text:
Why does the two-electron exchange integral in Hartree-Fock theory have positive values?
In Section 2.3.6 of Szabo & Ostlund's Modern Quantum Chemistry, the exchange integral has the form of
$$\int \mathrm{d}\mathbf{r}_1\,\mathrm{d}\mathbf{r}_2\, \psi_a^*(\mathbf{r}_1)
\psi_b(\mathbf{r}_1)\frac{1}{r_{12}}
\psi^*_a(\mathbf{r}_2) \psi_b(\mathbf{r}_2)
$$
or $\langle ab|ba\rangle$ in physicists' notation. According to this textbook (and many other books), $\langle ab|ba\rangle$ is positive. The conclusion seems obvious but I just cannot find a proof. Is there any simple reason that $\langle ab|ba\rangle$ must be positive?

The conclusion seems obvious but I just cannot find a proof.

No, I don't think it is obvious. At least I could not think about a simple algebraic proof that the exchange integral $\langle ab \vert ba \rangle$ defined as follows,
$$
\langle ab \vert ba \rangle :=
\iint \overline{\psi}_a(\vec{r}_{1}) \overline{\psi}_b(\vec{r}_{2}) r_{12}^{-1} \psi_b(\vec{r}_{1}) \psi_a(\vec{r}_{2}) \mathrm{d}\vec{r}_{1} \mathrm{d}\vec{r}_{2}
$$
is always positive. But I could think about relatively simple physical argument that proves that at least the sum of all the exchange integrals is positive. The argument is based on the very notion of a variation principle and goes as follows.

We know that exchange integrals arise only if we use an antisymmetric product of orbitals as our trial wave function. If we use just a simple product of them there will be just Coulomb integrals in the energy expression.
Now since the true wave function is antisymmetric, if we use an antisymmetric product of orbitals as our trial wave function instead of a simple product of them, we are guaranteed to get the lower energy since we use the qualitatively right wave function.
Exchange integrals enter the energy expression with the negative sign, thus, they have to be positive, otherwise we won't get the lower energy by using an antisymmetric product of orbitals instead of a simple product.

Again, it does not prove that each and every exchange integral is positive, only that their sum is.

Cool. If we apply this argument to a two-electron system, e.g. He or H2, then we can "prove" the exchange integral is positive?

@Molec, well, yes. :D

conceptually it makes sense. That means individual exchange term may be positive or negative but the sum must be positive as coulomb integral is obviously repulsive (or positive).

This is completely incorrect because point 2 is completely incorrect. The variational principle applies to only to actual solutions to the Hamiltonian. Arbitrary functions that are not can take any value. (In fact the 2RDM method converges to the correct answer from below, a sort of anti-variational principle, as the constraints of the true wave function are better satisfied)

@levineds, no, I dont think that you are correct. A Hartree product is basically a diagonal Slater determinant, so the space of Hartree products is just the subspace of the space of  Slater determinants, and, as a result, variationally you can't get energy lower with a Hartree product than with a Slater determinant. Can you quote any source which states otherwise?

The case of 2-e density matrices is a different story. With Hartee products there is no issue of $N$-representability: they are square-integrable functions of $N$-particles. Again, its not the whole space, but just the subspace of such functions, but they are certainly there, so you can't get energy below true energy varying orbitals in a Hartee product.

I agree the case of 2RDM is different for all the reasons you said. However, the point still stands that just because the true wavefunction is antisymm, an arbitrary antisymm trial wf does not mean you are any closer in energy (well, it does if you assume the result you are trying to prove). Your logic was: the real answer is antisymm, so an antisymm guess must be better in energy, but that does not follow directly. My point about the 2RDM was to rebut the point that saying, "here is my model and now I incorporate more correct physics which means my energy must be lower" is not valid.

@Wildcat Minor comment: some complex conjugations are missing in your exchange integral.

Pilar offers this explanation in his book Elementary Quantum Chemistry (paraphrased):
$\psi_a$ and $\psi_b$ are orthogonal so the function $f(\vec{r})=\psi_a(\vec{r})\psi_b(\vec{r})$ must have positive and negative regions with equal volume.
If $f(\vec{r}_1)$ and $f(\vec{r}_2)$ have the same sign, $f(\vec{r}_1)r^{-1}_{12}f(\vec{r}_2)$ will make a positive contribution to the integral
The smaller $r_{12}$ is, the more $f(\vec{r}_1)r^{-1}_{12}f(\vec{r}_2)$ will contribute to the integral
The smaller $r_{12}$ is, the more likely it is that $f(\vec{r}_1)$ and $f(\vec{r}_2)$ will have the same sign.
So positive contributions to the integral will be larger than the negative contributions, resulting in a positive exchange integral.

This answer is also not strictly correct. See Theo Jacobson's answer below for a counterexample to this argument. Points 3 and 4 require much more careful treatment to make this argument valid.

yes, $V$ probably has be to be some inverse function of $r$ ($V(r) = 1/r^x$ where $x > 0$) for this to work.

In a nut-shell: exchange integrals are two-electron integrals, and two-electron integrals yield positive values. Note that the "kind" or "meaning" of the input functions is irrelevant, because in practice, you will always have linear combinations of primitives, and in most cases gaussians. For the proof of the claim about positive values, I will defer to the experts, [1, HJO] who cite previous work. [2] As taken from the book:

The two-electron intergrals can be viewed as a matrix with the electron distributions [($\Omega_{ab}, \Omega_{cd}$)] as row and column labels [using AO labels $a,b,c,d$, see above]
  $$
g_{abcd} = \int \int \frac{\Omega_{ab}(\mathbf{r}_1) \Omega_{cd}(\mathbf{r}_2)}{r_{12}} \mathrm{d}\mathbf{r}_1 \mathrm{d}\mathbf{r}_2
$$
  Assuming that the orbitals are real, we shall demonstrate that this matrix is positive definite [2]. Let us consider the interaction between two electrons in the same distribution $\rho(\mathbf{r})$:
  $$
I[\rho] = \int \int \frac{\rho(\mathbf{r}_1) \rho(\mathbf{r}_2)}{r_{12}} \mathrm{d}\mathbf{r}_1 \mathrm{d}\mathbf{r}_2
$$
  Inserting the Fourier transform of the interaction operator
  $$
\frac{1}{r_{12}} = \frac{1}{2\pi^{2}} \int k^{-2} \exp[\mathrm{i}\mathbf{k} \cdot(\mathbf{r}_1 - \mathbf{r}_2)] \mathrm{d}\mathbf{k}
$$
  and carrying out the integration over the Cartesian coordinates, we obtain
  $$
I[\rho] = \frac{1}{2\pi^{2}} \int k^{-2} \vert \rho(\mathbf{k}) \vert^2 \mathrm{d}\mathbf{k} \quad\quad \text{(eq. 4)}
$$
  where we have introduced the distributions
  $$
\rho(\mathbf{k}) = \int \exp(-\mathrm{i}\mathbf{k}\cdot\mathbf{r}) \rho(\mathbf{r}) \mathrm{d}\mathbf{r}
$$
  Since the integrand in [(eq. 4)] is always positive or zero, we obtain the inequality
  $$
I[\rho] > 0
$$

HJO go on to expand the charge distribution $\rho$ in one-electron orbital distributions and get back to the original $g_{abcd}$, noting afterwards that two-electrons thus satisfy the conditions for inner products, in a metric defined by $r^{-1}_{12}$. Therefore, Schwarz-style inequalities hold and are used extensively in integral screening to throw out insignificant integrals before evaluating them.

[1] T Helgaker, P Jørgensen, J Olsen, Molecular Electronic-Structure Theory, Wiley (2002), p. 403f.
[2] CCJ Roothaan, Rev. Mod. Phys., 23, 69 (1951).

I think neither Wildcat's nor Jan Jensen's answers are satisfactory. In Wildcat's answer, point 2 essentially assumes the result. In Jan Jensen's answer, the role of the magnitude of $f(\vec r)$ is not controlled for, so the conclusion doesn't follow by the logic given.
I found myself at this page because the only proofs of positivity I could find refer to the particular form of the potential, which puzzled me. I thought perhaps the result should be true if we were to replace $1/r_{12}$ by any positive function that is monotonically decreasing as $r_{12}$ grows. However I've constructed a simple finite dimensional example that leads me to doubt that even that is true.
In the finite dimensional analogy, the possible points of space are replaced by just three points. Then consider the following for the analog of the potential $V$ and the function $f=\psi_a(i)\psi_b(i)$:
$$V=\left(\begin{array}{lll} 1 &1&1/2\\1&1&1\\1/2&1&1\end{array}\right),
\qquad f=\left(\begin{array}{l} 1 \\-2\\1\end{array}\right).$$
Note that this $f$ has the required property $\sum_i f_i = 0$, which follows from the orthogonality of the two wave functions (which could be taken as $(1,\pm\sqrt{2},1)$). Also, the potential has the property of decreasing away from the diagonal.  And, yet, we have
$f^T V f = -1 <0$. Also, while the potential $V$ above isn't strictly decreasing away from the diagonal, we could replace the off diagonal 1's by 0.9 and again we'd get a negative value, $f^T V f = -0.2 <0$. I think this counterexample shows that the explanations given above are not valid, and it strongly suggests that, to some degree, the specific form of $1/r_{12}$ is important for the result.

If you combine the fact that the exchange integral is a sum of terms denoted by $G^k$, with positive coefficients,1 and Racah's proof that the $G^k$ are positive,2 then the positivity of the exchange integral follows. That the exchange integral is unlikely to be negative was suggested by Bacher in 1933.3
References

Condon, E. U.; Shortley, G. H. The Theory of Atomic Spectra. Cambridge University Press, Cambridge, U.K., 1935; p 176.
Racah, G. Phys. Rev. 1942, 62 (9–10), 438–462. DOI:10.1103/PhysRev.62.438. The proof is on p 460.
Bacher, R. F. Phys. Rev. 1933, 43 (4), 264–269. DOI:10.1103/PhysRev.43.264.

I will ofer just a simple, qualitative, resoning.
In the expresion
 $\langle ab \vert ba \rangle :=
\iint \overline{\psi}_a(\vec{r}_{1}) \overline{\psi}_b(\vec{r}_{2}) r_{12}^{-1} \psi_b(\vec{r}_{1}) \psi_a(\vec{r}_{2}) \mathrm{d}\vec{r}_{1} \mathrm{d}\vec{r}_{2}$
The term
 $\overline{\psi}_a(\vec{r}_{1}) \psi_b(\vec{r}_{1})$
represents the "overlap" of  $\psi_a$ and $\psi_b$, while,
$\overline{\psi}_b(\vec{r}_{2})\psi_a(\vec{r}_{2}) $
also represents the "overlap" of  $\psi_a$ and $\psi_b$.
Thus, we get a square of the "overlap" of  $\psi_a$ and $\psi_b$, which has to be positive and smaller then Coulomb integral ("overlap" with itself).
Also, if you are wondering why it even has to be real,  $\overline{\psi}_a\psi_b$ and $\overline{\psi}_b\psi_a$ are Hermitian conjugates of each other.


text:
FT-IR background lines
I started doing IR spectroscopy a few months ago and have run into the following problem: Here's the IR background that everyone knows and loves:
CO2 peaks in the middle, water lines on either side. I get that. But what about those lines in the CH-stretch region? They are obviously very normal for an FT-IR background- everyone sees them, they're all over the internet. But WHAT ARE THEY???? No one knows- not even the IR spectroscopy veterans I've talked to. They're some kind of hydrocarbon, I guess, and they're not a gas because they are unaffected by purging. This is a spectrometer looking at no sample, with no windows, nothing, and fwiw this is using an MCT/A detector.
I feel like I should know this, but maybe I'm not using the right resources?
Thanks for any help!

