text:
Brass Instrument: Dezincification or just scrubbed off?
I do own a brass instrument (a trombone). The lacquer is damaged.
I asked how to deal with it here.
Now I learned that the brass contains zinc, and there is a corrosion process called "dezincification". Now they told me to ask here how to identify real dezincificated parts. Here are the pictures:
Main slide handle:

Bell "middle" section:

Bell inner rim:

The issue is: On trombones (other brass instruments) red rot (dezincification) usually only appears on parts where saliva touches the instrument, i.e. in the front parts. On a trombone that would be the slide. So it is kind of unlikely but not unheard of that this might be "red rot" or dezincified.
It might also just be lacquer which went off due to improper handling (like not cleaning water spots, showering/cleaning the trombone with way too hot water, etc.).
Actual questions

How do I identify dezincifiaction correctly? Can it be seen on those pictures? (most important question)
What do you think of chemical cleaning? It sounds dangerous to me. Isn’t it going to damage the instrument?
What do you think of cleaning such an instrument soap? Is this going to damage the lacquer or material?

I don't know enough to give a full answer but: 1. the cork barrel wear is lacquer only, 2. chemical cleaning should be safe, 3. the "shower technique" wouldn't cause this. The answer on the music post is excellent and should be followed. Find a reputable brass instrument (repair) shop, not just a music store, and talk to them. You probably don't need soap unless you delacquer the horn. IIRC lacquer on pro horns can be quite thin.

Brass Instrument: Dezincification or just scrubbed off?

How do I identify dezincifiaction correctly? Can it be seen on those pictures? (most important question)

Instruments are made from solid brass, this is required as a stiffer metal like brass is required to have good sound quality. Your brass was not scraped off and the spots are indeed a result of dezincification. Water could get here for a number of reasons from fog mist to moist hands, it does not have to be saliva only.

What do you think of chemical cleaning? It sounds dangerous to me. Isn’t it going to damage the instrument?

From what I can find chemical cleaning looks like they use a mild acid and chelatior (like citric acid or EDTA) to etch away some of the red rot. However the brass still has to be polished and the lacquer needs to be reapplied.

What do you think of cleaning such an instrument soap? Is this going to damage the lacquer or material?

Soap won't damage the lacquer, but scrubbing might. That said your red rot shows that the lacquer is already damaged to you might as well scrub/polish the spots and reapply lacquer.

It sure looks like dezincification to me. I have seen several cases in petrochemical service. It is fairly difficult to reproduce in realistic tests ( without unusual chemicals). It is surprising to see it in such ordinary environment. Zinc has been removed, so copper needs to be removed until you get down to unaltered brass . No easy chemical way to do this; grind,sand, buff, polish to restore the color ( then lacquer). It may not be worth the time or maybe there are instrument restorers that can do it. Try a few spots to determine how much work ( depth ) is needed. ............ If the manufacturer had used Inhibited Admiralty Brass ( 69 Cu : 30 Zn ,1 Sn , trace of As ) this would not have happened. They used Cartridge Brass ( 70 Cu : 30 Zn). Possibly it is a 60 Cu : 40 Zn brass , where again it would have been prevented with 1 % Sn. I am guessing more knowledgeable manufacturers use Admiralty or Navel Brass to prevent dezincification.

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text:
Separation of argon in Dewar's charcoal method
I was studying about the Dewar's charcoal method. It says in the book Modern Inorganic Chemistry by R. D. Madan on p. 1291 that at $\pu{-100 ^\circ C}$ Ar, Xe and Kr is adsorbed by the charcoal.
Then, if the charcoal is further cooled to $\pu{-180 ^\circ C}$, then Ar (boiling point $\pu{-186 ^\circ C}$) gets liquefied  and thus separated.
Xe (boiling point $\pu{-107 ^\circ C}$) and Kr (boiling point $\pu{-152 ^\circ C}$) should also be liquefied and separated. So why do they remain adsorbed?

Guess they are effectively solid by then.

Also, there's so little Kr, Xe (and Rn, hopefully!) that they're difficult to separate. Ar is the third-most common gas in air.

source: Stack Exchange

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sort: votes


text:
Deviation in compressibility factor in different gases
Why does the compressibility factor of gases like Carbon Dioxide and Nitrogen first decrease then increase with increase in pressure , but for Hydrogen and Helium it only increases with the increase in pressure ?

See my answer here, which addresses this question almost exactly. The only thing to add is the same reasoning I applied for $\ce{H2}$ also applies to helium. Note that for both molecules there is actually a minimum, but they are extremely tiny and not observable on the scale of the plots you're referring to.

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text:
How to do Lowdin symmetric orthonormalisation?
Löwdin symmetric orthonormalisation seems to be a common practice in quantum chemistry. I come from a different background though and have to understand it and possibly implement in a computer code. Here is the definition of the transformation;
$$
|{\phi'}\rangle = S^{-1/2}|\phi\rangle,
$$
where $S$ is the overlap matrix between the non-orthogonal basis,
$$
S = \left< \phi|\phi \right> .
$$

My first question is how to obtain $S^{-1/2}$ analytically?
Are there any parallel algorithms available to compute it numerically?

I would recommend checking out Szabo and Ostlund Modern Quantum Chemistry. Its a very affordable book that discusses a lot of the framework of computational chemistry.

Your first question has already been answered, but in words: to find $f(A)$ for some matrix $A$, you diagonalize it to obtain the eigenvalues $a$ and eigenvectors $U$, apply $f$ to the diagonalized matrix (the eigenvalues), then back-transform $f(a)$ using the eigenvectors to the original non-diagonal basis.
For the second question, yes. Most revolve around calling the LAPACK routine dsyev (real double-precision symmetric eigen decomposition) or its variants, ssyev for real single-precision, {c,z}heev for complex single- and double-precision Hermitian decomposition. The vast majority of quantum chemistry uses entirely real coefficients and double-precision.
Assume I have already compute the overlap matrix $S$ by some method, usually by calling a program's integral engine. Here is a sample implementation in Python using NumPy:
print("Overlap matrix")
print(S)

lam_s, l_s = np.linalg.eigh(S)
lam_s = lam_s * np.eye(len(lam_s))
lam_sqrt_inv = np.sqrt(np.linalg.inv(lam_s))
symm_orthog = np.dot(l_s, np.dot(lam_sqrt_inv, l_s.T))

print("Symmetric orthogonalization matrix")
print(symm_orthog)

From the documentation of numpy.linalg.eigh:

The eigenvalues/eigenvectors are computed using LAPACK routines _syevd, _heevd

Here is a sample implementation in C++ using Armadillo:
S.print("Overlap matrix");

arma::vec lam_s_vec;
arma::mat l_s;
arma::eig_sym(lam_s_vec, l_s, S);
arma::mat lam_s_mat = arma::diagmat(lam_s_vec);
arma::mat lam_sqrt_inv = arma::sqrt(arma::inv(lam_s_mat));
arma::mat symm_orthog = l_s * lam_sqrt_inv * l_s.t();

symm_orthog.print("Symmetric orthogonalization matrix");

Both working examples can be found here; check the Makefile for how to run.
In the Armadillo source tree, the files include/armadillo_bits/{def,wrapper}_lapack.hpp contain more information about which LAPACK routines are called for which types.
Regarding parallelization, your BLAS + LAPACK implementation (MKL, OpenBLAS, ATLAS, ...) is most likely threaded and can be controlled by <something>_NUM_THREADS=4, where <something> might be OMP, MKL, OpenBLAS, or possibly something else, but you should check the documentation. This means that as long as your environment is set up properly, math library calls with NumPy or C++ template libraries like Armadillo or Eigen will run in parallel without explicit OpenMP annotations or MPI code. For distributed parallelization (MPI), there is ScaLAPACK, which shares a similar interface to regular LAPACK.

Just to compare to a possible alternative, scipy.linalg.fractional_matrix_power(S, -0.5) gives the same result, but is much slower (~15x slower for a 24 x 24 overlap matrix) than your approach :)

You do not have to compute the inverse. The matrix is already diagonalized. Just compute the power of the vector with the eigenvalues and then construct the matrix with the vector as the main diagonal. That is, np.diag(lam_s**(-0.5)).

Also, symm_ortho = np.einsum("ik,kj,lj->il",lam_s_vec, lam_sqrt_inv, lam_s_vec) could be faster.

You obtain the diagonalization of $S$ as defined
$$
S = U \cdot s  \cdot U^{\dagger}
$$
(which holds for the case of symmetric $S$, which it is) and apply the power needed
$$
S^{-\frac{1}{2}} = U \cdot s^{-\frac{1}{2}}  \cdot U^{\dagger}
$$
For more background, try the application section of Wikipedia's article on diagonalizable matrices.
As far as understanding it goes, I'm afraid not much beyond doing some model calculations on 2D or 3D vectors can be done.
I do not have a strong background in parallel computing, but all operations involved are well-known and your linear algebra framework (BLAS, anyone?) will document how to run any of this in parallel and what storage may be needed for it. Note that, typically, diagonalization is not the most expensive step of a self-consistent field calculation (Hartree-Fock or DFT) anyway, so parallelization may not be critical for this step.

Hi!
I know that this was posted a while ago, but I would appreciate your help.
How did you know that you diagonalize S as UsU^T?

@JunJang That is the basic property of a diagonalizable, symmetric matrix. I would not know what information to give you besides the linked wiki article or to point you to linear algebra textbooks.

I see. Will definitely look into wiki/textbooks. May I ask you one more question? If you were to explain the lowdin orthogonalization to a 5th grade, how would you explain it? I am working in finance and worked on a project using the lowdin orthogonalization and have to present it to management. I am using this to identify the underlying uncorrelated components of the factors and maintains the interpretations of the original factors.. How would you describe the orthogonal transformation and the associated diagonalization step in layman's terms? Thank you very much!!!!

Not sure that can be done easily, any decent explanation would have to resort to Taylor series. There are illustrations on how vectors are changed by orthogonalization, but I cannot find a good one.

would there be any way to explain the symmetric orthogonalization to someone without sufficient mathematics background?

Just as a curiosity, this strategy (u, s, vh = np.linalg.svd(S); A = u @ np.diag(s**(-0.5)) @ vh) is ~2x faster than pentavalentcarbon's, even though both approaches are very similar. (I tested for a 24 x 24 overlap matrix.)


text:
How do antioxidants work?
A friend and I are doing a project that involves the process behind manufacturing Jet A-1 and specifically the specifications highlighted in DEF 91-91. I was interested to see the information at the bottom of page 8.

