text:
What is the structural formula of alkali hypohalite: MOX or MXO?
There seems to be an inconsistency in the formula. Somewhere it is written as MOX and in some places, it is written as MXO. This is what Google gives when you search sodium hypochlorite:

I think the formula of alkali hypohalite should be MOX because since the negative charge bears over the oxygen atom, it accepts the electron coming from the alkali metal ion and forms a duplet  thus forming ionic bond.

Is my assumption correct? What is the correct formula of alkali hypohalite?

I felt tempted to close this as a duplicate of the sodium acetate notation question, because of my answer there, but the question in itself does not ask the same thing, so I’ll just rewrite the relevant parts of my answer.
In short, there is no correct formula. Depending on which point you want to bring across, you can use either, and depending on the context, $\ce{ClNaO}$ may also be acceptable.

Structually, you are correct with your assumption; the oxygen bears the negative charge. But that doesn’t include any implication that the formula should be written that way; compare $\ce{SO4^2-}$ where the charge is also on the oxygen atoms and yet they come last.
In fact, that leads us to our first clue. By convention, oxygen atoms are typically written last; compare $\ce{ClO3- , ClO4- , BrO3-}$ and others. You could probably explain that with electronegativity, the convention being that electronegative elements come last. This assumption argues for $\ce{NaClO}$.
On the other hand, the corresponding acid $\ce{HOCl}$ is often understood as derived from water $\ce{HOH}$, replacing a hydrogen with a chlorine atom. From this starting point, it makes sense to write $\ce{NaOCl}$, because then you replaced the other hydrogen; this time with sodium. It also includes a resemblance to $\ce{NaOH}$.
Both ways can thus be supported by the reasons given above. In practice, I think I have always seen the substance being referred to as $\ce{NaOCl}$ but I could simply have ignored the occurances of $\ce{NaClO}$.

$\ce{ClNaO}$ is the Beilstein-type lookup formula and accepted whereever alphabetic ordering of elements is preferred/required.

Both formulae are correct. NaOX is correct structurally but preferred form is NaXO in correspondence with other series salts like Oxo-halate(III) NaXO2, Oxo-halate(V) NaXO3 and  Oxo-halate(VII) NaXO4.

I know both of them are correct but which one should be used? Explain with some reasoning. For ex- sodium chloride is also (incorrrectly) written as ClNa but that's not we use.


text:
Request for a graph of the combined probability density of the three p-orbitals
This started when i was wondering why p-orbitals have an unsymmetrical shape (which isn't clear to me yet). But then i came across Halliday/Resnick/Walker's Fundamentals of Physics. In it, they have said that the three p-orbitals show their individual shapes in presence of a magnetic field. But normally, in an isolated atom, the three p-orbitals have the same energy and are indistinguishable, and that their combined probability density is spherically symmetrical.
I am having trouble in visualizing this. The three dumbells do not look like they could complete a sphere! For example, what about the region in the direction of $\hat i + \hat j + \hat k$? Wouldn't gaps be left behind? Wouldn't this destroy the symmetry of the distribution?
A visual graph of the three separately and then the three together (like different coloured dots for different p-orbitals) would help. I have no resources or skills (not even remotely close!) to do it myself, so i am requesting this to the skilled people out here. Thank you very much!

I have no graph but if you mathematically add up the squares of the wavefunctions (to be more precise $\psi^*\psi$), you get a probability density that is independent of $\theta$ and $\phi$. The dumbbell diagrams are only pictorial representations of a concept (probability density function) that we cannot adequately depict in 2D. It is not surprising that when you add these limited diagrams you don't get the full truth.

The dumbbell pictures show where the probability density is greatest for each of the orbitals, but there is still some probability density in the area outside the dumbbells. When you add the orbitals together, the area inside each dumbbell has a large contribution from one orbital and near zero from the other two, while the "gaps" have a small contribution from each of the three orbitals, totalling the same amount.

To make orthocresol's comment more concrete, you can try to calculate $|Y_{1,0}|^2+|Y_{1,-1}|^2  + |Y_{1,1}|^2$, see this quantity is independent with $\theta,\phi$

@f'' Thank you for the comment! So this means the probability plot for the p-orbitals together would look like the s-orbitals with value of $n$ one less than their own? (That is to say, $3p$ orbitals would look like $2s$ orbitals and so on)

@orthocresol Hmm, that makes sense! Then what is the factor that separates them in the first place? Thank you for your comment!

What do you mean by separate?

@orthocresol I mean what factor causes p-orbital to be spherically unsymmetric?

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text:
pH of liquid hydrogen chloride vs hydrochloric acid
Is hydrogen chloride in liquid state acidic like hydrochloric acid?
I know that hydrochloric acid is hydrogen chloride dissolved in water but I was wondering if liquid hydrogen chloride was also an acidic substance.
Is this the same for other acids?

related http://chemistry.stackexchange.com/questions/53580/is-there-any-notion-of-ph-out-of-solution

The problem here is that the question is unintentionally loaded. As is, pH is a terrible measure of acidity or basicity for pure substances or when trying to compare solutions in different solvents. Furthermore, acidity and basicity are far more general concepts than pH. Better measures are pKa and proton affinity, but they too have their limitations.
That said, this is a clear-cut case. In truth, pure hydrogen chloride (gaseous or liquid) is a much stronger acid than aqueous hydrochloric acid. When hydrogen chloride is dissolved in water, it reacts quantitatively to produce hydronium ions $\ce{[H(H_2O)_{n}]^+}$. This is the actual acid present in hydrochloric acid solutions. The fact that this reaction between hydrogen chloride and water is quantitative, fast and very exothermic is a suggestion that the water is taming the acidity of the hydrogen chloride. This is a general phenomenon known as solvent levelling.
The exact factor by which the acidity of hydrogen chloride is reduced when dissolved in water is hard to measure precisely. There are attempts to form universal acidity/basicity scales, such as this article (overview here), which tries to define an absolute pH scale, as follows:

From the authors' data , gaseous hydrogen chloride at a pressure of $\mathrm{1\ bar}$ is an acid with an absolute pH ($\mathrm{pH_{abs}}$) of about 160, which is around 30 orders of magnitude stronger than even the most concentrated aqueous hydrochloric acid solution (pH in water approximately equal to -1), which would correspond to a $\mathrm{pH_{abs}}$ of about 190.

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text:
For what is reaction enthalpy more relevant than reaction free energy?
I know enthalpy is a component of the Gibb's Free Energy (along with entropy). For practical purposes, though, isn't Gibb's Free Energy what a chemist is actually interested in, as that determines the overall equilibrium of a reaction? What important detail does enthalpy inform us of and what applicable function does that detail serve?
I am asking this because I have a DFT calculation that yields both the enthalpy and free energy, and I see a paper citing the enthalpy, so I am wondering why it is being preferred over the free energy.

While a detailed answer to this question could go on at length, the major practical distinction is the following:

The enthalpy change of a reaction/process indicates how much heat input or output is involved with the reaction, with a negative $\Delta H_\mathrm{rxn}$ indicating heat output.
As you note, the Gibbs free energy change of a reaction/process indicates whether a reaction is thermodynamically favored or disfavored, with a negative $\Delta G_\mathrm{rxn}$ indicating a favorable reaction.

Note that, e.g., a reaction/process can have $\Delta H_\mathrm{rxn} > 0$ (endothermic), but also $\Delta G_\mathrm{rxn} < 0$ (thermodynamically favored).  That is, despite the reaction requiring heat input, it is still favored to occur.  This is because the change in entropy of the process, $\Delta S_\mathrm{rxn}$, is sufficiently large that the $T\Delta S_\mathrm{rxn}$ term is able to overcome the positive $\Delta H_\mathrm{rxn}$.  A common example is the endothermic dissolution of various salts, such as ammonium nitrate $\left(\ce{NH4NO3}\right)$.
Thus, the paper probably used the enthalpy results because they were comparing their calculations to experimentally-measured enthalpies (heats) of their process(es) of interest.

Hi, thank you very much for the response! Would you mind telling me why the heat of a reaction is useful? For example, the Gibb's Free Energy tells us whether a reaction is favorable or not. We can draw conclusions about the usability of the reaction compared to an alternative reaction, as we can derive the reaction metric K. What application does the enthalpy, or rather the heat, have, that is useful to a chemist in the lab? It seems like any experimentally useful conclusion we'd like to draw from it is unreliable because the other half of the picture, entropy, could alter the interpretation.

If you are designing a chemical reactor for a chemical processing plant, you need to perform a heat balance on the reactor (and all other equipment in the plant) so  that the temperatures into and out of the various pieces of equipment are satisfactory and so that you control the energy budget of the plant.  The energy budget is related to $.  Also, for the reactor, you need to know how much heat you need to supply or remove.  Plus, the temperature in the reactor controls the kinetics, which, in turn, determines the size of the reactor (equipment cost).

I see. That's exactly the kind of direction I'm going for with my question. So just to be clear, enthalpy is a metric more useful on an industrial scale than laboratory?  Any idea why a paper would cite the enthalpy over the free energy for a lab-scale reaction? I guess if someone told me the enthalpy for a lab reaction, I wouldn't know what to make of it and would prefer to know the free energy instead. I strongly suspect I am looking at this the wrong way somehow.

@Blaisem "Enthalpy is a metric more useful on an industrial scale than laboratory." No, this is incorrect; enthalpy and free energy are useful on both scales, but the manner in which each is useful at each scale may or may not be different, depending on the application. "Useful" itself is a problematically vague term here; without knowing the specific context of interest, it's difficult to say a priori which is more "useful".

@Blaisem "Would you mind telling me why the heat of a reaction is useful?" This is a fairly fundamental chemical concept; if you're unclear on it, you should consult an introductory chemistry textbook. It's certainly too broad to answer in these comments, and it may be too broad for Chem.SE in general.


text:
Is there any notion of pH out of solution?
For example, could one define a $\mathrm{pH}$ for pure acetic acid? It's a weak acid in water, but if someone handed you $1~\mathrm L$ of pure acetic acid, what would its $\mathrm{pH}$ be?

pH is usually for more or less diluted water solutions. But here is some insight http://chemistry.stackexchange.com/a/7000/9961

Is it? Strictly, it's just $-\log_{10}{\left[\ce{H+}\right]}$, right? Acetic acid may be a bad example, since I would think it wouldn't auto-ionize, but in general, if any solvated proton is present, I would think a $\mathrm{pH}$ could be defined, even if it wasn't particularly useful.

Edited, but it simply doesn't work the same in other situations.

@Brian It does auto-ionize, but in pure CH3COOH  will [CH3COOH2+]  be more important then concentration of acid itself? In what context?

H0 function is used is such cases.

This answer is also related: http://chemistry.stackexchange.com/a/5741/9961

The IUPAC definition of pH is:

The quantity pH is defined in terms of the activity of hydrogen(1+) ions (hydrogen ions) in solution:
$pH = − lg [a(H^+)] = − lg [m(H^+) γ_m (H^+) / m^⦵]$
where a$(\ce{H+})$ is the activity of hydrogen ion (hydrogen 1+) in aqueous solution, $(\ce{H+})$(aq), $γ_m(\ce{H+})$ is the activity coefficient of $\ce{H+}$(aq) (molality basis) at molality m$(\ce{H+})$, and m$^⦵ = 1$ mol kg $^{−1}$ is the standard molality.

