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Can I grow chlorapatite crystals at home?
Can I grow chlorapatite crystals, $\ce{Ca10(PO4)6(Cl)2}$, at home?
Is it as simple as ordering the elements online and throwing them into a beaker, and warming it up maybe?

I don't think you can buy the elements online, unless you happen to be in a lab at some reputable institution. Anyway, simply mixing the constituent elements wouldn't work at all.

Calcium is a food supplement. Chlorine is a pool chemical. And Phosphate is a fertilizer. I looked up how minerals form and it looks like they form by cooling down/evaporating...

why wont it work?

im totally clueless atm, any hint would be nice

Calcium may be a food supplement, but that does not mean you can buy elemental calcium (or, God forbid, add it to your food). There are more obstacles with other elements. Oxygen is a gas, so you can't simply throw it into a beaker. Ditto for chlorine, which is also pretty toxic. No, chemistry totally doesn't work like that.

Chlorine tablets that you buy do not contain elemental chlorine. I sincerely hope you do not come into contact with $\ce{Cl2}$. And yes, chemistry just doesn't work that way at all.

so what you guys are telling me is: there is no calcium in calcium food supplements,  there is no chlorine in pool chlorine,  and I cant add oxygen into a beaker. What a world, what a world. I better give up and go eat a peanut-butter and jelly sandwich. Thanks for the hints.

Why, of course there is calcium in calcium food supplements, only it is not in elemental form - that is, calcium is trapped in some compounds containing other elements you don't need in your apatite, and they won't just magically go away.

perhaps adding some sugar and spice will make everything nice?

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. We do very much appreciate, if you would limit posting questions to a single site. Cross-posting is not well received in our community.

beggars cant be choosers

It might be a lot harder than you think it is, because essentially every other element will influence crystal grow and purity. I found an interesting article by Esther García-Tuñón et. al., which is public access, outlining one of the possible routes to go. There might be more, but I am only using this as an illustrative example.
Here is the experimental procedure (shportened):

$\ce{CaCl2}$ [...] and β-TCP [$\ce{β-Ca3(PO4)2}$, [...]] powders were mechanically blended in different proportions in an agate mortar [...] for 12 min. Prior to mixing the $\ce{CaCl2}$ flux was dehydrated at 110 °C for 2 weeks. Thermo-gravimetric analysis [...] of the dehydrated flux showed that the remaining $\ce{H2O}$ percentage was 8wt.% and water is fully eliminated at 300 °C. After milling, the mixtures were uniaxially pressed at 60 MPa into cylindrical pellets (2.54 cm diameter) using a stainless steel dye. Five pellets (~12–15 g per batch) were placed inside a Platinum crucible and the growth process took place in a rapid melting furnace [...] at 1,100 °C for times ranging between 15 and 240 min. The heating rate was 10 °C min-1, and the cooling rates varied between 1 °C min-1 and quenching in air. [...] After cooling, the sample was washed, by rinsing in deionized water to dissolve the leftover flux. The crystals were vacuum-filtered and washing was repeated three times. Subsequently they were sieved trough a 63 μm mesh, and dried at 110 °C for 20 h.

As you see, the procedure requires high temperatures and pressures. If there was an do-it-at-home method I would assume they would have gone for that instead.
The reasons for that can be found in the discussion section.

Three different mixtures of CaCl2 and β-TCP (Fig. 1) were used to synthesize ClAp crystals. This system was selected to avoid the presence of additional species that could get incorporated into the ClAp structure, such as potassium or carbonates from the flux or the starting reagents [...].

Reference: Esther García-Tuñón, Ramiro Couceiro, Jaime Franco, Eduardo Saiz, and Francisco Guitián, Author manuscript, available in PMC 2013 Oct 1. PMCID: PMC3638812
Published in final edited form as:
J Mater Sci Mater Med. 2012, 23(10), 2471–2482.
Another possibility is the following paper, to which I do not have access. Here is the abstract in full:

The first hydrothermal growth of single crystals of chlorapatite is reported. Crystals grown from the system chlorapatite-HCl-H2O at 50 000 psi and pH = 1 with the growth zone at 465 °C and dissolution zone at 360 °C are found to be of high stoichiometry.

This seems also quite extreme to me.
Reference: A. Roufosse, M.L. Harvill, O.R. Gilliam, E. Kostiner, J. of Crystal Growth 1973, 19(3), 211-212.