Specifically:

Why are 2,6-ditertiary-butyl-phenol, 2,6 ditertiary-butyl-4-methyl-phenol and 2,4-dimethyl-6-tertiary-butyl-phenol used rather than "more common" antioxidants like ascorbic acid?
How do these antioxidants function and do they work any differently
to biochemical antioxidants?
What factors determine the effectiveness of an antioxidant?
Why are a mixture of chemicals used at different ratio's rather than just one. Do they work more/less effectively at different temperatures for example?

https://chemistry.stackexchange.com/questions/62729/what-makes-a-good-antioxidant

1) solubility. Typical liquid fuels consist entirely of hydrocarbons and are non-polar and thus cannot dissolve polar molecules like ascorbic acid.
2) The molecules mentioned in the provided quote can form stable radicals on loss of $\ce{OH}$ group hydrogen, i.e. act as a radical trap, transforming active radicals into non-active particles. It is a typical mechanism for antioxidants to work.
3) Strongly depends on the definition of effectiveness employed. Differen applications demand different properties.
4) Because in industry people typically do not bother with isolation of individual products if a mixture can work just as well. Sure, one can isolate individual products of phenol alkylation, but it is a tiresome work, so why bother, if the mixture works just fine?


text:
Chemical structure of xenon heptafluoride anion
When $\ce{XeF6}$ acts as a fluoride acceptor, $\ce{XeF7-}$ is formed. What is the hybridisation, geometry, and shape of $\ce{XeF7-}$?

Congratulations on your Yearling Badge!

About the geometry and shape:

Source: Housecroft, Sharpe, Inorganic Chemistry, Pearson Education Limited, 2005

As we know that XeF6 has a lone pair of electron then what happens to that lone pair of electron when XeF6 reacts with F-?


text:
Washing stainless steel separator from organic residue
In my lab we have a stainless steel separator for liquid products of one of our hi-pressure catalytic testing assembly. It is a steel cylinder with two short pipes and one another short pipe with a valve. The problem is, the separator cannot be disassembled and washed. It accumulated organic residues for unknown time, presumably years (it happened before it moved into our responsibility). Some residues were washed with acetone and other solvents, but we want it much cleaner.
Personally, I would love to threat it repeatedly with a strong oxidizer, but I'm afraid that stainless steel might disagree. Also, heating above 200-300 Celsius unfortunately is out of question.
The residue is from Fischer-Tropsch synthesis, so it probably consists of heavy hydrocarbons and possybly some oxygenates and products of their reactions.
Any suggestions?

Rather depends on the nature of the organic residues. Strong aq hydroxide soln?

Maybe sth that passivates steel?

@Waylander Added the infor

First try commercial cleaning solutions such as Alconox or Pierce RBS-35, with ultrasonic agitation (if you can sneak a transducer into the reactor).
Second, an oxidizer such as Godax Labs' Nochromix might work, but test it on an unimportant part of the reactor to see if it safe and effective. Read and understand its safety data sheet before use.
A last choice is chromic acid, $\ce{H2CrO4}$, which in some concentrations does not damage some types of stainless steel. N.B. Chromic acid is toxic, carcinogenic and harmful to the environment. With chlorides, it can produce toxic and volatile chromyl chloride. In high concentration it can cause ignition of materials. Check a material safety data sheet before trying it! Also prepare to dispose of this as hazardous waste.
Even if the reference states it is safe for the particular steel in use, try it on an unimportant area... perhaps with some of the organic "gunk", to see if the gunk is removed and the reactor is unharmed.


text:
Why is the excitation energy different from the energy difference between the orbitals involved in transition?
Why is the excitation energy different from the energy difference between the orbital from which a electron is excited to the orbital in which the electron is excited to? Consider that the transition mainly consists of the above specified 2 orbitals only. Do comment if I need to add additional information to the question.

Just look at the Hartree-Fock equations that define orbitals
$$
  \hat{F} \psi_{i}(1)
  =
  \varepsilon_{i} \psi_{i}(1) \, ,
  \quad i = 1, 2, \dotsc \, ,
$$
where $\hat{F}$, the Fock operator, is given by
$$
  \hat{F}
  =
  \hat{H}_{\mathrm{core}}
  +
  \sum\limits_{j=1}^{n}
  \big(
    \hat{J}_{j} - \hat{K}_{j}
  \big) \, .
$$
First, notice that summation in the series goes over the number of electrons, or, in other words, over all occupied orbitals. Secondly, note that $\hat{J}_{i} \psi_{i}(1) = \hat{K}_{i} \psi_{i}(1)$, and thus, the Fock operator for the $i$-th occupied orbital does not include the Coulomb and exchange contribution from the $i$-th occupied orbital itself. The so-called self-interaction is perfectly canceled in the HF model and this is a rather nice feature of this model.
It also means that an electron in each and every occupied orbital "feels" only $n - 1$ electrons in other occupied orbitals, while an electron in a virtual orbital "feels" all $n$ electrons in all occupied orbitals without any exceptions. As a result a virtual orbital is useful only for describing the system in which an electron was added to it but none of the occupied orbitals were depopulated, so that electron in the previously unoccupied orbital still "feels" all $n$ electrons.
Now think about exciting an electron from an occupied orbital $\psi_o$ to a virtual one $\psi_u$. Before excitation an electron in $\psi_u$ would "feel" all $n$ electrons, but after excitation it "feels" only $n - 1$ electrons since $\psi_u$ is not unoccupied anymore. That is clearly a contradiction. So, indeed, $\varepsilon_u - \varepsilon_o$ should not be used to calculate the excitation energy, because exciting an electron from $\psi_o$ to $\psi_u$ leaves behind a hole in $\psi_o$ and $\varepsilon_u$ describes the case when one puts an electron into $\psi_u$ but does not remove one from $\psi_o$.


text:
Reverse osmosis membrane used as an osmosis membrane
I recently came across the so-called 'thin-film composite membranes'. They are build up of layers with one of them being semipermeable. As I'm not well versed in chemistry, I'm unsure, if these membranes can be used for standard osmosis too, or if their permeability-property is reliant on the exerted pressure i.e., they allow the flow of water just in one direction.

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text:
Crystallization of zinc sulfate monohydrate
I'm currently running an electrolysis experiment and I have a few questions regarding crystallization of zinc sulfate monohydrate $\ce{ZnSO4 * H2O}$.
I have prepared a saturated solution of zinc sulfate, will it crystallize as monohydrate, or only heptahydrate $\ce{ZnSO4 * 7 H2O}$ will?
And if so, what is the reason behind it? Does it has to do with the monohydrate and heptahydrate part?

You need to crystallize monohydrate at elevated temperature, above 70 °C. Between 40 and 70 °C you obtain hexahydrate, below 40 °C – heptahydrate.

I have prepared a solution of ZnSO4.H2O (monohydrate) mixed with water.  I have saturated it enough I think. I saturated it and brought it to a boil. Now I'm waiting for it to cool without bothering it for a few days. Do you think after a few days crystals seeds will form? I have done it before with copper sulfate pentahydrate, the same process and it worked fine; crystals nucleated. I did the same process this time with ZnSO4.H2O. I'm hoping I'm on the right track.

There is no such thing as water solution of $\ce{ZnSO4.H2O}$. There is just water solution of $\ce{ZnSO4}$, period. You might have used solid $\ce{ZnSO4.H2O}$ to prepare it, but that's irrelevant. As soon as $\ce{ZnSO4}$ is dissolved, it has no idea that it comes from a monohydrate.

I'd use a warm water bath and a saturated at 75 - 80 °C solution.

@IvanNeretin From what I understood, it's about crystallizing the monohydrate, not keeping it in solution.

Yes, I'm trying to crystallize the monohydrate. I apologize for my english. If I leave the saturated solution of ZnSO4.H2O  for a few days without touching it, as I did with the CuSO4.5H2O, will it crystallize? Is there hope? or should I throw the solution away. Should I wait a few days?

I'm too lazy to search for a proper reference, but you can check German Wikipedia article (or Russian one) which says the same as my first comment.

I figure after a few days, ZnSO4.H2O solution will cool off (temperature drops and cools off to the desired temperature; 70 celcius) as you wrote, crystals might form. Otherwise I will restart again. Does this sound possible? Could? I brought the solution to extreme boil when saturating it.

No, you keep it above 70 °C as you grow the crystals. If you just let it cool off, you end up with heptahydrate. At least that's how I interpret that information.

@user50190 Yes, I got that. Your English is fine, but I must insist: there is no such thing as water solution of $\ce{ZnSO4.H2O}$, saturated or not. Ditto for any other hydrate. What will crystallize is determined by the temperature, as andselisk already explained.

TL; DR: obtaining a reasonably big single crystal of $\ce{ZnSO4 * H2O}$ seems to be a tricky task, whereas one can easily obtain $\ce{ZnSO4 * H2O}$ as a result of thermal dehydration of heptahydtare above $\pu{70 ^\circ C}$ in a form of melt, and above $\pu{120 ^\circ C}$ – as a polycrystalline solid material.

I finally got across to suitable references to back up my comments. In nature zinc sulfate monohydrate occurs as mineral gunningite, always in a form of efflorescence deposition on the surface of sphalerite and other minerals and doesn't form single crystals [1]:

Surface and ground waters carrying dissolved oxygen attack
  the sphalerite, yielding soluble sulphates of zinc [...]:
$$\ce{ZnS + 2 O2 -> ZnSO4}$$
[...] In the presence of abundant water all of the soluble salts are removed. If, however, the conditions are such that the supply of water is restricted or evaporation takes place, the metal-bearing solutions become supersaturated with the consequent precipitation of a number of hydrates of zinc, manganese, and iron of which gunningite, $\ce{ZnSO4 * H2O}$, is one.

Thermal dehydration among zinc sulfate hydrates has been studied with DTA in several publications. The most relevant and thorough investigation [2, p. 471] suggests the following scheme:
$$\ce{ZnSO4 * 7 H2O (s) ->[\pu{37 ^\circ C}] ZnSO4 * 6 H2O (s) ->[\pu{70 ^\circ C}] ZnSO4 * H2O (l) ->[\pu{120 ^\circ C}] ZnSO4 * H2O (s) ->[\pu{285 ^\circ C}] ZnSO4 (s)}$$
Original text in German:

Festgestellt werden vier endotherme Effekte: bei etwa $\pu{37 ^\circ C}$ verwandelt sich das Heptahydrat in Hexahydrat; bei etwa $\pu{70 ^\circ C}$ geht das Hexahydrat in Monohydrat über, wobei teilweise Schmelzen im Kristallwasser auftritt. Bei etwa $\pu{120 ^\circ C}$ siedet die Schmelze, Wasser scheidet
  sich ab, bis das feste Monohydrat zurückbleibt; bei etwa $\pu{285 ^\circ C}$ erfolgt der Übergang des Monohydrates in wasserfreies Salz.