So since the definition specifically refers to "aqueous solution", pH is undefined unless an aqueous solution is being considered.
There would be a p[H+] in pure acetic acid based upon the self dissociation of acetic acid, where [H+] is the concentration of hydrogen ions in acetic acid solution, but this is not pH according to the IUPAC definition.
According to Acid-Base Equilibria in Glacial Acetic Acid. III. Acidity Scale. Potentiometric Determination of Dissociation Constants of Acids, Bases and Salts J. Am. Chem. Soc., 1956, vol. 78, pages 2974–2979:

the autoprotolysis constant of acetic acid is calculated to be $3.5 \times 10^{-15}$ (pK = 14.45)

Therefore p[H+] = 7.2 for pure acetic acid.
This value is not a measure of acidity, but simply the concentration of solvated hydrogen ions in pure acetic acid.


text:
The most variant groups of periodic table
Is it okay to call group 11 (formerly 1B) a "group"? I meant that according to my knowledge "A group is a horizontal column consisting of elements with similar valencies and a continuous gradation in physical and chemical properties" but there is no gradual trend or we don't obtain similar reactions for same circumstances. Members do not show similar oxidation states and do not form analogous compounds. Electronegativity values and other bonding characteristics also deviate badly from one another. Are they in a group just because their atomic numbers allot their positions in same vertical column? Shouldn't 1B and other d-block "groups" (which show such variant properties) treated as "ungrouped region" or something like that?

Placing in groups is directly related only to electronic configuration, "results may vary".

I assume you meant "vertical column"?

Sure Group 11 is a group. And if you look in the right areas you see similarities between copper, silver and gold:
1)  Among the best electrical conductors even compared with other metals (http://eddy-current.com/conductivity-of-metals-sorted-by-resistivity/).
2)  Tend to prefer combining with "soft base" nonmetals:  Cu and Ag found as sulfides, gold is oxidized by $\ce{HNO_3}$ when a chloride source is added.
3)  Have some chemistry in the +1 oxidation state, especially wuth said soft-base nonmetals.

Conduction is somewhat physical quantity as I think and much more lanthanides tend to combine with soft base non-metals e.g iron. And iron and copper possess more similarity than copper and silver from this prospective.

Iron is found in nature primarily combined with oxygen (at least near the surface where it's chemically combined with anything at all).  It favors hard-base nonmetals much more than anything in Group 11.


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Coupling constant (Jab) - Why more unpaired electrons means a smaller coupling?
I hope this is not a "homework" question, but I'm having a bad time trying to figure this out.
Available literature proposes 3 equations to calculate the coupling constant during a Broken-Symmetry approach:
\begin{align*}
J(1) = \frac{-\left ( E_{HS}-E_{BS} \right )}{S_{max}^{2}} \\
J(2) = \frac{-\left ( E_{HS}-E_{BS} \right )}{S_{max}\left ( S_{max}+1 \right )} \\
J(3) = \frac{-\left ( E_{HS}-E_{BS} \right )}{\left \langle S_{HS}^{2} \right \rangle-\left \langle S_{BS}^{2} \right \rangle}
\end{align*}
I'm intrigued by the fact that, from the equations, the more the system have unpaired electrons, the minor will be $J_{ab}$. Why does this happen? Doesn't more unpaired electrons increase magnetic momenta (and an increase in magnetic coupling))?

What do you mean by minor and $J_{ab}$ (to which $J$ does this refer)?

I hope I have understood your question correctly.
For illustrative purposes, I will restrict myself to the Hartree-Fock picture. In a broken spin system, e.g., UHF, the spin state is not a good quantum number. That is to say that the UHF wavefunction need not be an eigenfunction of $S^2$ or $S_z$. For example two complexes with a single unpaired electron each will combine three singlet states (two ionic states and one open-shell) and three triplet states ($m_s\in \{-1,0,1\}$).
The open-shell singlet ($|\alpha\phi_1 \beta\phi_2|-|\alpha\phi_2 \beta\phi_1|$) and the $m_s=0$ triplet ($|\alpha \phi_1 \beta \phi_2|+|\alpha\phi_2 \beta\phi_1|$) states are not single Slater determinants. The singlet states need not be of the same energy, whereas the triplet states have to be degenerate for a Hamiltonian without explicitly spin-dependent components such as spin-orbit coupling. Hence, you have to accommodate for the multiple counting.
For higher spins, the degeneracy grows with $S_{max}$ and must be corrected accordingly. This does not imply that coupling necessarily decreases with higher spins.

Thank you for your answer, @Deathbreath.
I'm sorry if I haven't formulated my question in a clear way before (but you answered correctly). The doubt appeared when I was studying an isostructural coordination polymer of the kind [M(L)] (M is Co(II) or Co(II)). The comparison in energies from E(high spin) - E(singlet) is remarkable: Cobalt is 8,065 cm**-1 and copper is 0,804 cm**-1, but the coupling constants are very close: -0.9 for Co and -0.8 for Cu (with cobalt having more unpaired electrons).

So, from your answer, this difference in energy exists because if an atom has a higher multiplicity you still have to remember that the Hamiltonian is representing a "spin ladder" and not a single spin state, Correct?

I'm not sure what you mean by soon ladder, but the Hamiltonian commutes with $S^2$ and $S_z$, whereas the wave function is not a pure spin state.


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Oxidation states of iron in Roussin's salts

For preservation of meat, sodium nitrite is usually added and as a result $\ce{NO}$ is then formed. Consequently, $\ce{NO}$ reacts with the sulfur and iron atoms from decomposition of proteins, forming $\ce{[Fe4S3(NO)7]-}.$ X-ray crystallography shows that the complex anion has a structure as shown below:

a) Blacken all the circles corresponding to iron atoms and add symbols Fe(A),
  Fe(B), Fe(C) and Fe(D) beside the circles in the sequence of top → left → right.
b) The configuration of 3d electron shell of the iron atoms has been studied with modern structural analysis. Knowing that the mean oxidation number of the
  four iron atoms is $–0.5$, give their configurations of 3d shell, respectively.
  Assume that each iron atom adopt sp hybridization.
$\ce{[Fe4S3(NO)7]^{–}}$ anion can be reduced and a new complex $\ce{[Fe2S2(NO)4]^{2-}}$ is formed
c) Give the oxidation state of each iron atom with Arabic numerals.
d) $\ce{[Fe2S2(NO)4]2-}$ can be converted into $\ce{[Fe2(SCH3)2(NO)4]^{n}}$, a carcinogen. Which of the following three species is added to $\ce{[Fe2S2(NO)4]^{2-}}$ : $\ \ce{CH3+}$,$\ \ce{•CH3}$ or $\ \ce{CH3−}$?

My Attempt
a)

I don't have any real reason to why I chose those 4  atoms. I chose them since they were the 4 biggest atoms in the ring and made sense they would alternate with the sulfur atoms in the ring. Is there a more solid explanation?
b) Firstly, why do the iron atoms have a negative oxidation state, doesn't it 'give' its electrons to the sulfur and nitrogen atoms since it is more electropositive? I look up negative oxidation states of iron and I found out when bonded to carbonyl compounds, the iron atom usually has a negative oxidation state. So is the $\ce{NO}$ ligand in this case acting like $\ce{CO}$?
The answer is Fe(A) has 3d7 configuration while the others have a 3d9 configuration. Could someone please give an explanation for this.
c)

I have drawn what I think is the structure above, but I have the same problem as part b)
d) I have no idea. I am guessing the methyl cation since the iron atom has a negative oxidation state (I am guessing from part c) hence it will more likely react will a cation rather than an anion or radical.

Iron-sulphur complex is known to be multi-reference. It is not always clear-cut to assign oxidation state. I think this problem is not very suitable for homework and different people may have different answer..

@Rodriguez what do you mean by multi-reference?

Means there are several Slater determinants have similar weights. In another word, Hartree-Fock and many post-HF approaches are not quite suitable for this type of system.

@Rodriguez so you are saying that technically at a deeper, more technical level it is hard to assign its oxidation state. However is there a simple way (not necesarily to give the actual answer) of determining the oxidation states of each iron atom because there must be a way to do this question which doesn't require a lot of complicated calculations.

related https://en.wikipedia.org/wiki/Metal_nitrosyl_complex https://en.wikipedia.org/wiki/Roussin%27s_red_salt https://en.wikipedia.org/wiki/Roussin%27s_black_salt

There may be a simple way to determine the oxidation state in this system. However, I think the concept of oxidation state anyway has its limitation. In this case, we hit the boundary. Like to say, all models are wrong, some are useful.

I suppose it's asked about formal oxidation numbers of iron as there is not enough information to make assumptions about the ones that reflect electronic structure.
For instance, analysis of interatomic distances could point to a delocalized electron pair, but you are not given any linear or angular geometrical parameters to begin with.
It's only the symmetry that's evident from the crystal structure: the thioferrate cluster $\ce{[Fe4S3(NO)7]-}$ belongs to $C_\mathrm{3v}$ point group, hence the only possible arrangement of four iron atoms is a tetrahedral one.
Another conclusion that can be made is that three equatorial iron atoms are equivalent ($\ce{Fe(B)}$, $\ce{Fe(C)}$, $\ce{Fe(D)}$), and the last apical one ($\ce{Fe(A)}$) most likely deviates in terms of oxidation state.
Supposing formal oxidation number ($\text{O.N.}$) of three equivalent irons and the last one are $y$ and $x$, respectively, and the fact that the average $\text{O.N.}$ is $-0.5$, the following holds true:
$$x + 3y = -0.5\cdot 4 \implies y = -\frac{x + 2}{3}$$
As formal oxidation numbers are integers, there are several pairs giving the solution:
$$
\begin{array}{l|rrrrr}
\hline
x & \ldots & -5 & -2 & \color{green}{1} & 4 & 7 & \ldots \\
\hline
y & \ldots & 1 & 0 & \color{green}{-1} & -2 & -3 & \ldots\\
\hline
\end{array}
$$
among which the least "extreme" (e.g. the one with the absolute lowest negative $\text{O.N.}$) for a metal would be the $-1;+1$ pair; so that for $\overset{0}{\ce{Fe}}:~[\ce{Ar}]\,(\mathrm{3d})^6\,(\mathrm{4s})^2$ your answer gives two configurations, $(\mathrm{3d})^7$ for $\overset{+1}{\ce{Fe}}\ce{(A)}$ and $(\mathrm{3d})^9$ for $\overset{-1}{\ce{Fe}}\ce{(B,C,D)}$.

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Can an ion isotope exist?
Can an isotope have a charge, i.e. the number of neutrons and electrons differ in that atom?
If it does exists, is it correct to call it an ion isotope or isotope ion?

Yes and yes. Isotope is just an atom; it does everything that normal atom does.

In good old table salt, about 75% of the anions are $\ce{^{35}Cl^{-}}$ while 25% are $\ce{^{37}Cl^{-}}$. Nothing too special.