Would this be an artifact of the FT? The "beat" frequency difference between the large peaks at about 1500 and and 3600?

@MaxW- Interesting idea! I think though, that if it were the beat frequency of the water lines, then it would be affected by purging the air inside the spectrometer. What's more, the spectrometer can easily resolve all those little rotational states- so I think the difference frequencies should also show up as a comb of lines? The mystery lines are always broad and have no smaller scale structure.

The only other idea that I had was the beam splitter. Does it have two pieces of glass glued together? Some sort of organic goo would probably be used between the two pieces.

This Shimadzu page has separate IR spectra for H2O and CO2, neither of which shows that 2800-3000cm-1 feature. That is in the C-H stretch region so hydrocarbons sound like a fair guess. Possibly a film of oily contamination on the interferometer or detector? If it appears in every spectrum recorded on that instrument, then contamination is likely.

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text:
Will an alloy leaching in an acid act as both anode and cathode?
If an alloy is placed in an acid where selective leaching is occurring, will the alloy act as both anode and cathode?
If the alloy does act as both anode and cathode then will the potential difference be zero as the cathode and anode are shorted together?

Hello! Welcome to Chemistry.SE. Can you expand a bit, give us your line of thought? Do you have any theories or ideas behind what could be the answer?

Well lets say I have a piece of of iron and aluminum alloy immersed in acid. Then shouldn't it act as both cathode and anode since aluminum is oxidized and the hydrogen is reduced at the surface of the alloy?

Thinking about something like leaching zinc from brass? Zn and Cu both dissolve, but the Cu plates back.

Yes like leaching, but the piece of alloy will act as both cathode and anode?

@Mike - Let's try again. Leaching brass. There are two reactions happening in te solution. (1) Both Cu+2 and Zinc+2 are leached form brass. So the brass is acting like an anode. (2) Zinc+2 stays in solution, but the Cu+2 plates back. So in second reaction the brass is acting like a cathode. // The net result is that the brass has less zinc.

You can setup a situation where one metal is acting as a cathode and the other metal acting as an anode. There will be a small potential difference due to their differences in Fermi Levels much like a thermocouple but the alloy will be at an electrochemical potential that is constant.
This is not that different than a single metal(like a piece of iron) in solution. In the corrosion reaction, the counter reaction may happen somewhere else on the iron. So the stick of iron may be the anode and the cathode at the same time.

In an alloy, the atoms of the different metals are usually so homogenized that there is no automatic anode-cathode separation. A separation can develop, as in the brass example mentioned by MaxW.
And as Burak Ulgut notes, even a single metal can have anodic and cathodic sites. In the old days, the zinc-carbon battery https://en.wikipedia.org/wiki/Zinc–carbon_battery had a problem in that the zinc might develop cathodic sites that ran the battery down very quickly. This was solved by coating the zinc with a very tiny amount of mercury that had a high overvoltage for hydrogen evolution, yet allowed the zinc to move freely and be at the surface so as not to decrease the cell voltage appreciably. Purer zinc and perhaps the different medium of alkaline batteries has reduced the need for mercury, hence the frequent claim "Mercury-free!"
Electrode visualization can be achieved with a single metal or a combination:

https://tgreen230493.wordpress.com/2014/10/30/galvanic-corrosion-of-an-iron-nail/

http://icaseonline.net/parsel/www.parsel.uni-kiel.de/cms/fileadmin/parsel/Material/Hatfield/pdf/Sculpture-_teacher_notes.pdf
In the bottom picture, the iron bar on the left is wrapped with zinc, and becomes the cathode while the zinc becomes a sacrificial anode. The cathodic reaction is
1/2 O$_2$ + H$_2$O + 2 e$^-$ --> 2 OH$^-$
The buildup of OH$^-$ turns the originally neutral solution (perhaps activated with chloride ions) pink/red from phenolphthalein.
The iron bar on the right is wrapped with copper sheet and corrodes away. In some work I did a few years ago, I added K$_2$Fe(CN)$_6$, which also indicated the presence of the Fe$^{2+}$ ions coming off the bar. In that case, I thickened/gelled the solution with agar and added NaCl to accelerate corrosion. The intent was to show that addition of Ca(NO$_2$)$_2$ would stifle corrosion by oxidizing some Fe$^{2+}$ to Fe$^{3+}$ and precipitating Fe$_3$O$_4$ at the potential anode. It did and so became a useful corrosion inhibitor.
Not all metal corrosion reactions can be analyzed as if they were electrochemical. A piece of sodium does not usefully develop an electrochemical potential in an aqueous system: too rapid. The iron exhaust manifold on cars develops an oxide coating: no solvent, just metal and air.

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text:
Why are absolute enthalpies of reactants not available?
The standard enthalpy of reaction is given as $$\Delta_r H = \sum\nu\;  H_m(\text{products})- \sum\nu\;H_m(\text{reactants}).$$
Peter Atkins in his book Elements of Physical Chemistry then writes regarding this equation:

The problem with this eqn, is that we have no way of knowing the absolute enthalpies of the substances . To avoid this problem, we can imagine the reaction as taking place by an indirect route, in which the reactants are first broken down into the elements. [...]
With the introduction of standard enthalpies of formation, we can write $$\Delta_r H = \sum\nu\;  \Delta_f H(\text{products})- \sum\nu\;\Delta_f H(\text{reactants}).$$  The first term on the right is the enthalpy of formation of all the products from their elements; the second term on the right is the enthalpy of formation of all the reactants from their elements..the fact that the enthalpy is a state function means that a reaction enthalpy calculated this way is identical to the value that would be calculated from the former eqn, if absolute enthalpies were available.

I'm not understanding why we can't calculate the absolute enthalpies. After all, enthalpy of a substance at a certain pressure (which is constant as we are talking about standard state & that is 1 bar) is given by the defining equation:
$$H= U +pV$$.
Yes, can't we calculate the internal energy taking a certain state as the zero of the energy? Why isn't it possible to calculate the absolute enthalpy?

If you "take a certain state as the zero of the energy", then it is no longer an absolute internal energy, it is a relative internal energy. But I see what you mean. After all entropies are also defined to be 0 at absolute zero and we call those "absolute". I imagine it has more to do with the internal energy of a molecule not being easy to quantify. It is probably possible, but not worth the effort, which makes the values "not available".

@orthocresol: I do count it as an answer but let the question take some time more to see what others think.

Nah, it doesn't deserve to be an answer.

@orthocresol: It's your opinion but your argument has logic, after all.

Related: Why it is not possible to find absolute value of internal energy?


text:
What's the difference between Gibbs Energy and Work?
Gibbs energy is usually defined as the maximum useful work that a system can do on its surroundings. So what's the main difference between Gibbs energy and work?

Hey, if you do a search on chem.SE, you will see many other people have wondered about this before. See for example http://chemistry.stackexchange.com/q/32873/16683 (there are more).


text:
Why do some chemical reactions lead to less stable state?
If the universe is always trying to find a more stable energy state, then how do reactions take place that lead to a less stable state e.g. the formation of ammonia from extremely stable diatomic $\ce{N2}$ molecules?

Thermodynamics doesn't say that an "unfavorable reaction" can't take place. Thermodynamics says that if an unfavorable reaction takes place then it takes energy input. See http://pubs.acs.org/doi/abs/10.1021/jp209795t The "lighting bolt" on the reaction arrow in the diagram is a photon supplying the energy.

You have to consider the entire system to see whether a reaction is favorable. It is true that exothermic reactions lead to more "stable" products. But there are other factors that are in place. Pressure, heat, and work. Later on you will learn how these come into the picture. But for now, your understanding of stability seems to only encompass the energy of the bonds of the products. But that is not the case.  Also, you are wrong about the Haber process being endothermic. It is not.

@user97554 I have edited my answer accordingly. And thank you MaxW for the link. As I am currently teaching myself (not attending an courses on the subject), are there any wiki pages or reading material that you'd recommend that might expand my view to encompass variables like pressure, heat, and work? Thanks

@user3256725 - not having any idea of your level of understanding I wouldn't know what to recommend. I will say that thermodynamics is one of the last areas to study.


text:
Why isn't N2 a Lewis base?
Why is $\ce{N2}$ (dinitrogen) not a Lewis base even with two lone pairs?

Actually, it is, but a very weak one.
The lone electron pairs in dinitrogen are in $\mathrm{sp}$ hybridization state, and nitrogen is quite electronegative, so they are not easily donated. Still, if a Lewis acid is strong enough, dinitrogen may form a complex.
It is known that dinitrogen adsorbs on acidic $\ce{OH}$-groups in zeolites, forming complex with IR-active frequency. The nature of the complex may be debated, as no full proton transfer occurs, but the electron pairs are definitely involved.
Diazonium salts may be considered as a lewis-acid complex of a carbocation and dinitrogen molecule. While no direct synthetic route exists from dinitrogen to diazonium salt (because typically a stronger Lewis base is available), they typically decompose with formation of dinitrogen molecule.
Significant amount of dinitrogen complexes with transition elements are known. Typically they are obtained indirectly, by forming dinitrogen ligand in-place, though.
Finally, adsorption of dinitrogen on the surface of ammonia synthesis catalysts starts as a Lewis-interaction. Similarly, nitrogen adsorption on cations in zeolites is known.

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text:
Simple distillation & azeotropes
In my organic chemistry lab, we recently had an experiment where we were given a mixture of 50:50 ethanol & ethylene glycol. We were supposed to perform a simple distillation to calculate the percent recovery. Based on my readings, I found that the ethanol has a lower boiling point than ethylene glycol and concluded that ethanol would be the compound that would vaporize and transfer over to the cooled flask where it could condense back. At the end of the lab, we just weighed the distillate and the holdup.
My question was, how do you know if the simple distillation process was successful? I assumed that it was successful based on the fact that it produced a distillate, but I realized that can't be the only reason. The difference in boiling point was large enough to be performed as a simple distillation, but how can I conclude that it was overall successful?
Also, based on this, can I figure out if there is an azeotrope for the boiling points of the two compounds? I read that azeotropes "prefer" to be boiled into the gas phase together, but doesn't that imply that both compounds could have condensed, making the simple distillation process unsuccessful?

By definition an azetrope is a distillate that boils at a constant temperature, but is a mixture of two or more compounds. So you can't purify an azetrope by using more distillations.

Pure ethanol has a bp of 78.37 °C and pure ethylene glycol has a bp of 197.3 °C.  The bp of the azetrope could be (1) below pp of ethanol, or (2) above bp of ethylene glycol. In theory the azetrope cannot be between or equal to either bp. The problem with theory is experimental error.

@MaxW I'm not sure if it's an error on my part, the mixture began to produce the distillate around 67°C and began boiling around that time, too. Since that was lower than the bp of ethanol, would the mixture still be considered as an azetrope? Thank you!

Were you measuring the temperature in the liquid or in the gas phase? I didn't find any evidence of an azetrope between ethanol and ethylene glycol with a google search. I'd suspect that you were measuring the temperature in the distillate (in the gas phase). That is tricky since the thermometer has some thermal mass. It takes a little while for the gaseous distillate to heat it up.

@MaxW The thermometer was like this: http://www.docbrown.info/page12/gifs/distill.gif, so I assumed that it was measuring the temperature of distillate when it vaporized. I think it makes sense that that they are azeotropes based on what I found on Google, too, but I just wasn't sure how to back up that observation.

I'm confused. I found NO evidence of an azetrope between ethanol and ethylene glycol. I think you read the thermometer too early.

@MaxW I read the thermometer when it began to produce the drops of distillate, but it's definitely possible that I had human error! Thank you for taking your time to answer this :)

!! You're missing the point !! If the thermometer is dripping distillate then it is colder than the distillate. You need to wait until the thermometer stops dripping so that you know it is at equilibrium with the temperature of the distillate.

