text:
Estimating stability and mean life of anomalous tautomers
A pair of bases in the DNA, say, A-T, have a tautomerized form A*-T* (resulting from switching the sides of both protons along the hydrogen bridges).
I have studied how, by means of DFT computations, one can compute the energy difference between the normal and the anomalous forms. But, another point of interest is the stability of the tautomer.
I have read several times that those anomalous tautomers "decay" to the normal A-T form very quickly (much quicker than the time it lasts a round of DNA replication), which is interesting because it implies that tautomerization is not likely to be a source of spontaneous mutations.
What I would like to learn is how can one compute this expected life-time of the anomalous tautomer
I have an idea (which I don't know if it could be rigth or if maybe is completely worthless), which involves making a Born-Oppenheimer separation for the two protons (''after'' the DFT computation of the electronic ground state) and then study the non-adiabatic transitions between the different vibronic states of the couple of protons. I am happy to provide details, if someone whises so.
But perhaps "my method" is too messy, or naive (or both) and / or there are other standar approaches to compute approximately the expected mean life of anomalous tautomers such as the mentioned A*-T*.
I would be very grateful if someone could apport some insigth and/or references into how are these expected life-times calculated typically in computational chemistry!

I am by no means  an expert, but molecular dynamics simulations are sometimes used to study lifetimes. A lot of force fields exist, both for unreactive and reactive systems. Seeing as how tautomerism involves breaking and reforming of bonds, perhaps the ReaxFF could be fitted, validated, and then used?

find TS for tautomerizarion -> compute rate constant using TS theory -> compute half-life for tautomer

I had not heard of the ReaxFF, thanks for the note. @Jan Jense: could you elaborate further your answer, if you have time to spare?

Have a look at http://molecularmodelingbasics.blogspot.dk/2009/08/get-half-life-from-transition-state-to.html

Thank you very much, Jan Jensen. So, a first guess would be to compute the expected half-life from a reaction rate constant $k$ given by $k=\frac{\text{k}\text{B}T}{h}\exp\left( -\frac{\Delta F^{\ddagger}}{\text{k}\text{B}T} \right),$ where $\Delta F^{\ddagger}$ is the free-energy difference between the tauotomer and the transition state... even accepting this, a doubt remains: how do I "choose" a transition state? There a two coordinate reactions (there are two protons in hydrogen bonds) and thus tons of possible paths to the tautomer!

The protons might tunnel, so kinetics vs. barrier height might have a different relationship.

TLDR: With a quantum/semi-classical transition state theory.
I order to determine the lifetime of the tautomer, you need to know the rate constant for the decay from the tautomer to the conventional dimer.  This is a kinetics question rather than a thermodynamic question so you won't be able to answer this by comparing single point DFT energies.  The rate will also vary with temperature.
A complicating factor is that the nuclei moving are protons in hydrogen bonds.  Hydrogen bonds are frequently double well potentials with two minima, one near each end of the bond - in this case the tautomer and conventional dimer.  In these case the physical distance between the minima is small and the proton may tunnel through the potential barrier between the two minima - hence nuclear quantum effects are likely to be important, at least at low temperatures.
Methods that can calculate nuclear quantum effects can be very computational expensive and may require you to work in a reduced dimensional represention - only considering the two protons quantum mechanically and potentially simplifying the surrounding structure.
Two potential methods that could calculate a rate taking account of nuclear quantum effects are  semiclassical transtition state theory, or ring-polymer molecular dynamics which has been applied within pockets of larger systems such as proteins.

For more information about his class of reactions - often referred to as proton-coupled electron transfer - see this review.


text:
Reaction between sulfur and sodium hydroxide
Sulfur reacts with sodium hydroxide in the following way:
$\ce{3S + 6NaOH->2Na_2S + Na_2SO_3 + 3H_2O}$
It is a disproportionation reaction: sulfur's oxidation changes from zero to -2 in sodium sulfide and from zero to +4 in sodium sulfite.
But how to remember the products of this reaction? By rote memorization?
The preparation of sodium sulfide is apparent. Is there a way to understand why sodium sulfite forms? We have the hydroxide ion, and it somehow reacts with sulfur to yield the $\ce{SO3^{2-}}$ ion - but how?

Why, that's simple. Sulfide (a reduced form of $\ce{S}$) is apparent, you say. OK, then some $\ce{S}$ is going to be oxidized as well. What is the next stable oxidation state? It is +4. So sulfite it is. Then both react further with neutral $\ce{S}$, and things start to get hairy, with all them polysulfides and thiosulfates.

see this: https://chemistry.stackexchange.com/questions/34424/sodium-polysulfide-production-from-naoh-and-sulfur

According to webelements, Sulphur reacts with hot aqueous alkali hydroxide, to form trisulphide and thiosulphate species.
$$\ce{S8(s) + 6OH-(aq) → 2S3^2- + S2O3^2- + 3H2O(l)}$$
The primary source of this reaction might be from here:

Chemical reactions of sulfur compounds have been studied in equimolar $\ce{NaOH-H2O}$ melt at 100°C by voltammetry and UV spectrophotometry. Disproportionation of sulfur is fast and quantitative according to:
$$\ce{S8 + 6OH- -> 2S3^2- + S2O3^2- + 3H2O}$$
On the contrary, addition of sulfur to sulfide is not quantitative and
gives $\ce{S2^2-}$ besides prevailing sulfur
disproportionation.$\ce{SO3^2-}$ reacts with $\ce{S2^2-}$ and
$\ce{S3^2-}$ , while $\ce{S2O3^2-}$ decomposes $\ce{S^2−}$. A
pseudoequilibrium is observed:
$$\ce{S^2- + S2O3^2- <=> S2^2- + SO3^2-}$$
with concentration quotient Formula Q = $\ce{10^{-0.8}}$.

Also, reasonably stable sodium polysulfide ($\ce{Na2S_x}$) seems to form at high temperature. Check the references given in the researchgate site.

Sorry, please see my recent EDIT citing work published in Chemical Society of Japan,  which clearly puts holes into your reciting this very likely inaccurate (over simplified) chemistry. There may, indeed, be an apparent cited role for oxygen and even radicals! By the way, if you google this question, there is very little support for the above position, and more shocking our 2 answers comes up first!

The actual mechanics of the chemistry, I suspect, is advanced citing a recent (2019) reference noting some radical chemistry. My take is as follows:
First, heating in air (an O2 source) a boiling mix of NaOH and water in the presence of sulfur, I would argue may involve the reversal of the following radical reaction:
O2•- + OH• =  O2 + OH-
where explicitly the reverse reaction, which was actually proposed as early as 1963 by Schroeter in alkaline conditions (per this Google Book source), is:
O2 + OH- = O2•- + •OH  (Source above, Eq 5.98)
which apparently can be accomplished in the presence of high oxygen pressure, heat and alkaline conditions (as employed in oxygen-alkaline bleaching, see comments here. In the current text, the reaction equilibrium may be moved to the right with the consumption of formed radicals with say sulfur/sulfur compounds.
In particular, I suspect, the sulfur in heated water/OH- may acquire a charge upon heating and in accord with the electrostatic properties of a colloidal suspension resulting in the presence of some solvated electrons. Then, the formation of sulfur related radical(s):
S + e-(aq) = S•−
A path to the consumption of the highly reactive and non-selective hydroxyl radical (created above) could be:
S•− + OH• = S + OH-
One could also argue that the action of heat/light (see, for example, this Science Direct compilation) may reverse the above reaction leading to S•− radical.
Next, an interaction with oxygen from a 2019 semantic scholar article:
HS•/S•− + O2 --> SO2•- (+ H+)  (Source Page 7, Eq (7))
SO2•- + SO2•- -->  S2O4(2-)    (Same source, per comment)
Then, per Wikipedia on dithionite which notes that dithionite undergoes an acid hydrolytic disproportionation to thiosulfate and bisulfite (presence of CO2 may  assist):
2 S2O4(2-) + H2O --> S2O3(2-) + 2 HSO3-
[EDIT AUGUST 2021] Here is an interesting partial confirmation of my proposed mechanics. To quote the abstract from this work The Chemical Behavior of Low Valence Sulfur Compounds. I. Oxidation of Elemental Sulfur with Compressed Oxygen in Aqueous Ammonia Solution published in Bulletin of the Chemical Society of Japan, Volume 38, Issue 10, 1965:

The oxidation of sulfur with compressed oxygen proceeds through the following sequence of steps: (a) The disproportionation of sulfur to both pentasulfide and thiosulfate. (b) The oxidation of pentasulfide to thiosulfate. (c) The oxidation of thiosulfate to sulfate and sulfamate through a series of complicated intermediates.

Sulfamate forms only in the decomposition of trithionate.

The overall reaction of the oxidation is prompted by a rise in the temperature, an increase in the ammonia concentration, and the addition of transition metal ions, especially copper ions.

So, a confirmation from a respected journal of my claim of the role of oxygen forming the long known reported products: pentasulfide and thiosulfate.
Also, an interesting reference to an apparent fostering role of transition metal-based radicals, where the citation of Copper is especially interesting as it engages in REDOX reactions over a wider range than other transition metals (like Iron) creating active species.
Now, what could be a possible path, perhaps re-reading my suggested answer will assist.

Welcome to Chemistry SE! I've forgotten all about this reaction, but I hope that your answer will help others.

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text:
What is the oxidation state of copper and iron in chalcopyrite?
What is the oxidation state of Cu and Fe in chalcopyrite ($\ce{CuFeS2}$)?
Some websites tell that it is $\ce{Cu(I)Fe(III)S2}$, and others tell that it is $\ce{Cu(II)Fe(II)S2}$. Which is right?

A number of papers report that the formal valency states of chalcopyrite are best considered as $\ce{Cu+Fe^3+S2}$, based on various computational and spectroscopic evidence.
References:

Klekovkina, V. V.; Gainov, R. R.; Vagizov, F. G.; Dooglav, A. V.; Golovanevskiy, V. A.; Pen’kov, I. N. Oxidation and magnetic states of chalcopyrite CuFeS2: A first principles calculation. Opt. Spectrosc. 2014, 116 (6), 885–888. DOI: 10.1134/S0030400X14060149.
Pearce, C.; Pattrick, R.; Vaughan, D.; Henderson, C.; van der laan, G. Copper oxidation state in chalcopyrite: Mixed Cu d9 and d10 characteristics. Geochim. Cosmochim. Acta 2006, 70 (18), 4635–4642. DOI: 10.1016/j.gca.2006.05.017.
Goh, S. W.; Buckley, A. N.; Lamb, R. N.; Rosenberg, R. A.; Moran, D. The oxidation states of copper and iron in mineral sulfides, and the oxides formed on initial exposure of chalcopyrite and bornite to air. Geochim. Cosmochim. Acta 2006, 70 (9), 2210–2228. DOI: 10.1016/j.gca.2006.02.007.

One clue to the likely oxidation state is offered by some experiments I was doing in the chemistry laboratory when I was in graduate school.  Looking for a clue to how thiourea might alter copper ekectrodeposition, I tried adding this stuff in bulk to copper(II) salt solutions.