Short answer: Yes.
Longer answer:
Isotopes are atoms of the same element that have different numbers of neutrons in their nuclei. Everything else about them is the same. The numbers of protons and electrons in a neutral atom is governed by the atomic number.
So, for example, chlorine has two naturally occurring isotopes:
$$\begin{array}{|c|c|c|c|c|c|c|}\hline
\mathrm{isotope} & Z & A & p & n & e & \mathrm{abundance}\\ \hline
\ce{^35_17Cl} & 17 & 35 & 17 & 18 & 17 & 75.78\%\\
\ce{^37_17Cl} & 17 & 37 & 17 & 20 & 17 & 24.22\%\\ \hline
\end{array}$$
Where:
$$\begin{aligned}
&Z=\mathrm{atomic\ number}\\
&A = \mathrm{mass\ number}\\
&p = \mathrm{number\ of\ protons}\\
&n = \mathrm{number\ of\ neutrons}\\
&e = \mathrm{number\ of\ electrons}\\
\end{aligned}$$
Isotopes generally behave the same chemically1 since almost all chemistry is governed by the number of electrons ($Z$). Thus, anything that contains chloride ions contains a roughly 3:1 ratio of $\ce{^35Cl-}$ and $\ce{^37Cl-}$. We could call these ion isotopes or isotope ions, but in most all cases, we do not distinguish between the two in reactions or compounds involving chloride.
1 There is one caveat in the "isotopes behave the same chemically" rule of thumb that is worth mentioning. Since isotopes have different masses, they can produce kinetic isotope effects. Given the $KE=\frac{1}{2}mv^2$, two isotopes with different masses will move at different velocities and have the same kinetic energy. Heavier isotopes generally move more slowly.
This effect is most pronounced in hydrogen, where the second most common isotope $\ce{^2H}$ ($\ce{D}$) is twice as massive at the most common isotope $\ce{^1H}$ ($\ce{H}$). If two atoms have the same kinetic energy, then:
$$\begin{aligned}
m_1 v_1^2 &=m_2 v_2^2\\
\dfrac{m_1}{m_2} &=\dfrac{v_2^2}{v_1^2}\\
\sqrt{\dfrac{m_1}{m_2}} &=\dfrac{v_2}{v_1}
\end{aligned}$$
For the isotopes of hydrogen, the heavier isotope of hydrogen moves at only 70% of the velocity of the lighter isotope. This kinetic isotope effect can be used to probe reaction mechanisms. If the rate decreases dramatically when $\ce{H}$ is swapped for $\ce{D}$, then that hydrogen atom is involved in the rate determining step.
$$\begin{array}{|c|c|c|c|}\hline
\mathrm{isotope} & A & \mathrm{rel.\ mass} & \mathrm{rel.\ velocity}\\ \hline
\ce{^1H} & 1 & 1 & 1 \\
\ce{^2H} & 2 & 2 & \dfrac{1}{\sqrt{2}}\approx 0.707\\ \hline
\end{array}$$
For chlorine, the outcome is a little different. Since the ratio of the masses of the two isotopes is close to 1, the ratio of the velocities is also close to 1.
$$\begin{array}{|c|c|c|c|}\hline
\mathrm{isotope} & A & \mathrm{rel.\ mass} & \mathrm{rel.\ velocity}\\ \hline
\ce{^35Cl} & 35 & 1 & 1 \\
\ce{^37Cl} & 37 & 1.06 & 0.97\\ \hline
\end{array}$$

Hmm... The origin of kinetic isotope effects lies in potential energy, or zero-point energies, not kinetic energy. The relevant quantity is not the mass $m$ of the atom in question, but the reduced mass $\mu = m_1m_2/(m_1+m_2)$ of a bond being broken.

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text:
Molecular knots absolute configuration
How is the absolute configuration of chiral molecular knots determined? What rules should be applied? For centrochiral compounds, the CIP rules are applied, for axially chiral is described here and for planar chiral here. I can't find the system for assigning R/S (or possibly some other stereodescriptors) for molecular knots, e.g  the attached knot doesn't contain an Sn improper axis, therefore it is chiral.

I edited your question to improve it, but it would be still nice if you elaborated.

This is where topology\mathematics meets chemistry and becomes fairly complex fairly quickly.  Mislow, one of the great, modern-day stereochemists, got the ball rolling.  If you google "Mislow, trefoil chirality" you'll get a number of useful links, but again, it's involved.

I've read it a bit and encountered this interesting question: could these compounds be resolved?

The nomenclature of knot-like structures is not yet developed in IUPAC, let alone their stereochemistry.
However, there are some attempts in chemistry literature to describe the knots, drawing inspiration from math knot theory (however, even the detection of chiral knot is not yet completely resolved problem, apparently).
And sometimes there are contradictions. In TAUBER, Stephen J. Absolute Configuration and Chemical Topology. J. Res. Nat. Bur. Stand. A, 1963, 67: 591-599., you example knot's (let us just rotate him slightly in the paper plane by 60° clockwise or anticlockwise, does not matter, for "compatibility") intersection points are, according with the rule:

designated with plus signs at the intersection points:

and the enantiomeric one with minuses:

This notation is mentioned elsewhere, e.g. in SCHILL, Gottfried. Catenanes, rotaxanes, and knots. Elsevier, 2013.
However, in SAUVAGE, Jean-Pierre; DIETRICH-BUCHECKER, Christiane (ed.). Molecular catenanes, rotaxanes and knots: a journey through the world of molecular topology. John Wiley & Sons, 2008., in the chapter DNA Knots, your example knot is denoted in the opposite sense with minus signs,

and named "TREFOIL KNOT (−), 31−", and the opposite one as "TREFOIL KNOT (+), 31+".

I guess that the IUPAC Gold Book rules for helicity apply here.

The chirality of a helical, propeller or screw-shaped molecular entity. A right-handed helix is described as P (or plus), a left-handed one as M (or minus).

The above knotane is right-handed, so it should be P.
For more complex molecular knots such simple rules most likely will fail.

I agree that this system is applicable, but I would argue that the helicity here is left handed (M). Do you have any source applying the this system trefoil knot? Btw,  here is knots helicity boosted one level above :)

From both the image in the IUPAC Golden Book and from applying the right hand rule to the propeller (https://en.wikipedia.org/wiki/Right-hand_rule) I concluded that it should be P. However, I don't know any source where the system is applied to the molecular trefoil knot. Mathematically the trefoil knots are classified the other way round. So I might be wrong.

i think it could be designated P using both the knot intersection points method in my answer and helicity nomenclature when focusing on fan lobes, or in opposite sense M when focusing on the interior – the knot it can be threaded on the "anti-screw" (levorotatory helix)


text:
Predominance of III oxidation state for lanthanides
For most lanthanide metals, the stable oxidation state is III [*]. The general electronic structure is $$\ce{[Xe] 4f^{0}^{-14} 5s^2 5p^6 5d^{0}^{-1} 6s^2}\ \ [**].$$
Elements that have the d-electron are La, Ce, Gd, and Lu. Furthermore, the f-subshell is considered relatively stable in states $$f^0, f^7, \text{and} f^{14}.$$
We can conclude that La, Gd, and also Lu easily form $\ce{E^3+}$ ions. Yet, as would be predicted by this easy approach, we would also see $$\ce{Sm+, Tm+ (f^7 \ and \ f^14 ), \\ Pr^5+, Dy^5+ (f^0 \ and \ f^7)}.$$
This is definitely not the case. But the why? alludes me. According to Radiochemistry,

The 4f binding energy is so great that the remaining 4f-electrons are regarded as "core-like". [Italic added]

Could someone explain this further? [***]
Radiochemistry then goes on to assert,

Interelectronic repulsion in related not just to electron pairing, but also to [the] angular momentum of the electrons. [---]
e.g., in $\ce{Pr^2+ (4f^3)->Pr^3+ (4f^2)}$ ionisation removes repulsion between [electrons] of like rotation [---]
$\ce{Pm^2+ (4f^4)->Pm^3+ (4f^3)}$ ionisation removes repulsion between [electrons] of unlike rotation [---] [emphasis added]

What exactly is meant by "electrons of like/unlike rotation"? Plus it mentions some "quarter / three-quarter" rule which I possibly cannot rationalise for a total of 14 electrons.
Any help would be greatly appreciated.

[*] Ce, Pr, and Tb also have the oxidation state IV. Eu and Tm have the additional state II.
[**] Ordering changes with the number of electrons.
[***] I am, of course, aware that

each electron would be harder to remove as Culonic attraction increases with positive ionic charge,
lathanide contraction is a part of the explanation.

(Somewhat related: Plutonium having more oxidation states than samarium?)

EDIT
Every thanks to Brian for providing awesome title suggestions. These are the four new subquestions:

Why don't we see these lanthanide species?
Howcome orbitals become 'core-like' when electrons are removed?
What is meant by 'electrons of like/unlike rotation'?
What's up with this quarter / three-quarter rule?

Just an FYI for future chainposters, there's a sixty-second waiting period after asking two consecutive questions.

There are a lot of different questions in here; I see at least four. I recommend breaking out "Why not $\ce{Sm+}$, etc.?", "What does 'core-like' mean here?", "What is meant by 'electrons of like/unlike rotation'?", and "What's up with this 'quarter / three-quarter' rule?" into separate questions, perhaps edited to contain links among all of them.

@Brian: Are you sure? I thought it would be weird to post so many different questions in a short time. That's why I tried to find the best common ground :D. I will have some time tonight (or tomorrow, worst case scenario), and the corrections shall be made. Maybe not four questions but two or three ;-)

Yeah, the StackExchange approach favors a larger number of more narrowly scoped Questions. Ideally, any given Question will only contain a single question, about a single topic. Not always possible, but it's the guiding principle. The beauty of the way the site is coded is that when you link to other questions, they show up in the sidebar under "Linked", so it creates pretty strong connections between crosslinked Questions.

@Brian: done and done. Be sure to let me know if anything still needs tweaking.

On the whole, looks pretty good. Scope of the four subquestions is good, I think. In accord with the downvote it's gotten, the 'one-quarter / three-quarters' question is really unclear. See more comments there.


text:
Why is the Dulong–Petit law not applicable to the elements Be, B, C, or Si?
According to the Dulong–Petit law, the molar heat capacity of a solid element is approximately $3R$ (where $R$ is the gas constant $\pu{8.314 J K-1 mol-1}$.)
This is useful for calculating the atomic mass of an element when its specific heat capacity (i.e. heat capacity per unit mass) is given.
However, the Dulong–Petit law is said to not apply to the solid elements Be, B, C, and Si. Why is this so?

I refer to the same link and add a summary:
The deviation is quantum mechanical in nature. The Dulong–Petit Law is exact only if all vibrational modes are fully activated, in which case equipartition theory can be used. As every atom in a solid can be considered to be a three-dimensional harmonic oscillator, the contribution to the heat capacity is $3k_\mathrm{B}$ for one atom, or $3R$ for one mole.
However, this applies exactly only as the temperature goes to infinity. For finite temperatures we can still come close if the thermal energy is high enough versus the energy required to excite the vibrational modes:
$$k_\mathrm{B}T \gg \hbar\omega$$
For most elements – except the elements listed in the question – room temperature is high enough to come close to full excitation. For the elements in question, the vibrational frequency $\omega$ is large, particularly because of the strong bonds and relatively small masses ($\omega \propto \sqrt{k/m}$.)


text:
Moles of Diatomic Molecules
Does "Mole of Iodine" or any other element that normally appears as diatomic molecules refer to a mole of single atoms ($6.02\times 10^{23}$ iodine atoms) or a mole of the molecules ($6.02\times 10^{23}$ molecules of $I_2$)?
EDIT: If there is no specification of moles of atoms or moles of molecules, what do I do? Do I default to one of them?
e.g. in this question

For the combustion of sucrose:
$\ce{C12H22O11 + 12O2 -> 12CO2 + 11H2O}$
there are 10.0 g of sucrose and 10.0 g of oxygen reacting. Which is the limiting reagent?

It depends on context. Mole is simply a unit.

Thanks for the link, I've edited the question to clarify.

sigh If not specified otherwise it's about atoms and iodine is simply treated as element, not collection of molecules, as it doesn't have to be molecular.