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Are there any other (meta)stable oxygen molecules beyond O3?
Apart from Ozone, are there any longer oxygen molecules that are at least stable for short amounts of time? What configurations would that be?

According to my German version of Holleman-Wiberg: Inorganic Chemistry there is one additional, very short lived species, the so called tetraoxygen. A short article is also available on Wikipedia.
Wiberg describes it's synthesis from dioxygen in a mass-spectrometer employing the "neutralisation-reionisation-technique" (I am translating and paraphrasing at the same time, there might be something lost in translation.) It's stable time frame is in the micro second area. The decomposition barrier of $\ce{O4}$ is about 25 kJ/mol. The structure is not known, but most likely an aggregate of two oxygen molecules, i.e. $\ce{(O2)2}$. In lower concentrations $\ce{O4}$ might be found in liquid oxygen and they may be a template for the  ε-high pressure modification of solid oxygen forming at 0.6 Mbar.

Holleman, A., Wiberg, N., Wiberg, E., et al. (2008). Lehrbuch der Anorganischen Chemie. Berlin, Boston: De Gruyter. Kapitel XIII 1.1.4 Kurzlebige Sauerstoffspezies. S. 513f.  (in german)
Translation of the 101st edition available as Egon Wiberg and Nils Wiberg (2001) Holleman, Wiberg. Inorganic Chemistry. Academic Press.

According to wikipedia the epsilon phase of solid oxygen referred to in the answer has been shown by X-ray crystallography to contain O8 molecules. See https://en.wikipedia.org/wiki/Solid_oxygen

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Why isn't viscosity considered a property of solids?
Viscosity is defined as a measure of resistance to flow which arises dues to internal friction between layers of fluid as they slip past one another while liquid flows.
Why can't we say that viscosity is a property of solids as well? Except, it would then be said that solids have infinite viscosity.

Why, of course you can. It won't even be infinite, just very large. Think of the flowing glaciers or the Pitch drop experiment.

And to confuse the issue further, there are non-newtonian fluids, such as mayonnaise (which becomes less viscous on stirring), and "oobleck' (corn-starch in water, which behaves as a solid under pressure). And glasses, solids for all practical purposes, which never actually crystallize...

@IvanNeretin, That's the definition given in my book( NCERT Chemistry Class11)..so it got me wondering..     At the same time, the Pitch drop experiment , that you've mentioned, is for highly viscous  liquids, but ultimately not a solid.Right?

True, bitumen might be considered a highly viscous liquid, but what about ice? It is as solid as solid can be; it is crystalline. And yet glaciers flow to the ocean.

@IvanNeretin ,don't know much about the flowing glaciers, but isn't it something like ice flowing on water, instead of viscosity of ice itself

Well, it is definitely not that a whole glacier moves like one huge solid body sliding on the wet bedrock. No, its shape changes in the process; in fact, it looks mighty like a river, just an immensely slow one. Look up "ductile zone".

So what you are saying is, it is ice(some part of the glacier) moving over ice. But, if its shape changes, that means a part of the glacier moves over the melted ice, which is now water...so  basically,it is ice moving over water...

You seem to apply some kind of circular logic in assuming that ice can never be made ductile (despite the link above), hence if it moves, it must have melted. I don't think it works that way, but I don't know a great deal about glaciers either. So let's just say that the answer to your initial question is "yes", the solids do have viscosity; whether it is truly infinite or just extremely high (that is, infinite for most practical purposes) is not all that important.

The concept of deformations of solids under loading has a long history. It just isn't called 'viscosity' - instead this is just part of the mechanical properties of solids.
Relevant places to start include:
Wikipedia on Plasticity
Wikipedia on Deformation
The Non-Destructive Testing org on Elastic/Plastic Deformation
In the purely elastic regime, if you pull on a steel bar it will extend under the load, and will return to its original dimensions when the load is removed. If you apply too large of a load, it will plastically deform, so that the dimensions will not be the same when the load is removed. Under some circumstances even fairly small loads over long times can result in permanent deformation.
Physically, these can be a result of:

Dislocation formation and movement in crystals (e.g. slip bands)
Grain boundary motion
Creep (point defect formation and diffusion)
Crazing in plastics (formation of open volumes)
And a number of other, less common, mechanisms.

These topics are typically covered in metallurgy and materials science courses/textbooks. Again, even in amorphous solids, one would not tend to call it viscosity - that is reserved for liquids.