Copper sulfate gave a loss of color, blue -> colorless.

Copper chloride gave a white precipitate and then it dissolved, forming a colorless solution, when more thiourea was added.

Given that white or colorless copper compounds are typically copper(I), it appeared that thiourea reduces and complexes the copper, and in the sulfur-ligand environment provided by thiourea copper favors the +1 oxidation state.
A Cu(I)-Fe(III) model for chalcopyrite, where the metals are bonded to sulfur, would be quite consistent with such findings.

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text:
Does entropy depends on temperature?
Does the entropy of a system depends on the system's temperature?
I know that $\Delta S_{ext}=\frac{q_{rev}}{T}$ ($q_{rev}$ is the heat in a transformation if it is reversible).
My textbook says that the entropy of the exterior always depends on that formula (so it depends on the temperature).
But it says the entropy doesn't always depends on the formula, and we can only say the that in a transformation the entropy of a system can increase or decrease. So does it depend on the temperature?
Thanks!

Yes, entropy can depend on temperature (and often does). Examples include things like point defect populations in solids giving rise to configurational entropy, or molecule dissociation in the gas phase.

Entropy is a physical property of a material that, for a single phase substance of constant composition, can be expressed as a function of temperature and one other intensive physical property such as pressure or specific volume.  The equation that you wrote down is a relationship that we can use to determine the change in entropy of a substance from one thermodynamic equilibrium state to another, by devising a reversible process from the first state to the second state, and calculating the value of the integral of dq/T for that process.  All reversible processes between these same two states will give exactly the same value for the integral.

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text:
At what temperature does carbon dioxide decompose into carbon and oxygen?
If you were to super-heat a mole of carbon dioxide gas, at what point would it decompose into carbon and oxygen? If it decomposes into carbon monoxide and oxygen, at what point do those products decompose? I'm interested in the process from a hypothetical perspective, so energy is not a concern, nor is industrial feasibility.
I'm looking for the heat required for carbon dioxide to break down into atomic carbon and oxygen, if carbon monoxide is being formed then its not hot enough yet!

http://www.nist.gov/data/nsrds/NSRDS-NBS30.pdf

Disclaimer: I am too lazy to check whether the reaction $\ce{CO2(g) <=> CO(g) + 1/2 O2(g)}$ is more favourable.
There is no single tipping point where all carbon dioxide will suddenly be converted to carbon and oxygen. As you heat carbon dioxide, the percentage of oxygen present would gradually increase.
You're interested in the reaction
$$\ce{CO2(g) <=> C(graphite) + O2(g)}$$
for which, at $298~\mathrm{K}$, $\Delta_\mathrm{r} H^\circ = +393.51~\mathrm{kJ~mol^{-1}}$ and $\Delta_\mathrm{r} S^\circ = -2.862~\mathrm{J~K^{-1}~mol^{-1}}$ (data from Atkins, Physical Chemistry 10th ed., pp 974-7). The equilibrium constant for this reaction is remarkably simple:
$$K = \exp{\left(\frac{-\Delta_\mathrm{r}G^\circ}{RT}\right)} = \frac{a_{\ce{O2}}}{a_{\ce{CO2}}}$$
and if we assume ideality, then both activities are proportional to the partial pressures, which are in turn proportional to the number of moles of both substances:
$$K \approx \frac{n_{\ce{O2}}}{n_{\ce{CO2}}}$$
Therefore if $K = 100$, then you have a hundred times more oxygen in your system than carbon dioxide. If $K = 1$, then you have equal amounts. The pressure and volume of the system do not matter in this interpretation of the value of the equilibrium constant. Our goal is now to find out the variation of $K$ with temperature.

The variation of $\Delta_\mathrm{r} H^\circ$ with temperature is given by Kirchhoff's law, and to simplify things, let's assume that the heat capacities at constant pressure are independent of temperature over the temperature range we are interested in:
$$\begin{array}{cc}
\hline
\text{Compound} & C_{p,\mathrm{m}}^\circ\text{ / }\mathrm{J~K^{-1}~mol^{-1}} \\
\hline
\ce{C(graphite)} & 8.527 \\
\ce{O2(g)} & 29.355 \\
\ce{CO2(g)} & 37.11 \\ \hline
\end{array}$$
$$\Delta_\mathrm{r}H^\circ(T) = +393.51~\mathrm{kJ~mol^{-1}} + (0.77~\mathrm{J~K^{-1}~mol^{-1}})(T - 298~\mathrm{K})$$
One can derive a similar expression for the entropy change, which is "left as an exercise for the reader":
$$\Delta_\mathrm{r}S^\circ(T) = -2.862~\mathrm{J~K^{-1}~mol^{-1}} + (0.77~\mathrm{J~K^{-1}~mol^{-1}})\left[\ln{\left(\frac{T}{298~\mathrm{K}}\right)}\right]$$
All that remains is to calculate the Gibbs free energy change
$$\Delta_\mathrm{r}G^\circ = \Delta_\mathrm{r}H^\circ - T\Delta_\mathrm{r}S^\circ$$
and the equilibrium constant
$$K = \exp{\left(\frac{-\Delta_\mathrm{r}G^\circ}{RT}\right)}$$
You can do this easily with a spreadsheet. I've thrown together some very messy Mathematica code and I'm sure someone will come along and tell me that there's a much more concise way of doing it, but:
Manipulate[{Power[10,logT],Exp[((393510+(0.77*(Power[10,logT]-298)))-(Power[10,logT]*(-2.862+(0.77*(Log[Power[10,logT]/298])))))/(-8.31446*Power[10,logT])]},{logT,2.47422,12}]
The first number in the curly brackets gives you the value of $T$ in Kelvins, and the second gives you the value of $K$ at that temperature $T$. Here are some of the numbers:
$$\begin{array}{cc}
\hline
T\text{ / K} & K \\
\hline
298 & 7.513 \times 10^{-70} \\
10^3 & 2.073 \times 10^{-21} \\
10^4 & 0.007895 \\
10^5 & 0.6899 \\
10^6 & 1.307 \\
10^9 & 2.598 \\
10^{12} & 4.9257 \\
\hline
\end{array}$$
$K = 1$ when $T = 2.52 \times 10^5~\mathrm{K}$.

This is well above the temperature at which carbon vaporises ($4827~\mathrm{^\circ C}$ according to Wikipedia). If you redo the calculations using instead the equation
$$\ce{CO2(g) <=> C(g) + O2(g)}$$
then you get these new expressions (thermodynamic data again from Atkins):
$$\Delta_\mathrm{r}H^\circ(T) = +1110.19~\mathrm{kJ~mol^{-1}} + (13.083~\mathrm{J~K^{-1}~mol^{-1}})(T - 298~\mathrm{K})$$
$$\Delta_\mathrm{r}S^\circ(T) = +149.498~\mathrm{J~K^{-1}~mol^{-1}} + (13.083~\mathrm{J~K^{-1}~mol^{-1}})\left[\ln{\left(\frac{T}{298~\mathrm{K}}\right)}\right]$$
and much more favourable numbers:
$$\begin{array}{cc}
\hline
T\text{ / K} & K \\
\hline
298 & 1.637 \times 10^{-187} \\
1000 & 1.470 \times 10^{-50} \\
2000 & 3.417 \times 10^{-21} \\
5000 & 0.003130 \\
6000 & 0.351794 \\
7000 & 10.65 \\
10000 & 5594.31 \\
\hline
\end{array}$$
This time, $K = 1$ at $T = 6276~\mathrm{K}$. You can play around with the code here:
Manipulate[{T,Exp[((1110190+(13.083*(T-298)))-(T*(149.498+(13.083*(Log[T/298])))))/(-8.31446*T)]},{T,298,10000}]
Why does $K$ for this reaction increase so drastically with temperature? It is likely to be because of the entropy contribution: because the entropy change is much more positive, increasing the temperature makes the term $-T\Delta_\mathrm{r}S^\circ$ much more negative, lowering $\Delta_\mathrm{r}G^\circ$ and increasing $K$.

To all readers: Please tell me if there's an error somewhere, I wouldn't be surprised, the maths is rather fiddly.

Dissociation to CO is more favorable, but it later also dissociates.

@orthocresol when it is hot enough for atomic C to form, monoatomic O rather than diatomic O predominates

@DavePhD Thanks. I thought so too, but couldn't really be bothered to calculate it, maybe I'll do the maths later.

I wouldn't even mention temperatures like $10^5;\mathrm K$ and above. Anyway, by the time we get there, all chemistry is long gone, and the only real process that matters is the ionization of atoms.


text:
What is the colour of chromium(III) chloride solution?
I had a question asking about the colour of the solution of chromium(III) chloride in my exam, all my colleagues have different answers, I couldn't find any exact answer after searching!

Related: How hydrate isomerism results difference in colours?

It depends on the amount of time since the solution has been created, the temperature, and possibly present excess chloride ions.

wikipedia says it it dark green when exist as hydrate.

The color of the solution may vary as long it is not in equilibrium.
According Bjerrum and Bjerrum1 a $\ce{1\mathrm{M}}$ chromium (III) chloride solution in equilibrium at 25° C contains about 13% monochloro and 85% hexaqua ions.
The absorption spectrum published shows minima at $\ce{480 \mathrm{nm}}$ (blue) for the hexaqua and $\ce{510 \mathrm{nm}}$ (green) for the monochloro complex.

When we integrate over the spectrum we get the following colors.

However, it should be noted that the colors vary with spectral optical depth (concentration, thickness of the medium).
1 Bjerrum, M.J. and Bjerrum, J., 1990. Estimation of Small Stability Constants in Aqueous Solutions. The Chromium(III) Chloride System. Acta Chem. Scand. 44: 358-363.


text:
Why Heating turns Amorphous to Crystalline and How?
I found this answer here but iam still confused:
"Polymer chains are entangled and frozen in an amorphous state. The role of heating is to liberate these entangled chains, so some parts of chains are arranged in regular order, these regions are called crystalline regions. In between these ordered regions chains are randomly oriented."
But these regularly ordered portions are already found in amorphous solids in scattered form in between of which the arrangement is disordered. That's why they are defined as having SHORT RANGE ORDER..
Now what special happen when we heat them that they turn into Crystalline solid....

You allow them to diffuse around and find more highly ordered configurations that lower the free energy. Diffusion is (typically) an activated process, so you need to either lower the activation energy to move or increase the temperature.