When we ask, for example, "how many moles of Iodine (or Oxygen or Hydrogen) is contained in 100g?" It ALWAYS refers to the molecule (e.g. I2, O2, H2 etc). Therefore, the number of 6.02 x 10^23 refers to the number of molecules, not atomes of Iodine and therefore you actually have 12.04 x 10^23 atomes of Iodine contained in one mole of molecular Iodine. However, when you look at the periodic system, they always refer to atomic mass not molecular. In case of Oxygen, the atomic mass of 16 g/mol refers to 1 mol of O, not O2 as molecular form. Further, when you want to know mass of the Oxygen (gas form, O2) then: 2 (number of atoms) x 16 g/mol = 32 g/mol - this is the mass of 1 mol of O2 (Oxygen-diatomic form - gas present in the atmosphere).

Good practice is to nip the problem in the bud by precisely defining what you are counting.  Thus, properly, one mole of $\ce{I_2}$ molecules contains two moles of iodine atoms.


text:
Equilibrium Constant Explained
I'm not interested in a simple telling of what the properties of the equilibrium constant are and how to use them. I'm more hoping for an explanation of where it came from/how it might be possible to derive it.
My current guess is that it is a probability equation:
The numerator in the equation is proportional to the probability reactants meet and form products in the volume and the denominator is proportional to the probability products meet and form reactants in the volume. This ratio clearly must stay constant (unless the proportionality constants are changed via temperature or such).
Would anyone help?

Equilibrium is the state at which the rates of the forward and reverse reactions are equal.
$$\mathrm{rate_{forward}=rate_{reverse}}$$
Let's take a generic reaction:
$$\ce{A + B <=> C + D}$$
For simplicity's sake let's have both the forward and reverse reactions be concerts fundamental reactions. Then we can write the rate laws for the forward and reverse reactions:
$$\begin{aligned}
\mathrm{rate_{forward}}&=k_1 [\ce{A}][\ce{B}]\\
\mathrm{rate_{reverse}}&=k_{-1} [\ce{C}][\ce{D}]
\end{aligned}$$
If we set them equal to each other, we can rearrange the equation to get the rate constants on one side and the concentrations on the other.
$$\begin{aligned}
k_1 [\ce{A}][\ce{B}]&=k_{-1} [\ce{C}][\ce{D}]\\
\dfrac{k_1}{k_{-1}}&=\dfrac{[\ce{C}][\ce{D}]}{ [\ce{A}][\ce{B}]}
\end{aligned}$$
The ratio of rate constants is the equilibrium constant. Thus is derived the law of mass action.

I appreciate it, but now it reduces to instead understanding the rate laws for forward and reverse reactions. It is, for instance, possible to derive PV = nRT not through experiment, but through simple logic. Nevertheless thank you. It helped.

The rate laws arise from the frequency of molecular collisions of particular kind; that's quite obvious. This is a sound reasoning indeed, but there is more to it. Thermodynamics is kinda "more fundamental" than kinetics. Reactions may have complicated mechanisms and do not necessarily obey the simple rate laws. The equilibrium constant, though, is always the way it is, regardless of these details.

I believe my question is solved then. I'm aware that some things in thermodynamics are too complicated to understand simply and must be instead determined experimentally, but the equilibrium constant seemed to be the kind of thing which would have a simple explanation as to why it works. Now it makes sense.


text:
Electrolytic rust removal
I'm currently removing rust on some bike parts so decided to use electrolysis. It's working fine, but the solution has turned a deep/dark blue whilst in most online videos it turns orange with the rust. My sacrificial anodes aren't stainless or chromed, and I dont believe they're galvanized either.
Is the blue colour any cause for concern (am I making anything toxic accidentally)?

Probably just copper and/or nickel from the metal parts. I wouldn't be concerned but of course use gloves and proper PPE. Never touch or mess with something whose identity you're not confident of.

If your anode/cathode are from copper, then the blue chemical in the solution is copper chloride (CuCl2).  Copper chloride can be toxic. So you should have safety equipment.
As for rust removal in general, I'd recommend you to use Citric acid (Squeezing lemons for the lemon juice isn't efficient since it will contain 5-6% Citric acid - so its better to buy powdered Citric acid from a store).

I recommend using citric acid
Burn it with blowtorch later so it does not get rust.

TL;DR: Probably not.
However, it's always prudent to follow proper safety procedures when doing ANYTHING with chemicals. Given you're doing something with electrolysis, there's probably an acid involved, or, if not, some kind of ionic solution. If you're using an acid, I'd recommend keeping a neutralizer (baking soda) on hand, as well as wearing rubber gloves and some kind of goggles. A rubber apron would also be recommended, although, not everybody owns one of those. (And, to be fair, if you did, you'd look a little suspicious.)
However! Please list the specific metals and solutions you're using in this process. It'd help greatly in determining whether you're in any danger.


text:
Nomenclature of salts of oxyanions
I want to know what is the way to determine a compound's composition by just seeing the name. For example, if I see the name
"copper (II) sulfate"
How would I go about deducing that its formula is CuSO4? I know that copper(II) refers to the copper cation with a +2 charge. I know sulfur can have either a -2, +4, or +6 charge. And I know that oxygen has a -2 charge. I also know that an ending of "-ate" means a high amount of oxygen. But it doesn't specify exactly how many oxygen atoms I'll need. I'm new to studying chemistry so excuse me if this seems like a bad question.


text:
Nernst equation: half cell vs overall reaction
Can the Nernst equation always be applied to the overall redox reaction as opposed to only half cell reactions?
For example, in the following silver-silver chloride electrode: $$\ce{Ag (s)|AgCl (s)|Cl- (aq)}$$ rather than applying the Nernst eqution to the half cell reaction $$\ce{AgCl + e- -> Ag + Cl-}$$ can I apply it to $$\ce{AgCl -> Ag+ + Cl-}$$
The answer is the same for the question I was trying to solve (whether I use the overall reaction or the half-cell reaction), but I was wondering if this was just a coincidence or if it was the same.
If they are the same, are there any other factors I should consider such as whether or not the electrolyte fully dissociates in the given conditions?

$$\ce{AgCl -> Ag+ + Cl-}$$ is not a redox equation!

It is coincidence, since
$$\ce{AgCl -> Ag+ + Cl-}$$
is not a half cell reaction.
The silver chloride electrode functions as a redox electrode and the reaction is between the silver metal (Ag) and its salt — silver chloride (AgCl). There are two ways to represent the processes (half reactions) happening on the electrode:
$$\ce{AgCl + e- -> Ag + Cl-}$$
or
$$\ce{Ag+ + e- -> Ag}$$
$$\ce{AgCl -> Ag+ + Cl-}$$
And you should be applying Nernst equation to one of the two half cell reactions shown above.
The reason why you got the same result/answer is because the Nernst equation for silver chloride electrode has the dependence on the activity or effective concentration of only chloride-ions.
$$E = E^0 - \frac{\mathrm RT}{\mathrm F}\ln(a(\ce{Cl^-}))$$


text:
Titration of acid with base or base with acid: can I always use the same indicator?
If you are doing a titration with $\ce{NaOH}$ and $\ce{HCl}$, does it matter which one goes in the burette, and will this change the outcome of whether the solution in the beaker will turn pink due to phenolphthalein?

Possible duplicate of Question about titration theory

When phenolphthalein is used as an indicator, NaOH is taken in a burette because then the colour change is from colourless to pink. And this time, it will be easier to observe.
On the other hand, if HCl is taken in a burette, colour change will be from pink to colourless which can't be observed in a single drop.
When NaOH is titrated with HCl using methyl orange as an indicator, HCl is taken in a burette because then the colour at the end point changes from yellow to red which is easier to observe rather than that from red to yellow.

This is just for the sake of convenience of calculation..


text:
Does almond extract made from bitter almonds contain any cyanide?
NOTE: This question is not seeking medical advice. I am just trying to understand the chemistry behind why bitter almonds are poisonous but bitter almond extract (presumably) isn't.
A friend made pancakes and decided to add flavor by adding almond extract, the ingredients of which were "water, organic alcohol, glycerin, natural bitter almond oil". She added about a teaspoon.
Later she was feeling very nauseated and had a stomach ache. After a couple hours, the nausea went away followed by the stomach ache. Other people in the house ate the same pancakes and she was the only one who got sick. (And there similarly wasn't evidence that it could be food poisoning from the night before since she ate the same as others.)
(She does have an allergy to sunflower seeds and certain nuts, but she has no problem eating regular sweet almonds.)
I read that eating a small number (like 10) of bitter almonds can be fatal because of the cyanide. Does almond extract made from bitter almonds contain any cyanide, and could this be related to her getting sick? I tried to Google it and found lots of info about bitter almonds being dangerous, but nothing that mentioned problems eating a lot of almond extract.

If the extract were that poisonous, it probably wouldn't be allowed by FDA. However, I'd be unwilling to bet my life on it. Don't they print any warnings on the label, or at least something about amygdalin?

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Personal medical questions are off-topic on Chemistry. We can not safely answer questions for your specific situation and you should always consult a doctor for medical advice. That being said, it would probably be a good idea to formulate it more in general. I think the following article would be quite helpful to understand why bitter almond is toxic and what it means for the natural oil extract.

Related: How do I extract cyanide from apple seeds?

Crude extracts from bitter almonds or apricot seeds do contain amygdalin, which is hydrolyzed in the human body and releases cyanide.
Food-grade bitter almond extract does not contain amygdalin, because the crude extract is typically treated with a with a mixture of calcium hydroxide and $\ce{FeSO4}$. In this process, the amygdalin is decomposed and the cyanide is extracted in the form of $\ce{Ca2[Fe(CN)6]}$.
The typical flavour of the resulting product stems from benzaldehyde.


text:
Balancing using oxidation numbers
For the reaction
KCl + MnO2 + H2SO4 -> K2SO4 + MnSO4 + Cl2 + H2O
I have Mn going under oxidation from +4 to +2
I have Cl going under reduction from -1 to 0
Then after using my teachers method o multiplying the oxidation and reduction atoms I have
4KCl + 1MnO2 + H2SO4 -> K2SO4 + 1MnSO4 + 2Cl2 + H2O
However I'm not sure I'm on the right track, but if I am, what are the next steps to completely balance the equation?

I'd rather go with half-reactions, but if you insist, this way may work out as well.We don't touch the redox part anymore; now what elements are not balanced yet? Is the reaction balanced in S, for example? What can we do about it?

@Ivan Neretin I prefer half reactions as well, however teacher is marking it based on our ability to do it without it. Also what your saying is my method is correct, and all that's left is to balance the S O and H's?

Yeah, that's about right.

It is really easy to solve by the half reaction method, here is my answer:

Half reaction method
Global eq: KCl + MnO2 + H2SO4 -> K2SO4 + MnSO4 + Cl2 + H2O

Cathode: 2Cl-   --> Cl2 + 2e-

Anode: MnO2 + 4H+ + 2e- --> Mn+2 + 2H2O
So the final balanced eq is this one:

2KCl (ac)  + MnO2 (s) + 2H2SO4 (ac) --> K2SO4 (ac) + MnSO4 (ac) + Cl2 (g)  + 2H2O

Variations method
Without half-reaction method will be like this:
You have one Mn+4 reduced to Mn+2
That is 2e x 1 atom (ion indeed) so this is a net variation of 2
On the other hand you have Cl- wich oxidizes to form a Cl2 molecule so this is
1e x 2 atoms wich also is a net variation of 2
Now you use these numbers as the coefficient of the products that contain these atoms and you'll have
KCl (ac)  + MnO2 (s) + H2SO4 (ac) --> K2SO4 (ac) + 2MnSO4 (ac) + 2Cl2 (g)  + H2O
and considering this as valid just left to adjust the rest of the elements
and finally you'll have:
4KCl (ac)  + 2MnO2 (s) + 4H2SO4 (ac) --> 2K2SO4 (ac) + 2MnSO4 (ac) + 2Cl2 (g)  + 4H2O
Wich is also a valid adjust
I always recommend using half-reaction method or undetermined coefficient method instead. Good luck!!


text:
Does 6-methyl-2-(methylamino)pyrimidin-4-ol show aromaticity?