By definition an azetrope is a distillate that boils at a constant temperature, but is a mixture of two or more compounds. So you can't purify an azetrope by using more distillations.
Pure ethanol has a bp of 78.37 °C and pure ethylene glycol has a bp of 197.3 °C. The bp of the azetrope could be (1) below pp of ethanol, or (2) above bp of ethylene glycol. In theory the azetrope cannot be between or equal to either of the boiling points of the two pure chemicals. (The problem with theory is experimental error....)
I can find no evidence of an azetrope between ethanol and ethylene glycol using Google. So I very confident that no such azetrope exists.
You noted that the distillate temperature was about 67 °C as the pot started to boil. You also noted that you read the thermometer when it began to produce the drops of distillate.
Look at the drawing below.

The glassware setup has two thermometers.

Thermometer-1 is in the pot.
Thermometer-2 is in the pot.

Ideally the pot temperature and the distillate temperature would be the same. However when the solution starts to boil the distillate is just starting to contact the thermometer. The thermometers have some thermal mass, so it takes time to get them up to temperature.
The distillate thermometer is relatively slow to equilibrate. So long as it is dripping distillate solution then it is colder than the distillate and will not give an accurate temperature reading.
So the only explanation for this anomalous result is that you made a small mistake taking the reading.


text:
Chiral centers in albuterol-like compounds

My question is apart from the C in the straight chain with $\ce{-OH}$ group, should $\ce{N}$ be counted as chiral center since it also has tetrahedral geometry and all different substituents?
There are more such examples.

Also, check out these related questions: 1, 2, 3 and 4. Still leaving this open, because I’m not sure if one of these is an exact duplicate.

Inversion at nitrogen chiral centres proceeds rapidly, so the answer is no.

@orthocresol Would you explain, please?

Read those related questions.

You can surely count that nitrogen atom as a "chiral center" because it has four different groups attached to it.
However, the net optical activity due to that chiral center is zero because the two optical isomers from that nitrogen atom are rapidly interconverting with each other ("Amine inversion"). They always exist as a racemic mixture, and hence are optically inactive.


text:
Which of the following is a secondary alkyl halide
a $\ce{(CH3)2CHCH2CCl(CH3)2}$
b $\ce{(CH3)2CHCHClCH3}$
c $\ce{(CH3)2CClCH3}$
d $\ce{(CH3)2CHCH2Cl}$
This is just to check that I'm understanding the material correctly.
I'm thinking it's c because when drawing the line structure the alpha carbon is connected to two other carbons.

If I remember correctly, you're wrong: b is the secondary alkyl halide. c (as well as a) is the tertiary alkyl halide, and d is the primary one.

B is connected to one CH3 groups as well as a CH meaning it's connected to three carbons making it secondary, but C is set up the same way.

In b the carbon atom to which $\ce{Cl}$ is attached is indeed connected with $\ce{CH3}$ on the one side and $\ce{CH}$ (more precisely, $\ce{(CH3)2CH}$ on the other side. But this means that this carbon atom is connected to just two other $\ce{C}$ atoms. Where did you get 3?

Well, no. Here is the quote from Wikipedia:

From the structural perspective, haloalkanes can be classified
  according to the connectivity of the carbon atom to which the halogen
  is attached. In primary (1°) haloalkanes, the carbon that carries the
  halogen atom is only attached to one other alkyl group. An example is
  chloroethane ($\ce{CH3CH2Cl}$). In secondary (2°) haloalkanes, the carbon
  that carries the halogen atom has two $\ce{C–C}$ bonds. In tertiary (3°)
  haloalkanes, the carbon that carries the halogen atom has three $\ce{C–C}$
  bonds.

So, the presence of $\ce{CH2Cl}$, $\ce{CHCl}$, and $\ce{CCl}$ are the indicators of primary, secondary, and tertiary haloalkane respectively.
Here is b with the $\ce{C}$ atom to which $\ce{Cl}$ is attached as well as its two $\ce{C–C}$ bonds highlighted in red.

Thanks, I was drawing C wrong which is why I was seeing it as a secondary and not a tertiary.


text:
What is the difference between TIC and TICC?
I have picked up a textbook on mass spectrometry, and I am finding it did not clearly distinguish between total ion chromatogram (TIC) and total ion current chromatogram (TICC) in the section (1.3 Ion Chromatograms) I just read. I suspect there was some sort of typo, but I don't have the experience to tell yet.
Here is an abridged excerpt from the textbook that has confused me:

The total ion current (TIC) can either be measured by a hardware TIC
  monitor before mass analysis (nA to uA range), or its equivalent can
  be reconstructed or extracted after mass analysis. [...] Thus, the TIC
  represents a measure of the overall intensity  of ion production of
  ion production or of mass spectral output as a function of time,
  respectively. The TIC obtained by means of data reduction, [...], is
  termed total ion chromatogram (TIC). The term total ion current
  chromatogram (TICC) refers to a chromatogram obtained by plotting the
  total ion detected in each of a series of mass spectra recorded as a
  function of retention times of the chromatographically separated
  components of a mixture (which is essentially implicated by: TIC).

I would like to an explanation of the difference between total ion chromatogram (TIC), and total ion current chromatogram (TICC).

Ok, I had to read the actual text on pages 11 and 12 of the textbook link to understand.
TIC - In the olden days of mass spectrometers there was a way to measure the ion current from the sample (all masses). This total current is the current of all masses that is being injected into the mass spectrometer. Since the current in detector was also being measured, the spectra is plotted as the ratio of currents vs the mass (scan time).
TICC - The current from the sample is ignored and the spectra is just plotted as current in the detector vs mass (scan time).


text:
How does heat energy interact between two bodies of different heat capacities in order to attain thermal equilibrium?
Suppose there are two objects $A$ & $B$ of same mass; the former has specific heat capacity $C_A$ and the later has $C_B$ such that $C_A\gt C_B.$ Let $A$ be at temperature $T$ while $B$ be at $T+\Delta T.$ So, in order to attain thermal equilibrium, they have to attain the same temperature $T+ \frac{\Delta T}{2}.$ That means there would be a change of $\frac{\Delta T}{2}$ in both bodies. For that, $A$ must gain heat energy $Q_A= m\;C_A\cdot \frac{\Delta T}{2}$ while $B$ has to loss heat energy $Q_B=m\;C_B\cdot \frac{\Delta T}{2}.$ Now, $Q_A-Q_B\gt 0$ since $C_A\gt C_B.$
Then in order to attain the equilibrium temperature, $A$ has to get $Q_A$ energy which it receives partly from $B$ viz. $Q_B$ but where does the additional heat energy $Q_A-Q_B$ come from? It can't come from $B$ as that would deviate it from achieving the thermal temperature; $B$ can lose only $Q_B$ in order to attain $T+\frac{\Delta T}{2}.$ But where does $A$ get the extra energy $Q_A- Q_B$ from?
I'm confused as I thought so-far that when two bodies achieve thermal equilibrium, they do it by exchanging heat energy between themselves - the cooler body receives heat energy fro the hotter body till the equilibrium is reached.
But here, it seems that $A$ in order to achieve thermal equilibrium, needs energy $Q_A$ which cannot be provided by $B$ since the energy $Q_B$ it releases  is less than $Q_A.$ Where would then the extra energy $Q_A-Q_B$ come from?
Let's say, somehow $A$ gained that energy from the surroundings; however what would happen if they were isolated from the surroundings?
I don't know where I'm mistaking. Is it true that the hotter body itself provide the required energy to the cooler body to attain thermal equilibrium?
Please help.

According to your description and choice of symbols, $C_\mathrm A$ and $C_\mathrm B$ are heat capacities, not specific heat capacities.

$Q_A$ has to equal $Q_B$. Your mistake comes from the fact that at thermal equilibrium, the temperature of the two objects is not at $T+\Delta T/2$ if $C_A \neq C_B$.

@swenger:  I was guessing this reason but didn't have any logic to explain the point. Could you elaborate it  on why the thermal temperature shouldn't be $T+\frac{\Delta T}{2}?$

@Loong: Sorry for that.

Say the initial temperature of $A$ is $T_1$ and the initial temperature of $B$ is $T_2$. To determine the temperature ($T_f$) of $A$ and $B$ at equilibrium, we start with the equation $q_A=-q_B$, where $q_A$ and $q_B$ are the heat gained by $A$ and $B$ respectively as they reach equilibrium. We have $q_A=mc_A\Delta T_A=mc_A(T_f-T_1)$ and $q_B=mc_B\Delta T_B=mc_B(T_f-T_2)$. Thus the equation becomes $$mc_A(T_f-T_1)=-mc_B(T_f-T_2)$$ From here, we can solve for $T_f$. We can also note that if $c_A=c_B$, $T_f=\frac{T_1+T_2}{2}$ , which in your example is $T+\frac{\Delta T}{2}$. Otherwise, $T_f$ will not be the midpoint between $T_1$ and $T_2$.

You've pre-assumed that $q_A=-q_B.$ Is this just?

Since the system is isolated, $q_A=-q_B$ because of conservation of energy


text:
Obtaining magnesium oxide from magnesium hydroxide
What's the simplest and safest DIY way to obtain magnesium hydroxide to magnesium oxide?

Magnesium hydroxide decomposes at 332 °C (630 °F) :
$$\ce{Mg(OH)2_{(s)} -> MgO_{(s)} + H2O_{(g) }}$$
So heat the magnesium hydroxide in a covered crucible. The released water will cause the particles of magnesium hydroxide to "explode" so you want to do this covered.


text:
Electrolysis, Concentration of Electrolytes

So it says the Chlorine gets discharged as the concentration of chlorine ions are higher than OH ions, but why don't the concentration applies to Na+ ??

The real question is why they don't say $\ce{Cl-}$ gets discharged because of its position in the electrochemical series.

It is not only concentration - the potential is important too.  $\ce{Na+}$ is much harder to reduce than $\ce{H+}$, so despite the concentration, $\ce{H+}$ goes first. $\ce{Cl-}$ are easier to discharge than $\ce{OH-}$, and also more abundant, so the choice is obvious.

Really? But According to my book, it says OH- has a higher tendency to discharge when compared with Cl-, as shown in this picture..
https://sciencezoneja.files.wordpress.com/2013/05/electrochemical-series.jpg

$E^\circ(\ce{Cl2}/\ce{Cl-}) = +1.36~\mathrm{V}$ and $E^\circ(\ce{O2}/\ce{OH-}) =  +0.40~\mathrm{V}$. $\ce{OH-}$ is easier to oxidise than $\ce{Cl-}$. I think it is a matter of concentration. Related question: http://chemistry.stackexchange.com/questions/15435/why-is-chloride-oxidised-instead-of-water-in-copper-chloride-electrolysis Loong mentioned overpotential in that answer.

@orthocresol It is weird to see someone talk about $E^0$ and mention $\ce{OH-}$ at the same time … My source says $E^0 (\ce{O2}/\ce{H2O}) = 1,23~\mathrm{V}$. Which is rather close to chlorine’s. Remember that standard potentials are measured at pH 0. But you are correct and I was wrong with their order.

@Jan I agree that $E^\circ(\ce{O_2}/\ce{H2O})$ is much more appropriate. $E^\circ(\ce{O_2}/\ce{OH-})$ (or $E(\ce{O_2}/\ce{OH-})$) is measured at pH 14 which is obviously not the case for a concentrated salt solution.


text:
Complete Oxidation and Redox Equations for Sacrificial Galvanic Anode
I'm trying to get a more complete picture of how the chemistry works within a typical marine setup including a sacrificial anode. The setup consists of an aluminum hull, another metal (say a stainless steel propellor) and then a zinc sacrificial anode. We can ignore the steel propellor. What are the oxidation and redox equations for the aluminum and zinc? Seems like marine blogs don't care about this level of detail and chemistry courses don't use saltwater as the solution...
I'm really curious about what is going on at the aluminum cathode. I think the oxidation equation for the zinc is something to the effect of:
$$
\ce{Zn + something -> Zn^2+ +H_2}
$$
Therefore the zinc disappears. But what happens to the aluminum? There is no aluminum in solution to precipitate so where do the electrons go?