You've misunderstood the term "short range order". In a melt or glass, you (e.g. an atom) can be quite sure that the center of your next neighbours are no nearer than the sum of your and their atomic radius, and not further away than perhaps 20% more. Beyond your next and perhaps second-next neighbour, everything is completely random.


text:
How does the "methylamine test for sugars" work exactly?
There is a test that enables distinction of sugars between mono- and oligosachharides.
It consists letting methylamine react with an aqueous solution of the sugar under alkalic conditions and elevated (~70°) temperature.
Monosachharides always produce a yellow color; reducing oligosaccharides a red one and nonreducing sugars no coloration.
I wondered how this reaction works and of what "kind"/structure the resulting dyes might be.


text:
Why does Sulfur have 12 valence electrons in the Sulfate ion?
So I am completely and utterly confused about why sulfur has $12$ valence electrons. I understand in it's configuration, it has $2$ electrons for the $3\mathrm s$ subshell, and $4$ electrons from its $3\mathrm p$ subshell, and it can use the $3\mathrm d$ subshell since its in period $3$. Therefore it is able to exceed the octet, and form more than just $4$ covalent bonds.
But how exactly do you calculate the exact number of "$12$ electrons" in the valence electron, which thus allows it to participate in $6$ covalent bonds with oxygen in the sulfate ion, $\ce{SO4^2-}$? I can't seem to understand.
Thanks so much!

It doesn't and it was already discussed here.

related http://chemistry.stackexchange.com/questions/29142/hypervalency-and-the-octet-rule

more precisely I recommend this answer: http://chemistry.stackexchange.com/a/5242/9961

none of those answers explain how sulfur has specifically 12 valence electrons. I know it can exceed the octet, but how do we figure out, but HOW MUCH the octet is exceeded by?

Sulphur has six valence electrons. It is in group 16. I don’t get how you could assume any other valence electron count. Also, sulphur does not exceed the octet.

in the Sulfate ion it has 12 valence electrons though?

Then see this http://chemistry.stackexchange.com/questions/5964/why-does-sulfate-have-this-structure

The sulphate ion is made up of sulphur (6), four oxygens (4 · 6 = 24) and two additional electrons for the charge. Sulphur has six valence electrons, full stop.

Sulfur has 6 Valence electrons, 2 in the first shell, 8 in the second shell, and 6 in the outermost layer (third layer). They can determine the number of kernel electrons and the number of valence electrons due to the bonds they form, for example Sulfur is more likely to form ions with the Alkaline earth metals and form different covalent bonds. In the case of SO4, Sulfur is sharing electrons with 4 Oxygen atoms forming a covalent bond but the overall charge is not 0, it's -2, so it's not exceeding the octet law, it's just a polar covalent bond, this has to do with electronegativity but that's a different story.


text:
Separating Lipids from Milk
I am trying to develop a procedure to test the percent mass of lipids in (update: homogenized) milk. It doesn't need to be very accurate, just an interesting application of chemical principles. As I was looking on traditional ways this is tested, I came across the Babcock method, which involves pouring concentrated sulfuric acid into the milk, heating, and centrifuging the sample. According to Wikipedia, "key to this process is that everything in milk except the fat dissolves in sulfuric acid."
The lipids of are surrounded by what is similar to a cell membrane. The fat is an emulsion with these surfacants holding it in the water. The other components of milk are water, colloidal micelles, which hold other proteins (of no interest to me). My question is, how does sulfuric acid dissolve "everything else"? In a biological procedure I know of to break cell membranes, isopropyl alcohol is used. The cell membrane in that case is broken when the alcohol behaves as a surfactant toward individual triglycerides. However, sulfuric acid does not have a nonpolar region, so how does it work to dissolve the phospholipid membrane?
Many Internet resources on this topic are not written from a chemical perspective. In addition, they seem to confuse the lipid bundles with the other colloidal particles.
Also, why not just heat the milk strongly (such as with a Bunsen burner) to break the protein, or add electrolytes to it, as is a common way to separate colloids?

Do you want to analyze a fresh milk, or homogenized (i.e. almost all milk in stores) ?
In principle, the standard method is described at: https://en.wikipedia.org/wiki/Butyrometer

Milk is mostly water, and diluting it (roughly speaking) with sulfuric acid to a concentration of ~50% milk and 50% sulfuric acid will result in a very acidic, but still aqueous, solution.  We can examine the likely behavior of other milk components using basic chemical principles.

Sugars.  Milk is rich a variety of di- and oligo-saccharides such as the eponymous lactose, as well as many other more complex molecules:

[Free oligosaccharides are natural constituents of all placental mammals' milk. Human milk contains 7–12 g/L oligosaccharides, making the oligosaccharide fraction a major component of human milk. Compared with human milk, the concentration of oligosaccharides in the milk of the most relevant domestic mammals is smaller by a factor of 10 to 100.]

Fortunately, the effect of sulfuric acid on the sugars is easy to predict: they will simply stay dissolved in aqueous solution.  Hydrolysis of the sugars is possible, but will not happen to any great degree without heating or prolonged incubation.
Proteins.  Milk contains thousands of unique proteins.  However, by far the most predominant proteins are called caseins.  Caseins are relatively hydrophobic, proline-rich molecules and in milk are present as tiny aggregates -- a colloid -- of molecules.  Adding sulfuric acid will denature the casein.  As a result, the casein molecules are likely to stick to each other and form a centrifugable pellet that goes to the bottom of a centrifuge tube after spinning.  Proteins contain amino acids linked via amide bonds that are unlikely to hydrolyze in the acid, unless very high temperatures or very long incubation times are used.
Phospholipids.  Phospholipids in milk include glycerophospholipids and sphingolipids.  Polar lipids play an important role in stabilizing the triglyceride (fat) emulsions that keep fat in solution.  They do this by forming bilayers and other structures on the surface of triglyceride microdroplets.  To do this job, phospholipids have a phosphate head group attached to a fatty acid tail.  The fatty part of the phospholipid sticks to the triglyceride, and the polar head group interacts with aqueous solution.  The polarity of the head group is tunable by pH.  At neutral to mildly acidic pH, the phosphate head groups are (negatively) charged, greatly strengthening their propensity to interact with water and maintain emulsion stability.
Phospholipids may be partially hydrolyzed by the acid, but without very high temperatures or prolonged incubation, most molecules will remain intact.
The strong acid conditions that arise from adding $\ce{H2SO4}$ will protonate these head groups, neutralizing the negative charge and vastly weakening their emulsifying properties.
Triglycerides.  Triglycerides are the molecules that provide most of the fat content of milk.  They are triacylglycerols.  Unlike phospholipids, they have no polar head groups, and thus they do not dissolve in water.  They would "want" to form a separate, oily phase from the water (just like mixing vegetable oil and water), but the fact that they are usually bound in tiny droplets by phospholipids prevents this separation.  Because triglyceride oils are less dense than water, if they did ever get the chance to coalesce into large droplets, they would rise to the top of the milk.  This happens when cream naturally separates from fresh (raw) milk, but in homogenized milk, the droplets are artificially forced to much smaller sizes, effectively preventing the droplets from coalescing or rising.
Upond acidification, the loss of phospolipid surfactant activity breaks down the stability of the fatty droplets, allowing them to rapidly coalesce, so they rise to the top of the tube, a process that is only accelerated by centrifugation.
Hydrolysis of triglycerides in the acid is possible, but unlikely.  In addition to required prolonged incubation or high temperatures, keep in mind that the separated triglyceride phase is actually not exposed to the acidic solution very much at all.

Are you sure than strong sulphuric acid would not hydrolyse the triglycerides?

So why use sulfuric acid rather than any other acid? If it's dilute, it doesn't exhibit any distinguishable qualities.

It's probably the cheapest acid, and was almost certainly the cheapest widely available acid when the Babcock method was initially developed.

When you add sulfuric acid, the phospholipid are hydrolysed into glycerophosphoric acid and fatty acid, therefore dissolving the fat globules membrane and freeing the tryacylglycerol.
Furthermore adding H2SO4 to milk is very exothermic (which help with the whole  dissolution of the solid non fat) because it react with the Lactose (milk sugar) with a dehydration reaction.

(i) I don't completely understand what you mean by your first sentence.  (ii) Even if I understood it, I'm not sure that it's true.  Would you be able to expand your answer, providing citations or links to evidence to back up your statements?  In particular, why is the ester bond in phospholipid less stable than the ester bond in triacylglycerol?


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Why does milk clot while making yogurt?
Why does milk clot when it becomes yogurt? Is it only because of the decrease in the $\mathrm{pH}$ of milk? If it does so, what is happening when I increase the $\mathrm{pH}$ level? And by decreasing the $\mathrm{pH}$ level can I clot any type of liquid like milk?

The clotting in a yoghurt is indeed caused by the lactic bacteria producing lactic acid and therefore lowering the pH of the milk. However in this particular case (yoghurt making) this is not due to the denaturation of the milk protein, they are in fact intact.
When the pH drop in the milk, it get near the isoelectric point (IEP) of the casein (milk protein) therefore making them partialy neutral (or completly if the pH reach it).
It weaken the electrostatic repulsion between the casein.
Consequently when the pH get closer to the IEP, casein solubility decrease and often precipitate.
And this is reversible, if you add caustic soda or another base, the yoghurt or any sour milk will go back to being liquid (well not perfect in some case because the bacteria often produce exopolysaccharide which thicken the milk).

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

Do casin milk protein repulse to each other?

Yes when the milk is at its normal pH (around 6.7) casein are charged negatively  and repulse each other.

The clotting that you see is caused by denatured proteins. You can denature proteins physically and chemically. You can use acids or bases (chemical). Also, you can heat it up shake the protein (physical). The $\mathrm{pH}$ of yogurt is different from milk because of fermentation. Fermentation causes production of lactic acid in milk. Which is responsible for the change in $\mathrm{pH}$ (yogurt $\mathrm{pH}$ 4.0 vs milk pH 6.5).


text:
Dipole-dipole interaction
So, dipole-dipole interaction exists between two polar molecules? A polar molecule consists of two ions and different charges (one is higher)?

Atoms in polar molecules do not necessarily have to have ionic bond - it can be covalent. It happens because different atoms have different electronegativity - some want electrons more than others. In water oxygen is more electronegative than hydrogen, therefore it pulls the electrons closer to itself and makes the water molecule polar. Thus water has partially negative and partially positive sides which could make dipole-dipole interaction with other molecules.
See this video on intermolecular forces.


text:
Table of expanded electronic structures of atoms
Where can I find the table containing expanded electronic configurations for atoms in the periodic table, showing how outer electrons are distributed between different atomic orbitals?
I really just need to know the expanded electronic configurations of sulfur S and tungsten W.
So, for sulfur, amongst others, there are $p_x%$, $p_y$, $p_z$ atomic orbitals. And for tungsten W, amongst others, there are $d_{xy}, d_{xz}, d_{yx}, d_{x^2-y^2}, d_{z^2}$ atomic orbitals. I need to know how exactly electrons are distributed.
Does anyone have any idea where I can find the information or how I can derive it?

They are degenerated for partial-filling case..

tungsten, sulfur?

The information you ask for does not exist, and not because of our limited understanding of nature.

Thanks @Rodriguez! You are right, I missed this moment...

This is one of those questions where I have to ask: Why? What for? Do you really get what you want with the information you requested?  You don't need to answer this comment, but maybe think about it for a second.