And in geology ; salt domes are the result of solid flow. And in oil/gas wells, flow of rock must be handled by high compression strength casing. And in extreme flow situation ,two concentric casings are used with cement between them.  Different types of solid rock flow at different rates.

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Estimate Partition Function from a simulation
I have a system and I've carried out a long molecular dynamics simulation over it. I would like to estimate the partition function $Z.$ Theoretically, one would compute:
$$Z=\dfrac{1}{N!h^{3N}}\int \exp \left(-\beta\frac{-p^2}{2m} \right) \exp\left(-\beta V(r) \right) \ dp \ dr, $$
but if the system I'm dealing with is, say, a protein in water or something like that, this integral is of course absolutely untractable. When thinking of how to estimate the partition function of susch a system from a molecular dynamics simulation, I guess the first thing that comes to mind is the following: Consider all the snapshots obtained during your simulation, label them with the index $j$ and call $E_j$ to the energy of each snapshot. For giant thermalized systems, I guess we can neglect the probability of a concrete configuration repeating, so I believe a nice try would be:
$$\tilde{Z}=\frac{dV_{\text{ps}}}{N!h^{3N}}\sum_{j} \exp\left( -\beta E_j \right),$$
where $dV_{\text{pc}}$ is the volume of small cube in phase space (its volume would have to be adjusted manually depending oh how many snapshots one has taken).
My question, briefly put, is, how can I estimate the error of estimating $Z$ by $\tilde{Z}$?
It seems intuitive that typically $\tilde{Z}$ should be a good aproximation to $Z,$ but one can easilly think of situations where this isn't true: Imagine that my initial configuration lies in, so as to speak, a valley in the energetic landscape, surrounded by a chain of "high mountains" (high-energy configuration) and that at the other side of these "mountains" lie low-energy configuration. It is clear that in such a case there could be a considerable error when estimating $Z$ from $\tilde{Z}.$ But since one does not know the aspect of the energetic landscape beforehand, it seems difficult to deal with this.
The question stops here, but just to provide context I'll explain briefly what I'm thinking about: I want to study a chemical reaction, and to that end I would like to know the free energy profile along a certain reaction coordinate. For that, an Umbrella Sampling technique is used: you drag the reaction coordinate along the reaction path, forcing it to be at certain positions at different steps (otherwhise, the simulation wouldn't take you over the energetic barriers in short simulation times). I am aware of several methods to compute free energies, like FEP or WHAM, which is quite sophisticated, but I was guessing that if the technique I mentioned above to estimate the partition function $Z$ was correct, then the goal of getting the free energy profile would be fairly simple: you just drag along the reaction coordinate, perform molecular dynamics simulations for different restricted values of the reaction coordinate (you force the reaction coordinate to be at a concrete position by means of a harmonic potential), and at each value of the reaction coordinate you estimate the partition function as above. That's the idea and that's why I am interested in having some idea on methods to estimate the error of approximating $Z$ by $\tilde{Z}.$
Note: this question was originally in Physics.stackexchange. It didn't receive any attention at all and I asked for it to be migrated to Chemistry.stackexchange, but the flag was ignored, so I've decided to delete my question on Physics.stacexchange and post it here

What you're saying does make sense. You can, in fact directly estimate the partition function using nested sampling, with error bounds. See for example this paper http://journals.aps.org/prb/abstract/10.1103/PhysRevB.93.174108 . In your case without PBC it's actually even simpler. Also check out Livia's blog here https://liviabpartay.wordpress.com/publications/

There's a deeper problem, in my opinion: your quantity \bar{Z} will not go to the partition function in the limit of infinite sampling. The problem is that your configurations j are already drawn from a Boltzmann distribution (if your simulation is at a defined T), and your expression for \bar{Z} then weights these Boltzmann-distributed energies by a second Boltzmann factor.
Getting the partition function (or configuration integral) from a thermalized simulation is not easy, and hence is not typically attempted.
You might be interested in a discussion of this in Andrew Leach's book, Eqs 6.21-6.28, in the 2001 edition.

Yes, this is precisely what I was worried about. I understand you. Weighting the energies with the Boltzmann factors would make sense if the different snapshots were uniform in the configuration space, but the sampling is already biased towars more probable regions.  So, to estimate the partition function, I would need the density of states... Is the Wand and Landau algorithm the best way to do that?