Does this structure show aromaticity? Or does this not work with nitrogen?

Mm, oxygen radicals... is that an OH group?

Yes, sorry my bad. That's an OH group

What's wrong about nitrogen? Pyridine is aromatic, so is pyrimidine, hence so should be your compound, once you fix that issue with O.

Each Nitrogen has 5 valence electrons. They are using 2 of those in sigma bonds to their neighbouring atoms, 2 in an sp2 lone pair in the plane of the ring, leaving only 1 electron to contribute to the delocalised pi system.
Each carbon is also contributing 1 electron to the delocalised pi system, and has 3 involved in sigma bonds.
This means each member of your 6 membered ring is contributing 1 electron, 6 in total. This means it is aromatic, as all 3 bonding orbitals are fully occupied.


text:
Significant Figures Calculations
I am confused regarding significant figures. Consider "Calculate the volume of 2.8 kg of liquid with density of 1.11 g/cm^3"
Then,
Volume = 2.8 kg * (cm^3 / 1.11 g) = 2522.522 (where 522 is recurring).
Since the original figure (2.8 kg) has only two significant figures, I thought the answer should be 2.5 kg.
The textbook says the answer is 2.523 kg, which is four significant figures.
Which is correct?

Unless greater precision is indicated by adding terminal zeroes to the mass measurement (2.800 kg) you should assume only two significant digits.  Note also that even if the mass is reported to high accuracy the density is given with only three significant digits.  The book is wrong to ask for four significant digits in the answer.

I thought it was obvious - but am just starting to learn significant figures so was not sure.
Thanks.

And then also, the answer should be given as $2.5 \cdot 10^{3}$.

Your answer of $2500$ and the book's answer of $2523$ both have four significant figures.
If you want to keep your answer to the same degree of accuracy as the given data with the least number of significant figures, i.e. $2.8$ kg, it is best to express your answer as $2.5\times10^{3}$.

Please, add the units in order to make your answer more accurate.


text:
Redox: Oxidation and Reduction
I don't get the difference between the gain or loss of electrons in a reaction and changes in oxidation state. Based on my notes and some research:

Gain or loss of electrons

A substance is oxidised if it loses electrons in a reaction.
A substance is reduced if it gains electrons in a reaction.

Changes in oxidation state

A substance is oxidised if the oxidation state of the element increases.
A substance is reduced if the oxidation state of the element decreases.

Whenever I see a chemical equation, I am confused as I do not know if it is a gain or loss of electrons or changes in oxidation state. I have an assignment to do which involves redox reaction, but it is the holidays and I can't contact my teacher. Could someone explain so that I could understand? For example,
$$\ce{2 Ca (s) + O2 (g) → 2CaO (s)}$$
I thought there is a change in oxidation state as on the reactants side, the oxidation state of both $\ce{Ca}$ and $\ce{O2}$ are $0$. As for the product, the oxidation state for $\ce{Ca}$ is $+2$ and $\ce{O}$ is $-2$. Therefore, $\ce{Ca}$ is oxidised and $\ce{O2}$ is reduced.
But one of the questions on my worksheet asked about electron transfer for this chemical equation, and I do not understand why it is a electron transfer.

Look this as:
$$\ce{2Ca + O2 -> 2CaO}\qquad\text{(complete reaction)}$$
then we can this split into two sub-reactions:
$$\ce{Ca - 2 e- -> \overset{+2}{Ca}}\tag{1}$$
Here "$+2$" means that $\ce{Ca}$ lost $2$ electrons (or two negative units) and therefore it was oxidized. Since it lost something negative it has a higher number of protons (positive particles) remained. In other words the atoms is now positively charged. The initial state was $0$ (the same numbers of protons and electrons, or positive and negative particles/units) and it was increased by loosing two electrons (the number increased).
$$\ce{O + 2 e- -> \overset{-2}{O}}\tag{2}$$
The oxygen got two electrons or two negative particles. Now oxygen has two negative particles more than positive particles (protons) and therefore it is $-2$. In other words, the number (oxydation state) decreased ($0 \to -2$).

You have an electron transfer because on order for the reaction to take place must exist an exchange of electrons from one atom to the other. Basically, this means that electrons are shared between both atoms to form a bond. In your case you'll have that $\ce{Ca}$ loses two $\ce{e-}$ and oxygen gains two $\ce{e-}$ because it's more electronegative than $\ce{Ca}$. This can be represented as follows:
$$
\begin{align}
\ce{Ca &-> Ca^2+ + 2 e-}\\
\ce{O2 + 4 e- &-> 2 O^2-}
\end{align}
$$
This is not the real process, I mean is all more complex than that, but it will give you and idea about what I mean.


text:
What are polydisulfide polymers?
What are polydisulfide polymers? How are these polymers applicable in energy sectors such as rechargeable batteries?

http://onlinelibrary.wiley.com/doi/10.1002/pat.1992.220030803/abstract

Disulfide batteries use a redox shuttle between disulfide bonds ($\ce{R-SS-R}$) and two thiolates ($\ce{R-S-}$), e.g.:

Often the counter-ions are $\ce{Li+}$ but this is hardly a requirement. Since two electrons can be transferred in the charging / discharging cycle, the theoretical energy density is very high.
One downside (e.g., in my past research) is that you're breaking the polymer bonds during discharge, creating oligomers and monomers that might be more soluble in the electrolyte than the original polymer. This might cause problems with cyclability, since your electroactive material might leave the electrode.


text:
Correlating the decrease in orbital radius and penetrating power with the increase in l value
I have read that s-orbitals have a stronger penetration effect as compared to p-orbitals and p-orbitals have a stronger penetration effect as compared to d-orbitals, etc.
Therefore, the electrons in an s-orbital are closer to the nucleus of the atom and their ionisation energies are greater than for an electron in a p-orbital. (Example: This is what we use to explain why the energy of ionisation decreases from $\ce{Be}$ to $\ce{B}$).
Everything made perfect sense to me until I read:

The orbital radius slightly decreases with an increase of $l$ value.

Since s-orbitals penetrate more, shouldn't that mean that being closer to the nucleus, they have smaller radius as compared to an orbital with higher $l$ value, say, the p-orbital?
How can we justify this? Or are these two pieces of information not related in this way?

tl;dr
The statement that "the orbital radius slightly decreases with the increase of $l$ value" is true only for orbitals with the same value of $n$ (shell). The penetration of the nucleus by an electron is measured by the relative electron density, which depends on both shell ($n$) and subshell ($l$) of an electron in an atom.
What is Orbital Penetration:

Penetration describes the proximity of electrons in an orbital
  to the nucleus. Electrons which experience greater penetration
  experience less shielding and therefore experience a larger Effective
  Nuclear Charge ($Z_\text{eff}$), but shield other electrons more effectively. Electrons in different orbitals have different
  wavefunctions and therefore different distributions around the
  nucleus. However, contrary to what many think, penetration is not the
  outer electrons penetrating through the shield of the core electrons.
  It is actually just how well the electrons feel the nucleus. This is
  similar to the idea of outer electrons penetrating, but not the same.
  They are not the same because the core electrons have more penetration
  than the outer electrons since they (the core electrons) feel the
  strongest pull.

Clarification
As you can see from the definition above, the penetration of a nucleus by an electron is measured by the relative electron density near the nucleus of an atom. It is essentially, how effectively electrons can get close to the nucleus.  For instance, below are the electron probability densities for s orbitals and a p orbital.

Nodes are regions of zero electron probability (white areas),
The orange color corresponds to regions of space where the phase of the wave function is positive, and the blue color corresponds to regions of space where the phase of the wave function is negative.

In a multi-electron system, the penetration of the nucleus by an electron is measured by the relative electron density near the nucleus of an atom for each shell and subshell of an electron. For example, the 2s electron is penetrating the nucleus of an atom more than the 2p electron, because the 2s has more electron density near the nucleus than the 2p electron. The penetration power of an electron, in a multi-electron atom, is dependent on the values of both the shell (n) and subshell (l) of an electron in an atom.
For the same shell value (n) the penetrating power of an electron follows this trend in subshells:
$$\ce{s > p > d > f}$$
And for different values of shell (n) and subshell (l), penetrating power of an electron follows this trend:
$$\ce{1s > 2s > 2p > 3s > 3p > 4s > 3d > 4p > 5s > 4d > 5p > 6s > 4f ...}$$


text:
Return time after perturbation in a chemostat
I'm wondering how I can derive the return time of the culture cell concentration back to a steady-state in a chemostat after a small perturbation of the bulk culture cell concentration.
More specific, I have the following situation. I have a very small chemostat with a working volume of about 200 µL. Influx is achieved via a very small needle which drops fluid in the reactor. At this small scale, the influx cannot be seen as a continuous stream, but rather as points in time when the next drop is added. Consequently, when a new droplet is added, the culture concentration in he chemostat drops slightly, which is seen as a perturbation from the steady-state situation. Now, I want to determine how long it takes for the system to recover and again arrive at the steady-state situation.
The chemostat is assumed to be perfectly mixed, and the working volume of the reactor is constant over time (efflux is removed by overflow at a certain level in the reactor).
Any answers or suggestions how to calculate this are very welcome!


text:
What reactions produce lots of Helium?
I'm working on a pet project and I'll need some helium. I was wondering if there were any reactions that I could perform at home to produce helium cheaper than it would be to buy helium from someone else.
Anyone know of such reaction off the top of their head?

Helium is not known to react yet, to the best of my knowledge.

Do you have ready access to a star? http://www.universetoday.com/75803/how-does-the-sun-produce-energy/

The US is effectively dumping helium so the price is depressed against the real cost. It's more cost effective and safer to buy a small cylinder e.g. for filling balloons at home

It has been reported that the reaction of needles with toy balloons may yield vast amounts of helium.

Helium doesn't form any compounds, being a noble gas. That is, you can't extract it from any reaction, because you don't have anything that it reacted with already. Also, it would've probably taken much more energy to form that helium compound than it would've to actually extracted it from the atmosphere.
Now, you can produce helium, technically. That's what an alpha particle is: A loose helium nucleus bouncing about. But mucking around with radioactive things is dangerous and can be quite expensive.
As a note, you could probably extract it yourself if you have a natural gas well that you can access directly and a way to separate it out with fractional distillation. Almost all the gasses will condense out before helium does, which means you'll end up with very pure helium left behind, with a minor impurity of neon. This process is dangerous, however, as you'll both be working with very cold temperatures and have liquid methane running amok.
You could probably get a tank from it from a chemistry supply store somewhere. (Party supply stores add oxygen to prevent people from suffocating off of their tanks. The impurity is not insignificant.)
TL;DR: No, unless you're willing to irradiate or freeze yourself.