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There are no wavefunctions in DFT
I have seen this phrase several times across DFT textbooks. However, I am not sure if it still holds. Was there a change or a theorem that proved it otherwise? Several programs display wavefunctions for DFT calculations. For example, Quantum Espresso has the option to collect wave functions. Some of the people that I have to talked to mention the Kohn Sham orbitals as if they were a real physical thing. Shouldn't it not be possible to obtain the wavefunction for a certain state? For example, I thought it was not possible to tell that a Kohn Sham orbital belongs to a wavefunction near the surface of a material.
I know that the Kohn Sham orbitals lack a physical meaning. They are simply wavefunctions that are used to obtain the correct density. They do not have an analogue of Koopmans' Theorem. In addition, "the exact wave function of the target system is not available in density functional theory" (Koch, 2001).
So why do people need/use wavefunctions (after all most DFT programs can group such wavefunctions)? I thought that they were Kohn Sham orbitals. Therefore, there should be seldom any use for them.

semantic problem..

I recommend this question: KS orbitals have at least interpretational value. Additionally the mapping between electron density and wave function in bijective. That is what the Hohenberg-Kohn theorems show. To say that there are no wave functions in DFT seems to be a bit incomplete. (It would be nice if you could extend your quote with a complete citation.)

@Martin-マーチン The quote is from "A Chemist's Guide to Density Functional Theory." In the section "Do We Know the Ground State Wave Function in DFT?" they mention:
"We just have to look at all the wavefunctions associated with the ground state density and select that one for which the energy is is lowest. Of course, like so many results presented in this chapter, this one is also absolutely usless in real applications. We have no access to all these wave funcstions and thus, in real life there is now way whatsoever to identify the correct wave function associated with a particular density."

http://as.wiley.com/WileyCDA/WileyTitle/productCd-3527303723.html

There are no wavefunctions in DFT.

I don't like that "wave function" is used here in the plural form and I feel like it reflects OP's misconception. The right way to say this is

There is no wave function in DFT.

See, there exist only one wave function, the function that describes the state of a system in question and can be used to calculate its properties. When it comes to a many-electron system, the wave function is a function that describes the state of all the electrons present in the system interacting with each other. This is the wave function people have in their minds when they say the it is absent in DFT. And that is true: the whole point of DFT is to get rid of this function as a description of state since it is too complex to work with and difficult to visualise.

But that is not the whole story. In practise the Kohn-Sham DFT is used which introduces a fictious system of non-interacting electrons. Each electron in the fictious system is described by a Kohn-Sham orbital, a single-particle wave function, while the whole fictious system is described by the Kohn–Sham wave function, a Slater determinant constructed from a set of Kohn-Sham orbitals. Still there is no wave function in KS-DFT in the above mentioned sense:

KS orbitals don't qualify as the wave function just because they do not describe the state of a whole system;
the Kohn–Sham wave function doesn't qualify because it describes a different system (fictious non-interacting rather than real interacting).

So, the KS wave function is not the wave function, since it doesn't describe the (real) system. You can't use it to calculate the properties of the (real) system, you have to use the electron density to do so. In this sence, the statement that there is no wave function in DFT holds true for KS-DFT as well.
Can the KS wave functions (both single- and many-particle one) be somehow useful in any other way is a different story. See, for instance, the discussion of the usefulness of KS-orbitals here.

Section 2.(xii) of Autschbach, J Chem Educ 89: 1032 (2012), doi:10.1021/ed200673w points out some of the meaning that can be found in the KS orbitals.

To address the question whether "there is a a wave funtion in Kohn-Sham theory", one has to decide first whether there is a wave function or not in Hartree Fock theory.
I argue below that the answer will be the same in both cases.
In HF theory, we seek the noninteracting wave function (Slater determinant) for which the expectation value of the interacting Hamiltonian H is minimum. Obviously this wave function is only an approximation and it is very different from the true ground state interacting wavefunction Ψ.
In KS theory, there is a similar point of view using just wave functions and the Rayleigh Ritz variational principle and never mentioning the density.
Let us consider the class of effective Hamiltonians, Hv=T+V, with kinetic energy T and an effective local potential V but with no interaction. Let us denote by Ev the ground state energy of Hv. There is an infinity of effective Hamiltonians Hv (differing by V) and we can ask which one of these effective Hamiltonians Hv adopts the true interacting ground state Ψ optimally as its approximate ground state.
Mathematically, we can use the variational principle: <Ψ|Hv|Ψ>-Ev > 0.
On the l.h.s. the wave function Ψ is given, so it makes sense to minimise the l.h.s. over the effective potential V to find that which makes the l.h.s. minimum. It turns out that the minimizing effective potential is the Kohn-Sham potential. So, from this point of view the ground state of the minimizing Hv (i.e. the exact KS wave funtion) is a meaningful approximate wave function for Ψ, in the same way that the HF wave function is a meaningful approximate wave function for Ψ. See the reference for more details: https://journals.aps.org/pra/abstract/10.1103/PhysRevA.83.040502


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How does a copper piece added in nitric acid change after nitrogen dioxide starts appearing?
There's an experiment I have no way of performing but want to know its results. The instructions (in short) are:

Dip a piece of copper in a glass of concentrated nitric acid, $\ce{HNO3}$, using a pincette. When a brown gas starts forming, take out the piece of copper and rinse it with water. Right after that dip it into $\ce{HgNO3}$ solution. Leave it in there for some time and then take it out of the solution. Rinse it with water and wipe it with a napkin.
How does the piece of copper look?

What I've worked out is that the first part can be written as:
$$\ce{Cu (s) + 4HNO3 (aq) -> Cu(NO3)2 (aq) + 2NO2 (g) + 2H2O(l)}$$
and that $\ce{NO2}$ is the brown gas.
What I'm  not sure about is why exactly I'm dipping the copper piece into a solution of $\ce{HgNO3}$ following the above reaction. And why am I rinsing it with water before dipping it into the $\ce{HgNO3}$ solution?

@student States of aggregation should not be subscripted, it is not wrong, but the recommendations (Sec. 2.1.) are different.

@Martin-マーチン Thank you, will bear it in mind.

Metals often have corrosion on their surface, such as oxides.  By dipping in nitric acid and then rinsing with water you would achieve a true copper metal surface free of oxide or other corrosion products.

You will likely be creating a mercury-copper amalgam similar to this


text:
Difference between water solubilities of phenol and cyclohexanol?
The solubility of phenol is 83 g/L, while that of cyclohexanol is 36 g/L. Both of them have a hydroxy group, and can form hydrogen bonds with water. Also, the hydrophobic part of phenol, as well as cyclohexanol is almost the same. Then what is the reason for phenol's better water solubility in water compared to cyclohexanol?

Phenol's $\ce{OH}$ is more polar (which is also manifested in its acidity).

The aromatic component draws significant electron density away from the hydroxy group. It is an example of conjugation. One of the Lewis structures involve a double-bonded oxygen.

@IvanNeretin Hmm, could you elaborate a little more on the reason for the higher polarity of Phenol's $\ce{O-H}$ bond you suggested. I don't see a particular link between bond polarity and acidity as my view of the higher acidity of Phenol is that the higher (relative) stability of the phenoxide ion due to resonance stabilisation of the negative charge (which is not present in the cyclohexoxide ion) leads to a an equilibrium that is shifted a bit more to the side of dissociated species compared to the dissociation equilibrium of cyclohexanol and thus to a higher acidity.

@Philipp It is all the same; if you have resonance structures that stabilize phenoxide ion by taking its negative charge and distributing it someplace else, then you also have the same structures in protonated form with (+) on oxygen. Or, as YunfeiMa put it, the aromatic ring draws electron density away from oxygen, thus effectively making it more electronegative.

And how 'exactly' does higher bond polarity make up to a better hydrogen bonding?

In addition to the acidity difference described in the comments, there is a more subtle effect involving conjugation in the undissociated phenol.  When the phenyl group withdraws some electron density from the π-donating hydroxyl group, it gains some negative charge and so is better able to attract the hydrogen ends of the water dipoles.  Ultimately this is a manifestation of the polarizability of the delocalized π electrons.


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Is it feasible to make dendritic crystals at home?
I've become interested in the forms of dendritic metal crystals from an aesthetic perspective, and I'm interested in whether it is feasible (from a cost, safety, resources) perspective to create metal crystals of this sort at home?
Would the energy involved, or the cost of materials, the difficulty of controlling conditions sufficiently, or the general complexity of techniques be prohibitive to a non-expert/ someone outside an institution? Are there some metals, or metal compounds (e.g. those with lower melting points - lead?) for which this might be easier?

The single displacement reaction between silver nitrate and metallic copper will yield dendritic crystal growth of metallic silver. Normally, a copper wire is placed into silver nitrate solution. The reaction that occurs is:
$$2AgNO_3(aq) + Cu(s) \rightarrow Cu(NO_3)_2(aq) + Ag(s)$$


text:
Reactivity towards SN1 reaction
In the following compounds , we have to find order of compounds of their reactivity towards SN1 reaction.
According to me , we should compare carbo cation stability which is formed as an intermediate.
So the 4th compound (tertiary butyl chloride)  should be in top in order of reactivity.
But how to compare other three compounds.

@Freddy the 1,2,3 compounds are 1°, now how to compare.

First of all, why the carbocation of fourth one is more stable. The major factor is hyperconjugation. Greater the number of hydrogen atoms on beta-carbons, greater is the stability achieved, as there are more resonance structures possible.
Therefore, 3rd one is least stable among. Likewise, you can easily point out that 1st one is more stable than the 2nd one.
Addition: After reading comments and other posts, I would like to share an image.

But in third after carbocation rearrangement (methyl shift) it will become mote stable.

yes. however that requires bond breaking.

The (4th) would be the first.
Then the third (explanation ) since methide shift is a preliminary property of a carbocation(C+) therefore the carbocation would do the same and become a 3 degree hense would be second.
Simillar is for the secon one here hydride shift comes into effect and thus carbocation becomes 2 degree
FINAL ANS.4>3>2>1.

Bond breaking is involved in alkyl or hydride shift, that means one more extra step. Thus reactivity is also affected.

THE shift is a step definately but it  is a fast step because carbocation approach to a more stable state the RDS(rate determining step) in all cases would be the formation of carbocation initially .

The shift requires breaking of C-H bond and it requires more energy and hence requires more time....

We ahould not see the thermodynamics when there is a posibility of 1->2 or 2->3 degree carbocation formation i have been taught this in my coaching

so you think there is no relation between thermodynamics and kinetics? Then why more stable tertiary carbocation is formed easily? I suggest you to think rather following what is taught in coaching.

I agree with adianadiadi, but the book answers according to mukhar jain. This shakes my concepts completely!


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Can ethanol alone be used as Fuel?
Ethanol is a clean fuel and added as an additive to petrol.
Why can't ethanol itself be used as fuel?

It can. E-85 is ethanol.