The tricky part is that monohydrate can be isolated in crystal form, but it initially isolates as a melt, and only above $\pu{120 ^\circ C}$ makes a polycrystalline solid phase (occurs as white powder or granules).
Aqueous solubility $\omega_2$ for zinc sulfate $\ce{ZnSO4}$ in mass% of a solute ($\omega_2 = m_2/(m_1 + m_2)$, where $m_2$ is the mass of solute and $m_1$ the mass of water) as a function of temperature ($T$) from CRC Handbook [3, p. 5-172]:
\begin{array}{r|rrrrrrrr}
T,\,\pu{^\circ C} & 0 & 10 & 20 & 25 & 30 & 40 & 50 & 60 & 70 & 80 & 90 & 100 \\
\hline
\omega_2,~\% & 29.1 & 32.0 & 35.0 & 36.6 & 38.2 & 41.3 & 43.0 & 42.1 & 41.0 & 39.9 & 38.8 & 37.6
\end{array}
With this in mind, what can be suggested is to prepare a saturated at $\pu{70 ^\circ C}$ solution, and then slowly increase the temperature (e.g. using a thermostat), hoping that zinc sulfate monohydrate starts to crystallize (using the decrease in solubility starting from $\pu{60 ^\circ C}$).
According to [4, entry 3575, p. 470], solubility of $\ce{ZnSO4 * H2O}$ also decreases in this interval: $\pu{101 g}/\pu{100 g}~\ce{H2O}$ at $\pu{70 ^\circ C}$, and $\pu{87 g}/\pu{100 g}~\ce{H2O}$ at $\pu{105 ^\circ C}$. The difference is rather small, so I wouldn't expect rapidly-grown large crystals.
References

Jambor, J. L.; Boyle, R. W. The Canadian Mineralogist 1962, 7 (2), 209–218.
Balarew, C.; Trendafelov, D.; Gerganova, M. Monatshefte für Chemie 1971, 102 (2), 465–473. DOI 10.1007/BF00909340 (in German).
Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC handbook of chemistry and physics: a ready-reference book of chemical and physical data.; 2017; Vol. 97.
Perry, D. L. Handbook of inorganic compounds, 2nd ed.; Taylor & Francis: Boca Raton, 2011. ISBN 978-1-4398-1461-1.


text:
Can buffers regulate the pH, or do they just smooth out pH bumps?
Hydrogen carbonate can react with hydrogen ions to create carbonic acid which degrades into water and carbon dioxide that can be exhaled or excreted otherwise. That lowers the pH, yes. However, in order to create new hydrogen carbonate ions, the reverse pathway takes place, and pH is raised. So, this will help smooth out the bumps in the pH, but can it somehow regulate it long-term?

This question is better suited for chemistry.


text:
What is the correct unit for the equilibrium constant?
I have the following equation
$$K_a = \dfrac{c\alpha^2}{1-\alpha},$$
with units $c= \pu{mol/cm^3}$ and $\alpha$ is the same. I'm not sure how to get the correct units for $K_a$. What I have come up with so far is
$$K_a= \pu{mol^2/cm^6}.$$

This looks like Ostwald’s dilution law written for simple binary electrolyte $\ce{AB <=> A- + B+}$, so $[K_\mathrm{a}] = [C] = \pu{M}, (\pu{mol L-1}, \pu{mol m-3})$. For more complex dissociation you need to reconstruct the formula, so there is no unified unit for $K_\mathrm{a}$.

Ideally, you would write the equilibrium constant using normalised unitless concentrations so $K_\mathrm a$ has no unit at all. (If it had, you couldn’t logarithmatise it.)

Andselisk correctly identified the law of dilution and the name Ostwald is often connected with it.
$$K_\text{dissociation} = \frac{\alpha^2}{1-\alpha}\cdot c$$
However, the degree of dissociation is $\alpha$ and has "no" unit, i.e. dimensionless quantity. Therefore the unit for the equilibrium constant is that of a concentration, in SI that would be $\pu{mol m-3}$.
It is derived from the Law of mass action; the derivation can be found on Wikipedia and various other sources.

In principle there is no fixed unit for the (generalised) equilibrium constant, as it is simply defined as a product of the involved quantities according to the IUPAC gold book.

Equilibrium Constant
  Quantity characterizing the equilibrium of a chemical reaction and defined by an expression of the type $$K_x = \Pi_B x_B^{\nu_B},$$ where $\nu_B$ is the stoichiometric number of a reactant (negative) or product (positive) for the reaction and $x$ stands for a quantity which can be the equilibrium value either of pressure, fugacity, amount concentration, amount fraction, molality, relative activity or reciprocal absolute activity defining the pressure based, fugacity based, concentration based, amount fraction based, molality based, relative activity based or standard equilibrium constant (then denoted $K^\circ$ ), respectively.

The standard equilibrium constant is always dimensionless, as it is defined differently (IUPAC gold book).

Standard Equilibrium Constant $K$, $K^\circ$
  (Synonym: thermodynamic equilibrium constant)
  Quantity defined by $$K^\circ = \exp\left\{-\frac{\Delta_rG^\circ}{\mathcal{R}T}\right\}$$  where $\Delta_rG^\circ$  is the standard reaction Gibbs energy, $\mathcal{R}$  the gas constant and $T$  the thermodynamic temperature. Some chemists prefer the name thermodynamic equilibrium constant and the symbol $K$.

Good answer. (+1) One problem with the 'equilibrium constant': unlike the standard equilibrium constant it is not constant at fixed temperatures, and will vary with composition. Only in ideal approximations will the quantity $K_x$ be truly constant at equilibrium. In this way the naming 'equilibrium constant' may be a bit misleading. doi: 10.1351/pac199466030533 This is more of a comment on IUPAC's recommendation than your post.

By definition, the equilibrium constant has no units, as we're supposed to be using active masses instead of the molarity/ concentrations of the respective substances.
Active mass is a unit-less ratio of the mass reacting to the total mass present. Ideally, the equilibrium constant therefore doesn't have a unit. However, due to physical limitations, the exact active mass is difficult to calculate. Therefore, we use the concept of concentration instead, which is basically same as the active mass.
However, concentrations have units of $\pu{mol/cm3}$. This results in equilibrium constants having various different units depending on the equivalents of reactants in the balanced equation.
In your equation, the units that you calculated are perfectly right, but as far as I know, the $\alpha$ variable is a unit-less ratio, called the degree of dissociation, which is mathematically equal to:
$$\frac{\text{amount of substance dissociated}}{\text{total amount of substance present}}$$ in case of Ostwald's dilution law. The actual units would be, as andselisk said, just $\text{mol/cm$^3$}$.

There is absolutely nothing wrong with expressing the same thing in different words. Some people might prefer your version. Therefore I have edited your version and undeleted it. (I can because I am a moderator. :D) One point though: Please note that the proper term for "(number of) moles" is amount of substance. The former would be the same as referring to the mass as "(number of) kilograms".


text:
pH of amphiprotic species
I was asked to find the pH of a diprotic species $\ce{BHA}$ where $B$ is the cation and $\ce{HA}$ is the anion given the first and second dissociation constant of acid $\ce{H_2A}$ and the base dissociation constant .
My book gives answer as $$\ce{[H+]}=\sqrt{K_{a_1} \left(\frac{K_w}{K_b}-K_{a_2}\right)}$$
I tried writing down the expression for the first and second dissociation constant and the base dissociation.Then I wrote mass balance as
$$\ce{[HA-]_{initial}=[HA-] + [H2A] + [A^2-]}$$
However I couldn't proceed after this. Any ideas.

How many unknowns and how many equations do you have?

@IvanNeretin I have found out one more equation as $$\ce{H+}=[A^{2-}]-[HA^-]$$ However there are more unknowns and I couldn't solve them

My question was by no means rhetorical: how many unknowns and how many equations do you have?

You may have more luck breaking it down into the individual, sequential ionizations.   Keeping an the actual chemistry in mind is always useful.  As an aside, of the initial concentration is large enough, you can often disregard the contribution of the second ionization to the total pH of the solution.


text:
Effect of direct exposure to sunlight on various drugs
Today, we were discussing optical activity in class and our teacher said that some drugs are kept in dark coloured bottles to avoid direct exposure to sunlight. Same is the case with some other drugs in which it's written that the direct exposure to sunlight should be avoided.
The reason he gave for the above was that the one form of isomer can get converted to other form in direct sunlight.
But as far as i know, photolytic decomposition can happen. Is the conversion of d to l isomer and vice versa possible in sunlight?

Cis/trans isomerisation can occur.

I wonder if there is some confusion here about language.  Optical activity refers to the rotation of plan-polarized light, where one enantiomer rotates the light in one direction and the other rotates the light to the same degree in the opposite direction.  This is a separate phenomenon from photosensitivity, where a compound can decompose or isomerize (or otherwise be affected) by exposure to light.

@MichaelLauman - no, I was actually asking about the conversion of one enantiomer to the other form in presence of sunlight. Not exactly about photosensitivity or optical activity but just wanted to know if the enantiomeric conversion is possible or not(in sunlight's presence).

Conversion of D to L isomer or vice versa is about the last thing to worry about in this situation, even if it may be facilitated by sunlight to a significant extent (which I doubt). Also, storage of chemicals in dark coloured bottles is by no means limited to compounds that have optical isomers, or even to drugs in general. Other examples are numerous; permanganate solutions and nitric acid come to mind.
Photons of visible light have pretty decent energy of a few eV. That's way higher than the energy of thermal motion at room temperature. When a molecule is hit by a photon, it is hit hard. If a molecule is fragile or strained in any way, it will probably break. Photons come in droves, and after a while your precious organic sample turns into a brown goo. (It would probably do so even in the dark, but that would take a bit longer.)
That's what's wrong about sunlight in the first place.

So dextro form can be converted to laveo form or vice versa by exposure to sunlight?

Just to strengthen the answer. This is not specific to drug and conversion between stereoisomers is, IF it happens to be, the last to worry about.

@anamikaSingh I already said that in my answer, but if that's too long for you, here is a brief summary: no.

@IvanNeretin - i read the answer but at that time i got confused btw the dextro(d) and laveo(l) and the D and L system used for naming the optical isomers. As in ,i got confused in the representation.  Sorry for that.  Anyways, thanks for clearing up the matter!  =)

Isomerism is only one (and probably a minor) pathway.
Many substances have chromophores that absorb light. Absorption leads to a high energy intermediate that is more reactive and able to access degradation pathways that were unavailable from the ground state. This is different from thermal excitation because the energizing photon carries a lot more energy and directly promotes an electronic transition between two states.

@Zhe - Just asking, in photosystems PS-I and PS-II , the central 'chlorophyll a' molecule  can be called a chromophore?

Yeah, it absorbs light. It's still pretty destructive. If I remember correctly, PS-II has to be rebuilt every 20 minutes or something like that. @anamikaSingh

@Zhe- I Didn't know that they are regenerated at that rate! Thanks!

@anamikaSingh I think I heard Dan Nocera say that during a talk once, but that was a long time ago. Please double check before quoting.


text:
What chemical is safe for humans in low doses, and can be tracked by a satillite
I'm looking for a chemical that can be injected into the human body and be tracked by satellite that will also be safe to absorb and have no ill or lasting affects. Can you point me in the right direction?

I guess water will do.

If you think you can inject a small amount of an inert chemical into a specific human and then track them from orbit, you really haven't put any effort into the problem.

Haha, thanks for the laugh man, but tracking anyone from satellite isn't exactly easy and targeting specific chemical, identifying person from such distance is complete nonsense. Human's chemical signature is scent.