I think you would have better throughput if you used  the fusion of deuterium and tritium, though at a greater capital cost.

True, but it'd also be much more expensive then simply using fractional distillation of ambient air.

The above answer is correct in that helium does not form any chemical compound. İt is formed in alpha decay. İn some minerals, this happens and the alpha particles are trapped in the mineral matrix. Dissolving this matrix releases the gas. I think the mineral was called clevite and the 'reaction' was noted by one of the early workers on noble gases.

That's right: there are monazite, thorianite and other $\alpha$-radioactive minerals. Also, there are compounds like $\ce{He@C_{60}}$. Many people would argue these are "not real", or "not chemical" compounds. But anyway, you react them with something and they release helium, so there are such reactions, after all. This is not cheap at all, though.

Short of alpha decay (nuclear fission, radioactive decay), you're out of luck getting it from a chemical reaction. Most of the helium found on earth is recovered from natural gas extraction. Unless you know of a geothermal engineer willing to give you some on the cheap from a new deep well, you're best bet is to go to a chemical supply company or other traditional means of purchase. Best of luck.


text:
Convert pg/mL to/from pmol/L?
In this article, 148 pmol/L is said to be equivalent to 200 pg/mL.
But this unit conversion site says that 200 pg/mL are equivalent to ~734 pmol/L, and this one gives 44 pmol/L for the same value.
What am I missing here? What is the correct way to perform this conversion, and why are different sites giving back different values?

Simple: your article is discussing vitamin B-12 whereas the first calculator you link is for estradiol, a compound with a very different molar mass.  Your second one works fine. You just have to use the Vitamin B-12 converter.

I didn't notice there was a specific calculator for B12 in the second site. Thank you A.K.!

Your question is how to do the conversion of the units of mole/liter to gram/liter.
Mole/liter, written as mol/L, is an amount (in units of mole) per liter.
To do this conversion, we need to know the convert mole to gram or vice versa.
To this end, we need to use the physical property of molar mass in units of gram/mole. This is the mass per mole of substance, in your case the substance is Vitamin B-12.
According to Wikipedia, the molar mass of Vitamin B-12 is 1355.37 g/mol.
$$\frac{200\times 10^{-12} \frac{g}{mL}}{1355.37 \frac{g}{mol}}=0.146\times10^{-12}\frac{mol}{mL}=146\times10^{-12}\frac{mol}{L}=146\frac{pmol}{L}$$


text:
To Debunk the 9/11 conspiracy, what is the science behind melting of the towers architecture?
I am wondering what the chemical validity is of jet fuel actually melting the steel beams that held up the twin towers during the September attacks.

I think this question is better suited for Skeptics.SE which has a related question and answer.

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

Related: How quickly will steel lose structural strength from a heat flash?

Jet Fuel type A (Jet A) burns at approximately 2,230 °C. 0.2% carbon steel has a maximum forging temperature (i.e., the point at which it'll melt) at about 2410 °C, while 1.5% carbon steel is at about 1920 °C. This also doesn't take into account effects that happen with how aerosolized jet fuel will burn, which is, much hotter, and strange effects that'll happen as everything else has it's surface area very spontaneously increased, subjected to a large amount of kinetic energy at once, and then lit on fire.
Not to mention everything else burning at the same time. Finely shredded aluminum is very flammable.
Also: As a final note: The steel didn't actually have to melt, persay. It only had to weaken enough that the mechanical forces of the skyscraper that it was supporting became too great for the softened steel, at which point it'll all collapse in a chain reaction.
TL;DR: Conspiracy theories are BS.

Also, thermite reaction.

Doubtful, but possible. The grain size needed would be difficult to attain.

You've got ALL facts and numbers wrong. "Jet fuel" does not have such a defined temperatue when burning, that depends a lot on circumstances. The aerosol created on impact burns away immediately, consuming all the oxygen, leaving the bulk of the fuel to just burn steadily with air flowing in from outside.  Iron melts at 1536°C, steel slightly below, depending on carbon content etc. Steel is forged at 950-1250°C. To ignite aluminium, you have to powder it. How should a plane crash do that, esp. when it melts at 930°C?

Karl: There's many different factors. I took the value of the maximum temperature of Jet Fuel Composition A, used in the United States since the 1960s, as a reference. From there, you don't need to melt the steel, like I said. In fact, jet fuel probably can't melt steel beams. All it needs to do is threaten the structural integrity of the beam, and the whole thing would come down around it. Also, I specifically mentioned that a termite reaction or any kind of explosive reaction involving aluminum is extremely unlikely. Also, there's a very strong possibility that the aerosolized jet fuel burnt

at a much higher temperature. This is a well observed phenomenon, and is the primary mechanism for fuel-air explosives.


text:
Reaction of Iron(III) with hydroxide ion
The observations:
1) Treatment of 1 mL of 0.05 M $\ce{Fe(NH4)SO4}$ solution with excess 1 M $\ce{NaOH}$ results in the formation of a rust-colored red-brown iron-based solution with many suspended particles in the solution. These particles match the description of $\ce{Fe(OH2)3(OH)3}$. This is the vial on the right in the picture.
2) Treatment of 1 mL of 0.05 M $\ce{Fe(NH4)SO4}$ solution with excess 6 M $\ce{NaOH}$ results in the formation of a clear but colored solution with no suspended particles in the solution or any sort of precipitate. The yellow-brown colored solution matches the description $\ce{Fe(OH2)4(OH)2+}$. This is the vial on the left.

Why is it that treatment with more concentrated base results only two apparent deprotonations rather than three?
More observations (after about 10 minutes):
1) The 1 mL of 0.05 M $\ce{Fe(NH4)SO4}$ solution treated with excess 6 M $\ce{NaOH}$ has developed suspended particles. This is the vial on the left.
2) The 1 mL of 0.05 M $\ce{Fe(NH4)SO4}$ solution treated with excess 1 M $\ce{NaOH}$ has started to become clear (the suspended particles are disappearing). It seems like much of the particles have settled on the bottom. This is the vial on the right.

It seems that the 6 M $\ce{NaOH}$ is eventually "getting the job done" by going through with the third deprotonation of $\ce{Fe(OH2)6^{3+}}$. I don't think that the 1 M $\ce{NaOH}$ is deprotonating $\ce{Fe(OH2)3(OH)3}$.
Still, why is it that treatment of iron(III) ions with concentrated base does not result in the formation of $\ce{Fe(OH2)3(OH)3}$ immediately but treatment with less concentrated base does?
Even more observations:
1) Treatment of 10 mL of 0.05 M $\ce{Fe(NH4)SO4}$ with excess 6 M NaOH results in immediate and obvious $\ce{Fe(OH2)3(OH)3}$ formation. This seems to support the nucleation explanation; the higher solution volume means lower concentration and lower supersaturation and therefore fewer and larger crystals of $\ce{Fe(OH2)3(OH)3}$ being formed.
Possible explanations I came up with:
1) I ran the results past a grad student and he said his best guess is that the excess $\ce{HO-}$ present stabilizes $\ce{Fe(OH2)4(OH)2+}$. He suggested a conductivity test to make sure that there are $\ce{Fe(OH2)4(OH)2+}$ ions in the clear solution. However, I did not have a conductivity meter handy.
2) A problem in the lab manual from an unrelated section notes that strongly electrolytic solutions, such as saturated $\ce{H4NCl}$ solution, can cause dispersion rather than dissolution of $\ce{Al(OH2)3(OH)3}$ in solution. A 6 M $\ce{NaOH}$ solution is strongly electrolytic, and as a result, the ions may be keeping the $\ce{Fe(OH2)3(OH)3}$ from clumping together.
What are your thoughts?

I wouldn't expect the reactions to yield different products but rather different particle size. The higher concentrated $\ce{NaOH}$ should hereby produce smaller particles†.
The "solution" in the left vial is probably rather a colloid of microscopically dispersed iron(III) hydroxide. It should show the Tyndall effect if you send a beam of light through it. However, the particles will aggregate over time to form larger particles that also precipitate, as seen in the second image.

† Higher concentration means higher supersaturation. Higher supersaturation leads to a higher nucleation rate, so more and thus smaller crystals are being formed.

why would the particles start out as microscopically dispersed?

@Dissenter They always do, but kinetics of aggregation may vary.

@Mithoron - makes sense. There was a question in the lab manual that talked about strong electrolytic solutions possibly causing metal hydroxides to "disperse" instead of dissolve. 6 M NaOH is a very strong electrolytic solution, and that may be keeping the metal hydroxide particles that form from clumping together.

@Dissenter Precicipate rarely 'aggregates', aging of precipitate includes dissolution of smaller particles and growth of larger ones. Obviously, strongly basic solution prevents dissolution of iron hydroxide.

@permeakra - why don't they clump together?

@aventurin what do you think about the anionic stabilization hypothesis?

It is known that colloid iron oxy-hydroxide can stabilize by adsorption of protons on the surface of the particles. They repel each other when charged. But I would expect less positive charge at higher pH. The sign of the charge can be determined experimentally by electrophoresis in a U-shaped tube. However it may be difficult to conduct with high concentrations where the particles coagulate fast and then sink to the ground.

@aventurin - by iron oxy-hydroxide do you mean FeO(OH) and related compounds?

Yes, this somehow describes the stoichiometry of the precipitate.


text:
Proof of buffers inability to maintain exact same pH
I was wondering if the following is a legitimate proof of the statement above. Assume we have
HA -> A- + H+
where [HA] + [A-] is our buffer.
If we add a small amount of H+ one of two things will happen:
1) The H+ will not react with A- ...thereby raising the pH so we are done.
2) The H+ will react with A- ...converting to [HA]
looking at the HH equation we know that
pH = pKa + log(ratio)
since ratio_before != ratio_after
pH_after = pKa + log(ratio_after) and pH_before = pKa + log(ratio_before)
so therefore
pH_after - log(ratio_after) = pH_before -log(ratio_before)
Now assume pH_after == pH_before.... we get -log(ratio_after) = -log(ratio_before) which is a contradiction of above, so pH_after != pH_before.
What are your thoughts?

Not sure if I'm reading maths, programming, or chemistry... Anyway, yes. Buffers are not intended to maintain the exact same pH and indeed they cannot. They are designed to maintain the pH within a small range acceptable for whatever system you are studying.

I'm curious though, is 2) really true? If you add H+ to a buffer system will all the H+ react with A-?

Your reasoning is right, buffers can't keep pH exactly the same.

It is not an all or nothing thing. Not ALL of the H+ will react but it does not invalidate your proof.

But when doing buffer pH change problems we assume that all the H+ reacts ... Is that just to simplify the problem?

A buffer is a solution that can resist pH change upon the addition of
  an acidic or basic components. It is able to neutralize small amounts
  of added acid or base, thus maintaining the pH of the solution
  relatively stable.