@YunfeiMa E-85 is an advertising slogan for 51% to 83% ethanol.  http://www.afdc.energy.gov/uploads/publication/ethanol_handbook.pdf

Ethanol clearly can be used as a fuel.
But there are several reasons why it isn't.
First is simple economics. Ethanol costs more to produce than gasoline: oversimplifying only slightly, fermentation is more complicated than simple distillation. Those countries that do use mixtures of ethanol in gasoline do so for "environmental" reasons (which are not always well thought through as being made from plants doesn't make something good if associated production costs are high or alternative uses are better). And "environmental" causes are a good excuse to subsidise favoured groups, like US or Brazilian farmers. In the absence of subsidy, there would be very little use of ethanol as a fuel.
But those are not the only reasons. Ethanol carries less energy than the same volume of gasoline. In other words your mileage will be lower; your fuel efficiency will be lower.
Worse, though, ethanol doesn't have exactly the same characteristics as gasoline. Engines may need to be adjusted so the alternative fuels burn equally well, which is annoying but fixable easily in electronically controlled engines. In addition, Ethanol may be more corrosive to some of the components used in fuel pipes (nothing like as bad as methanol, but clearly worse than gasoline). Ethanol is also much more hygroscopic than gasoline, absorbing significant amounts of water from air. This is another potential cause of contamination in fuel systems and may make corrosion worse.
In short, there are no fundamental reasons why ethanol can't be used. But there are a range of practical issues that make it less economical or inconvenient.

But one of the most important reasons is probably that petrol was much, much more easily accessable when motor vehicles started becoming a thing in the early 20th century and therefore 100 years of engine optimation went into petrol engines. But +1 anyhow for a nice summary!

Pure ethanol is used as a fuel in South America. Some vehicles can actually produce more power from ethanol than petrol.

The argument that ethanol by itself carries inherently less energy, and hence, as a potential fuel source is perhaps not a good power source, I would view as a likely misleading statement, given how it has been actually employed in practice.
In fuel engines, Ethanol is employed as a stand-alone fuel for combustion with Nitrous oxide (N2O). Further, research has even been performed on its use as a prospective rocket fuel (see,  Design and Testing of a Liquid Nitrous Oxide and Fueled Rocket Engine).  To quote:

A small-scale, bi-propellant, liquid fueled rocket engine and supporting test infrastructure were designed and constructed at the Energetic Materials Research and Testing Center (EMRTC). This facility was used to evaluate liquid nitrous oxide and ethanol as potential rocket propellants. Thrust and pressure measurements along with high-speed digital imaging of the rocket exhaust plume were made. This experimental data was used for validation of a computational model developed of the rocket engine tested.

Futher comments provided by another source:

Under certain conditions more energy is produced by the decomposition than necessary to reach the decomposition temperature -> run away reaction!
$\ce{N2O(g) —> N2(g) + ½O2(g) + 82 kJ/mol}$

Of import by the same source, there are reported explosion of a N2O hybrid motor, to quote:

Incident 4
Design flaw of the injector caused a contamination of the N2O channels with alcohol. “Water hammering“ in the N2O ignited the mixture and destroyed injector and valves.

The author cites a possible explosion trigger from so-called ‘water hammering’ caused by cavitation of N2O combined with organic contamination (including Ethanol) resulting in a  “Dieseling” based ignition scenario.
So a limiting factor on implementing Ethanol/N2O fuel engines is likely the need for research to manage safety issues and little to due with a low energetics argument (actually, more like the exact opposite).

Your file on EMRTC is corrupted...

It's not used widely mainly because of production cost and energy density that is lower than petrol.
for petrol 34.2 MJ/L
for ethanol fuel (E100) 20.9 MJ/L.

This low energy density is the reason why ethanol will never be able to be used in a plane. If you want ethanol to replace kerosene in a usual jet, you would have to fill the plane with all the ethanol you are to burn when flying. And there will not give any place for the crew and the passengers.


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Scavenging Xenon
This may well be off topic - if so I will remove it or the mods can.
We will be using a good few litres of Xenon by running it through a new analytical instrument. Given it's price (at a guess, around $15 per litre at STP) is there any standard method of recycling it? I really do not want to vent it to atmosphere.

Googling for 'xenon recycling system' comes up with multiple hits from several different areas (EUV litho systems, plasma processing, human anesthesia), so it is clearly possible. BOC Edwards appears to market (or did market) a turnkey recycling system. Whether it is worth it in your setup is a question only you can answer.

@JonCuster I'm starting to think "balloons"!

What is the Xe used for (carrier gas?), and how pure does it have to be? What else is going through the system? For example, you might get away with a LN2-cooled sorption pump to trap the Xenon, then every once in a while purge it through perhaps an appropriate molecular sieve?

@JonCuster We are going to measure the speed of sound in Xenon as part of a calibration test. I'm starting to think it might be too expensive. Flow is 100 ccm ( $1.50 per minute for several hours)

I would suggest SF6 but I'm not sure it is any cheaper. Do you have to flow it, or backfill and valve off?

@JonCuster I was going to say "flow" but I think maybe just filling the apparatus might be sufficient. Esp given the budget. We use flow as part of a continuous measurement process where we expect composition to change, but if it is a pure gas maybe no need to flow.


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How to convert SPC files to ASCII without original spectroscopy software?
IR and Raman spectroscopy are typically saved in *.SPC files (chemical/x-galactic-spc), which are binary files. Collected data is save to SPC files by default, but can be exported to txt (ASCII) files.
I now have original SPC files and the original spectroscopy software is not available. Is it possible to somehow read those files with some other (generic?) software?

If you don't mind Python on the command line, there's an spc converter on Github.
While the trial version of the GRAMS Suite 9.2 is available with a 30 days license, it seemingly doesn't work with any data but the samples provided with it and therefore isn't suitable for your purpose :(

A quick search turned up a number of useful tools. Wikipedia suggests:

Essential FTIR - which advertises batch conversion of SPC to other formats.
SPC2CSV - from AnalyzeIQ to convert to ASCII CSV format
Spekwin32 - general spectra software, including batch conversion
EPA's ShowSPC - not clear if it allows saving (or if it even runs on more recent OS)

If you use R for data analysis, take a look at the R package hyperSpec. Example:
install.packages('hyperSpec')
library('hyperSpec')
spectrum <- read.spc('<your spc file>')
...
plot(spectrum)

You can also use Sys.glob() function for wildcard search of your files, then read all of them with read.spc(), and call collapse() to build a single hyperSpec class with all data points together.


text:
Why Balmer series exist?
I was reading about Bohr's atomic model in our textbook when came up with this question:
Since the electron in hydrogen atom is most stable in the first orbit, why should it descend from higher orbits to any orbit except the first orbit(that means balmer series for example)?

https://en.wikipedia.org/wiki/Emission_spectrum and this one too http://www.avogadro.co.uk/light/bohr/spectra.htm

It says "This is as a result of an electron 'falling' from a higher energy level to one of lower energy." What I want to know is why"a lower energy level" and not the ground state as it is the most stable?

@Shadock The references cited here are more relevant http://www.ioffe.ru/astro/QC/CMBR/sp_tr.html Transition to the lowest energy is most probable, but others have a non-zero probability.  I think QED is needed to explain properly.

I see...; thanks, the tables are although well clarifying.

@DavePhD thank you for the link. What means QED?

@Shadock quantum electrodynamics

@orthocresol DavePhD ask him before if he knows that. If it is too complicated for the person who ask the question he will never accept your answer...

I think a less sophisticated explanation might suffice. Yes, sometimes an electron would fall to an intermediate state and remain there for a few long microseconds before falling further to the ground state; so what? Who said that all things must go straight to the ground state? If that were the case, all organic life (including us) would die and decompose instantly.

Depending on the higher energy wave function quantum numbers, decay directly to the ground state may not be allowable (at least in the dipole approximation).

@IvanNeretin maybe without QED like this http://farside.ph.utexas.edu/teaching/qmech/Quantum/node119.html#e3.115 ?

or with qed https://www.photonics.ethz.ch/fileadmin/user_upload/Courses/NanoOptics/app_b.pdf

The entire process has two parts: Excitation and relaxation. The first is clear enough, we excite an electron into a higher energy level.
Now this electron wants to fall back down. There is no roadsign or intrinsic selection rule that requires the electron to immediately go back to the ground state. Any state in a lower shell will be less energetic and by chance electron 1 happens to stop in state A while electron 2 of a different atoms stops in state B. Any electron that is not in a ground state will remain there — probably for a duration of about a few femtoseconds or less — and then relax further. The universe is not deterministic so nothing is telling the electron it has to take the fastest path.
Note that this gedankenexperiment does not make any prediction which path is most likely. You would need more theory (that DavePhD is likely to give) to explain that.

Well, for example if the H atom is in the 3s or any other s state, it will  need to go to an intermediate state before going to the ground state, for electric dipole radiation.  The 2s lifetime is therefore infinite for electric dipole radiation.  Generally, the other lifetimes are in the nanoseconds. http://www.physik.uni-wuerzburg.de/EP4/teaching/WS2012_13/KM_ueb/PhysRev.148.1.pdf


text:
Clear definition of a 'non-oxidizing acid'?
From the Beryllium article in Wikipedia:

Beryllium dissolves readily in non-oxidizing acids, such as $\ce{HCl}$ and diluted $\ce{H2SO4}$, but not in nitric acid or water as this forms the oxide.

I think it would be helpful to have a definition of 'non-oxidizing acid'. Is it an established term in chemistry? Am I right to understand that it is an acid that can only act through its $\ce{H+}$ moiety, and not by the reduction of its conjugate base part (as in the reaction between $\ce{HNO3}$ and $\ce{Cu}$)?
There's a little obscure page titled oxidizing acid on Wikipedia.
The trouble is, until I came across this term, I imagined all acids to be oxidizing by definition. It's not very intuitive.

P.S.
Related: Weird Wikipedia Section on Oxidizing Acids

Yeah, basically that's it.

"All acids are oxidising by definition" - this is not true - an acid (as far as Arrhenius was concerned) is simply something that releases H+ into solution. It may or may not then go on to oxidise something. HCl + NaOH is not a redox reaction; HCl is an acid but it is not oxidising anything here. HCl + Na is a redox reaction, and yes H+ is acting as an oxidising agent, but that is not because HCl is an acid. It is because Na is so easy to oxidise that H+ can oxidise it.

Hydroiodic acid is certainly not oxidizing.

A non-oxidising acid is an acid in which there is no stronger oxidant than $\ce{H+}$ at $\mathrm{pH}\ 0$.
Or, in a slightly different wording, a non-oxidising acid would be one that cannot dissolve any noble metal at $1~\mathrm{M}$ concentration, but should (in the absence of effects like overpotential) dissolve any non-noble metal at that concentration.
Nitric acid has the strong oxidation agent $\ce{NO3-}$ present, which will reduce to $\ce{NO}$ or $\ce{NO2}$ depending on the concentration. It is able to oxidise noble metals such as silver.
Hydrochloric acid, on the other hand, has no other oxidising agent. The chloride ion is already fully reduced. Therefore, it is not able to oxidise noble metals such as copper.
Some acids are more strongly oxidising if they are concentrated. Sulphuric acid is one. Only in higher concentrations will reduction of sulphate to $\ce{SO2}$ take place and make the acid oxidising. In lower concentrations, only $\ce{H+}$ takes part in oxidations.