What you are showing is a half-cell reaction. In "theory" the reaction shouldn't take place at all. The boat would just have some potential relative to the water. In reality all sorts of things are going on. But rather than have some other metal oxidize into the water via corrosion, the zinc oxidation occurs. Hence its name - a sacrificial anode.

Are you saying that no reaction should occur with the Aluminum? The aluminum cathode must still be shedding electrons somehow.

The point with Zn being a sacrificial anode is that the zinc should oxidize before any other metal. So the Zn protects the aluminum, and the stainless steel propeller and so on...

My understanding is that if the only metal was Aluminum there wouldn't be much of an issue. It would oxidize, but slowly since the negative charge would build up on the Aluminum. However, somehow the existence of a cathode helps in releasing the electrons back into the solution, thereby greatly speeding up the oxidation. Am I understanding that correctly?

Aluminum is sort of a funny metal. Normally when a metal oxidizes (rusts) the rust has a greater volume than the metal so the rust flakes off exposing more fresh metal. Aluminum oxide should coat the aluminum metal and "seal" the metal surface. But you can scratch the aluminum and so on...


text:
Continuously replenished reaction
Substrate ($m$) concentration in a reaction scheme is governed by the equation $$\frac{dm}{dt} = -\dfrac{r_1c_1m}{1+\dfrac{r_1}{r_2}m + \dfrac{r_1r_3}{r_2r_4}m^2}$$ where $c_1$ is the initial enzyme concentration and each $r_i$ is the reaction rate of the reactions present within this reaction scheme (which are positive). Now additionally suppose the substrate is being continuously replenished by a reaction $$\ce{A->[r_5]S,}$$ where $A$ has constant concentration $a_0$. Modify $dm/dt$ and show there are two distinct positive equilibria $0<m_1<m_2$ if $r_5<r_5^C$ where $r_5^C$ is a critical value to be determined.

With this information we have
$$\frac{dm}{dt} = -\dfrac{r_1c_1m}{1+\dfrac{r_1}{r_2}m + \dfrac{r_1r_3}{r_2r_4}m^2}+r_5a_0.$$
Equilibria when $dm/dt=0$. We rearrange this to produce the quadratic
$$r_2r_4r_5a_0 + r_1r_4(r_5a_0 - r_2c_1)m + r_1r_3r_5a_0m^2=0$$
So the equilibria occur at
$$m_{1,2} = \dfrac{r_1r_4(r_2c_1-r_5a_0)\pm\sqrt{r_1^2r_4^2(r_5a_0-r_2c_1)^2-4r_1r_2r_3r_4r_5^2a_0^2}}{2r_1r_3r_5a_0}$$
Now I need to justify these are positive and find the critical value for $r_5$ but I have no idea how to continue here.

If i am following you, $m$ is the substrate concentration. So $m = r_5a_0$ which is substituted into your first equation.

$m$ is the substrate concentration but I have not made that substitution

How did you get from your equation 1 to your equation 3 ?!?

I have added $r_5a_0$ to the rate of change for concentration of the substrate as it is being increased by $a_0$ at reaction rate $r_5$. Is this incorrect?

You have $\ce{A->[r_5]S}$ which means the concentration of S is given by $r_5$ times the concentration of A. In the first sentence you wrote "Substrate ($m$) concentration" so $m$ is the concentration of S.

Sorry for having mislead you. I totally misinterpreted the part "Now additionally suppose the substrate is being continuously replenished by a reaction..." and my answer was totally wrong so I deleted it. No sense confusing some other poor soul with nonsense.

I can't believe the way we solved the equation for the discriminant equal to zero.  You and I both did the same thing.  All we really needed to do is solve $R^2-\lambda = 0$ by writing $R^2=\lambda$, and taking the square root of both sides.  The resulting equation is then linear in r5.
Anyway, if I did the algebra correctly, I get the double root as:
$$m_1=m_2=\sqrt{\frac{r_2r_4}{r_1r_3}}$$
So, I guess one has to conclude that, in order to guarantee that the roots are  different, we have to take as the critical value of $r_5$ the value that we determined by setting the discriminant equal to zero.

When you square root both sides why do we take $R>0$, i.e. $R = +\sqrt{\lambda}$ over $R = -\sqrt{\lambda}$?

For the same reason we said before.  That's the only way to guarantee a positive r5.

The first step is to set the discriminant equal to zero, and thereby solve for the critical value of $r_5$. When I did this, I got $$r_5a_0=\frac{r_2c_1}{1+2\sqrt{\frac{r_2r_3}{r_1r_4}}}$$.  The next step is to substitute this into the equation for m, and find the two  corresponding equal values of m.

Actually it's not.  All that's required is a little more mathematical manipulation.  Factor out the $r_2c_1$ and, in the denominator, use the identity $a^2-b^2=(a+b)(a-b)$

Yes.  I left out the 2.  Thanks.  I'm going to go back and correct it.   We choose the negative numerator because that choice will guarantee that the answer is always positive.

Also replace all the $\ge$ signs with $>$ signs.

I like what you did much better than what I was suggesting.  Very nice reasoning.

Which would be the correct answer in the end then?

Your result is the correct bound.  My result is consistent with your more general relationship, but it has no deeper meaning than that.  All it does it tell you at what value of r5 the two roots are equal.

@ChesterMiller - I assume you started with  $$\frac{dm}{dt} = -\dfrac{r_1c_1m}{1+\dfrac{r_1}{r_2}m + \dfrac{r_1r_3}{r_2r_4}m^2}+r_5a_0.$$

Yes, that is the starting equation.


text:
London Forces and Liquifying Noble Gases
From my book, it says that 'without van der Waals (London Dispersion) forces, it would be impossible to liquefy noble gases.'
Why is this the case?

a noble gas can be liquefied due to instantaneous dipole induce dipole interaction and here london force provide dipole which can be considered as induce dipole

I'd say that because noble gases are non-polar atoms, the only force that applies to them is the London force.
If you want a liquid phase, you need to have forces between the atoms / molecules that constitute this liquid (forces too strong would lead to solid state).
So because London force is the only one that applies on noble gases molecules, they are necessary to liquify to gases.

I think this is half the answer. The Van Der Waahl's forces can't be the only forces because the noble gases will also form solids. So there must be something else. In solids the atoms are close enough to actually have have some weak chemical bonding (overlapping of atomic orbitals). In a liquid the atoms are moving around too much for such weak molecular bonding to occur. So in a liquid the Van Der Waahl's forces provide the necessary interatomic attractive force. Ar melts at 83.81 K and boils at 87.302 K.


text:
Can we add an electron to an atom?
Say that we have an atom which is inert, or stable. How would one go about adding one electron. Is it possible?

Yes. With an extra-electron such an atom/molecule would be "unstable". How long it would last depends on the conditions. In a gas ions can last for quite a while. In solids or liquids losing the extra electron would be almost instantaneous.

Negative ions are not that hard to make for many elements. Some will auto-ionize fairly quickly, some are reasonably stable...

F-center defects in crystals are due to unpaired electrons and can be caused by ionizing radiation; these defects can accumulate over thousands of years. This can be used to find the age of minerals with retroactive inclusions, or to intentionally color gems.
An electron gun, as in a cathode ray tube (CRT), will also do the job. Older televisions had CRT's, and to prevent electrons building up on the front of the tube, they were siphoned off to ground. "Static" electricity on the faceplate of a CRT can be felt; it is due to the build-up of electrons.
Of course, one could question if electrons in the above examples are bound to a particular atom, are interstitial, or are free to move, but as mentioned above by @MaxW, an ionized gas (plasma) does have particles with excess electrons, as well as those with a deficiency. In interstellar space, an ion could last for years.
Also, radical anions are easily produced, for example the reduction of naphthalene by an alkali metal.


text:
Mass of sugar that was added to the water (Raoult's law)

Water has a vapour pressure of $15.5~\mathrm{mmHg}$ at $25~\mathrm{^\circ C}$. An unknown amount of the non-volatile sugar glucose ($\ce{C6H12O6}$) is added to $1.31~\mathrm{kg}$ of water and the vapour pressure of the system is found to decrease by $0.13~\mathrm{mmHg}$.  What mass of sugar, in grams, was added to the water?

I'm a little bit confused about this problem.
Here it says that after the sugar was added to water, vapour pressure of the system is found to decrease by $0.13~\mathrm{mmHg}$.
Does it mean that before sugar was added, there was only one element $\ce{H2O}$?
And I used Raoult's law:
$$P_i = x_iP_i^*$$
where $P_i$ is the vapour pressure of the solvent, $x_i$ is the mole fraction of the solvent, and $P_i^*$ is the vapour pressure of the pure solvent.
So can I use this to find the partial pressure before the sugar was added?
I know that $1.31~\mathrm{kg}$ of water is $72.78~\mathrm{mol}$.
And if my assumption is right and before sugar was added there was only water, will the mole fraction before sugar was added just be $72.78$ since there was only $\ce{H2O}$?

Note that the question contains a (not relevant) flaw: The vapour pressure of water at $T=25\ \mathrm{^\circ C}$ actually is $p=23.8\ \mathrm{mmHg}$, not $p=15.5\ \mathrm{mmHg}$. The value $p=15.5\ \mathrm{mmHg}$ corresponds to the vapour pressure of water at $T=18\ \mathrm{^\circ C}$.

Your question

Here it says that after the sugar was added to water, vapour pressure of the system is found to decrease by $0.13~\mathrm{mmHg}$.
Does it mean that before sugar was added, there was only one element $\ce{H2O}$?

One component, not element. Water isn't an element. But yes. The question does actually say it: "sugar is added to water".

I used Raoult's law: [...] So can I use this to find the partial pressure before the sugar was added?

Why do you want to find the partial pressure before it was added? It's already given to you. When there's only one component in the system, the partial pressure is the same as the total pressure, which is the vapour pressure of pure water: $15.5~\mathrm{mmHg}$.
In fact, the partial pressure after it was added is also given to you, it's just $(15.5 - 0.13)~\mathrm{mmHg}$. You already have all the pressure data you need. This question is giving you partial pressure data and asking you to find the mole fraction.

I know that $1.31~\mathrm{kg}$ of water is $72.78~\mathrm{mol}$.

Yes. Although if you use a more accurate molar mass for water, you will get a more accurate answer.

And if my assumption is right and before sugar was added there was only water, will the mole fraction before sugar was added just be $72.78$ since there was only $\ce{H2O}$?

No, the mole fraction would be $1$. The mole fraction of a component is the number of moles of that component, divided by the total number of moles.