I am sure OP doesn’t want to use this to any evil plan

I think that no such chemical would exist.
Assuming passive detection
The chemical would need to decay and emit electromagnetic radiation. Human tissue would need to be transparent for that radiation. Looking at the electromagnetic spectrum, we are not transparent for large parts of it. One would go to low-frequency/low-energy (radio waves) or high-frequency/high-energy (gamma rays). The latter causes cancer. The former suffers from the fact that chemical/photochemical processes that release energy do so at higher, i.e. too high, energies (we emit infrared partly because that's our temperature, partly because that is what comes off some chemical reactions in our body).
Assuming active detection
So the satellite would emit some form of electromagnetic radiation, which would detect the chemical. The same problems as with passive detection apply. Nuclear magnetic resonance and electron spin resonance are the only spectroscopic techniques that use radiation in that region, and they require strong magnetic fields. In theory, you could use the earth's magnetic field instead (its not constant from place to place) - then you are at tiny energies and you still need to irradiate the whole earth, detect the tiny echo, which would have to be unique and would have awful signal/noise ratio.
So, yeah, I think that no such chemical would exist. You are better off implanting a radio transmitter in the subject's phone or head.


text:
How to tell between carbon powder or graphite powder?
I have a black powder that I cannot reliably tell just by looking or touching whether it is graphite powder or charcoal powder.
Are there any simple tests that I can perform to tell whether it is one or the other?
What I have tried:

Adding it to water, it sinks to the bottom and leaves a slight oily looking film on the top. Upon stirring, it darkens the water and then gradually sinks back to the bottom after some time.
I tried writing some thick lines on a piece of paper with a graphite pencil and testing for conductivity (high resistance).
Same high resistance with the powder (granted this is very unrepeatable)
Rubbing it into paper it has a much blacker colour than the graphite (which to me looks like a really dark gray). However, I don't know if that is a reliable indicator.

I've also added a picture, if it helps:

EDIT:
Ok, I've made a paste using isopropyl alcohol and the carbon powder and made a small "stick" or rod of the stuff. I have now been able to achieve solid continuity at around 2 centimeters in distance and the resistance dropped from the high 40 - 50 kOhms for the loose powder, to the low 10s (ie 40 Ohms). If this helps anyone.

Well, x-ray crystallography would work, but I suppose you're looking for some other way ;)

Besides, I'm inclined to believe that a given sample of charcoal is not 100% amorphous carbon, and that a given sample of graphite is not 100% crystalline either. So asking  whether this sample is "graphite" or "charcoal" poses other issues...

Yeah, I guess a question about the particular alotrope of carbon you have in your hand is perhaps much for "simple experiments".

No idea beside conductivity test . If you have access to Raman it would be rather simple to discriminate between mostly charcoal or mostly graphite.

Ok, when I test for conductivity, I get a good reading from the pencil led, max 30 Ohm. However, when I test for conductivity for the lines drawn on the paper, I don't get the beep since it's like 50kOhm resistance. Same for the powder, I get high resistance since it is not exactly a cohesive structure if you will.

Short answer: You can't reliably tell the "difference" (by any simple means)

Longer version:
First of all, the graphite you were taught about at school is a pedagogical construct. Any macroscopic sample of "real" graphite would contain both the layered honeycomb (crystalline) bits as well as amorphous bits of carbon (in addition to God-knows-what bunch of other substances that managed to wiggle their way into your sample).
What about charcoal then? It's amorphous carbon with significant quantities of several kinds of organic compounds (you can't expect them all to decompose to carbon...), depending on the source.
Trying to class your sample (which clearly has other stuff in it... I can see white/gray specks on it in the first picture ;P) as "graphite" or "charcoal" is an effort in futility; neither "real" graphite nor charcoal have a sharply defined constitution.
Simple tests of the physical nature of the powder would not be able to accurately distinguish it from being "mostly graphite" or "mostly charcoal", mainly, because it's a powder. Had you had the sample before it was pulverized you could (easily) tell the difference; the qualities that most people attribute to graphite (luster, slippery-ness) aren't easily discernible when you turn it into a powder and then compare it with powdered charcoal.

About the tests you conducted:
Your attempt to subject the sample to various (physical) tests is laudable (more users ought to do that before posting "Identify this thingy"-type questions here... it may give us something to work with). However those tests, apart from sorely lacking in some sort of mechanism/procedure to cut down on sources of error, don't really provide any meaningful data that you or me or anyone could use to determine what your sample is (largely) composed of.
Let's break down what you did

Adding it to water, it sinks to the bottom and leaves a slight oily looking film on the top. Upon stirring, it darkens the water and then gradually sinks back to the bottom after some time.

The film isn't really a giveaway. Even particles denser than water are known to "float" in these circumstances... well, they aren't technically "floating", it's the surface tension of the water that's holding it up. This is easily disrupted by stirring/agitation, and as you've noticed, it'll sink.
Graphites typically have have a much higher density (slight variation depending on purity and crystal defects) than water, so pieces of "graphite" would sink.
Blocks of charcoal typically have lower densities (varies depending on type/source of charcoal) than water and hence floats, but that changes when you pulverize them (and destroy most of the air cavities in the process).
Conclusion: No useful data.

I tried writing some thick lines on a piece of paper with a graphite pencil and testing for conductivity (high resistance).
Same high resistance with the powder.

That was a lot of effort... for nothing (you have my deepest sympathies).
Okay, I think I see your reasoning here:
"Graphite's pretty conductive (low-ish resistance) and charcoal is a poor conductor (very high resistance), so doesn't this test tell me something?"
Nope.
Thing about pure graphite (of the "Holy cow! A non-metal that conduct electricity!!!" fame) is that it only conducts (really well) along a particular plane; try another plane perpendicular to this and you'll find it to be a thousand times less conductive ( see answer to this question).
Even if your powder is (largely) graphite, surely the individual powder particles aren't going to align themselves into an extended plane between your terminals. They'll be oriented randomly, so expect the high resistance.
You first drew a line on a piece of paper using a pencil and measured its resistance to create a baseline to work up from. Smart. Just two things though:

1) Pencil lead isn't graphite (neither is it "lead" for that matter ;P); it's a mixture of graphite and clay. So unless you know for sure that the impurities in your sample are clay, this doesn't really serve as a good baseline. Besides, the graphite particles aren't all aligned in that plane I was talking about.
2) Your makeshift resistor ("thick line on a piece of paper") is a poor standard for another reason: dimensions. Your line has no uniform dimensions (length, breadth or even depth). If you need resistance from resistivity of a given material, or the other way around, you need the sample's dimensions. Neither your line nor your powder have fixed dimensions in this test.

Trying to figure out what your sample is by comparing its resistivity with that of another substance is a really neat idea, so don't get me wrong; the issue here is (1) the design of the experiment, (2) of the two substances you're trying to compare (graphite and charcoal), the latter's electrical properties are altered because it's been powdered and (3) your "control-test/sample" (the line-resistor) is neither graphite nor charcoal, making it a rather pointless "control sample".
Conclusion: No useful data.

Rubbing it into paper it has a much blacker colour than the graphite (which to me looks like a really dark gray). However, I don't know if that is a reliable indicator.

I don't need to tell you how poor a test this is (you already seem to doubt the validity of the test yourself)
Conslusion: No useful data.

EDIT: Ok, I've made a paste using Isopropyl Alcohol and the carbon powder and made a small "stick" or rod of the stuff. I have now been able to achieve solid continuity at around 2 centimeters in distance and the resistance dropped from the high 40 - 50 kOhms for the loose powder, to the low 10s (ie 40 Ohms). If this helps anyone.

Same argument as before: This test is pointless because your sample doesn't have fixed, uniform dimensions. Nor do the particles (if they're graphite; doesn't matter if it's charcoal) align themselves in extended planes.
As for the drop in resistance, I reckon it's because the sample isn't completely dry. The "isopropanol" you get at the drugstore is at least one-quarter water (not pure isopropanol). After you mixed it with your sample, a ton of electrolytes would've dissolved in the water (pretty sure your sample has loads of stuff that isn't carbon/ carbon-based). When you measured the resistance across this stick of yours, all that water hadn't evaporated by then. This is probably the (main) reason for the drop.
Conclusion: No useful data.

Interesting question by the way, you have my upvote ;)

Even bigger question: is there any point on the continuum where there is a clear dividing line between highly impure charcoal and relatively pure graphite when both exist as fine powders? I'm not sure there is.

@paracetamol - Thanks for all your input. I will definitely accept this answer as it does an awesome job of clarifying my muddied water. As for the isopropanol, I used tech grade (99.1%).

The first thing I would try would be to tie a length of thread to the ceiling and attach a non-magnetic sample holder to the free end. Put your sample in the holder and see how much you can deflect it from vertical with a strong pyramid magnet such as https://supermagnetman.com/collections/neodymium/products/pyr1000n . (If it's inappropriate to link to vendor sites please feel free to edit this out.)
Graphite is one of the more diamagnetic substances (although far short of superconductors), so it should deflect more than an equivalent mass of charcoal. For comparison you could get some pyrolytic graphite, a 1 in X 1 in X 6 mm block is about $50 I think, break (or cleave) a little chunk off and grind it up so you would have a good graphite standard for comparison.
They are fun toys to have around and worth it not just for the sake of this experiment.


text:
Ar atomic radius Vs. Mg2+, Al3+ and F-
I find it difficult to understand why the Atomic radius of $\ce{Ar}$ is lower than the ionic radius of $\ce{Mg^2+}$ and $\ce{F-}$, and higher than $\ce{Al^3+}$. According to the tables I've seen:
$$\ce{F- = 119 pm > Mg^2+ = 86 pm > Ar = 71 pm > Al^3+ = 68 pm}$$
Shouldn't Ar be larger than all three since it has a third electron shell, and they only have two? From what I was taught, the number of electron shells is more important then the number of protons in the atom. What differs so much between $\ce{Mg^2+}$ and $\ce{Al^3+}$ that $\ce{Ar}$ atomic radius is right between them?
Is there a way to predict this order, just by looking at the periodic table?

So-called ionic and atomic radii  aren't directly comparable, nor they give exact values  of sizes of anything.

I think I get that. though my professor did want me to find a way and compare between these 4 atoms, hence the question

related https://chemistry.stackexchange.com/questions/34555/why-are-there-drastic-changes-sometimes-in-radius-of-isoelectronic-ions https://chemistry.stackexchange.com/questions/66485/why-isnt-the-ionic-radius-of-mg2-bigger-than-f


text:
Hartree-Fock dipole moment
I read this sentence in a paper (J. Phys. Chem. A 2014, 118 (20), 3678–3687. DOI: 10.1021/jp502475e):

Calculations using Hartree–Fock theory systematically overestimated
  dipole moments and underestimate polarizabilities.

Is there any intuitive explanation that the HF approximation affects dipole moments and polarizabilities differently?

Perhaps this can be understood by considering that HF theory has exchange in it, so electrons of the same spin will tend to be separated. This leads to so called Fermi heaps. On the other hand, HF lacks Coulomb correlation, so these Fermi heaps will tend to be too large as Coulomb correlation will spread these charges out. So, there is a greater separation of charge which explains the overestimated dipoles. Then, to some extent the system would be "pre-polarized", so the system should be less polarizable. This is all very heuristic, however, so there may be more solid reasons.

