boiling their water before ingesting it and the online articles I've found on this seem very subjective.

Perhaps they just like tea.

Chlorinated water is safe and might be the biggest public health improvement in human history. Before it existed, water was often contaminated with dangerous microbial pathogens like cholera. There were only two ways to guarantee pathogen-free water: mix in alcohol or boil the water. Both kill bugs and both are no longer needed if your water is chlorinated.

For countries with an average temperature < 12°C or so, chlorinating is only necessary if your water infrastructure is in bad shape. In Germany it is never done (and forbidden) except as emergency measure after pipe bursts etc.

@matt_black Chlorinated water tastes bad. I would pick any (safe) alternative if I have the chance.

Chlorination (free or combined) is not very effective against some pathogens such as cryptosporidium, so it is not a "do-all" treatment option.

What do you mean by "safer"? If the water is pure to start, there's no point in boiling it. For mildly contaminated water, chlorination is inexpensive, but can form some toxic halogenated carbon compounds. Boiling is very energy intensive, and may require use of fossil fuels (unless you use a solar water heater). Both eliminate most microorganisms, though there are a few, mostly nonpathogenic ones, such as Deinococcus radioduran and others found at undersea volcanic vents, that can survive radiation, heat and other chemicals. Bacteria have even been found living in the common biocide, povidone-iodine! See this source for more information on biocides.
Better alternatives might be ultraviolet sterilization or ozonation.

Yes, but a boiling alert for eating/drinking is often issued by water companies when contamination of the treated water supply is suspected. It is effective.

True, and it is more effective than halogenation or ozonation when water has suspended particulates. It's just not efficient.

Boiling water will kill potentially dangerous microorganisms, such as the cholera bacterium.
See "Water Disinfection for International and Wilderness Travelers" by Backer et al.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the [help].

Filtration is not a clear term when used together with water. It could mean anything from getting some macroscopic debris out to a bacteria-withholding microfilter. The most common usages would probably be said microfilter and the filters common in most areas of Germany which reduce water hardness. While the former may be safe, the latter isn’t. Without more information on the filtering technique, we cannot say whether it is safe or not.
Chlorination will pretty reproducibly kill each and every microorganism. However, it cannot destroy all toxic compounds and not do anything with respect to toxic heavy metals.
Boiling water will also pretty reproducibly kill all microorganisms that have been known to populate human water supplies. Yes T. aquaticus etc. etc, but I don’t think that they will ever get the chance to enter our pipes. It may also destroy some toxic chemicals, but less. And it also cannot do anything with respect to toxic heavy metals.
So boiling water is not safer than chlorination, but it isn’t really less safe either. (If you have chemicals in there that survive chlorination or boiling, you have a very serious problem anyway and probably shouldn’t even shower in the water.) However, boiling has a significant advantage, because chlorine residues in water taste bad.


text:
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?

related http://chemistry.stackexchange.com/questions/69/splitting-of-multiplets-in-%c2%b9h-nmr-spectra

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.

The first important point to note is that magnetically equivalent nuclei do in fact couple to each other, however no splitting is observed in the spectrum. The second point is that chemically equivalent, but magnetically non-equivalent, nuclei couple to each other, and this coupling is observable in the NMR spectrum.
Spin coupling comes from a magnetic interaction between nuclear spins transmitted through the bonding electrons. Signals observed in the NMR spectrum are a transition between energy levels of allowed spin states. When two nuclei 'couple', the energy levels are stabilised or destabilsed slightly based on the relative orientations of the nuclear moments, so that (for a doublet) one transition is now δ+J/2, and the other transition is δ-J/2. These two transitions constitute the two lines of the doublet signal. When two equivalent nuclei couple, transitions between energy levels do not change because the interactions between the nuclear moments are the same, as are all other contributing factors such as the Fermi contact. As long as the transitions remain the same, all possible transitions will be equivalent.
The diagram below hopefully explains it a little clearer. The middle energy levels are for two non-coupled spins. Spin A (shown in red) has two possible transitions, both of which are equivalent. When the two spins are coupled, the energy levels are stabilised/destabilised as shown on the right. The transitions for spin A are now no longer equivalent, and they will appear as two lines (doublet). On the left, the equivalent nuclei have an overall change in energy level, but the transition between the levels remains the same, hence the observed line is still a singlet.

When there are two distinct nuclei (The AX case) and there is spin-spin interaction ($J$ coupling) between them (in addition to chemical shift) then fine structure can be observed in the nmr spectrum. The first diagram shows the energy levels and how they interact. Note how the $J$ coupling moves the levels up and down in addition to chemical shift.