The answer
The vapour pressure of pure water, which is $P_{\ce{H2O}}^*$ in Raoult's law, is $15.5~\mathrm{mmHg}$.
After you add a certain amount of sugar, the vapour pressure is decreased to $P_{\ce{H2O}} = 15.37~\mathrm{mmHg}$.
So, if you just plug these numbers into Raoult's law, you find:
$$\begin{align}
x_{\ce{H2O}} &= \frac{P_{\ce{H2O}}}{P_{\ce{H2O}}^*} \\
&= \frac{15.37~\mathrm{mmHg}}{15.5~\mathrm{mmHg}} \\
&= 0.9916
\end{align}$$
$x_{\ce{H2O}}$ is the mole fraction of water after sugar is added. So:
$$\begin{align}
x_{\ce{H2O}} &= \frac{n_{\ce{H2O}}}{n_\mathrm{total}} \\
0.9916 &= \frac{n_{\ce{H2O}}}{n_{\ce{H2O}} + n_\mathrm{glucose}}
\end{align}$$
The number of moles of water is:
$$\frac{1310~\mathrm{g}}{18.02~\mathrm{g~mol^{-1}}} = 72.70~\mathrm{mol}$$
This is what I said earlier about using a more accurate value of the molar mass. So:
$$0.9916 = \frac{72.70~\mathrm{mol}}{72.70~\mathrm{mol} + n_\mathrm{glucose}}$$
Rearranging:
$$n_\mathrm{glucose} = 0.6149~\mathrm{mol}$$
The molar mass of glucose is $180.16~\mathrm{g~mol^{-1}}$. So:
$$\begin{align}
m_\mathrm{glucose} &= (0.6149~\mathrm{mol})(180.16~\mathrm{g~mol^{-1}}) \\
&= 110.8~\mathrm{g}
\end{align}$$

thank you very much ! the reason i could not solve this question, because i did not know that when there is only one component in the system, the partial pressure is the same as the total pressure.

@student123 No problem. You might wish to read up more about vapour pressure, why it exists, how it is measured etc.  - it is a confusing concept especially the first time you see it - it certainly was for me.


text:
What is the mass fraction of C2HCl3 in the liquid phase?
Vapour-liquid equilibrium of a two-component ideal solution of trichloroethene (C2HCl3) and trichloromethane (CHCl3) is established at 25 °C. The mole fraction of CHCl3 in the vapour phase is 1. What is the mass fraction of C2HCl3 in the liquid phase?
The vapor pressure of trichloroethene and trichloromethane at 25 °C are:
Pvap, C2HCl3 = 73.0 mmHg
Pvap, CHCl3 = 199.1 mm Hg
I'm confused about this problem, if the mole fraction of one element equals to 1, would not be the mole fraction of another element be 0? Can someone advise me on how to approach this problem?

"I'm confused about this problem, if the mole fraction of one element equals to 1, would not be the mole fraction of another element be 0?" // Yes.

The system is supposed to be "ideal." What does that mean to mixture in vapor vs mixture in liquid? // Hint you need the boiling point of the two compounds.

@MaxW but i don't get it, if the mole fraction of another compound is 0, wouldn't that mean that the moles of this compound is 0? The only way to get mole fraction to be 0, is to have 0 in the numerator which is the number of moles of the compound.

Yes that is correct. What happens when the temperature changes for this system?

@MaxW so, at boiling temperatures  the total pressure of the system will be the same as the partial pressure of the CHCL3? And then, from here should i use PV = nRT to find the moles of CHCL3 in order to find the mass fraction of C2HCL3. But then, i don't know the volume and if the number of moles of C2HCL3 is zero, wouldn't that mean that the mass fraction of it is zero?

No. This is an "ideal system." Think of what happens as temperature changes. Start at 100 degrees and goto 200.

Don't think of the absolute vapor pressure, think of mole fraction in the vapor.

I'm sorry. I mislead you. Temperature changes, and the boiling points of the liquid have nothing to do with this problem as I erroneously commented.
The correct approach is to simply apply Raoult's law. For some temperature T, let

$p^*_i =$  partial pressure of some pure compound $i$
    $x_i =$ mole fraction of the compound $i$ in solution.

then let $p_i$ be the partial pressure of compound i above a liquid mixture. For an ideal system then:

$p_i = p^*_ix_i$

The problem states "The mole fraction of $\ce{CHCl3}$ in the vapor phase is 1."
As you deduced, since there is no $\ce{C2HCl3}$ in the vapor phase there can't be any $\ce{C2HCl3}$ in the liquid phase either.


text:
Cracking of dicyclopentadiene - was my simple Liebig condenser distillation setup adequate?
I recently did a distillation of dicyclopentadiene (bp 166 °C) to get cyclopentadiene (bp 40 °C). The setup as instructed by my supervisor consisted of a simple water-cooled Liebig condenser. Oil bath temp. at 180 °C, receiving flask held at -78 °C by a dry ice-acetone bath:

Everything seemed to work fine, and I got plenty of cyclopentadiene after distilling for about 4 hours.
Now (writing the report) I went hunting for some procedure references, and the first hit was this one (http://www.orgsyn.org/demo.aspx?prep=CV4P0238):

My interpretation of these instructions look like this:

My questions (too many I know, sorry):

Is my interpretation of that setup correct?
What is the Friedrichs condenser for?
Why should the water temp. be 50 °C? (and how important is it to get this temp. correct?)
Where would I attach the "calcium chloride drying tube" and what is it anyway, what's it for and how important is it to have one?
Other procedures I found attached Nitrogen to the setup. Why would I do that and how important is it?
How would you interpret "efficient water-cooled condenser"?
Finally, can you put me at ease and tell me that my simple distillation setup was just fine?

I remember that one time I used cyclopentadiene. We just took the bottle whose contents were solid wax (dimer) and put it into the rotavap’s water bath at $40~\mathrm{^\circ C}$ for the reverse reaction to happen … It was liquid (monomer) after a few minutes … You had to be quick, though, with measuring and adding because it would turn solid again quickly.

The melting point of dicyclopentadiene is around 20 - 30 °C. So to me that sounds like you just used the liquid form of dicyclopentadiene, not cyclopentadiene.

Hm, checking Wikipedia it sounds like you are right and our TA had no clue about the differences either xD. But then we were creating a metallocene so probably the equilibrium amount of monomer was enough (we used it as a solvent iirc).

Your setup probably works perfectly well. the extra complexity in the sketch looks like it is intended to improve the purity of the product (by avoiding too much contamination with the dimer). This often doesn't matter much so is often avoided to keep things simple.

The Friedrichs condenser is for refluxing dicyclopentadiene (bp ~170 C). It's kept at 50 C to let the cyclopentadiene (bp ~40 C) pass without condensing.

I see. Since I had my oil bath temp. at 180 °C, I ran the risk of distilling some "uncracked" dicyclopentadienes together with the cyclopentadiene. I speculate that it probably worked because the head temperature in all likelihood was below the bp of dicyclopentadiene (can't say for sure, as I didn't measure temperature there). - Edit: looking at my own foto, I notice that I actually did have a thermometer there, but I never read it :-(

That setup in the drawing is way over the top, and you'll never get a drop of destillate with it. The Cyclopendadiene will fully condense in the bridge and all flows back into the Friedrichs. Ten degrees temperature difference to the dew point will never get it over 20 centimeters of glass tube.
However you should attach a drying tube to the exit near the receiving flask to avoid condensing water into it, especially if you do that nonsense with the dry ice/acetone. Typical for organic chemists, I'm sorry to say. A simple ice bath would help as much. Just keep the receiving flask as cold as the water in the cooler.
And 180 dC is a bit much, I'd just stay under the boiling point of the dimer. I remember adding some iron powder was supposed to speed up the retro-Diels-Alder reaction, i guess by lowering the effective concentration of the monomer through complexion.

I agree the setup is a bit much. The way I viewed the above diagram the receiving flask is on the left, hence the Friedrich's condenser is used to rapidly condense the hot, cracked cyclopentadiene rapidly. As such it gets the cold water, and the other condenser gets the hot water to help the reflux. Its a good idea to keep it cold at this point since low temp slows the dimerization. When Ive done it in the past I've had my setup under flowing N2 to keep it dry, but the drying tube could work depending on how pure you need it. Also I would freeze the collected cyclopentadiene during collection and storage. Things typically react slower in the solid phase. J. Chem. Educ., 1987, 64 (10), p 898

You've got it wrong. Read the description above the sketch: "technical Dicyclopentadiene ...placed in two-necked flask equipped with thermometer and Friedrichs condenser (..50°C)". Also why would you put a thermometer in the receiving flask?

Your right I just looked at the picture and wondered why someone would put a thermometer above the condenser on the receiving flask side. Perhaps a thermometer In the receiving flask is to make sure the cracked product stays sufficiently cooled.

You put a thermometer in the educt to make sure you're not overheating it, most importantly. Does that publication you cite give a rate constant for the dimerisation (and at what temperature)? I somehow doubt that extra cooling is really necessary.


text:
Dissociation of zinc anode during electrolysis
Example electrolysis setting:
Cathode: copper
Anode: zinc
Electrolyte: dilute aqueous sodium chloride [$\ce{NaCl(aq)}$]

During electrolysis, zinc is more reactive than copper, hence electron flows from zinc anode to copper electrode.
At the zinc anode:
$\ce{Zn(s)->Zn^2+ (aq) + 2e-}$
Q: What is this process known as?
("Dissociation" is usually applied for salts that dissolve to form mobile ions, and "discharge" is usually used for ions in the electrolyte itself that gain/lose electrons. Hence, what would a proper term be?)
When the zinc anode "dissociates" in the  electrolyte, $\ce{Zn^2+}$ ions are formed.
Q: Do the $\ce{Zn^2+}$ ions remain "isolated"?
(Because $\ce{H+}$ions are selectively discharged at the copper cathode, this leaves the $\ce{Zn^2+}$ions  $\ce{Na^+}$ions are bonded to  $\ce{Cl-}$ions, so does the  $\ce{Zn^2+}$ions form just move into the electrolyte just like that, with an unstable electronic configuration?)

Q: What is this process known as?

This is a particular case of dissolution. Call it electrolytic dissolution, if you want to be more specific. The word "dissociation" certainly does not apply here. It is used when talking about salts and other ionic compounds.

Q: Do the $\ce{Zn^2+}$ ions remain "isolated"?

Well, $\ce{Zn^2+}$ just goes into the solution and gets solvated (hydrated), that is, surrounded by interacting water molecules, pretty much like any other ion. Some people in some contexts would probably call these ions "free", though they are surely not as free as in vacuum. The same applies to the $\ce{Na+}$ and $\ce{Cl-}$ ions out there, and no, they are not linked to each other in any way except for electrostatic interaction. Have you heard of dissociation? That's what it is about. A $\ce{Cl-}$ ion does not "know" that it "belongs" to $\ce{Na+}$ and not $\ce{Zn^2+}$. They all just coexist there together, probably with some other ions like $\ce{OH-}$. As the process goes on, I expect some $\ce{Zn(OH)2}$ precipitate to appear.

The process in the particular case of dissolution is also known as oxidation.


text:
Reactions of acidified substances
Case study: acidified potassium manganate(VII)
Does "acidified" suggest that the substance contains $\ce{H+}$ions?
If so, will acidified $\ce{KMnO4}$ react with a base, such as ammonia?

Yes it will react, but it is the $\ce{H+}$ that is reacting, not the $\ce{KMnO4}$.

Permanganates, like KMnO4 are powerful oxidizers, but their redox potential and thus reactions are pH dependent. Lowering the pH by adding an acid increases the oxidizing power of permanganates.
Acidified KMnO4 most likely means a solution of KMnO4, with some sulfuric acid added. Since the solution is acidic it will react with bases like ammonia, but that is not the intended reaction of such a solution. It is used to oxidize a wide range of organic and inorganic chemicals, for example toluene to benzoic acid.


text:
Why do unboiled eggs last longer than boiled eggs?
Just like the title says, I watched this video https://youtu.be/CHMY4G9gTPA and then became intrigued.

An eggshell is a complex structure.  From Wikipedia:

Boiling the egg removes the waxy cuticle from the outside of the egg, dissolves a small but not insignificant amount of calcium carbonate from the shell, damages the protein matrix that holds calcium carbonate crystals in place in the shell, and can disrupt or destroy the two shell membranes.  All of these contribute to allowing bacteria a means to enter the egg and spoil it.
Note: Though calcium carbonate is actually less soluble in hot water vs cold, the rate of dissolution is substantially greater.  Unless the egg is boiled in a saturated solution of calcium carbonate, it will lose material to the solution when boiled, and any pores in the shell become enlarged as a result.