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Role of NaOH in synthesis of Dithiocarbamate?
What is the role of NaOH in a given reaction. whether It act as a base to removal  proton from amine or just to provide Na ion for the synthesis of sod. dithiocarbamate?

Proposed mechanism

$\ce{NaOH}$ is used as a base to remove the amine proton, which improves the nucleophilic power of the molecule. This way the addition to the double bond in $\ce{CS2}$ is favoured.

NaOH can't deprotonate amine.

@Mithoron to an appreciable extent, maybe, but it's not like it can't happen at all. This seems like the most likely mechanism to me.

@Mithoron But once the dithiohemicarbamate is formed, sodium hydroxide can remove the proton on the nitrogen and prevent the return reaction. (Sometimes just the order of steps matters.)

@Mithoron, I agree with your opinion, I think NaOH only act as source of Na ion,

@Jan, i proposed a mechanism and edited my question


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where will be highly electronegetive atom placed?
consider a molecule PCl3F2 here F is more electronegetive than Cl, which of them should occupy axial position??when i searched on internet i found these two images , they increased my confusion instead of solving my problem.can anyone breifly explain me the which one would be on axial position(F or Cl).

also http://chemistry.stackexchange.com/questions/45058/in-pcl2f3-why-are-two-cl-atoms-not-on-axial-position

http://pubs.acs.org/doi/abs/10.1021/ic50022a021?src=recsys


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Multiplicity of orbitals in a magnetic field
Solving past questions preparing for a test, I am stuck at number 11. Is multiplicity the ml value

$m_l$ is the magnetic quantum number, and multiplicity refers to the number of allowed values that $m_l$ can be.

I found strange the text "in a magnetic field".

@user1420303 Why? In an atom, an applied electric field will only shift the energy levels, not split them. You need an applied magnetic field to do that.

I agree with your answer.  But the number of allowed values for $m_l$ (for some $l$) is 2$l$+1 when no magnetic field is applied. As the typical hydrogen atom case. In presence of an external magnetic field there would be further lost of degeneracy. That is why I do not get why to state "in a magnetic field" in 11.
I would not say that there is not split of energy levels in presence of an electric field (for example, crystal field theory).

I was overthinking it; just because the energy does or doesn't change has no affect on the allowed values. I was also only thinking of the electric dipole approximation between p orbitals. But yes, it is more general than "in a magnetic field".


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Balancing the reaction equation for the oxidation of lead(II) sulfide with ozone
While I was reading about the chemical reactions of ozone, I came across this reaction.
$$\ce{PbS + 4O3 -> PbSO4 + 4O2}$$
Though the reaction seems well balanced, but when I attempted to balance it myself, I did this as $$\ce{PbS + 2O3 -> PbSO4 + O2}$$
I understand that there must be some reason that the 1st reaction is given in the book. I'm just not able to figure out why.
Can somebody give an explanation as to why is it so?

This is a semi-standard example of why 'naive' balancing does not work.
Ozone is a source of atomic oxygen, producing free oxygen molecules. So, in 'mild' conditions only one oxygen per ozone molecule would react as a strong oxidizer, and the remaining molecular oxygen would require elevated temperatures to react. Thus the equation would be
$\ce{PbS + 4O3 -> PbSO4 + 4 O2}$
Similarly, in excess of reducer at elevated temperature (though why would you want it with ozone is another matter) all atoms in ozone would react. Thus, the reaction should be
$\ce{3PbS + 4O3 -> 3PbSO4 }$

Because remember ozone always gives nascent oxygen, that O single atom, which is a  very strong oxidizing agent. Your mistake here is that you are involving all 3 oxygen atoms in the reaction which isn't the case.


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Copper sulphate, Benedict's Reagent and albumin
When my student's add Benedict's Reagent to albumin and heat it, the solution turns purple (as if testing for proteins).  Is it possible that the copper sulphate in the Benedict's is reacting with the protein in the albumin?

Yes, it's most likely the same reaction as in the Biuret test that detects peptide bonds.

Suggest above comment be moved to answer.