I originally wrote this as a comment, but I will post it as an answer as I'm fairly certain this is most of the answer.
In general, there are two distinct types of electron correlation which can be described. Exchange correlation, which results from the indistinguishability of electrons, and Coulomb correlation which correlates the motions of electrons by considering individual repulsions between electrons.
Hartree-Fock (HF) theory accounts for the exchange interaction, but does not have Coulomb correlation as HF is a mean-field theory. That is, each electron experiences the average potential of the other electrons.
So, there is a sort of imbalance here from the beginning. It is known that the presence of this exchange interaction, which only happens for electrons of the same spin, results in two phenomena called Fermi holes and Fermi heaps. The references in that Wikipedia page are quite good. Essentially, a Fermi heap results when you have electrons of opposite spin. Because they must have a spatially symmetric wavefunction, the ends up being a greater probability of finding both electrons at the same location. This sounds silly, but read the references in the wikipedia page and you'll see what I mean. A fermi hole results when you have two electrons of the same spin. These must have an antisymmetric spatial wavefunction, so there is no probability of finding them in the same location.
If we think about these two effects in terms of electrical repulsion, the Fermi hole is a lower energy configuration than the Fermi heap. Because we almost always have both spin-up and spin-down electrons in multi-electron systems, both of these phenomena will take place, and the overall electron density will be some kind of average of all of these. (This is only really a conceptual framework.) If we take a simple case of a homonuclear diatomic, then these heaps and holes will all average out and we will get a symmetric electron density. If, on the other hand, we take a heteronuclear diatomic, then we see that there will be a tendency for these Fermi heaps to form over the atom with the larger nuclear charge just based on Coulombic attraction.
From here, I believe it is fairly clear that Coulomb correlation will tend to make the effect of these Fermi heaps smaller, and hence the overall separation of charge will decrease rather than increase, which explains why HF tends to overestimate dipole moments. By the same reasoning, the electron density is already polarized due to this heaping effect, so the system is not likely to be as polarizable as if this electron density were allowed to relax as is the case when Coulomb correlation is included.
This is all a very conceptual way of thinking about this, but I don't know that the math describing HF will be all that much more useful.

I've written an introductory bit about the quantum mechanics of holes and heaps, if it is useful to anybody reading this: https://chemistry.stackexchange.com/a/58652/16683

jheindel has given an interesting and valid perspective on the issue, but I believe that one does not necessarily have to go holes and heaps to understand the situation if configuration interaction (CI) will do.
The explanation that I have been given comes down to: restricted HF yields wave-functions that have a too high portion of ionic character. This ionic character also manifests itself in the dipole moments. The overly ionic character is known to cause the absurd dissociation curves of RHF. Note that unrestricted HF can partially repair this, at the cost of introducing spin contamination. At typical bond distances, UHF will collapse to the RHF solution anyway.
For an extensive discussion, I suggest: F. Jensen, Introduction to Computational Chemistry, Wiley, 2nd Edition (2007), Chapter 4.3: Illustrating how [Configuration Interaction] Accounts for Electron Correlation, and the RHF Dissociation Problem. Of that discussion, I will write up the key points here.
Let's assume the following model system: $\ce{H2}$, minimal basis set. RHF will yield the following ground state wave-function (and since it's RHF; this is true for all nucleus-nucleus distances):
$$
\Phi_0 = [\chi_A(1) + \chi_B(1)][\chi_A(2) + \chi_B(2)] \\
 = \underbrace{\chi_A(1)\chi_A(2) + \chi_B(1)\chi_B(2)}_{\mathrm{Ionic \ Part}} + \underbrace{\chi_A(1)\chi_B(2) + \chi_B(1)\chi_A(2)}_{\mathrm{Covalent \ Part}}
$$
$\chi$ are the AOs. Atom/AO labels $A,B$. Electron labels $1,2$, where given and hopefully obvious otherwise. This is after the spin functions have been integrated out/away and the LCAO ansatz has been used. The ionic part refers to a $\ce{H+}\ce{H-}/\ce{H-}\ce{H+}$ situation, whereas the covalent part looks like two hydrogen atoms next to each other.
Several other (excited) configurations exist, but only one of them is relevant (due to symmetry and spin considerations), which is:
$$
\Phi_1 = [\chi_A(1) - \chi_B(1)][\chi_A(2) - \chi_B(2)] \\
 = {\chi_A\chi_A + \chi_B\chi_B} - {\chi_A\chi_B - \chi_B\chi_A}
$$
They can be combined by linear combination (obeying normalization) as in:
$$
\Phi_{\mathrm{CI}} = c_0 \Phi_0 + c_1 \Phi_1
$$
which allows for the down-scaling of the ionic term as appropriate for the nucleus-nucleus distance.
Jensen goes on to note that the too-high dissociation energy of RHF manifests itself also in short bond lengths and high vibrational frequencies
because the potential hypersurface is distorted. The too-high ionic character finally leads to high dipole moments and atomic charges (as in Mulliken charges etc.).
Note: I am not bored enough for quantitative analysis yet.

I like this answer as I believe it says mathematically what my answer says conceptually. I've never been a big fan of the holes and heaps thing, but I believe the lack of Coulomb correlation explains also why the coefficient on the "ionic" term would be too large.


text:
Can you explain how we get gold(II) sulfide?
Can you explain how we get gold(II) sulfide, when gold has oxidation numbers of 1 and 3?

One solution would be if you have two gold atoms at +1 and +3, they average to +2.

Then again, it is probably closer to interstitial compounds like $\ce{Fe3C}$ or $\ce{WC}$ than to actual ionic sulfides like $\ce{CaS}$.

Brauer's Handbook of Preparative Inorganic Chemistry ([1, p. 1062] and references therein) lists the following method for gold(II) sulfide synthesis:

$$\ce{$\underset{1/10}{}$\qquad$\underset{242.7}{\ce{8 AuCl3}}$ + $\underset{20.2\,l.}{\ce{9 H2S}}$ + $\underset{7.2}{\ce{4 H2O}}$ = $\underset{183.3}{\ce{8 AuS}}$ + $\underset{87.5}{\ce{24 HCl}}$ + $\underset{9.8}{\ce{H2SO4}}$}$$
A neutral 1-3% solution of $\ce{AuCl3}$ is precipitated in the cold (the temperature must not exceed $\pu{40 ^\circ C}$) with $\ce{H2S}$ or an alkali sulfide. The precipitate is filtered, thoroughly washed with water, and
  treated with alcohol, anhydrous ether, $\ce{CS2}$ and again with ether.
  The product is dried at $\pu{130 ^\circ C}$ over $\ce{P2O5}$.
Alternate method: A solution of $\ce{AuCl3}$ is added in drops to an
  aqueous solution of sodium dithiosulfatoaurate(I) [...].
Properties: Formula weight 229.07. Deep black. Insoluble in water and acids; soluble in aqua regia and solutions of potassium cyanide and alkali polysulfides. Resistant to KOH in the cold, decomposes after prolonged boiling, liberating gold. Thermal decomposition begins at $\pu{140 ^\circ C}$.

Reference

Handbook of Preparative Inorganic Chemistry, 2nd ed.; Brauer, G., Ed.; Academic Press: New York; London, 1965; Vol. 2.

There is a bit of information about gold monosulfide($\ce{AuS}$) that can be found here:

Hydrogen sulphide or an alkali-metal sulphide precipitates the
  monosulphide from solutions of auric chloride, and it is also produced
  by the action of hydrogen sulphide on sodium aurothiosulphate. It is a
  black substance, insoluble in acids except aqua regia, but soluble in
  alkali-metal sulphides.


text:
Reusable and safe endothermic reaction
I am looking for a reusable endothermic reaction that is safe to both touch and handle, it needs to be able to be stored in a jacket and light enough to do physical activity. Any suggestions?

https://chemistry.stackexchange.com/questions/6568/homemade-reactor-for-water-cooling

There are many endothermic physical processes such as melting ice or sublimation of solid carbon dioxide $\ce{CO2}$. One should also mention thermoelectric cooling effect (Peltier device), which is is rarely effective in practice, yet still an option.
A suitable endothermic chemical reaction would probably be an electrolytic dissociation of certain salts, among which dissolving ammonium nitrate $\ce{NH4NO3}$ in water is probably one of the most well-known and effective examples.

@Jan I like to see the interaction between the solvent and the compound being dissolved as mostly of a chemical nature, but yes, there is no there is no clear line inbetween. Solvation ability is guided by the nature of the chemical bonds, and the solvent's dielectric constant is also related to its composition and chemical properties.


text:
Identification of an aromatic compound for a quiz
I need help identifying this structure for a puzzle. Original text was blurry, so a white overlay was added for readability, though some fragments might be different. Any help or suggestions are appreciated.

I suppose the SO3 groups make it water-soluble. Does not really look like a dye (yet could be), I would rather bet on pharmaceutical.

I would expect neither Ne (Na only), nor 8 (3 only) for the marked fragments on the given structure. Both groups are most likely just sodium sulfonate $\ce{-SO3^-Na+}$ fragments. The name generated by MarvinSketch for the given structure is disodium 4‐(3‐carbamoyl‐5‐acetamidobenzamido)benzene‐1,3‐disulfonate:

using this in a substructure search in chemspider gives this neat looking thing : http://www.chemspider.com/Chemical-Structure.9564201.html I guess this image could be part of that.

@gilleain Good find, I guess your comment might as well be an answer:)


text:
Weight percentage of a component in gaseous mixture

Given a mixture of $\ce{He}$ and $\ce{O2}$ with density $\pu{0.47 g/L}$, temperature $\pu{298.15 K}$ and pressure $\pu{0.948 atm}$. What is the weight % of $\ce{He}$?

I get stuck in the formula
$$\frac{m(\ce{He})}{m_\mathrm{tot}} = \frac{p(\ce{He})}{p_\mathrm{tot}} \cdot \frac{4}{32}$$
I don't seem to have enough information to calculate $p(\ce{He})$.

@Vedant The homework tag is deprecated, please do not use it.