Not to forget the hole if you prick the shell where the air pocket is...

It isn't just the membrane. My recollection is that boiling an egg will  inevitably lead to at least small cracks in the shell, which will rip the membrane.

In addition to Jason answer: egg is a live organism with repair and immunity capabilities. Small punctures in egg membrane may be healed, and bacteria are most likely to get killed by white blood cells using antibodies or other techniques. Boiling kills everything, including egg immunity system - existing antibodies are denaturated (they are protein too) and there are no white cells to produce fresh ones.

I don't think eggs have white blood cells.


text:
Nucleophilicity vs Basicity
Can something be both a nucleophile and a brønsted base?
Also , I'm actually quite confused about the difference between these two terms.

by definition a nucleophile, is a lone pair donor. phile meaning 'loving'. i.e its attracted to species which are electron deficient. the term is used to describe a species in an organic mechanism.a  bronsted base, is a proton accepter. Mostly in organic chemistry, you draw curly arrows, going from the nucleophile to electrophiles,most usually a electron deficient carbon. so it acts more as a Lewis base. which is a lone pair donor. however there are protons floating around in most organic reactions, also very electronegative atoms such the halogens(leaving groups). which pick up protons so i guess by defintion you could say they act as bonsted bases as well as nucleophiles. However the term nucleophile is mostly reserved for the species acting on the electrophile in an organic mechanism.

authors: Завьялов Виктор


text:
Base catalyzed rearrangement of an alkyl substituted cyclopent-2-enone
I have been asked to provide a mechanism for the following transformation:

I don't see any viable way to make this happen. All I can see that is possible is the formation of an acetal or a Michael addition. Neither of which seem useful. Can you help?

I believe that the mechanism for rearrangement lies in a retro-aldol condensation (technically a Michael attack as a first step) followed by a new aldol condensation; both catalysed by the excess base which can both catalyse the mechanism and be the Michael attacker in step 1. My version of the mechanism is shown here:

One should be able to support the mechanism by labelling either of the ketone carbons with $\ce{^{13}C}$. (Oxygen labelling would not work since in the hydroxide conditions the oxygen will exchange rapidly via a hydrate intermediate.)
As I forgot to say and as Orthocresol correctly noted: Every step is to be seen as being in equilibrium and the overall equilibrium is shifted towards the thermodynamically most stable product: the tetrasubstituted double bond.


text:
Ionisation Energy and Evidence of Subshells
In Chemistry Electron classes, I was told that first ionisation energy gives evidence of electron shells.

Now is this true due to the steep increase shown on a first ionisation graph when the previous shell is emptied, or am I wrong completely?
Thanks, Dave.

I imagine you are correct. If you go on to study chemistry at a more advanced level you will learn more evidence for the notion of electron shells, but for now, you are right.

Let's start with the most basic planetary model. If the planetary model were right then copper could have any ionization energy. Since that isn't the case there must be a more complex explanation. If you then plot first ionization emery vs atomic number then you can see that there is some sort of complex pattern.  The pattern is due to the electronic shell structure. The rub here is that the shell structure is too complex to deduce from the first ionization energies alone. When you figure out the shell structure though, the first ionization energies fit.


text:
How much worse is methane than CO2 as greenhouse gas?
I can't find a coherent answer to this question. The numbers are all over the place.

25 times greater
84 times more potent

The reason I'm looking is because I can't find a reason in physics why this would be so*:

There is nothing “special” about methane that makes it much powerful than $\ce{CO2}$ (link to source)

What makes $\ce{CH4}$ absorb more infra-red than $\ce{CO2}$? What experiments have shown this?
Methane absorbs infra-red at different bandwidths than $\ce{CO2}$ (see here). Does the earth radiate more heat in those bandwidths?

The short answer is that all such "facts" are dependent on the planetary climate model which is being used. Such models are very complex. The sort of situation where you push something in here and it bulges out over there. //  In general most of the heat the earth transmits (via electromagnetic radiation, i.e. light) to space is done via infrared radiation. So having strong IR absorbers in the atmosphere stops infrared light from earth from getting into space.

See also [EarthScience.SE] Stack Exchange.

The reason for the discrepency is that the excess methane is removed from the atmosphere much quicker - about 12 years for methane compared to centuries for CO2. Source 1 says methane is "25x over 100 years", source 2 says "84x over 20 years". These are both true, it's just a difference of the timescale they're considering.

Let's take a look at this graph just so we have a reference.

This shows the IR absorption bands of various gases found in the atmosphere. For this answer, let's focus in on the absorption bands of $\ce{H2O}, {CO_2}$, and $\ce{CH_4}$.
First, the answer to why methane is considered a worse greenhouse gas than carbon dioxide is largely because methane absorbs in regions where carbon dioxide does not absorb at all. That is, methane absorbs different wavelengths of light than carbon dioxide.
Another thing to note is that methane's absorption peaks overlap with some of the absorption peaks of water vapor. This is an interesting thing to note because the strength of methane's absorptions in those overlapping regions are much stronger than the absorptions of water.
So, we have two things which contribute to why methane is sometimes called a worse greenhouse gas than carbon dioxide. First, it absorbs in areas where $\ce{CO2}$ does not absorb. Second, these absorptions are very strong.
Something you might be interested in, however, is the fact that there are some people who make the argument that methane essentially doesn't matter as a greenhouse gas. This would be due to the overlap in absorption between water vapor and methane. For instance, methane has an average concentration of 1.7 ppm (parts per million) in earth's atmosphere. Compare this with the fact that the earth's atmosphere can be composed of 1% water vapor at sea level and .4% across the whole atmosphere on average.
Thus, the amount of water vapor in the atmosphere is much more than that of methane at any given time. So, it can be argued that we do not need care much about methane as a greenhouse has because water vapor will do all the IR absorption that methane would have done.
I don't know enough about atmospheric chemistry to really say if I agree or disagree with those claims. One thing I do know, however, which is interesting is the fact that the idea of global warming would seem to present more of a problem with water vapor than methane or carbon dioxide because as temperature increases, the ability of the atmosphere to hold water vapor increases which would accelerate the warming from water vapor, which would increase the amount of water vapor the atmosphere can hold...
Also, I believe that the metric which says methane is more dangerous than carbon dioxide is a comparison which does not account for other things which might absorb that light. In other words, the figure you quote do not account for the fact that water vapor might make methane irrelevant as a greenhouse gas.
Interesting Read: http://www.nasa.gov/topics/earth/features/vapor_warming.html


text:
Can I freeze water at 100 °C if I add 28 grams NaCl in 100 grams water?
Can I freeze water at 100 °C if I add 28 grams NaCl in 100 grams water at 20 °C?
I found this diagram that says, if I have 28.05 wt% NaCl in water, then the equilibrium temperature between two phases (ice/liquid) will be at 100 °C. Am I understanding it right? I also found this solubility diagram which says that the solubility is ok for 28 grams at 20 °C.

I am sorry but I don't see how the diagram you linked implies that. If anything, dissolving salt in water should lower the freezing point to below $0~\mathrm{^\circ C}$. https://en.wikipedia.org/wiki/Freezing-point_depression

@orthocresol What you say is happening only when the wt% is below 26,28%. After this limit the freezing point is increased dramaticly. Look at this diagram  too. link

It is not a freezing point. It is the point of equilibrium between the liquid and solid.

@Ivan Neretin What do you mean as solid? Like a crystal?

If you read the note below the table, it says: "$T_\mathrm{eq}$ is the equilibrium temperature between two phases: ice/liquid solution for $T_\mathrm{eq} < 0 - 0.1$ and $\ce{NaCl}$/liquid solution for $T_\mathrm{eq}$ above $0.1~\mathrm{^\circ C}$". So, your equilibrium at $100~\mathrm{^\circ C}$ is not a water/ice equilibrium.

If you look at the phase diagram, it is an equilibrium between "liquid solution" and "liquid solution + $\ce{NaCl (s)}$". So, what the table is actually telling you is the solubility of $\ce{NaCl}$ in water (when it is insoluble, it will precipitate out, giving you the saturated "liquid solution" as well as solid $\ce{NaCl}$).

Yes, like a crystal. Only it's not an ice crystal.


text:
Why does stale bread go soft in the microwave?
It's fairly well known that stale bread will go soft in the microwave. It makes hard bread edible (albeit a little bit chewy sometimes.)
It's obvious that moisture is being released from somewhere in the bread, and this is what causes the softness.
But the bread doesn't blacken as it would if the carbohydrates were decomposing to carbon, so where (chemically) does this moisture come from?

Poked around a bit and I think the answer is the gluten content. Know when you cut into a cooked steak how juice will ooze out? That is because cooking denatures the proteins in the steak. Microwaving does the same thing to bread. It denatures (removes water) from the gluten and the rest of the bread absorbs the moisture.

I'd say the residual water acts a plasticiser. When quickly heating the bread, it goes above Tg, becoming soft. If that's true, it should become stiff again quickly when cooling. Does it?

Another possibility: fresh bread contains water, we can see that once water evaporates it gets hard. Does some water remains? Probably, but this might not be the only reason why it softens when heating it up. When you heat the 'dry' bread up it seems reasonable that some component in the mixture fuses and because of it, it softens. Probably water helps too...https://chemistry.stackexchange.com/questions/96459/why-does-butter-get-soft-with-heat-but-not-ice?rq=1 seems relevant

@MaxW 5 years later, but the same happens to gluten free breads as well for whatever it's worth

I believe this has to do with its moisture content. While certainly dry, it still has some moisture, and what little moisture is left gets excited by the microwaves, heats up (the heat helps it soften too) and steams the bread essentially. I think that MaxW's explanation of glutens for the question 'Where does the moisture come from?' is a likely explanation.
Note that if the bread is sufficiently dry enough, this little microwave trick wont help. It is usually advised that you add a few drops of water to the bread before heating up in the microwave in this way. This is to ensure the bread doesn't actually become harder in the microwave, as it boils the water and will drive it off if microwaved in excess.
Also, note that bread WILL become blackened, charred and start smoking if left in the microwave long enough. I did this by typing 30 minutes instead of 30 seconds. It started smoking around 5 minutes.


text:
Periodic table- quantum numbers
I have come across many questions like: "if electron had 3 spins (-1/2,0,+12) then what change will be there in the periodic table?", also sometimes " if the capacity of each orbital becomes 5 then in what period or group will the particular element be?" So what is the relation between quantum no. and the periodic table?