Buffers do not maintain exact same pH, but keep it relatively stable i.e. close to the original value.
log(ratio) << pKa, therefore, pH changes will be minimal, unless substantial amount of acid has been added to the solution.
For instance, when the ratio between the conjugate base/acid is equal to 1, the pH = pKa.
If the ratio between the two is 0.10, the pH drops by 1 unit from pKa since log (0.10) = -1.
If a ratio increases to a value of 10, then the pH increases by 1 unit since log (10) = 1.
As you can see, changing ratio by 10 times, affects pH value only by 1 unit.


text:
Why don't gases of elements with negative electron affinities exist as ions in nature?
Fluorine has an electron affinity of about -300 kJ/mol. Lithium has an electron affinity of about 60 kJ/mol.
As I understand it when $\ce{Li^-}$ loses an electron about 60 kJ/mol is released but when $\ce{F}$ gains an electron about 300 kJ/mol is released.
Furthermore, when $\ce{Li^+}$ gains an electron about 520 kJ/mol is released because that is the first ionization energy of lithium and when $\ce{F^+}$ gains an electron about 1700 kJ/mol of energy is released because that is the first ionization energy of fluorine.
Shouldn't the cases that release the most energy be $\ce{Li}$ and $\ce{F^-}$ and so shouldn't those be the states that gases of those elements occupy?
Is it the case that $\ce{2F^-}$ is less stable than $\ce{F2}$ or something? Or is it the case that $\ce{F^-}$ could only exist in a highly artificial situation such as a magnetic contraption that trapped ions?
Why don't gases of elements with negative electron affinities exist as ions in nature?

Well, you need something to give it the electron to begin with. Nature doesn't have free electrons floating around for fluorine to pick up. If you put fluorine anywhere near sodium or anything that is willing to give up an electron, it will obviously take the offer.

It totally does not work like that. You can't say that $\ce{2F^-}$ is less stable (or more stable) than $\ce{F2}$; you can't compare them at all. People would say "B is more stable than A" when there is a spontaneous reaction $\ce{A\to B}$, but this is not the case here. A reaction must be balanced in charge and in all elements, otherwise it is just a bunch of letters that makes no sense. There is no reaction $\ce{F2<->2F^-}$ (either way), much like there is no reaction $\ce{F2->Au}$.
There is indeed a reaction $\ce{F2 + 2e- -> 2F^-}$, and it is quite favorable energetically (as you already know), but that's another story. You need the electrons for it, and they are just not there. Fluorine can't wrestle those electrons from another fluorine, as explained by IT Tsoi. And if you have lithium around, why, of course they would react to form $\ce{LiF}$, but that's a chemical compound, not just $\ce F$ by itself; also, at room temperature it is solid.
There is more to it. A gas (as well as a liquid or solid, for that matter) formed of negative ions alone is very much impossible. The ions will repel with such force that no material in the world will hold them even for a second. (I mean macroscopic amount, not a thin trickle of ions inside a mass spectrometer.) So you can't have fluorine in this state even if you have a stash of electrons which you are willing to share. $\ce{F-}$ ions do exist, of course, and there is nothing exotic about it (arguably, almost all fluorine in the world is found in this form), but they exist as a part of various ionic compounds, and not by themselves.

So consider a vessel wherein which we have Fluorine atoms (F) in gaseous state.
Now, they have seven outer shell electrons and would really love to get the 8th one for an octet. What would have to happen is:
$$\ce{2F -> F+ +F-}$$
It's not possible for both to become fluoride ions because there would be no charge balance.
The total change in energy is then given by:
$$\Delta E = -300 + 1700 = + 1400\ \mathrm{kJ/mol}$$
So it is not energetically favorable for one Fluorine atom to completely give up its electron in favor of the other. Hence, they prefer to share one electron each in a covalent bond. And this results in the release of $156.9\ \mathrm{kJ/mol}$ of energy.


text:
Calculate the pH of Na2CO3
Given information:
$K_\mathrm{a}(\ce{H2CO3}) = 4.3 \times 10^{-7}$,
$K_\mathrm{a}(\ce{HCO3-}) = 4.8 \times 10^{-11}$,
${{\ce{[Na2CO3]}=0.20\ \mathrm{M}}}$
This was a question in an exam and I'm struggling to understand how to solve it. I understand that $\ce{Na}$ is soluble in water, so $\ce{[CO3^{2-}]}=0.20\ \mathrm{M}$. I also know how $\ce{H2CO3}$ dissociates:
$$\ce{H2CO3 <=> H+ +HCO3- <=> 2H+ +CO3^{2-}}$$
When I try solve it using $K_\mathrm{a}=\frac{[P]}{[R]}$ I get:
$$\frac{\ce{[H+]}^2}{\ce{[H2CO3]}}=1.032 \times10^{-16}$$
Past here I'm not sure. I've been told to find $K_\mathrm{b}$ because it's a base but I'm not sure why that would make a difference. Please could someone make a worked solution or give me some hints?

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Some users updated your post with chemistry markup. If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details.

It suffices to consider only the reaction
$$\ce{CO3^{2-} + H2O <=> HCO3- + OH-}$$
with $k_{B_1} = 2.1 \cdot 10^{-4}$, since the reaction
$$\ce{HCO3- + H2O <=> H2CO3 + OH-}$$
with $k_{B_2} = 2.3 \cdot 10^{-8}$ will only contribute a small amount of $\ce{OH-}$.
$$k_{B_1} = \frac{[\mathrm{{HCO_{3}}^{-}}] [\mathrm{OH^-}]}{[\mathrm{{CO_{3}}^{2-}}]} = \frac{x \cdot x}{c - x} \approx \frac{x \cdot x}{c}$$
$$x = [\mathrm{OH^-}] = \sqrt{k_{B_1} \cdot c} = 6.455 \cdot10^{-3}$$
which gives a pH of 11.8.


text:
Hydrogenation of alkenes
How many structural isomeric alkenes ($\ce{C_6H_{10}}$) can be hydrogenated to give methylcyclopentane? Would there be three? I drew out what I thought was correct but I just want to confirm that there are three.

There are 4 structural isomeric alkenes ($\ce{C_6H_{10}}$) that can be hydrogenated to give methylcyclopentane.
1-Methycyclopentene

3-Methylcyclopentene

4-Methylcyclopentene

Methylenecyclopentane


text:
how does incomplete combustion of methane produce soot?
Ususally the incomplete combustion is formed this way:
hydrocarbon + oxygen-> carbon + carbon monoxide + water.
what about methane?

Methane is no different from other hydrocarbons in this respect.

The incomplete combustion of methane can result in a slew of partially oxidized products, including $\ce{CO}$, but also methanol, formic acid, formaldehyde, and higher hydrocarbons also.
However,the oxidation of methane to soot is generally given as the following:
$$\ce{CH4(g) + O2(g) → C(s) + 2H2O(g)}$$

fascinating! Is there a source for the first sentence?

It is not necessary that every hydrocarbon's incomplete combustion produces carbon.
Methane has a high calorific value, so there are less chances that it will undergo incomplete combustion. Even if it goes,it will follow the reaction
2CH4 + 3O2 ----> 2CO + 4H2O

True, methane produces less soot than other hydrocarbons; I attribute this to the absence of $\ce{C-C}$ bonds in it. Calorific value has little to do with completeness of combustion. Your reaction is of course possible, but what if we just don't have enough oxygen even for that?

Calorific value has a role in completeness of reaction. As if there was limited amount of air and there was also a substance with low calorific value, Methane would burn faster than that, therefore reducing the amount of O2 left for that other substance.  And actually, we see these reactions at macro level. A reaction is never completed (in real world). We consider it to be completed as the reactant unreacted is very negligible. And so O2 would be available for combustion.

OK, but what if we just mix methane with oxygen, 1 to 1, no other substances?


text:
How does entanglement explain myoglobin's preference for O₂?
I was reading this article (full PDF here) about the myoglobin's unusual preference for O2  than CO, in which it was written:

It turns out that some electrons in the myoglobin involved in binding CO and O2  exhibit a strong ‘entanglement’ effect, which means that their motion cannot be described independently. The all-important strength of this effect is primarily controlled by a property of quantum mechanics (Hund’s exchange) that has been traditionally neglected in such simulations; the team now believe that classical electric repulsion effects are far less important in determining which of CO and O2  is more energetically favourable for binding.

But I cannot understand how entanglement explains this phenomenon. Has it been concluded from this (quantum mechanical) perspective? If yes, can anyone explain how?

For the convenience of anyone who wishes to answer, the paper linked in the article is here: http://www.pnas.org/content/111/16/5790.full.pdf

So, there are two effects at play here which they account for. One they denote by a $\textbf{U}$ and the other by a $\textbf{J}$. $\textbf{U}$ represents many-body coulomb repulsions between the electrons. That is, they are accounting not just for two electrons repelling each other, but up to three or more. $\textbf{J}$ represents valence fluctuations in spin. Specifically, iron in hemes are in a high-spin triplet state. What they are saying then is that it is necessary to include spin fluctuations which result in changes in the magnetic moment of the system in order to properly capture the binding of ligands. If you look at figure 4 of this paper, you'll find that they have included a comparison of spin-fluctuations in $\ce{MbO2}$ and $\ce{MbCO}$ which are used as model systems and known to be analogous in ligand binding to hemes. Basically what this means is that the increased spin-fluctuations cause $\ce{O2}$ to bind more strongly, but this has basically no effect on $\ce{CO}$. They describe these effects in the following quote:

In this case, the effect of J on the
  binding energy of O2 may be regarded as a balance between two
  competing effects. The charge analysis (Fig. 3) reveals that
  metal-to-ligand charge transfer is higher for J =0 eV, which is
  expected to enhance ligand–protein interactions for small values
  of J. However, NBO analysis reveals a larger ligand-to-metal
  back charge transfer for J = 0.7 eV, which is consistent with the
  increased occupancy of the Fe $\ce{d3z^2−r^2}$ orbital (Table 1), and is
  expected to cause variational energetic lowering at higher values
  of J due to electronic delocalization.

I won't pretend to fully understand their argument (frankly it seems to me like they are waving their hands at a solution and are basically saying this is some complicated effect having to do with spntaneously changing magnetic moments).
So, in that paper they only ascribe the effect to strong electron correlation, many-body coulomb effects, and spin-fluctuations.
So, they definitely use the word entanglement but don't define exactly what they mean or even use it as an explanatory device. I found in another one of their papers though, Weber, C., O’Regan, D. D., Hine, N. D., Littlewood, P. B., Kotliar, G., & Payne, M. C. (2013). Importance of many-body effects in the kernel of hemoglobin for ligand binding. Physical review letters, 110(10), 106402., where they do actually define what they mean by entanglement and it depends on these parameters $\textbf{U}$ and $\textbf{J}$. You can find their discussion of this on page 11 in an appendix.
Their use of the word entanglement is derived from the Von Neumann Entropy which can be used as a measure of the amount of entanglement in a given system. This is because if two particles are entangled in some way, their entropy will be lower than if the two particles were not entangled. Note that this use of the word of entanglement is related to the usual descriptions of entanglement, but it is more of a measure of the average of spontaneous entanglement and disentanglement. That is, particles can become entangled due to some way in which they interact, but almost immediately interact with something else that breaks their entanglement so that it's very rare to find systems where this has any meaningful effect. This spontaneous entanglement in these heme systems is a result of the many-body effects which I've been describing. This manifests itself in the entropy they calculate. This highlights the importance of using dynamical mean-field theory (which they use) as follows:

With this definition of the entropy, suitable for DFT
  calculations, we found that L = 0.757 in FeP-p and
  L = 0.765 in Fep-d. The entropy at the DFT level
  is hence much smaller than the DFT+DMFT entropy,
  which is in the range Λ = 2.5 − 4.2. This is explained by
  the absence of the many-body excitations induced by the
  correlations (U and J) which contribute significantly to
  the entropy.

There's no doubt they aren't being very forthcoming in their terminology or definitions, but they are also doing some very complicated and interesting stuff.