There is no lack of data. Keep in mind that usually you are allowed to approximate a given gas mixture using ideal gas law:
$$pV = nRT = \frac{mRT}{M} \tag{1}\label{eq:1}$$
On the other hand, density by definition is
$$\rho = \frac{m}{V} \tag{2}\label{eq:2}$$
Combining \eqref{eq:1} and \eqref{eq:2}, and applying the equation to a binary mixture of gases ($\bar{\rho}$ and $\bar{M}$ are average density and average molecular weight, respectively):
$$\bar{\rho} = \frac{m_\mathrm{tot}}{V_\mathrm{tot}} = \frac{p_\mathrm{tot} \bar{M}}{RT} \tag{3}\label{eq:3}$$
From \eqref{eq:3} one can determine average molecular weight of the mixture. Remember to use SI units: $\bar{\rho} = \pu{0.47 g L-1} = \pu{0.47 kg m-3}$; $p_\mathrm{tot} = \pu{0.948 atm} = \pu{96056 Pa}$:
\begin{align}
\bar{M} &= \frac{\bar{\rho} RT}{p_\mathrm{tot}} \\ &= \frac{\pu{0.74 kg m-3} \cdot \pu{8.314 J mol-1 K-1} \cdot \pu{298.15 K}}{\pu{96056 Pa}} \\ &= \pu{1.213e-2 kg mol-1} \\ &= \pu{12.13 g mol-1} \tag{4}\label{eq:4}
\end{align}
At the same time average molecular weight of a binary mixture is
$$\bar{M} = \omega_1 M_1 + (1 - \omega_1)M_2 \tag{5}\label{eq:5}$$
Equating \eqref{eq:4} and \eqref{eq:5}, and solving the obtained equation for $\omega_1$ gives the answer ($M_1 = M(\ce{He}) = \pu{4.00 g mol-1}$; $M_2 = M(\ce{O2}) = \pu{31.99 g mol-1}$).

 $$4.00\omega_1 + 31.99 - 31.99\omega_1 = 12.13 \implies \omega_1 = 0.71\, \text{or}\,71\%$$

Pro tip: $$a=b\tag{a}\label{eq:a}$$ [...] $\eqref{eq:a}$ Only automatic enumeration does not work. $$a=b\tag{a}\label{eq:a}$$ See equation $\eqref{eq:a}$.


text:
Phases of Atomic or Molecular Orbitals
What does phase mean in orbitals?
I know that phases are separated by nodes. They are in some way related to wavefunctions, I can't understand, how? How can wavefunctions be negative, since they are complex numbers (as I know)?

Phase doesn't mean much; the difference in it does. BTW, I vaguely remember hearing that the eigenfunctions for a bounded state can always be chosen so as not to be complex.

This post of mine may shed some light on your question: http://www.sciencemadness.org/talk/viewthread.php?tid=62973&page=10#pid419076

Complex Numbers
As you point out, general a wavefunction if complex-valued.  A complex number and be described by two real numbers and hence are of depicted on the 2D complex plane (sometimes called the argand plane).
Numbers in the complex plane can be expressed in multiple forms.  The one usually taught first is as the sum of a real part and imaginary part:
$$\begin{align} z = x + iy\;{} & ; x,y \in \mathbb{R}\\
\Re(z) = x\; {} & ; \Im(z) = y \\
|z| = {} & \sqrt{x^2+y^2}
\end{align}$$
Alternatively it can be described by a magnitude ($r$) and a phase ($\phi$):
$$\begin{align}
z = r e^{i\phi}\; {} & ;r\in\mathbb{R}^{+0}, \phi\in[0:2\pi)\\
\Re(z) = r \cos(\phi) \; {} &; \Im(z) = r \sin(\phi) \\
|z|{} & = r
\end{align}$$

Two signals or functions with the same magnitude are said to be perfectly/exactly 'out of phase' if there phases differ by $\pi$ as $e^{i \pi}=-1, and perfectly 'in phase' if they have the same phase.
Orbitals
Now we can apply this to some orbitals.  Consider a cross-section of the 2p$_{\text{x}}$ orbital in the $x$-$y$ plane through $z=0$:
$$\psi_{2px}(x,y,z) = r(x,y,z) e^{i \phi(x,y,z)} \propto x \exp\left(-\sqrt{x^2+y^2+z^2}\right)$$

The intensity of the colour represents magnitude, $r(x,y,z)$, of the wavefunction and the colour itself represents the phase - red for $\phi=0$, blue for $\phi=\pi$, with a colourless node separating the two lobes.
As the probability density is equal to the square modulus of the wavefunction: $$\rho(x,y,z) = |\psi(x,y,z)|^2$$
we can multiply the wavefunction by a constant phase factor $e^{i\theta}$ and still represent the same particle:
Consider:
$$\begin{align}
\tilde{\psi} {} & = \psi e^{i\theta} \\
\to \tilde{\rho} {} & = |\tilde{\psi}|^2 \\
& = |\psi|^2 e^{i\theta} e^{-i\theta} \\
& = |\psi|^2 \\
& = \rho
\end{align}$$
So both $\psi$ and $\tilde{\psi}$ are valid wavefunctions with either one is, and the wavefunction can only be defined up to multiplicative phase factor, so we can talk about absolute phase, but we can talk about relative phase and phase differences.
Consider the out of phase 2p$_{\text{x}}$ orbital:
$$\tilde{\psi}_{2px}(x,y,z) = \psi_{2px}(x,y,z)e^{i\pi} = - \psi_{2px}(x,y,z) $$

Molecular Orbitals
As we just showed they both have the same probability density, but when we begin to combine them, the phases dictate how much the wavefunctions constructively/destructively interfere.  We can combine linear combinations of atomic orbitals to make molecular orbitals.
Consider adding together two adjacent wavefunctions with the same phase on different atoms:

The adjacent regions of different phase cancel one another out.  When the probability density is examined:

we can see than the chance of the particle being between the two atoms is very low and it is localised around each nuclei.  This would be an anti-bonding molecular orbital.
On the other hand, adding together two adjacent wavefunctions with the opposite phase on different atoms:

The adjacent regions of similar phase add together.  When the probability density is examined:

we can see than the chance of the particle being between the two atoms is very much higher and it is localised in a bond.  This would be a bonding molecular orbital.
This is why we talk about in phase and out of phase orbital when talking about bonding - the regions of in phase overlap of wavefunctions increases the probability that a particle will be found in there, and out of phase overlap decreases the probability.
The physical result is the wide variety of bonding phenomena we see in nature

covalent bonds: local in phase overlap
anti-bonding orbitals: local out of phase overlap
conjugation: extended in phase overlap etc...


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Why does hydrochloric acid not damage the human stomach?
Hydrochloric acid ($\ce{HCl}$) is found in the stomach, so why it doesn't damage the stomach despite that it is considered to be one of the most dangerous acids?

Short answer: the stomach has a lining and is constantly renewing it.

Of course it does. How do you think stomach ulcer and gastritis come about? (Sure, there are other reasons and counter-reasons.)

This question already has an answer here on Bio.SE: https://biology.stackexchange.com/questions/6820/how-does-hcl-not-burn-our-stomach

I'm closing this question as off-topic because it is not fundamentally about chemistry.

I am voting to reopen this question, because it does address chemistry, even though it is chemistry in living species.


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Group 15 Ionisation Energies
I had a little confusion regarding ionisation energies.
I get that in general, the second ionisation energy > first ionisation energy, due to increased effective nuclear charge on the second electron.
However, in the case of an element of group 15, say phosphorus, shouldn't it be the other way around because of the half-filled orbital configuration ([Ne]3s23p3, for phosphorus)? Won't it be harder to ionise the first electron because of increased stability?

You're right in saying that the ionisation energy of phosphorus should be greater than that of sulphur due to its half-filled p subshell. This is one of the many exceptions... There's also the second group alkaline earth metals whose ionisation energy is greater compared to that of the thirteenth group elements, due to a fully filled s subshell...

Periodic trends are just that, trends.   They give you a general idea of what to expect, but each case needs to be examined individually.

This is a common misconception if you don’t think through the concept of ionisation energy enough. By your proposal, the second ionisation energy of certain elements (e.g. phosphorus) should be lower than the first. But that is never the case; all second ionisation energies are higher than the first.
The first ionisation energy corresponds to the following reaction:
$$\ce{P(g) -> P+ (g) + e-}\tag{1}$$
The second ionisation energy corresponds to this reaction:
$$\ce{P+(g) -> P^2+ (g) + e-}\tag{2}$$
It is very simple to see why the second ionisation energy must be greater than the first. While in the first, you are separating a positive and a negative charge, which already requires force (and thus energy), when moving on to the second ionisation you are removing a negative charge from an already positively charged body. Since electromagnetic attraction depends on the value of the charge, the second electron is drawn to the doubly-ionised ion twice as strongly as the first.
Therefore, there is no element whose second ionisation energy is lower than the first.


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Why does orthocresol behave as a skin irritant?
o-Cresol or ortho-Cresol or 2-methylphenol is a constituent of tobacco smoke, and, like most other phenols, behaves as a skin irritant, causing a burning sensation when it comes in direct contact, or upon inhaling it in.
What is the biochemical reaction that causes our skin to produce such a reaction to orthocresol? Is there any correlation with orthocresol being an acidic compound? Can this be explained by +I effect or Hyperconjugation effect of the $\ce{-CH3}$ group?
Unlike Vinegar or Alcohol, which have different effects (vinegar feels like water to me, and alcohol has a cooling sensation), why do cresols and phenols in general have a stingy sensation on touching?

There's high probability that any random organic liquid would be irritant, that's nothing special at all. Check out specific data and compare with others.

It is the phenol moiety of ortho-cresol that causes this. Phenol is classified as toxic because of the way it can penetrate the skin and cause damage.
In its neutral state, the phenol molecule is a slighlty polar but generally organic compound. Since our skin is designed to mainly keep water and water-soluble compounds out, it is lipophilic; this in turn helps lipophilic compounds to penetrate it and be absorbed by the body. As soon as phenol has been absorbed, it is again in an aquaeous environment. Here, phenol’s acidic properties ($\mathrm pK_\mathrm a = 9.95$) start playing a role: it can be deprotonated to the phenolate anion. This is hydrophilic and so will be able to dissolve in the water inside your body. If any membrane needs to be crossed, just add a proton to regenerate the lipophilic free phenol.
Another way in which phenol can often react is the single electron transfer mechanism to generate a phenol radical which can then cause havoc in cells. And finally, phenol should also be able to intercalate into DNA further causing damage.
So the stingy sensation is caused by the slight burns that you encounter once phenol has penetrated your skin. since neither ethanol nor vinegar can penetrate your skin as well as phenols can (they are too polar), these react differently and give you a different sensation.


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Why does a protein crystallize with a specific salt but not with another?
So I am conducting an experiment and I do not understand this super fundamental part.
I know salting out works via solubility, and I know crystallization works via ordered structures. But could someone explain to my why one salt allows a crystal to form, but another salt doesn't. A component of one of my salts (Mg) has been found to be important to my protein in optimization and activity, but why would another cation with the same anion (I know the anion is not the problem) discourage the formation of crystals. If this is a clearly stupid question I apologize.
EDIT:looking for theory. I have already crystallized everything, and have CRYSTALS. Will be performing x-ray crystallography.This post was more to get me on the right track in thinking.

I'm not getting it, clarify your question.

why would a salt, such as NaCl allow crystallization with a protein, but another salt like CaCl have no crystal formed. If both are salts playing on the solubility of the protein, and interacting more or less the same, what is the difference. @Mithoron

Precipitation isn't equal crystallisation. Good crystals of proteins are often very difficult to make. There may be specific number of Mg cations usually bound to specific protein, but it has not much of a connection with what happens if solution make the protein bind much more of them. I think you should read much more about this. https://en.wikipedia.org/wiki/Protein_precipitation may be a start.