Related: http://chemistry.stackexchange.com/q/28921/7951

If electrons could have $m_s = -1/2, 0,$ and $1/2$, then the entire structure of the Periodic Table would be different. The s-block would have 3 elements instead of the 2 that we see now, because each s orbital can accommodate 3 electrons, and similarly the p-block would have 9 elements instead of 6.
It is not possible to say, for example, "sodium would be in so-and-so position in the new Periodic Table" because in such a universe, the element sodium would not even exist the way it does in our universe. The new Periodic Table would not have an element with the electronic configuration $\mathrm{(1s)^2(2s)^2(2p)^6(3s)^1}$. What I am basically saying is, there is no one-to-one correspondence between the elements in our universe and the elements in such a hypothetical universe (mathematically speaking, you can, but it would not make any chemical sense). Nevertheless, you could say that sodium is defined to be the element with 11 protons and 11 electrons. In that case, it would have a configuration of $\mathrm{(1s)^3(2s)^3(2p)^5}$ and it would be the fifth element in the Period 2 p-block.
Actually, there is no evidence that the aufbau principle should still hold true in such a universe: the large repulsion that arises from putting 3 electrons into the same orbital might well make filling the 2s orbital before the 1s orbital more favourable. So, our hypothetical "sodium" might have a weird electronic configuration of $\mathrm{(1s)^2(2s)^2(2p)^7}$. The Periodic Table is useful partly because it groups elements with similar electron configurations - which implies similar chemical properties - together. However, if these elements do not obey the aufbau principle, then there is no guarantee that the elements in the same group should have similar chemical properties, and therefore no guarantee that a Periodic Table in such a hypothetical universe would even be useful.
I should end off by saying that the laws of quantum mechanics stipulate that the different values of $m_s$ must increase in steps of $1$. Therefore, it is not actually possible that $m_s$ could ever take on values of $-1/2, 0$, and $1/2$, since these increase in steps of $1/2$. You might say: "Ah! But we could say, what happens if $m_s = -1, 0, 1$." Well, in that case, you have much more to worry about than just the orbital occupancy. Without going into too much detail, the Pauli exclusion principle would no longer apply to electrons, and you would have more problems than just "one orbital can fit 3 electrons".


text:
Finding the pH among HF and KOH
Okay I ran into this question in homework.
A 100.0 ml sample of 0.20M HF is titrated with 0.10 M KOH. Determine the pH of the solution after adding 200 ml of KOH. The Ka of HF is 3.5 x 10-4?
Oh yeah its multiple choice

3.46
10.54
8.14
9.62
7.00

I've found answers to similar questions.Here
But I come into a different problem with this question. The HF and the KOH cancel out each other for they have the same amount of moles
I assume what ever we're looking at on the other side will have 0.02 moles and is an acid

HF -> 0.1L * 0.2M = 0.02 mol
KOH -> 0.2L * 0.1M = 0.02 mol
HF-KOH = 0

I beleive this means I can't use the hasselbach equation so I did this:
(M of ???) = 0.02 mol / 0.3 L = 0.667*10^(-2)
3.5*10^(-4) = x^(2) / (0.667*10^(-2))
sqrt(3.5*10^(-4)*(0.667*10^(-2))) = x

so there is another way of solving it
I have had some feed back on this reaction: "You assume the first reaction goes to completion when the Ka is not that high"
If it doesn't go to completion what do I do?!
I really just need to see how this works out and know why things are happening

I would recommend you stick to the assumption that the reaction is complete first, before going on to consider the incomplete case. If the reaction is complete, what products are you going to form?

I'm sorry but I wasn't sure really either FH2 + KO or H2O + FK not really sure actually

Which of HF and KOH is the base? Which is the acid?

HF is the acid KOH is the base

Uhhuh. So the acid is a proton donor right? And the base is a proton acceptor right? What species will they form? Remember KOH exists as $\ce{K+}$ and $\ce{OH-}$ in solution.

So H2O(which is neutral) + FK

Yes. And as you said, the number of moles of each is equal, so you now essentially have a solution of the weak base, $\ce{F-}$. Calculate the concentration of $\ce{F-}$ and from there calculate the pH.

I hate stupid questions where you can know too much for your own good.  0.20 m HF is +/- 0.005 molar. The !@#$%^& pH could be "all of the above".

@orthocresol and what I got for the - log(F-) Which F- should be (0.02 mol / 0.3 L) didn't appear on the answer list and i thought this might have led me to the pOH since we are left with a base. Which didn't show up either. Where does pKa/Ka show up in this solution?

@MaxW  also what...?

It has to do with significant figures. In chemistry when we write 0.20 when mean 0.20 +/- 0.005. If the solution were 0.200 molar then that would be +/- 0.0005. Since both the acid and base solution were +/- 0.005 molar, all of the ph values listed could be possible answers. In other words the acid could have been 0.203863859462929428 molar which rounds to 0.20.

@RyanHenry Please go and look up "pH of a weak base" on SE, or the Internet in general. There is no point in calculating $-\log_{10}[\ce{F-}]$ because that does not correspond to either pH or pOH. Instead, you need to find out how to determine $[\ce{OH-}]$ from $[\ce{F-}]$ and from there you can determine pOH and pH.

In other words you notion "The HF and the KOH cancel out each other for they have the same amount of moles" is right. Write the balanced chemical equation. What do you have on the right hand side? What happens to the products?

Okay Thank you all I finally got it. so K is just thrown out the window. and we are left with the equation (F- + H2O <-> FH+ OH-) So i take the Ka to make Kb by (Kw/Ka) and my concentration of F- and OH- was (0.02/0.3) which I plugged into sqrt((Kb)*(M of OH- or F-)) took the -log of that got the pOH and then converted it back to pH (14 - pH) One final thing why did I K out the window does that mean its neutral. Oh and the answers 8.14

Yes. After balancing the chemical reaction the question is basically "What is the pH of a 0.0667 molar KF solution?"

@RyanHenry the K+ ions do not affect the pH... at least, probably not enough for you to notice.

From the comments:

[OP] Okay Thank you all I finally got it. so K is just thrown out the window. and we are left with the equation (F- + H2O <-> FH + OH-) So i take the Ka to make Kb by (Kw/Ka) and my concentration of F- and OH- was (0.02/0.3) which I plugged into sqrt((Kb)*(M of OH- or F-)) took the -log of that got the pOH and then converted it back to pH (14 - pH) One final thing why did I K out the window does that mean its neutral. Oh and the answers 8.14

Paraphrased with paragraphs:
A strong base ($\ce{KOH}$) reacts with a weak acid ($\ce{HF}$) at stoichiometric ratio:
$$\ce{KOH + HF -> H2O + F- + K+}$$
The major species is fluoride, a weak base with pKb = 14 - pKa, where pKa is that of hydrofluoric acid. The potassium ion is a spectator. To find the pH, use your favorite strategy for a pure weak base. The result is pH = 8.14.
As MaxW pointed out in the comments, this relies on getting the stoichiometric ratio just right. If $\ce{KOH}$ is even in slight excess (let's say one tenth of a millimole per liter too much), the pH will be strongly basic (pH 10). If you err in the other direction, you will have an $\ce{HF, F-}$ buffer with an acidic pH.


text:
Reference request for properties of amorphous titanium carbide (a-TiC)
I assume through some side reaction I'm getting $\ce{TiC}$ in my $\ce{TiO2}$ samples after calcination and I need some basic information about the properties of amorphous $\ce{TiC}$.
Is it paramagnetic or are there paramagnetic centers in the powder, i.e. $\ce{Ti (III)}$ species? What is the colour?

https://en.wikipedia.org/wiki/Titanium_carbide


text:
Calculate the concentration of the deprotonated adipic acid anion

Adipic acid is a diprotic acid having $K_\mathrm {a1}=3.9\times10−5$ and $K_\mathrm {a2}=3.9\times10−6$. A saturated solution of adipic acid is about $\pu{0.10 M}~\ce{HOOC(CH2)4COOH}$. Calculate the concentration of the $\ce{−OOC(CH2)4COO−}$ in this solution. Express your answer to two significant figures and include the appropriate units

Lets represent adipic acid as $\ce{H2A}$ for the sake of simplicity.
$$\ce{H2A <=> H+ + HA-}\qquad   K_{\mathrm a1} = 3.9 \times 10^{-5}$$
$$\begin{array}{lccc}
\text{Initial}\quad &0.1 \quad &&0 \quad &&0\\
\text{Change}\quad &0.1-x \quad &&+x \quad &&+x\\
\text{Eq}\quad &0.1-x \quad &&x \quad &&x
\end{array}$$
$$3.9 \times 10^{-5} = x\times x/0.1-x$$
$x^2 = 3.9 \times 10^{-5} \times 0.1$ (neglecting $x$ in nominator because $x\ll0.1$)
$x = x_1 = 0.00197\ \mathrm M$
Let's designate $x$ as $x_1$. $\ce{HA-}$ dissociates again as
$$\ce{HA- <=> H+ + A-}\qquad K_{\mathrm a2} = 3.9 \times 10^{-6}$$
$$\begin{array}{lccc}
\text{Initial}\quad &0.00197 \quad &&0 \quad &&0\\
\text{Change}\quad &0.00197 -x \quad &&+x \quad &&+x\\
\text{Eq}\quad &0.00197 -x \quad &&x \quad &&x
\end{array}$$
$$3.9 \times 10^{-6} = x^2/0.00197$$
$x^2 = 0.00197 \times 3.9 \times 10^{-6}$ as shown above
$x = x_2 = 8.77 \times 10^{-5}$   (designate $x$ as $x_2$)
Concentration of $\ce{HOOC(CH2)4COOH}$ ($\ce{H2A}$) $= 0.1 - x_1 = 0.098\ \mathrm M$
But that was wrong. I then  $x_1 - x_2 = 0.00197 - 8.77 \times 10^{-5} = 0.00188\ \mathrm M$ but that answer is wrong.

I'll follow your lead and represent adipic acid as $\ce{H2A}$ for the sake of simplicity.
Equations we know:
(I) o.100 = $\ce{[H2A] + [HA^{-}] + [A^{2-}]}$
(II) $\dfrac{\ce{[H^+] [HA^{-}]}}{\ce{[H2A]}} = 3.9 * 10^{−5}$
(III) $\dfrac{\ce{[H^+] [A^{-2}]}}{\ce{[HA^{-}]}} = 3.9 * 10^{−6}$
(IV) $\ce{[H^+]} = \ce{[HA^{-}] + 2[A^{2-}]}$
There are four equations in four unknowns so the system does have a solution. I could find any way to simply all the equations into anything which seems reasonable for an exact solution.
The way to solve this problem is to solve for the first ionization and assume that the second ionization has an insignificant contribution. A check can be done with equation (IV) to verify that  $\ce{[HA^{-}] \gg 2[A^{2-}]}$.
I am going to show an excessive number of significant figures in the answers so that various methods of calculation can be compared. The acid constants are only given to 2 significant figures so that is the limiting number of significant figures that can justified.
Quick and dirty
Let $\ce{[H^+] = [HA^{-}]}$ and let $\ce{[HA]} = 0.100$
then from (II)
$\ce{[H^{+}]^2 = [HA^{-}]^2 =} \sqrt{(0.100)(3.9E-5)}$
$\ce{[H^{+}] = [HA^{-}] =} 1.97E-3$
thus $[\ce{HA}] \approx 0.1 - 0.00197 = 0.09803$
from equation (III)
(V) $\ce{[A^{-2}]} = \dfrac{\ce{[HA^{-}]}}{\ce{[H^{+}]}}3.9E-6 = 3.9E-6$
So indeed $\ce{[HA^{-}] \gg [A^{-2}]}$
This method can be made somewhat better by noting
                  x*(0.1/0.10197)
   HA    0.10000  0.09807
   HA-   0.00197  0.00193
         -------  -------
         0.10197  0.10000