Jan, are there any cases where the oxidising power of the anion and the H+ (in an acid with acidic hydrogens) are roughly or exactly equal? Such acids would oxidise with both the proton and the anion, and would be of an intermediate class of acids between oxidising and non-oxidising.

Also, isn't pH=0 and concentration=1M only for strong acids?

@harry Good point. The standard potential assumes a $\pu{1M}$ concentration of all dissolved ions (including $\ce{H+}$ which maybe doesn't work as well with weak acids. I would have to research what is done in those cases but I am too lazy. As for oxidising anions, I am not aware off the top of my head of any anion that would have an equal oxidising power, but do check the electrochemical series.

As an add-on, I'll post an excerpt from the 6th edition of "Chemistry: The Molecular Nature of Matter" by Jespersen et al.
Kudos to Freddy who posted it in the chatroom.

Basically my answer but backed by a textbook ;)

@Jan - a lot of good stuff gets dissociated in the chat, why not bring it here, by a salt bridge, as it were; lest the half-cells become unbalanced.

I thought this would be more of a metallic bridge but yeah. Not even complaining. Upvoted it towards sportsmanship ;)

There is really nothing more to say than what's in the Wikipedia article referred to by the OP and the comments.  "Nonoxiding acids" are those acids that react with metals by displacement of hydrogen, such as $\ce{HCl}$ reacting with zinc.  "Oxidizing acids" like $\ce{HNO3}$ contain more strongly oxidizing components that react instead of the solvated $\ce{H^+}$, or possibly convert wiuld-be displaced hydrogen back to solvated $\ce{H^+}$.

you can, and should, use $\ce{}$: $\ce{H2O + H+ -> H3O+}$ gives $\ce{H2O + H+ -> H3O+}$ Package documentation

I went with "solvated $H^+$" because the concept could be applied to nonaqueous solvents.

It was just an example of the formatting syntax, I was not commenting on any of the things you wrote

Ok, point taken.


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Transition metals have a half filled s band in their metallic states
I was reading this "Chemical bonding at surfaces and interfaces" book. Here in page 69, they argued that interactions of transition metals to s band of adsorbate would be same because transition metals have half filled s band at their metallic state. My confusion is in electronic arrangement of transition metals I am seeing $s^2$ except for atoms having $d^5$ or $d^{10}$. Even $d^{10}$ is actually d block metals not transition metals. What could be the possible explanation of this line?


text:
Hydrazine vs 1,1-dimethylhydrazine for rocket fuel?
The question in my textbook is as follows:

Hydrazine and 1,1-dimethylhydrazine both react spontaneously with O2
  and can be used as rocket fuels.
$$\ce{ N2H4(l) + O2(g) -> N2(g) + 2H2O(g)}$$  $$\ce{N2H2(CH3)2(l) + 4O2(g) -> 2CO2(g) + 4H2O(g) + N2(g)}$$
The molar enthalpy of formation of N2H4(l) is +50.6 kJ/mol, and that
  of N2H2(CH3)2(l) is +48.9kJ/mol. Use these values, with other $\Delta fH^{\circ}$ values , to decide whether the reaction of hydrazine or 1,1-dimethylhydrazine with oxygen provides more energy per gram.

My attempt at the solution:
Hydrazine reaction:
$\Delta fH^{\circ}\ \ce{N2H4(l)} = +50.6kJ/mol$
$\Delta fH^{\circ}\ \ce{H2O(g)} = -241.83kJ/mol$
By Hess's law of summation
$$\Delta rH^{\circ}=\sum\Delta fH^{\circ} products -\sum\Delta fH^{\circ} reactants  $$
$$ = [ 0 + 2(-241.83kJ)] - [(50.6) + 0] $$
$$ = -483.66 - 50.6 $$
$$ = -534.26 kJ/mol $$
1,1-dimethylhydrazine reaction:
$\Delta fH^{\circ}\ \ce{N2H2(CH3)2(l)} = +48.9kJ/mol$
$\Delta fH^{\circ}\ \ce{H2O(g)} = -241.83kJ/mol$
$\Delta fH^{\circ}\ \ce{CO2(g)} = --393.509kJ/mol$
By Hess's law of summation
$$\Delta rH^{\circ}=\sum\Delta fH^{\circ} products -\sum\Delta fH^{\circ} reactants  $$
$$ = [ 2(-393.509kJ/mol) + 4(-241.83kJ/mol) + 0] - [48.9kJ/mol + 0] $$
$$ = -483.66kJ/mol - 50.6kJ/mol $$
$$ = -1803.238 kJ/mol $$
Now, my first reaction is to say that the 1,1-dimethylhydrazine provides more energy, as it evolves 1803.238 kJ/mol of energy as heat for the reaction vs the 534.26kJ/mol evolved by the Hydrazine. I have a strong suspicion that this is wrong however based on the fact that a quick google search reveals that the Hydrazine is indeed used in rockets. The questions does say " energy per gram " so i also tried working out the number of moles in 1g of both Hydrazine and 1,1-dimethylhydrazine and then working out how much energy per mole equivalent to 1g but 1,1-dimethylhydrazine still evolves more energy when i do those calculations.
I unfortunately don't have an answer to this problem so i'd really appreciate any feedback. Thanks in advance!

Nitpicking 1) Your second reaction is missing $\ce{2CO2}$ 2) Enthalpies of formation and enthalpies of reaction are in kJ/mol not kJ 3) Your value of the enthalpy of formation of CO2 isn't right 4) The third last line of the equation is wrong. Be careful with your copy pastes

Furthermore the fact that N2H4 is used as rocket fuel is hardly a sign that you are wrong. If the choice of fuel was solely based on the amount of energy released per gram, then all our vehicles would be running on nuclear fusion right now. By the way, N2H2Me2 is also used as a rocket fuel.

@orthocresol thank you - have corrected. That is true, i suppose there are actually many factors to consider such as cost of production, density of propellant etc. Thanks for the link, very interesting.


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Precipitation occurred when adding vanilla extract to oak extract. What caused this?
Sorry if this is a very obscure question, but I don't know where to ask it.
120ml of oak extract, made by soaking oak chips in ethanol, was added to a flask. At this point, the solution was very dark with zero precipitate. 13ml of vanilla extract, consisting of ethanol, water, vanilla, and corn sugar were added to the solution. This extract was also free of precipitate before being added. Immediately upon adding the vanilla extract, a large amount of precipitate fell out the solution. Can someone please explain what happened, and what fell out of solution?

Well, unless you want exact composition of precipitate, then it's rather simple imo - one extract was on almost pure ethanol other on diluted so you basically added water and lowered solubilities of most of compounds in this solution.

The oak extract was 95%, and the vanilla 35%. That would have only dropped it to a 89.1% strength. The amount of precipitate (it's about half the volume of the solution) made me think the small dilution wasn't the cause.

Also some of vanilla extract may precipitated as diff was much bigger for it. Main thing seems to be how much extracted stuff was it those extracts.


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Hybrid orbitals in molecular orbitals
In our textbook there are two topics that I can't make connection between:

hybrid orbitals which (as I understand) are used to demonstrate the spacial shape of molecules like CH4 with VSEPR while taking atomic orbitals in account.
molecular orbitals which are an accurate way to look at covalent bounds.

But the problem is: In molecular orbitals we start talking about "s, p, ..." orbitals again and not hybrid orbitals "sp, sp2, sp4, ...".
Why is it like that?

What context are you talking about? For example, in arenes you talk about unhybridized p orbitals forming pi complexes and sp2 orbitals forming sigma bonds.

Traditionally, hybrid orbital belongs to valence bond theory, which in-general adopts non-orthogonal basis; molecular orbital is another scheme, which adopts orthogonal basis. Sometimes, localized molecular orbital refers to hybridization, e.g. natural hybrid orbital .

I'm really suspicious of your statement; what is the ground behind this? I really am not getting that. Just keeping your notions and believes aside for a second, please carefully read ortho's answer:http://chemistry.stackexchange.com/a/40635/5764  and then decide the fate of your thinking.


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Chromatography: Why does the insoluble compound get pulled along?
In paper chromatography, we see that a paper with an ink spot on it get dissolved in the liquid and travels up with water due to capillary action. Then the compounds with more affinity to the paper get deposited on the paper. My question is why do the molecules with more affinity to the paper move with the liquid in the first place and then ultimately settle down after some time?

Chromatography only works when the affinity for the paper vs. affinity for the water is in a dynamic equilibrium.  Suppose there are three compounds in the ink, called $\ce{I}$, $\ce{N}$, and $\ce{K}$.  Lets call the  spots on the paper fiber that bind ink molecules as $\ce{P}$.  Then let's write the dyanmic equilibria this way:
$$\ce{I(aq) + P <=>> PI} $$
$$\ce{N(aq) + P <=> PN} $$
$$\ce{K(aq) + P <<=> PK} $$
This means that any given $\ce{I}$ molecule will be absorbing to paper, then becoming unstuck from the paper, then absorbing to a (probably different) site on the paper, then desorbing again and returning to aqueous solution.  The same is true of $\ce{N}$ and $\ce{K}$.
There are two further facts that we need to realize to understand the chromatography.  The first is that the equilibria are balanced at different points.  This is why the size of the arrows in the reactions I drew above are different.  As drawn, component $\ce{I}$ has stronger affinity for the paper and a lesser affinity for water, as indicated by the longer rightward top arrow and the shorter bottom left arrow.  The reverse is true for component $\ce{K}$, and $\ce{N}$ is somewhere in between.
The second fact is that the water is moving due to the capillarity you mentioned. So when the compounds are in aqueous solution, they will flow with the water.  When they are stuck to the paper, they will not move.  Since all of the molecules are very rapidly transitioning between paper-bound and water-"bound" states, they will all move some of the time, and be still another part of the time.  The key is that the different components spend, on average, different fractions of time moving vs. being still.  This makes their average speed different.
As for the question of why they eventually come to rest, this is because the capillarity-driven flow eventually runs out of steam (i.e. gravitational forces compete with the capillary forces after the fluid rises far enough), and the water itself stops flowing very much.  Thus the bands of separated molecules eventually settle into a relatively fixed position.
So to answer your title question, truly insoluble compounds would not be separated by chromatography.  Compounds need to be soluble to at least a degree in order to be separated.

That would mean that the compounds which are a little soluble in water would move along with water. Why would a substance with greater affinity to the paper move with the solvent altogether? Shouldn't they all keep sticking to the paper on account of being more attracted to the paper? @Curt F.

Also, as the water ascends, the insoluble compounds of water are not able to keep pace with water's speed whereas the soluble compounds move higher up because of the kinetic energy provided by water. Now as the portion of water level that is "moving" exceeds the insoluble compounds, the insoluble compounds are left in the portion of water which is not really moving, thus their kinetic energy further decreases and they eventually come to a stop.- does this explanation make sense? @Curt F.