I understand this; however, I have already formed the crystals and photographed them and will be using them in xray crystallography this weekend. They are not just in phase change or precipitates, but rather beautiful large crystals. My question was more on theory, as to why one of my plates has crystals using one salt at different concentrations and the other (with a different salt) has nothing

That's better, edit this into your question, still you should know more about this before experiment. It may save you much time. As I mentioned, it's not just salting out, and even if it was, MgCl2 solution would have different ionic strength and pH then say, of MgHPO4. It would be good if you mentioned what was this not unimportant counterion.

sorry I was trying to be as vague as possible intentionally to understand this amongst all salts but okay I am using 3 chloride salts: NaCl, MgCl2 and CaCl2 and three sulfate salts: NaSO4, MgSO4, and CaSO4. But you did just make me realize I needed to consider the pH of the SALTS on the protein(and not just my buffers which were sodium phosphate, and acetic acid), I don't know why that escaped my mind;however still fails to explain why one nacl and mgcl2(both at pH 7 have two different effects) And yes I realize now I should have thought more about this before... @Mithoron

It would be a lie to postulate we knew much about the formation of protein crystals. Yes, we know the basic theory — that we need a supersaturated solution in which the protein is not denatured and which would then slowly evaporate until hopefully a single crystal starts growing — but we don’t have the slightest of clues of what exactly helps and what doesn’t help.
This is why protein crystallography is still trial and error (and nothing else). You try thousands of different conditions in an automated manner, select those with promising results for further examination and finally — if and only if you are lucky — end up with crystals suitable for X-ray diffraction.
Therefore, we don’t know and nobody knows why magnesium chloride might help the crystallisation of your protein but sodium chloride, calcium chloride, aluminium chloride or magnesium bromide don’t.
We don’t know and nobody knows why a $\pu{1.3M}$ solution of magnesium chloride might help the crystallisation but a $\pu{1.4M}$ or a $\pu{1.5M}$ solution don’t.
We don’t know and nobody knows why $\pu{23^\circ C}$ might be a good temperature for crystallisation but $\pu{21^\circ C}$ and $\pu{25^\circ C}$ aren’t.
And so on and so forth.

Thank you so much! I was thinking that this was known based on how my lab coordinator was asking me questions.

Well, some things are known, and this rather looks like a extended ranty comment then proper answer. Even https://en.wikipedia.org/wiki/Crystallization_adjutant shows it's not exactly mysterious magic.

@Mithoron Please state where in that article you read something about a known mechanism with the one exception of micellar substances to stabilise membrane protein crystals.


text:
Is this definition of the verb oxidize correct?
UPDATED VERSION:
There are a dozen definitions of the verb oxidize in various dictionaries. My question is whether the definition in the end of this post holds true. I have added the words in bold.
And if we think of oxidation as "electron theft", would the two pairs of three sentences match as to who's getting "robbed"?
Oxygen oxidizes Sodium.
Sodium is oxidized by Oxygen.
Sodium oxidizes (intransitive).
Oliver robs Sophie.
Sophie is robbed by Oliver.
Sophie robs.
"Undergo or cause to undergo a reaction in which electrons are lost to oxygen or another species"

I'm voting to close this question as off-topic because the definition is actually quite clear; the problem may be more likely to be solved on [ELL.SE]

Found this sort of unclear use in French chem slang too. Where the equivalent of oxidize may refer to what has been oxidized or to what indeed oxidizes a reducent species.

Why are you trying to learn chemistry from a dictionary? Anyway from the language point this is perfectly clear and also chemically correct. When I say "Fred made Joe hit Bill", you also know who is hurt, right? Don't let yourself be confused by the chemical verb. It's still english, not chemlish.

@Alchimista, exactly! When the dictionary definition says that to "oxidize" is to undergo a reaction where electrons are lost to another species, it doesn't say who's doing the losing. It could very well be oxygen that comes in contact with another species and does the losing.

@Karl. The dictionary definion left out who's doing what. I was wondering if the definition would still be correct if I inserted the part in bold? If you say "Fred made Joe hit Bill" you've introduced three characters. I wish there were "charcters" in the dictinoary definition.

@Gelb No it doesn't, please read properly. Definition of to cut: split into pieces by use of a knive or scissors. Get my point? You don't need paper or string or carrot to define "to cut". And yes, the other thing, that "looses electrons to oxygen or another species" could also be another oxygen. Oxygen oxidises an oxide to peroxide for example.

@Gelb "I was wondering if the definition would still be correct if I inserted the part in bold?" Why didn't you say so in your original question?

@Karl. Thanks. Cutting is easy to visualize. Let's say Joe cut Bill in the face; no doubt who's doing what. But if ordinary oxygen and sodium atoms get together and oxidize, by the words it sounds like oxidize belongs to oxygen. Alas, I see "curse of knowledge" made me not emphasize what you quoted in the original question. I see it would've been much clearer that way.

In everyday language "oxidation" is used to describe a reaction with oxygen.
This definition, however, is not scientifically correct.
The more correct definition would be "A reaction in which an atom loses an electron" or "A reaction in which the oxidation state of an atom increases".
This means that in a reaction between hydrogen and chloric gas, which results in HCl (hydrogen chloride) is also an oxidation, because the hydrogen in HCl has a partial positive charge it did not have before, which counts as "losing an electron". This means that an oxidation does not require oxygen to be involved.
You might consider looking up how to determine the oxidation state of an atom because it makes deciding if an oxidation takes place easier.

Thanks, GreenSmurf. So, the definition I suggested was correct. When oxidation occurs then electrons can be lost to oxygen, but they needn't be. So "lost to oxygen or another species" would we correct.


text:
Final Temperature of Water

The problem statement, all variables and given/known data:

A certain mass m of water in a copper pot is heated over a stove that releases  Q joules of heat to heat the water. Later the same mass m of water in an aluminum pot is heated over another stove that releases the same amount of heat  Q in order to heat the water. Then the water in which pot will have the higher final equilibrium temperature?

Relevant equations:
Q = mc∆T
The attempt at a solution:
I think the final temperature of the water in the metal pot with the least specific heat will be higher because that pot can transfer heat more easily. Is this correct?

(Edit) I believe you heat both of the pots for the same amount of time (not long enough to boil) and then stop the stove. Then wait until the equilibrium temperature for the metal and water is reached and compare it for both of them.

Well it depends on when you stop your experiment. In both cases water will boil at the same temperature but it will not take the same time. But here final as no sense, you need to be more precise.

I believe you heat both of the pots for the same amount of time (not long enough to boil) and then stop the stove. Then wait until the equilibrium temperature for the metal and water is reached and compare it for both of them.

@Hexacoordinate-C "Final" means "when both plates have given off the amount of heat Q". That's what the problem states, can't get any more precise than that.

You are right of course, but not perfectly in your reasoning.
The transfer of heat is not the point, because the question already says it is $Q$ in both cases, not $0.9 Q$ in one and $1.2 Q$ in the other. Pot and water have the same temperature ("equillibrium"). That's of course nonsense (bottom of pot would obviously be hotter than the rest in reality) but you have to stick to the made-up facts your teacher gave you.
Both plates have given off the same amount of heat(energy), so the pot with less heat capacity is hotter.
Note: This is a Gedankenexperiment. Don't make the error to think up an actual setup with a stopwatch, thermometer and power meter. For example the plate would also give off heat to it's stand and surrounding air, how could you measure that?


text:
Is this model to explain children about oxidation and fire correct?
I am teaching two young boys (elementary school) some basics about chemistry, but I have to learn as I go myself.
Is this a viable model to explain oxidation and fire?
When we light up the fireplace, oxygen "steals" electrons from the wood, which is carbon. But carbon won't let oxygen just leave with its electron(s), so this theft results in them clinging together, and we call that Co2.
At the moment they cling together, their tight grip gives off heat. We call this an exothermic reaction. This word may not be easy to remember, but think of "exhaust pipe" (going out, so "exo") and "thermos" (it stores something hot). So when you hear exothermic reaction you know it's a reaction that gives off heat.
And how much energy is required to get oxygen and carbon to let go of their tight grip? Exactly the same amout of thermic energy they gave off when they first clung together!

Why don you need that much to talk about electron in the first place? We had basic chemistry in elementary school, and no one was particularly puzzled by the basic concepts.

This interview with Feynman pretty much does the job.  Since I've got nothing else to add to it, I've posted it here as a comment (instead of a full-fledged answer). O:)

@Greg. I've purchased a chemistry set for kids and I am going over everything with them. To get through it the concept of the electron must be introduced.

@paracetamol. When you say you have nothing else to add, I'm assuming there were no blatant errors in my model? I just made it up, and I'm a complete neophyte in chemistry. Thanks a lot for the video!

related https://chemistry.stackexchange.com/questions/1254/what-are-the-chemical-reactions-behind-fire

@Gelb You don’t need to introduce all the concepts, electrons, oxidation, bonds, heat etc in the same time just because it was said in a box

I feel it's not exactly important if it's "correct". Everything in our description of real world is just a model, illusion that we truly understand, how things work. More to the point, there no "electron grabbing" and there's no reason to approximate wood with pure carbon, if you can simply tell about burning pure carbon...

@Gelb Ah no, I meant I had nothing else to add to the video. Also, I'd go with what Greg said: You don't have to introduce everything to them at once... take it slow ;) [As for your "model", nope, no blatant errors]

There is no need to even mention esothermic

@paracetamol. Thanks a million. You also gave me some outstanding entertainment on YouTube.

@Mithoron. That's true, burning pure carbon would be easier. But I will talk about it when we put logs of wood in the fireplace. If I had some coal lying around, I suppose we would have used that. But when you say "there no "electron gramming"", what you do mean by that Mithoron? Is there are a blatant error in the model after all? I like the proverb "a model is a lie that helps you see the truth", but I don't want to tell blatant lies to the kids.

Oh, well... maybe you could put it like that, but it's quite metaphorical. Also as wood is made of carbohydrates (besides other stuff) you could say that carbon is oxidated and water "liberated" from structure.

On the other hand it seems to be the clue of your other question, grabbing, losing, stealing etc. might be tolerable as metaphor, but are no good  for defining this process.

@Mithoron It was Karl who told me in the comments that oxidize behaves like other verbs, and layed out the example "Joe hit Bill". So therefore I just updated the oxidize question. But I am still none-the-wiser with regards to how the verb "oxidize" behaves. I thought the chemistry forum knew more about grammar than the grammar forum knew about chemistry. So, it would be here the expertise was. Alchimista told he was a bit befuddled by it himself in French. Anyway, thanks for your comments. They have made a big difference.


text:
Crystal field splitting in complexes other than octahedral or tetrahedral
Is there a methodology to developing the crystal field splitting diagram for various crystal fields?  I know the octahedral and tetrahedral diagrams, but if I'm asked to draw something like square pyramidal, trigonal prismatic, etc. that I haven't memorized, then is there a process I can go through?  I searched Google, but am having little luck finding anything that shows the process.
Thank you!

If you know the process to create the octahedral and tetrahedral crystal fields, why can’t you apply that same process to other fields?

I don't know the process, I just have them memorized.  Memorization without the process is quite honestly worthless at this point.