Now there are selection rules that allow the radio frequency radiation to couple different levels together and so produce a spectrum. The rule is that the $m_z$ quantum number (variously called magnetic, or azimuthal or projection  quantum number) has to change by $+1$ or $-1$. This means that in the first diagram only levels in which one alpha changes to beta or vice versa, as shown by the vertical arrows, are allowed transitions. (The selection rule occurs because the photon (even if at radio frequency) has one unit of angular momentum and total angular momentum is conserved.)
In the equivalent nuclei case (called A2) the interaction between spins is still present but because of the magnetically identical nuclei the spin states are not either symmetrical or anti symmetrical to exchange of nuclei and a linear combination has to be made. This is shown on the left of the figure below. The reason that the splitting in energy levels is not observed is that selection rules make the transitions unobservable

An alternative answer using the product operator formalism:
The fundamental equation describing NMR quantum mechanically (neglecting relaxation) is the Liouville–von Neumann Equation (in frequency units, i.e. setting $\hbar = 1$):
$$\frac{\mathrm d}{\mathrm dt}\hat{\rho}=-\mathrm i[\hat{H},\hat{\rho}]$$
For a 2-spin system with identical chemical shifts and a coupling
$$\hat{H}=\Omega\hat{I}_{\!1z}+\Omega\hat{I}_{\!2z}+J(\vec{\hat{I}}_{\!1}\cdot\vec{\hat{I}}_{\!2})$$
After a 90 degree pulse on both nuclei the density matrix $\hat{\rho}$ is of the form $\hat{I}_{\!1\chi} + \hat{I}_{\!2\chi}$, where $\chi = x$ or $y$. Now it is a bit lengthy but easy to show that
$$[\hat{I}_{\!1\chi}+\hat{I}_{\!2\chi},J(\vec{\hat{I}}_{\!1}\cdot\vec{\hat{I}}_{\!2})]=0, \quad \text{with } \chi=x,y,z$$
meaning that the coupling Hamiltonian does not influence the signal after a 90 degree pulse, as $[\hat{H}_\text{coupling},\hat{\rho}] = 0$.
I don't know, but maybe it might be possible to generate a coherence on one of the spins only, but I doubt it. It's of course not possible with a single pulse only, since that would always generate coherences on both spins.


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How to determine if a given reaction requires heating to initiate?
At room temperature, some reaction propagate merely upon intermixing the reagents (e.g. autoignition of triethylborane), while others require additional heat to initiate (e.g. crystallization of calcium chloride).
How can I determine if a given chemical reaction would require heating above room temperature to initiate? And to what temperature?

The ease of initiation of a reaction depends on the activation energy required. In order for reactions to proceed (at a reasonable rate), the forces binding the individual reactants must be overcome, which may require energy, solvent, catalyst or other input.
For example, $\ce{H2O2}$ is quite stable at room temperature, but exposure to horseradish peroxidase, silver oxide or other catalyst causes rapid decomposition. Iron filings do not oxidize quickly, but apply a bit of salt water and they rust so quickly they make an effective heater. $\ce{NI3}$ has an extremely low activation requirement!

In your H2O2 example it seems that the thermal energy at room temperature is enough to overcome the activation energy. Was there a way for you to know this without trying out in an experiment?

You need to study the reaction kinetics as a function of the species concentrations and temperature.  To quantify the behavior, you need to know the reaction mechanism, and then quantify this mathematically in terms of reaction rate expressions and reaction rate constants.  Experimental measurements then establish the values of the reaction rate constants as a function of temperature.


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Ester foam insert and isopropyl alcohol
I recently purchased a used set of lens for my telescope. They came in a nice aluminum case with foam inserts, but the foam has an odor. I am currently using just the sun to try to rid the foam of the musty odor. It's slow going. I was wondering if there would be any adverse interaction if I sprayed the ester foam inserts with isopropyl alcohol? I am hesitant to spray anything on the foam for odor removal, for fear any lingering residue may harm the lenses in the long term.
The foam looks like a gray ester poly foam. Bottom is solid, top is convoluted.
Thank you.

I would just try soap and water first. If you want to try any kind of solvent like alcohols, you really needn't be concerned with lingering smell, but I would be concerned with the (slight) possibility of destroying the foam.
If soap and water doesn't work, just try placing a drop of isopropanol on both the top and bottom of the insert and let it dry. If there is any change in the consistency or feel to it I would stop there. If not, then progress to cleaning a small area, and if that does no damage go ahead and soak the whole thing if needed.


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Safe pH levels for blood
What exactly happens to blood cells when a persons blood is to alkaline?
I know that if your pH level goes below 6.8 or above 7.8 you will die. I have read that if you go below 6.8 your cells will stop functioning. Is the same true if you go above? If so what causes the cells to stop functioning?

I should be asked on Biology SE not here.

I'm voting to close this question as off-topic because it's better suited for biology.SE.

While there are a lot of different things that can go wrong if your blood's pH changes too much, in this answer I am going to focus on one of the most chemistry relevant issues with a large change in pH: the denaturing of proteins. Your body is 70% proteins by mass, and they are responsible for functions as diverse as catalyzing reactions to structural support. While many factors influence a protein's function, one of the most important aspects of a protein is its three dimensional shape. Proteins are folded into a shape so that they can optimally preform their function. If a protein's three dimensional structure is distorted, it will normally lose its functionality. The process of a protein losing its three dimensional structure is called denaturing, and it can be caused by a variety of factors including changes in temperature and pH. Once a protein has been denatured, the process becomes irreversible and the protein loses its functions. One of the biggest risks of a change in your blood's pH is that your proteins will start to denature, causing devastating effects on your body's function. Either an increase or a decrease in your body's pH can cause proteins to denature.