Interesting answer, +1! "Basically what this means is that the increased spin-fluctuations cause O2 to bind more strongly, but this has basically no effect on CO." and "This spontaneous entanglement in these heme systems is a result of the many-body effects which I've been describing." Can you add some more details on these points? :)

The first quote box is about all I can offer in regards to the preference for $O_2$. They are saying that when spin-fluctuations are allowed to occur, the charge-transfer from ligand to metal is increased and this increase compensates for the metal-to-ligand charge transfer which is lost due to spin-fluctuations. If you wish, you can regard spin-fluctuations as a mechanism by which the electrons can be occupied, and simply because this can happen, it does to some extent, which weakens the metal-to-ligand charge transfer, but oxygen then donates more charge so it all works out.

Note, however, that they are choosing the value of $\textbf{J}$ such that it matches experiment. One can always choose a set of parameters that match experiment, but that doesn't make it right. Nonetheless, it seems they have good physical reason for doing this. As far as the entanglement goes, I don't what to add besides that particles are always being entangled, but this entanglement is disrupted by random collisions and other things. This can be thought of as the environment "making an observation" on the entangled pair of particles.

Nice points! Just one more thing, please add them into your answer ;)

More of a comment, but it is too long for the comment format.
I scanned the article. I'm not qualified to evaluate their QM model. I do wonder about their use of the word "entanglement" however. From wikipedia:

Quantum entanglement is a physical phenomenon that occurs when pairs or groups of particles are generated or interact in ways such that the quantum state of each particle cannot be described independently of the others, even when the particles are separated by a large distance.

The phrase "separated by a large distance" in Wikipedia's definition is an important qualifier to me. To me it means I'm measuring a atom/molecule on this side of the lab and comparing it to a atom/molecule on that side of the lab. But for some reason due to how I am making the measurements the two atoms/molecules have correlated quantum states.
If you do a "full" MO analysis on the molecule then you have to consider how all the parts interact. Chemists typically just do a MO on the specific chemical bonds between two parts of the interacting molecules because the approach is simpler. That doesn't seem to work for myoglobin's unusual preference for O2 over CO.
Look at the blue box at the end of the paper. Instead of "quantum entanglement" I think "quantum interactions" would be more appropriate wording.

Well, I think entanglement would fit there. Its just that many people (except Wikipedia) don't define quantum entanglement properly. And since (I suppose) the electrons are farther than in an atom, so only entanglement and tunneling (maybe 1-2 more things) would be applicable in the case.

The point here is not separated by large distance but even when separated by large distance, in a way meaning that distance does not play much role here.

Is not 'entanglement' a bit of a buzz word at present? If the coupling between electrons is stronger than the combined disruptive (de-phasing) effects of electric & magnetic fields  of all nearby atoms then it would seem that their behaviour should be correlated.  How 'correlation' and 'entanglement' differ is not clear, nor is it clear (in the paper to me at least) how Hund's coupling differs from the more familiar antiferromagnetic or exchange coupling, but I'm not an expert in this.

@porphyrin and how would coupling between electrons be stronger than overall disruptive effects? If not entanglement, then some other effect must be responsible for this, and thats what I want to know.

As I understand it the reason that its hard to detect entanglement (other than in photons) is that because of disruptive 'noise' from surrounding atoms that any coherence  lasts only for a few hundred femtoseconds at most, e.g as in the initial entangled state ( exciton) in energy transfer in antenna of photosynthetic organisms.  In general the size of the electron's interaction depends on  its causes, so could cover a wide range of energies. I don't know enough about the details to comment in more specific terms.


text:
What is the meaning of [2,1-c][1,4] in pyrrolo[2,1-c][1,4]benzodiazepine?
Pyrrolobenzodiazepines are a group of DNA-alkylating natural products. What is the exact meaning of the [2,1-c][1,4] part in the full descriptive name pyrrolo[2,1-c][1,4]benzodiazepine?
![Pyrrolo2,1-c][1,4]benzodiazepine
General structure of pyrrolo[2,1-c][1,4]benzodiazepines. The 2-3 double bond is not present in all natural products.
Thanks to Loong I may be able to figure it out.
1,4-Benzodiazepine is pretty trivial, as [1,4] defines the position of the nitrogens like this:

For clarification [1,5] could look like 1H-1,5-benzodiazepine:

[2,1-c] is answered in What does the "h" "k" and "(5.2.2.02,6)" mean in dibenzo(h,k)-4-oxatricyclo(5.2.2.02,6)undec-8,10- diene-3, 5-dione?

According to the current version of Nomenclature of Organic Chemistry –
IUPAC Recommendations and Preferred Names 2013 (Blue Book), isomers of
fused polycyclic ring systems are distinguished by lettering each
peripheral side of the parent component using the italic letters a, b,
c, etc., beginning with a for the side numbered ‘1,2’, b for ‘2,3’,
etc.

So the letters would be:

So pyrrolo[2,1-c] would be attached to the c-side.
I still need clarification in one point though:
Does the direction 2,1 (NOT 1,2) show the direction of the attachment?

Related: Numbering in chlordiazepoxide and What does the “h” “k” and “(5.2.2.02,6)” mean in Dibenzo(h,k)-4-oxatricyclo(5.2.2.02,6)undec-8,10- diene-3, 5-dione

Yes, the order of the numbers indicates the relative orientation of the fused ring system. The direction of the numbers of the attached component (e.g. ‘1,2’ or ‘2,1’) corresponds to the direction of the lettering of the parent component (i.e. a, b, c, etc.).
For example:

The correspondig Subsection P-25.3.1.3 in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), reads as follows:

(…)
  Isomers are distinguished by lettering, continuously, each peripheral side of the parent component (…) using the italic letters a, b, c, etc., beginning with a for the side numbered ‘1,2’, b for ‘2,3’ etc. To the letter as early in the alphabet as possible that denotes the side where the fusion occurs are prefixed, if necessary, the numbers of the positions of attachment of the other component. These numbers are chosen to be as low as is consistent with the numbering of the compound and their order conforms to the direction of lettering of the parent component. (…)


text:
Solubility of silver acetate in the presence of nitric acid

What will be the effect on the solubility of $0.1\ \mathrm{mol}$ of $\ce{CH3COOAg}$ in presence of $1\ \mathrm{l}$ of $0.1~\mathrm{M}\ \ce{HNO3}$ solution?
$$K_\mathrm{sp} (\ce{CH3COOAg}) = 10^{-8}\\
K_\mathrm{a} (\ce{CH3COOH}) = 10^{-5}$$

I know that silver acetate will react with nitric acid to form silver nitrate and acetic acid. So the solubility of $\ce{CH3COOAg}$ will increase. But in the end, I am not able to find out the final solubility (as concentration of $\ce{H+}$ and $\ce{NO3-}$ is coming different).
Maybe I have an error somewhere?

Interestingly, Ag+ is much more strongly positive than H+, and -COO- more strongly negative than NO3-.  Unless acetic acid splits out of solution (unlikely), what you will have is AcAg solvated to some degree in aqueous solution.  When reacting metallic silver with HNO3, some acid oxidizes silver 0 to silver +1, hence AgNO3.

The process you are trying to investigate is:
$$\ce{AgOOCCH3(s) + HNO3(aq) <=> AgNO3 (aq) + CH3COOH (Aq)}$$
with the net ionic equation:
$$\ce{AgOOCCH3(s) + H+(aq) <=> Ag+(aq) + CH3COOH(aq)}$$
The equilibrium constant is
$$K_c = \dfrac{[\ce{Ag+}][\ce{CH3COOH}]}{[\ce{H+}]}$$
Note that this is law of mass action can be derived from the $K_a$ and the $K_{sp}$ you have:
$$\begin{aligned}
K_{a}&=\dfrac{[\ce{H+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]}\\
K_{sp}&=[\ce{Ag+}][\ce{CH3COO-}]\\
K_c &= \dfrac{K_{sp}}{K_a}=\dfrac{[\ce{Ag+}][\ce{CH3COOH}]}{[\ce{H+}]}
\end{aligned}$$
The problem is now an ICE (Initial-change-equilibrium) problem.
$$\begin{array}{|c|c|c|c|} \hline
\ & [\ce{H+}] & [\ce{Ag+}] & [\ce{CH3COOH}]\\ \hline
I & 0.1\ \mathrm{M} & 0\ \mathrm{M} & 0\ \mathrm{M}\\
C & -x & +x & +x \\
E & 0.1 -x \ \mathrm{M} & x \ \mathrm{M} & x \ \mathrm{M}\\ \hline
\end{array}$$
Plug values into the law of mass action, and solve for $x$, the concentration of $\ce{Ag+}$, and then you can determine the solubility.

Just one thing, i got x=10^-2. So solubility is 10^-6 or 10^-2 ?

Solubility is the amount that dissolved per unit volume.


text:
Amide from acyl azide
There are several reports of synthesis of amines from acyl azides via Staudinger reaction with triphenylphosphine.
Are there any reports of Staudinger reactions of acyl azides to amides? If not, are there any reports on reduction of acyl azides to amides? I have found a paper (http://www.tandfonline.com/doi/abs/10.1081/SCC-120014032) that talks about the reduction of acyl azides to amides using zinc and ammonium chloride, but I am looking for some other reactions as well.


text:
Calculate the rate of decomposition of H2O2 at 25 °C when [H2O2] is exactly 10.0 mmol/L
I'm really confused?
$$\ce{2H2O2 -> 2H2O + O2}$$
$$\begin{array}{ccc}\hline
\text{Experiment no.} & c(\ce{H2O2}) & \text{rate} \\
\hline
1 & 14.45 & 3.35 \times 10^{-3}\\
2 & 28.90 & 6.65 \times 10^{-3}\\
3 & 57.80 & 13.35 \times 10^{-3}\\
\hline\end{array}$$
So from previous parts of the question, I found the order of reaction (1st), the rate equation for the reaction ($r = k[\ce{A}]$), the rate constant ($2.3 \times 10^{-4}\ \mathrm{s}^{-1}$, which is correct according to the answers), and now I need to find the rate of decomposition of $\ce{H2O2}$ when $\ce{[H2O2]}$ is exactly $10.0\ \mathrm{mmol\ L^{-1}}$.
I thought it's just $2.3 \times 10^{-4} \times 10.0$, which is $2.3 \times 10^{-3}$, but the answers provided is double that (so $4.6 \times 10 ^{-3}$), so I must have gotten the rate equation wrong?

It is a matter of definition:
The rate of the reaction is defined as: $$r=-\frac{1}{2}\frac{d[\ce{H2O2}]}{dt}$$
The rate of the decomposition of $\ce{H2O2}$ is defined as:$$r_{\mathrm{decomp}}=-\frac{d[\ce{H2O2}]}{dt}= 2r$$

so if I'm calculating for k, would I use r = k[H2O2]?

Yes. Your calculation of k is quite right.

so if, for example, the rate was r = (-2)*(d[H2O2]/dt), would rate of decomposition be (1/2)*r? am I understanding this relationship correctly?

In the expression of the reaction rate on devise by the stoichiometric coefficients of each reactant or product.

The rate of decomposition of a reactant or the rate of formation of a product is given without dividing by the stoichiometric coefficients .

The correct answer depends on what who wrote the problem had in mind about the values of rates  in the  last column in the data table:
If he considers the  rate is  rate of decomposition of $\ce{H2O2}$      , then  the answer is $2.3\times 10^{−3}$
If he considers the  rate is  rate of formation of $\ce{O2}$, then  the answer is $4.6\times10^{−3}$,
How we answer if we write the balanced equation in this way:

$\ce{H2O2 -> H2O + 1/2 O2}$