Yes, memorisation without knowing the process is always worthless; especially in this case where memorising requires a lot of brain capacity to do while deriving would be a very simple job.

Yes, the methodology is the same that is used to derive the crystal field theory a priori. Your course should have taught you that when it introduced crystal field theory, otherwise it is not worth its money.
Crystal field theory assumes that the ligands will approach the central metal in a certain manner and that these ligands will be point-shaped negative charges. Before the ligands approach, all orbitals of the metal’s same subshell will be degenerate, i.e. have the same energy. Because these orbitals have an orientation in space (e.g. the $\mathrm d_{z^2}$ orbital can be seen as ‘pointing’ its lobes in the direction of all the coordinate axes), the approach of the ligands will cause interactions between them and the negatively charged electrons that make up the orbitals. Therefore, any orbital that is pointed towards the approaching ligand will be destabilised (and all others stabilised to a certain degree).
In the octahedral case, the ligands approach from $\pm x,\pm y$ and $\pm z$. I have already pointed out that the $\mathrm d_{z^2}$ orbital would be pointing towards them; the $\mathrm d_{x^2-y^2}$ orbital does, too. Therefore, these two are destabilised while the other three orbitals $\mathrm d_{xy}, \mathrm d_{xz}$ and $\mathrm d_{yz}$ are stabilised.
Applying this general principle to other geometries is left as an exercise to the reader.

Thank you!  I went back through my notes and my professor never did explain how to derive the splitting diagrams, but between you and a textbook I found, I was able to learn.  Many thanks!


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Using Avogadro's law to find gas volumes

An important step in the preparation of sulfuric acid involves the production of sulfur trioxide ($\ce{SO3}$) from sulfur dioxide ($\ce{SO2}$) and oxygen.
$$\ce{2 SO2 (g) + O2 (g) -> 2 SO3 (g)}$$
If all volumes are measured at the same temperature and pressure, calculate the volume of sulfur trioxide produced when $\pu{25 mL}$ of sulfur dioxide is placed in a container with $\pu{8 mL}$ of oxygen. What will be the total volume of all gases remaining in the container?

I seem to keep getting the wrong answer for the question, and I'm not totally sure why. I assumed I needed to use Avogadro's law, given that temperature and pressure are constant. So I used:
$$\frac{V_1}{n_1} = \frac{V_2}{n_2}$$
I determined $V_1$, $n_1$ the following way:
$$V_1 = \pu{8 mL} + \pu{25 mL} = \pu{33 mL}$$
$$n_1 = \pu{2 mol} + \pu{1 mol} = \pu{3 mol}$$
From the equation I assumed $n_2 = \pu{2 mol}$. This gave me
$$\frac{\pu{33 mL}}{\pu{3 mol}} = \frac{V_2}{\pu{2 mol}} \implies V_2 = \pu{22 mL}$$
The provided answer says $\pu{16 mL}$, and I am just unsure of how to alter the method.

Here's a hint: if I increased the volume of sulfur dioxide to 1000mL, but kept the amount of oxygen the same (8mL), would you expect there to be much more sulfur trioxide produced? Why or why not?

@chipbuster I would assume that not much more would be produced, because it would be limited by the smaller amount of oxygen??

@uyujc Absolutely, so why don't you determine the limiting reagent first, and then apply the same law again? $$\frac{V_i}{n_i} = \text{const}$$

@andselisk Ah! Thank you!! I got the first bit. Would you be able to help with how to go about finding the total volume of all gases remaining?? I know there would be 16 mL of sulfur trioxide (produced from the reaction), and would it be right to assume all the oxygen has been used up??

@uyujc Well, assuming the reaction is complete and irreversible,  in the system you have some leftovers of sulfur dioxide $\ce{SO2}$ and the product, sulfur trioxide $\ce{SO3}$ (as you mentioned, all oxygen is used). All you have to do is to find how much $\ce{SO2}$ is left and add $\pu{16 mL}$ of $\ce{SO3}$ to it. Keep in mind you can post an answer to your own question, this is considered a good practice here:)


text:
Why do the Ions Swap in double displacement?
I'm curious but in a double displacement reaction I know that the anions are swapped around and normally a precipitate is formed, but why do the ions swap? Has it got something to do with electrical charges?

They don’t really swap; the swapping of ions is a way to simplify things for the beginner — and I simplification I do not necessarily consider to be necessary or good.
What actually happens: the ions in a double displacement reaction are typically in solution. In solution, the salts are no longer salts but separated into their ions. For example with sodium chloride:
$$\ce{NaCl(s) ->[H2O] Na+ (aq) + Cl- (aq)}\tag{1}$$
I could draw the same thing for silver nitrate:
$$\ce{AgNO3 (s) ->[H2O] Ag+ (aq) + NO3- (aq)}\tag{2}$$
Now if these two solutions are brought together, we initially have all four ions swimming around. But silver and chloride will form a badly soluble solid under these conditions. So, as soon as silver ions and chloride ions meet, the reverse reaction will happen:
$$\ce{Ag+ (aq) + Cl- (aq) ->[H2O] AgCl (s) v}\tag{3}$$
In the solution there are still the nitrate ions and the sodium ions which just weren’t affected by this process and see no need to take part.
Formally, considering your reactants and products, it seems as if the ions had been swapped. But that is not really the case; the final solution only contains those ions that did not precipitate as a solid together.


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Translucifying chemical?
I was recently shown an object and asked if it was a chestnut. It looked like a regular horse chestnut except instead of an opaque brown surface, the entire nut was slightly translucent, in similarity to dark amber, and with an almost oily texture to the surface.
Is there a chemical that this supposed chestnut could have possibly been dipped in, to give it this translucence?
Apologies if this question is incorrectly tagged, if so, re-direction most appreciated.

It's hard to tell without a photo. Can it be just a core of a chestnut in epoxy resin imitating the shell? There is a recent trend of making low-cost jewelry with environmental and natural object captured with epoxide.

3D printed plastic chestnut? Saw it open and put it under a microscope, then you'll know. It might be possible to chemically modify an actual chestnut, but not by just dipping it into sth., more like boiling it in different chemicals at elevated pressure for days.


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Why droplets of water and ethyl alcohol solution freeze weirdly?
I'm doing a project where I freeze water drops and solutions with water and ethyl alcohol. It's a phenomena where drops create a conical shape at the end of their freezing. But with ethyl drops, with concentrations between 10% and 20%, they act weird. Most of them have a conical shape, but that shape, compared to shape of water, isn't symmetrical or they have weirdly shaped fragments on their tip. (not the usual fragments appearing after the drop freezes completely)
Any information about heat transfer of ethyl alcohol or about ethyl in general  would help. Thank you.
This is what I'm expecting it to look.

Sometimes this happens

Note: The temperature of the metal where I put my droplet on is about -40 degrees celcius. This rarely happens on -15 degrees.
The drops on the pictures are the same:
1) they contain 10 microliters
2) the temperature is -40 degrees, and it's constant
3) both of the drops have 17% of ethyl alcohol
I assume all forces that affect on the drops are the same, so there must be something in the structure of ethanol that I don't know about.

When your solution freezes, it is water that actually freezes, and the rest of liquid gets enriched with ethanol. To freeze it completely, you would have to go way below -40.

I have considered that solution (no pun intended). The freezing point of 17% ethanol is between -4 and -9 degrees celcius.  link @IvanNeretin

That's irrelevant. Solutions are not like pure compounds. They start freezing at one temperature (that's what your link gives) and do not freeze completely until another temperature, sometimes quite a while lower.


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Why is viciously shaking a jar of water work in the thermodynamic sense?
According to thermodynamics, internal energy can only be increased either by doing work on the system or heating the system. Work done on the system is defined by $$W=-P_\text{external}\times\Delta V$$
So when a jar of water is viciously shaken, the temperature of the water inside is increased. How is work done correlates over here?

This can be called "work" but it is different from pressure-volume work which requires change in volume.

There are many ways to do "work", i.e. transfer energy. If you put a resistor in a sealed, and allow electricity through, it will also get hot. In 1798, Benjamin Thomson showed that kinetic energy can be changed to thermal energy... but the details are boring.


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How to calculate he equilibrium constant change for acid-base reactions in a different solvent?
So we know that $K_\mathrm w$ is $10^{-14}$. But what if I am working with non-aqueous systems? Suppose I were to add LDA to liquid ammonia, or dissolve HCl in concentrated sulfuric acid. How would I calculate the percentage of dissociation? I have some inkling of how to do this but I don't really see discussions online and want to make sure my procedure is correct. Furthermore, I don't know of what good keywords to find the self-dissociation constant of ammonia or sulfuric acid are and this is really frustrating.
To emphasize: I am looking for solutions for non-aqueous systems.

Different solvent basically means different everything. The new constants can't be calculated easily.

To emphasize Ivan's point, there is no conversion formula for converting equilibrium constants in aqueous solutions to equilibrium constants in liquid ammonia. You need a whole different set of experimentally determined constants.

I'm not exactly getting the point of question here. There's no problem with getting Kw of ammonia or H2SO4 but strength of acids and bases varies heavily depending on solvent.

For all solvents that can dissociate into ions, a dissociation constant can indeed be measured. Expect them to be unpredictable in each direction.
Furthermore, the $\mathrm pK_\mathrm a$ values you know are typically those measured in water, although some organic chemists rely more heavily on the DMSO tables e.g. the one by Evans. Just a quick look at the first column shows you that there is no way you can predict the $\mathrm pK_\mathrm a$ value in one solvent from the one in a different solvent:
$$\begin{array}{lcc}\hline
\text{compound} & \mathrm pK_\mathrm a(\ce{H2O}) & \mathrm pK_\mathrm a(\ce{DMSO}) \\ \hline
\ce{HF} & 3.17 & 15\phantom{.0}\\
\ce{HCN} & 9.4\phantom{0} & 12.9\\ \hline\end{array}$$
While $\ce{HF}$ is the stronger acid in water, $\ce{HCN}$ is the stronger acid in $\ce{DMSO}$.
Couple this with a different dissociation constant means that you can calculate nothing ab initio.

This probably should've been asked in chat, but still: do you know a good source besides Evans' DMSO table for $\mathrm{p}K_\mathrm{a}$ values in other solvents (primarily high-polar aprotic ones), where all the data would be accumulated together and citeable?

@andselisk I only know Evans and the Bordwell data referenced therein. Both are only water and DMSO (Bordwell also a few gas phase values). Sorry I can’t be of help.


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Elegant solutions to non-conjugate buffer systems: ammonium carbonate, pyridinium dihydrogen phosphate, lysine and glutamic acid etc
Okay, this has really been frustrating me as a tutor because there is such few materials online on what to do in this case (besides use a brute-force method with systems of equations). Almost every pedagogical resource assumes a weak base and strong acid, or weak acid and strong base, and stops there. Oh, sometimes they will cover ammonium acetate, which is honestly cheating because pKa + pKb of both species add up to 14, so it's really a conjugate system in disguise. Is there a pedagogical resource for these general situations? I also realize my knowledge is lacking because I only was taught the system of equations method as a backup.