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why are amines more water soluble than alcohol for similiar molecular weight?
kindly someone please explain the reason behind this behaviour despite the fact that oxygen is more electronegative and forms stronger hydrogen bond.

Possible duplicate of Solubility of n-butanol and n-butanamine (however, no answers yet)

More soluble in what? Water?

yes they are more soluble in water

@Loong ,yes this is what I am talking about. 1 comment in that question's discussion mentions about H-bonding but i think both amine( 2H +1 LP) and alcohol (1H+ 2LP) will be able to form 3 H-bonds. Correct me if I am wrong

I think it may be due to van der waals force (v.w.f). For both alcohols and amines the v.w.f acting between them and water molecules is keesom force i.e. dipole dipole interaction. In such case v.w.f is directly proportional to molecular mass , but if the molecular mass is also same then v.w.f is directly proportional to surface area of molecule which is inversely proportional to branching. Hence since amines have more branching than alcohol they have less surface area and hence less van der waals force. Due to weaker v.w.f amines are less soluble than alcohol.
Please correct me if I am wrong.

Please someone comment on my answer. Am I right or wrong


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Measuring the Kc of equilibrium between NO2 and N2O4
I'm trying to measure the $K_c$ of the equilibrium reaction
$$\ce{2 NO2 (g) <=> N2O4 (g)}$$
in different temperatures. How would I do this do this in a normal school laboratory? I was thinking of shining (laser) light through a test tube that contained $\ce{NO2 (g)}$ and looking at the intensity of light emitted. $\ce{NO2 (g)}$ is dark-reddish brown and $\ce{N2O4 (g)}$ is colorless. That means that the more to the right the equation is, the more light will be emitted.
However, I don't think this would give me accurate values for $K_c$. Are there any other methods that I could use to do this which preferably don't require advanced equipment (which I don't have)?

Nice question! You've got me thinking.  So you know, I took the LaTeX out of the title because it makes things look messy in searches.

@Brian Oh, thanks! I didn't realize that, but it makes sense. I'll keep that in mind for the future! Also, thanks for correcting the equation and tags – that was a silly mistake on my part :)

No prob - welcome to the site!

If you have an equipment, that allows you to measure pressure change during reaction, you can easily compute partial pressures and equilibrium constant.

@VadimShkaberda wow, that's a great idea! I can measure pressure, and I probably could use these calculate that and try to find $K_c$ that way. I'll look into that!

Skel, can you add a summary of what equipment you do have available? Otherwise, it's really hard to narrow down possible approaches.

I agree with @Brian, it is really not possible to answer this without the knowledge of your equipment. Also note, that you are probably trying to measure $K_p$ (with partial pressures) as you have gasses.

@Brian hey! Actually, I decided to try to use a pressure meter that we have, but it unfortunately wasn't accurate enough (the pressure seemed to fluctuate the whole time). I wanted to use the partial pressures method, but maybe I just don't have equipment accurate enough for it. That's why I decided to try another experiment, like perhaps finding the solubility curve of a certain substance in a certain solvent. If you want me to, though, I can list the equipment that I have! I'm really on the fence with the Kc experiment :)

Skel, it's up to you if you still want an answer to the question..! Just edit the list of equipment into the question, and it'll trigger a reopen vote.

@hBy2Py Thanks! I decided not to, though. But thanks for helping! :)


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Equilibrium constants and
I have a few questions about this two relations:
We know that:
$K_p$ = $e^{\frac{-\Delta G^0}{RT}}$ $(1)$
and
$K_c = K_p (RT)^{\Delta n}$ $(2)$
Ok now I'm solving this problem about solubility:

The solubility product $K_s$ of $Ni(OH)_2$ is $6.0\times10^{-16}$ at $T=25ºC$
Calculate $\Delta G^0$ of this reaction at $25ºC$

I checked my teacher's resolution and he is using $(1)$ to directly calculate $\Delta G^0$? But is that correct since $K_s$ is equivalent to $K_c$ and not $K_p$?
But I also thought about converting $K_s$ to $K_p$ using $(2)$ but does that make sense since we are not dealing with gases?
Can someone please clarify me what's the best process to follow here?
Thanks!

http://chemistry.stackexchange.com/a/40586/16683

Also your equation for $K_c$ depends on the ideal gas law.  It isn't true for non-ideal gases.  (And solubility equilibria happen in liquids, and liquids are definitely not ideal gases...)


text:
Is there a difference between NaCl or table salt (processed) and salt from the Himalayas?
So, I was doing some research on bonding and electronegativity and one of the examples used was NaCl (Sodium Chloride) which is salt. We don't buy processed salt, rather we buy salt from the Himalayas.

Apparently this is better salt because it is not processed, made, etc.
But, is there a difference between NaCl and Himalayan salt in regards to their molecular structure?

Exactly why would processing something to make it consistently pure and free from potentially dangerous ingredients make it worse?

@matt_black "better" can mean a lot of things. It could have better taste, less environmental impact, etc. (I don't know if it does, just giving examples). It's not clear to me what OP meant by it.

Typically, the people making claims like "Apparently this is better salt because it is not processed, made, etc." are the ones selling the stuff. A little skepticism is warranted.

I'm guessing one important difference is the price.

@matt_black processing or purification can add impurities of chemicals used to purify the desired chemical, e.g. organic solvents, metallic catalysts etc. So it can happen that you removed one impurity and added the other. Of course, I don't want to get into regulations of food processing here, but it is a thing to consider

Since it's  a different colour, what makes you doubt for an instant that it's different? Clearly there's something in those crystals other than a regular NaCl lattice, so when you say "in regards to their molecular structure", what are you really asking about?

@matt_black, I think your question was answered by user2829 -

@CarlosCarlsen Sure, processing can introduce impurities, but food products are highly regulated and usually don't. The claim that it is better because it isn't processed is ridiculous as many natural things contain dangerous or poisonous ingredients.

@matt_black - Right, but in the case of Himalayan salt, usually there isnt anything poisonous.

@CarlosCarlsen Possibly, but that isn't a given and you can't be sure without analysis. Being natural is irrelevant and some natural salts clearly contain heavy metal contaminants as DavePhD points out.

No salt will be pure NaCl.  Each will have some degree of other elements.  The fact that the salt isn't white confirms there are other elements present.
See Analysis of Gourmet Salts for the
Presence of Heavy Metals which investigates 14 salts including two Himalayan salts.
See especially "Table 3. Comparison of toxic elements in Table Salt".
A "Himalayan Pink Fine Mineral Salt" was found to have the highest level of cadmium and tied for first for the highest level of nickel.
NaCl isn't molecular, but instead is a lattice of Na+ and Cl- ions.  Other ions can lie within this lattice as impurities.

Wait, so Himalayan salt is more toxic than processed salt?

@VPrime not necessarily, but if it isn't white or colorless it's obvious that there is other stuff in there.  The reference says in the toxic elements section "highest concentration of
elements were found in the darker or deeply colored
salts".

"No salt will be pure NaCl." - Do you mean nothing in the grocery store is pure NaCl?

@Era It's somewhere between 97% and 99.7% pure.  We have another question about that http://chemistry.stackexchange.com/questions/24816/how-pure-nacl-is-typical-table-salt-how-is-it-purified

@Era Salt for human consumption (in the developed world, at least) is also deliberately idodised.  The grocery store salt you're talking about has iodine "impurities" added deliberately, in addition to the general truism that nothing's 100% pure.

@HopelessN00b not really, "the World Health Organization’s (WHO) European Region has been identified as having the lowest coverage of salt iodization of all the regions." http://www.who.int/nutrition/publications/micronutrients/iodine_deficiency/9789241593960/en/

@HopelessN00b Kosher salt might be almost pure. It at least usually does not contain intentionally added iodine, but it could contain other impurities.

@WilliamKappler Pickling salt is also generally iodine-free.  You can get uniodized salt, yes, but at least in North America, (and some, but not all, European countries, apparently) it's something you have to look for.

Iodine is added to our salt because it is a necessary nutrient for our bodies, but in some parts of the world there is no natural or sufficient source of Iodine. Same way, in some other parts of the world you can find table salts with added iron.

@DavePhD - so you are saying that NaCl is more 'pure' than himalayan salt and therefore would have less elements? If so, would it be plausible to say that Himalayan salt is therefore more beneficial than NaCl?
P.S. Are you also saying that the color of the salt gives a general idea of how many elements there are in the salt? E.G. Himalayan salt is pink, thus would it be logical to assume that Himalayan elements > than those in NaCl?

@CarlosCarlsen I'm not saying salt is automatically less pure because it comes from the Himalayas.  That's a huge region and there are probably lots of different mines with each mine having different salt, even different parts of the same mine would have differenced.  Heck, even the individual grains in the photo have visible differences from each other.  Unless you actually analyze the salt, who knows what's in it?

@DavePhD, right. I am asking my questions in a more general sense.
Is NaCl more 'pure' - can you derive to this conclusion from its color?
Would you say that Himalayan salt (in general) would have more elements in it and could you say that its color can give you that impression?

@CarlosCarlsen “NaCl” refers a molecule composed of one sodium atom, and chlorine atom, and nothing else. A substance is “purely” NaCl when it consists of only this type of molecule, and nothing else. So when you ask “is NaCl more ‘pure’?” the answer is, by definition, yes—you have defined the substance as being of a particular type of molecule, and have indicated that this is all it is composed of. But you seem to be using “NaCl” to mean “processed table salt” which is not quite true: processed table salt is mostly NaCl, but nothing in practice is 100% pure, plus many places add iodine.

@CarlosCarlsen So table salt is mostly NaCl, possibly with some iodine, and then very small amounts (0.3% - 3% to use DavePhD’s numbers) of whatever else. It is therefore 97% “pure NaCl.” And since NaCl is white, something that is 97% NaCl is probably white (and is in the case of table salt). Something like the Himalayan salt, that is mostly NaCl but visibly not white, must have some larger amount of something else to give it that color: it must be less “purely NaCl” than table salt. So yes, the color of the Himalayan salt indicates that there is more than just NaCl molecules in it.

@KRyan Ahhh, I see. Thanks a lot for your clarifications - I can better understand this now!

@CarlosCarlsen Sure, you could say that, though it’s almost a circular statement: the purest form of NaCl is NaCl itself. Well, yes. Processed table salt, at least when it isn’t iodized or “iron-ized” or whatever, is “supposed to be” NaCl, so it’s 100% pure when only NaCl molecules are present, at which point we could accurately call it just NaCl instead of table salt (which in reality is a more complex mixture of mostly NaCl but also some other things). But if we were talking about water, that’s “supposed to be” H₂O, so adding NaCl makes it less pure.