Cleaner solution
The Quick and dirty is a bit dodgy since $\ce{[H2A]}$ was assigned a value of 0.100 to calculate, then reassigned a value of 0.09803. Since the pK constant is only given to two significant figures this isn't really wrong, but a more consistent approach can be easily calculated.
Let $\ce{[H^+] = [HA^{-}]} = x$ and let $\ce{[HA]} = 0.100 - x$, and also let's use $k_2$. Substituting into equation (III):
$x^2 = (0.100 -x)k_2$
$x^2 + k_2x -0.1k_2$
This is a quadratic in x so roots are:
$x = \dfrac{-k_2 \pm \sqrt{k_2^2 - 4(1)(-0.1k_2)}}{2(1)}$
The negative root doesn't work so the positive one is the only one. Also $k_2^2$ is so small that it can be neglected so;
$x = \dfrac{-k_2 + \sqrt{0.4k_2}}{2} =0.00186$
$\ce{[H^+] = [HA^{-}]} = 0.00186$
$\ce{[HA]} = 0.100 - 0.00186 = 0.09814 $
Again from equation (V)
$\ce{[A^{-2}]} = \dfrac{\ce{[HA^{-}]}}{\ce{[H^{+}]}}3.9E-6 = 3.9E-6$
Exact Solution
Solve equation (II) for $\ce{[HA^{-}]}$ in terms of $\ce{[H2A]}$ and $\ce{[H^+]}$ to get:
$\ce{[HA^{-}]} = \dfrac{k_1\ce{[H2A]}}{\ce{[H^+]}}$
and multiply (II) and (III) to solve for $\ce{[HA^{2-}]}$ in terms of $\ce{[H2A]}$ and $\ce{[H^+]}$ to get:
$\ce{[A^{2-}]} = \dfrac{k_1k_2\ce{[H2A]}}{\ce{[H^+]^2}}$
Both of the above can be substituted into the left hand side of the following three equations to get the values shown on the right.
$\dfrac{\ce{[H2A]}}{\ce{[H2A] + [H2A^{-}] + [A^{2-}]}} = \dfrac{\ce{[H^+]^2}}{\ce{[H^+]^2 + k_1[H^+] + k_1k_2}}$
$\dfrac{\ce{[HA^{-}]}}{\ce{[H2A] + [H2A^{-}] + [A^{2-}]}} = \dfrac{\ce{k1[H^+]}}{\ce{[H^+]^2 + k_1[H^+] + k_1k_2}} $
$\dfrac{\ce{[A^{2-}]}}{\ce{[H2A] + [H2A^{-}] + [A^{2-}]}} = \dfrac{\ce{k_1k_2[H^+]}}{\ce{[H^+]^2 + k_1[H^+] + k_1k_2}} $
Being a glutton for punishment I created a spreadsheet with the calculations. After some manual manipulation I was able to determine the "exact solution" to a totally ridiculous number of decimal places.
$\ce{[\ce{H+}]} \text{ }= 0.00195932639309$
$\ce{[\ce{H2A}]}~~= 0.09804455814272$
$\ce{[\ce{HA-}]} ~= 0.00195155732146$
$\ce{[\ce{A^{2-}}]}~~= 0.00000388453582$
$\text{sum} \quad = 0.10000000000000$
$\ce{2[\ce{A^{2-}}] = [\ce{H+}] - [\ce{HA-}]} = 0.00000776907164$
$\ce{K_{a1}}$ from data = $3.9E-5$
$\ce{K_{a2}}$ from data = $3.9E-5$
$\ce{[H^+]} = \ce{[HA^{-}] + 2[A^{2-}]} = 0.00195932639309$


text:
What is the meaning of the prefix "per-"?
The prefix "per-" appears in the names of many compounds, such as in perchlorate, persulfate, peroxide. What does it mean?

See my comment in the thread under my answer here.

Per- can mean a number of different things, depending on the context.
It can mean that there is a peroxi bond(oxygen-oxygen bond) in the system, persulfate is a good example of this.
More generally it tends to mean that the compound is the most completely altered product of a certain(possibly hypothetical) reaction. For example perchlorates and permanganates are the most completely oxygenated compounds of chlorine and manganese, respectively. This also describes another common use of per-, perhalogenated compounds. In that context, it usually means that all hydrogens present in the parent compound, that were bonded to a carbon, have been replaced with a halogen, for example perfluoroethylene (C2F4, also known as tetrafluoroethylene, the monomer of Teflon(PTFE))

According to current IUPAC recommendations, the prefix "per-" is no longer recommended for organic compounds in which all substitutable hydrogen atoms are completely substituted (e.g. perfluoroheptane).

@Loong are there substituting recommendations or must one say octadecafluoroheptane?

@Jan The preferred IUPAC name for perfluoroheptane $(\ce{C7F16})$ is hexadecafluoroheptane. Note that all locants are omitted, i.e. the name is not 1,1,1,2,2,3,3,4,4,5,5,6,6,7,7,7-hexadecafluoroheptane.

I came here looking for this compound - perfluorobenzene - which has all its hydrogens replaced by a halogen, am commenting if anyone else googles this as well.


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Does the 'X' of Grignard's reagent react?
Till whatever I have studied in organic chemistry, Grignard's reagent has been the most helpful thing everywhere, whether is it preparation or reaction of organic compounds. But I haven't ever seen the Halogen part of RMgX reacting with any compound. Does the Halogen ever react in any reaction?

Maybe it does, it's just those reactions are not that helpful, so nobody uses them.

Lets look at Grignard's reaction route. Key step is nucleophilic addition of R. And MgX is removed with acid on the next step. Nobody cares about X - if you want MgX2 in solution you just add MgX2. The main reason to use Grignard's reagent is nucleophilic addition of R.


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Separating Enantiomers
I understand that you can react enantiomers with a chiral molecule to resolve the enantiomers of interest into diastereomers.
However I do not understand how exactly the added chiral molecule can make the enantiomers of interest into diastereomers.
Can someone please explain the process/mechanism ?
For example: here is a picture from my textbook.

How are those two salts Diastereomers, even though they are mirror images of each other.

The ways are probably numerous. One mechanism that I remember has the chiral counterpart on a column. So you would basically do like an ion exchange. You'd then flush the column to release the other enantiomer. See: https://en.wikipedia.org/wiki/Chiral_resolution

Are you sure they are mirror images? Look at the salt as a whole, not the individual ions.

The example on UC Davis ChemWiki is better, as they also offer the actual physical data with the compounds: http://chemwiki.ucdavis.edu/Organic_Chemistry/Fundamentals/Isomerism_in_Organic_Compounds/Diastereomers_and_Optical_Resolution

Enantiomers are mirror reflections, they have same physical and chemical properties. This like left and right hand.
Diastereomers have different physical and chemical properties. You can separate them on chromatography column or by crystallization.
In your example upper and bottom products are not mirror reflections.
Mirror relection of upper product looks like:

Just play with the 3D models in any drawing software to get the point.

Ah. Thanks. So the anion salt has to be taken into account as well. I was under the impression that you would only have to look a the cation.


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Can one component of a mixture penetrate the skin if the whole mixture cannot?
There's a general rule that says that compounds under 500 dalton are able to penetrate the skin without any special vehicle (such as ethanol).
I've started using castor oil, which contains different acids that are believed to be good for the skin (not discussing this here now). The oil is very thick, and each acid has different molecular weight. I'm mostly interested in ricinoleic acid, which has about 300 dalton.
If I were to apply this bare oil on my skin, would it be a correct to think that some acids would penetrate the skin and some others wouldn't (because of the 500 dalton rule), assuming this oil didn't have any special treatment (except, well, the process of extracting the oil itself)?

Well yes. Simplest example: An aquaeous phenol solution which will penetrate the skin even though water won’t.

@Jan then it may be possible that it takes more time to absorb the ricinoleic acid (to stick with the example) due to the whole mixture being thicker and mixed with some other components, right?

I can’t really comment on what might be absorbed and what might not; that’s more of a pharmacist’s question than a chemist’s one; sorry =C


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Is iron the most stable element in the periodic table?
According to the binding energy per nucleon vs mass number graph, it is observed that iron-56 has the maximum value of binding energy per nucleon ($\pu{8.75 MeV}$). It means that iron-56  is the most efficiently bound nucleus meaning that it has the least average mass per nucleon. This is the approximate basic reason why iron and nickel are very common metals in planetary cores since they are produced profusely as end products in supernovae and in the final stages of silicon burning in the stars. So, in one word, iron is quite stable.

But, what about helium and other noble gases? They are considered the most stable elements in the whole periodic table. But their binding energy per nucleon value is less than iron-56. So, they are not stable as iron-56. Is that true? Is iron the most stable element in the periodic table both structurally and chemically?

Chemical stability and the stability of the isotopes are two completely different topics. First is about the stability of the electronic shells and the second is about the stability of nucleus.

re: " This is the approximate basic reason why iron and nickel are very common metals in planetary cores, since they are produced profusely as end products in supernovae and in the final stages of silicon burning in the stars. " // That just isn't true. When the earth was very young it was molten. As it began to cool, then various phases started to separate. Fe/Ni formed a phase and since they are much denser that crustal compounds like SiO2 the Fe/Ni core "sank" to the center of the earth.

Similar question from Physics StackExchange: Why is iron the peak of the binding energy curve?

Iron-56 has the lowest mass per nucleon, nickel-62 the highest binding energy per nucleon.

Yes, $^{56}\ce{Fe}$ has the most stable nucleus, and $\ce{He}$ is the most chemically inert element. These are different and unrelated qualities, pretty much like physical fitness and intelligence in a man. As for structural stability, there is no such thing in chemistry (there is one in architecture and another in mathematics, but those are out of scope of this question).

What about the gold atom? I thought it's also very stable and hard to destroy.

No more so than any other non-radioactive atom.

In the far future, most stars will degenerate into iron-56 stars specifically because it is the most stable element in the universe

@cowlinator iron stars are unstable to core collapse, become neutron stars, black holes, or possibly strange matter stars.

@DavePhD  Neutron matter, strange matter, and black hole matter are not elements, so it seems irrelevant.

No, nickel-62 is the most stable on a binding energy per nucleon basis.  Fe-58 is second and Fe-56 is third.
See Fewell, M. P., "The Atomic Nuclide with the Highest Mean Binding Energy", Am. J. Phys., vol. 63, pages 653-658.

The other half of this is a star burning the atoms via fusion. I'm too lazy to look up all the reactions, but the burning is exothermic to Fe and then becomes endothermic. It has to do with the mixture of isotopes created as the star goes from element to element.

@MaxW, nickel-56 is the last energetically favorable fusion product in silicon burning (in a series of fusions with alpha particles).  The nickel-56 then decays to iron-56.  https://en.wikipedia.org/wiki/Silicon-burning_process

Nickel-62, Iron-58 and Iron-56 are the most tightly bound nuclei, but Iron-56 is the most stable nucleus.  The stability of Iron-56 results from the fact that an Iron-56 nucleus has a diameter about equal to the range of the nuclear force.  In an Iron-56 nucleus every nucleon is attracting every other nucleon.

Well, Iron has the most stable isotope(Iron-56), it has a large half-life which means it takes a fairly large amount of time to decay into the half of what was initially there. Noble gases have a low reactivity, which means a low electronegativity and a high ionization energy. Noble Gases should be identified as inert and not stable. They are not the same.

Large? As far as we know it has no half-life, unless proton itself decays.

Every element known to the humans have a half-life.  Iron - 56 is stable but that's an iron isotope. Isotopes can be stable but there are many isotopes of iron that are radioactive and decay. For example, Iron-72 has a half-life of around 150 nanoseconds, while Iron-60 has a half-life of  2.62 X 10^6 which is a fairly LARGE amount of time.


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Concentration of solutions
I'm stuck with this problem. If I have 200 grams of a solution at 30% how much water should I add so that the concentration becomes 25%?

This is a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed.

The answer is that for a simple dilution the following formula applies:
$$c_1m_1 = c_2m_2$$ $$ m_2 = \frac{c_1m_1}{c_2} = \frac{(200g)(30\text{%})}{20\text{%}} = 240g$$
Therefore the mass to add is $(240g - 200g) = 40g$ of $\ce{H2O}$ (which is 40 ml of $\ce{H2O}$).

I would +1 if you added units wherever it was necessary.

Agreed - A unit analysis is always right to be sure you're getting what you think you are getting.

It's a homework question - they shouldn't be answered with exact calculation

There is a rule for. In german wikipedia there is a article for 'Mischkreuz', I don't know the equivalent in english, maybe the 'cross over rule'.
If you have two solutions with different concentrations and have to mix up a concentration in the middle. The difference between the start concentrations and the final concentration gives the volume ratio of the input solutions.
In your example:
You need 25 parts of your concentrated solution. (difference: 25% - 0%)
And 5 parts of water. (difference: 30% - 25%)