You've already fallen to the marketing. Take a step back and ask yourself - how would a processed X be better then natural X? In many cases, you'll find advantages and disadvantages to both approaches. And don't forget that "natural" can kill you just as easily as "industrial" - you don't even have to go to outright poisions like death caps. A lot of the food we eat all the time is poisonous without proper processing - potatoes or rhubarb, for example. In other cases, we're outright trying to get those poisons in our food - chocolate, coffee, tobacco; of course, we're  calculating how much of the poison is appropriate to get the desired behaviour (good taste, "drug effect") while limiting damage. Humans are actually pretty tolerant to huge amounts of different poisons - the ones that are left are usually very serious indeed :)
But you can be sure that "molecular structure of NaCl" will always be the same - it's entirely determined by the laws of physics. There's been some crackpottery about ideas like "water memory" or "changing angle between atoms in water" - that's just pure hogwash playing on people who didn't pick up much of Chemistry in high school, so I feel I have to stress that. What might be different is composition (impurities in addition to the NaCl) and crystal structure (pure NaCl in a well made crystal is very optically transparent, the pink colour of Himalayian salt is due to impurities).
Processed salt in particular is something humans have done for thousands of years (in fact, the ancient Greeks and Arabs burned oil for salt processing :)). The technologically simplest way to make salt is by letting salt water evaporate - this was commonly used in places where sunlight and other power sources (mostly rock oil) were plentiful. However, salt water is far from being pure NaCl indeed - there's plenty of other salts involved, as well as various other contaminants. These contribute to the colour and taste of the salt produced.
The purest "natural" salts were formed when whole seas evaporated - in Europe, this is mostly the Middle-terranean, of course. That was the source of many salt deposits that were extensively mined in Europe. Since such huge volumes and areas evaporated, the different salts precipitated out of the solution at different times and in different places, so you'll find quite pure "rock salt" formations - 85% and upwards is quite typical.
"Refined" salt is produced by purifying salt sources - the resulting salt might be up to about 99.75% NaCl or so. The source for the refining might be anything from sea water to rock salt to "whatever salty thing came by".
So unless the refining process itself introduces harmful impurities, refined salt will be much more pure NaCl. Is that a good thing? I don't know. There's certainly nothing inherently good or bad about purity - perhaps the other impurities are useful (iodine); perhaps they are harmful (kidney stones fun :P). And that's what you should think about when considering marketing claims such as "Himalayian salt is better because it's not processed". Processing is generally a good thing - you probably prefer your meat cooked, and your vegetables devoid of E. Coli, don't you? Just like home processing, industrial processing can make things better or worse, or a mixture of both. That's just the way the world works :)
If you find that you like the taste of Himalayian salt, that's the impurities talking (assuming you can actually tell the difference from a placebo). Nothing inherently bad about that, enjoy yourself ;) And don't forget that it can also be human-made for a good bit of profit :P


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Measuring sweetness
According to my textbook, sucralose is about 500 times sweeter than sucrose. How is this measured? Does one have to use blind taste tests, or is there a quantitative chemical way of measuring sweetness?

Sweetness isn't a chemical property and it's perception may vary, even extremely.

According to Ullmann’s Encyclopedia of Industrial Chemistry, the determination of sweetness intensities requires sensory tests. Therefore, panellists are used, which have to be adapted to the test procedures according to their sensory abilities. The results have to be evaluated on a statistical basis because of individual variations in sweetness perception.
The sweetness intensity factor (or sweetness power) is the factor by which an sucrose solution with equal sweetness is more concentrated than a solution of the sweetener. However, such sweetness intensities depend on the sweetener concentration and the sucrose concentration used for comparison. Commonly used values are often based on comparison with sucrose solutions with a concentration of 0.1 mol/l or similar.


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Equilibrium constant in a reaction with stoichiometric coefficient 1/2
I calculated the equilibrium constant for the follow reaction at 298 K.
$$\ce{CH4(g) + 1/2 S8 (s) <=> CS2 (l) + 2 H2S(g)}$$
I got $1.65 \times 10^{-9}$. However my professor wrote the answer as $2.66 \times 10^{-18}$ (so the square vale of my value). And I wonder why? I checked my answer 3 times already and I get to the same result. Is it because of the coefficient 1/2? Do we need to treat our equation with integer coefficients?

We need more information! What are the partial pressures of the gases at equilibrium? You can ignore solids and liquids as they have activities of 1. $K= (\ce{H2S})^2/(\ce{CH4})$ Does this tally with your result?

There is no more information. Except a table of thermodynamics standard values. $K_p$ is calculated using $K_p = e^{\frac{-\Delta G^0}{RT}}$

There is that method too! But yes in general you would use lowest integer ratios to calculate this and that would lead to a doubling of $\Delta_rG^O$ and hence the square of the $K_p$ value. Although I would say in the above case you were asked for the $K$ value for the given reaction with that stoichiometry and I would award you the marks in an exam situation. If the question is ambiguous or not precisely stated then the professor is at fault. Now, however you probably have learned more from this exercise than if you got it right first time!

Check this question for further clarification...http://chemistry.stackexchange.com/questions/9034/why-does-the-equilibrium-constant-change-on-scaling-the-stoichiometric-coefficie?rq=1


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Can an ether be selectively hydrolysed in the presence of an ester?
What reaction, or protection/deprotection can be used to transform the ether group of a molecule  that also has an ester compound into an alcohol while keeping the  ester unchanged? For example:

The usual way of removing ethers, by acid hydrolysis, would also hydrolyse the ester. Could that preserve the ester under certain conditions?

There are certain ethers which are somewhat easier to remove than others. If you stipulate that the ether must have a CH2 group immediately adjacent to oxygen, then a couple of possibilities, off the top of my head, would be:

Benzyl ethers (R' = Ph), or substituted derivatives thereof (e.g. PMB, R' = 4-methoxyphenyl). These are typically removed with catalytic hydrogenation, or under reductive Birch-type conditions. A related choice is the benzyloxymethyl group (R' = OBn), which also comes off with hydrogenation. Electron-rich benzyl ethers can be removed by oxidation (e.g. DDQ, or CAN).
Allyl ethers (R' = CH=CH2) can be deprotected with Pd(0) or Ni(0) in Tsuji–Trost style chemistry.

If you don't need a CH2 group next to oxygen, then there are many more choices:

Silyl ethers can be removed with fluoride ion.
THP ethers can be cleaved under very mildly acidic conditions (a weak acid such as PPTS is sufficient). Esters usually need slightly more forcing conditions to be hydrolysed.

However, in general, a random ether (e.g. R = Et) would likely prove difficult to deprotect in a selective fashion. You could try using BBr3, but it might end up being very messy. Otherwise, ordinary ethers generally require very forcing conditions to be removed. There is a review on methods of effecting ether cleavage here: Tetrahedron 2005, 61 (33), 7833–7863 and a number of ways to cleave simple alkyl ethers are listed, but many of these are very harsh and are probably not very general.


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Interpretation of an H NMR

I am fairly new to NMR and require assistance to analyze and interpret the NMR I have provided (sorry for it being vertical). This is an unknown compound that I have derived during one of my lab classes and it would help me a lot to know and understand what compound this is.
This is what I have interpreted so far:
The compound should have between 12 - 13 hydrogens, probably 13 if there is an alcohol involved and no saturated bonding.
the multiple peaks at slightly below 4 ppm can be a ring but also can be a $\ce{CH2}$ near an oxygen and also bonded to a $\ce{CH3}$ group.
The multiple peaks at around 1.5 ppm are also $\ce{CH2}$ group bonded to another $\ce{CH2}$ group and further away from any oxygen
The triplet at slightly below 1 ppm is a $\ce{CH3}$ group attached to a $\ce{CH2}$ group
However, my interpretations don't seem to add up and I am still figuring out the double peak.

Hi and welcome to Chemistry SE. We are a supportive community who are happy to help you with problems you might have, however we are not a question and answer service. Without providing additional information showing what you have already deduced from your own interpretations, and specifically what you don't understand (rather than just 'what is my molecule'), it is very likely that this question will be closed under our homework policy.

You may need a few more pieces of information such as the molecular formula. If someone drops a spectrum on you without telling you what it is, it's pretty hard to offer any concrete advice.

@ITTsoi generally yes, but not in this case. This is a very simple molecule which is very easily recognisable.

This appears to be a homework question, please share your thoughts and attempts towards the solution. You might want to start an summarise the signals and their structures and coupling constants. You can also take a look at typical shifts for proton NMR. You can [edit] your question to include these efforts and it then can be reopened.

@long true, but always good to show your working from scratch with as much evidence, especially for beginners to help build the correlations and familiarity.

@long My first pass gives a 'no-way-is-this-being-asked-on-stack' result.

How did you arrive at thirteen hydrogens? Saturated compounds are typically $\ce{C_{$n$}H_{2$n$ + 2}}$

@jan, I meant to say that there can be 13 hydrogens if there is an alcohol attached and no double bonds. The reason I am inferring alcohol is because my professor "tipped" me toward there being a tertiary alcohol somewhere.

No, a saturated monoalcohol would follow the sum formula $\ce{C_{$n$}H_{2$n$+2}O}$.

@jan, Yes thank you! I forgot about the hydrogen on the alcohol itself facepalm

You have now given yourself some information to go on with, and indeed have identified most of the important fragments. Now, the tricky bit is putting them all together! Let's work through what you've got.
Let's first assume that the sample is pure, and the integrals are correct to within reasonable error. This then gives you 5 obvious environments with ratio 3:3:4:1:1. Environments with 4 equivalent protons are not that common, so we might suspect there is some overlap here, and that we have more than one group. So, I count 12 protons.
The peak at 0.9 integrates to 3 - it is most likely a methyl group, and is closely resembling a triplet, so we can make the guess that it is next to a CH2. (It is in fact not a true triplet, and is immediately recognisable as a terminal methyl for a longer alkyl chain - see this link for more information). But for now let's call it a triplet.
At 1.2, we also have a -CH3 (integrates as 3), which is a doublet, so must be next to a -CH.
The multiplet mess around 1.4 must contain at least one -CH2 (to couple to the methyl at 0.9), which will leave us with another -CH2, or perhaps two -CH groups.
That broad peak at 1.72 is characteristic of an alcohol peak. It is also indicative of a sample which, if not run almost neat, is very concentrated and in a dry solvent. Its integral is not quite correct, but probably because of some excess water in the sample. But it is broad, and in the correct region for an alcohol. You could confirm this by doing a D2O shake, and seeing this peak disappear.
And, at 3.8 you have a -CH which is coupled to more than one thing. You have reasonably proposed that it is adjacent to a -CH3 and -CH2- Its chemical shift suggests that it is near some electronegative group or deshielding group. Knowing you have an alcohol might help narrow the possibilities here. The chemical shift is not indicative of a ring on its own. An alkyl ring would still be to higher field.
So, let's summarise what you know:

CH3CH2-
CH3-CH(OH)-CH2-

I think there is perhaps only one way to connect these fragments....
The key to these types of questions is that once you propose a structure, you can then go back and rationalise what you see to test the structure. Sometimes, you might have to come up with several structures before you get the correct one. But it usually best to start with the simplest, most probable and start eliminating from that.
If you still have problems with understanding any of this, ask some more, and we can go from there.

Thank you a lot, long. This really helped me understanding the peaks much better.  And I think the compound is named 2-Pentanol?

Yes, my question has been answered very well.