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Experimentally proving the mechanism and rate laws of SN1 & SN2 reactions
I have been given the task of experimentally proving the rate laws and mechanisms of $\mathrm{S_N1}$ and $\mathrm{S_N2}$ reactions. Unfortunately, the only equipment I can use is limited to analytical equipment that is needed to complete A-Level and IGCSE chemistry courses. So, I cannot use NMR machines, mass spectroscopy/spectrometry, etc.
It's obviously very easy to look at a nucleophilic reaction and see how fast it goes (using $\ce{AgNO3}$), but I do not know how I would go about quantifying this, i.e. actually numerically measuring the rate so that I can use the data I collect in a rate law equation.
Furthermore, I do not know how I would definitively prove the mechanism of $\mathrm{S_N1}$ and $\mathrm{S_N2}$ reactions, through experiment. I originally, thought of measuring the rate of reactions with different primary, secondary and tertiary haloalkanes and try and use the rates I get to prove that they are undergoing $\mathrm{S_N1}$ or $\mathrm{S_N2}$, and go from there. But I don't think that would be deemed as definitive proof that what is going on is definitely, without a doubt, the exact $\mathrm{S_N1}$/$\mathrm{S_N2}$ mechanism.

Can you please clarify what analytical equipment you can use? That significantly changes the possibility of the ways to address the question.

I did an experiment of this type using titration as an undergrad. Set up the reaction with varying concentrations of hydroxide nucleophile and halide. Titrated after a few methods and used the method of initial rates.

@JM Of course. I'm currently at a secondary school, doing A-Level and Further Chemistry and so the equipment I can use is any equipment in the Dept. Prep Room. This includes, all analytical equipment that is needed to complete A-Level and IGCSE chemistry courses. So, overall, not much I'm afraid. I'm sorry that I cannot be more specific than that; I would be able to elaborate more, after I have visited my Chemistry Department tomorrow, if you need more specifics.


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Composites of Immiscible Solids - Aluminum and Lead Example
The $\ce{Al}$-$\ce{Pb}$ (aluminium-lead) phase diagram clearly depicts that when cooled from the liquid state, aluminium and lead are not miscible in the solid state, i.e. the crystal structure of only one of the components prevails while the other remains a liquid or does not interact in any way with the other in its own solidification process.
What does this have to do - if anything - with the fact that a solid combination of $\ce{Al}$ and $\ce{Pb}$ is not classified as a composite?

You are correct that solid Al (fcc) and Pb (fcc) are (surprisingly for both being fcc) not miscible in each other (well, there is always some miscibility as driven by entropy). Heck, there is even an miscibility gap in the liquid phase as well. Al and Pb really don't like each other.
So, as you cool the liquid from above 1418C (the top of the liquid miscibility gap), you will phase separate in the liquid phase. At the aluminum melting point, the aluminum will solidify, and the Pb will remain liquid. When the Pb melting temperature is reached, the Pb will solidify. The exact specifics of the solidification will depend on how it is done (the kinetics), not the thermodynamics. This means you can't predict the final micro-structure without  details of the process.
The final result will be solid Al and Pb, likely intermixed. It is not an alloy, but a two-phase mixture. I would not call it a composite as I would generally reserve that name for actual composite materials.


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Identify two amines
I was reading a random text about the surface chemistry of YBaCuO superconductors. In the text, there was a table. The table had a list of several chemical compounds, including these two:

Alkyl Amine: FcC(O)NH(CH2)4NH2
Aryl Amine: p-Fc-C6H4-NH2

Two amines. After extensive search, I was unable to find a structural formula for it (the closest I've found was this). I even tried to draw all structural formulas I could think off in a piece of paper. But I couldn't draw a single one satisfying the second one (the easiest probably).
Can you identify these amines? What's their structural formula?

What is 'Fc'? Or is that part of the question?

@gilleain Its a good question. I have no idea what Fc means. I assume it most be something from coordination chemistry. In my searchs, I simply ignored the Fc- and the p-Fc-.

Fc = ferrocenyl; and p-Fc = para-ferrocenyl.

The proposed strucutre of these compound are, correct me if i am wrong.

$\ce{Fc}$ stands for ferrocenyl.
So the first compound is an amide of ferrocenecarboxylic acid and tetramethylene diamine, and the second compound is p-ferrocenyl aniline.


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Electrostatic potential map and local ionization map
What's the difference between electrostatic potential map and local ionization map?
I think that the main difference is that the electrostatic potential map doesn't take into account resonance (delocalization), but only electronegativity, therefore nucleophilicity or electrophilicity of a species is much better estimated by using the local ionization map. Am I wrong?

Can you provide a reference for the local ionization map?

Huh? I've just started using Spartan and I couldn't find a neat explanation in the manual.

Ok, it's from Spartan; that's what I meant by "reference". I've never heard of it before, but now I have a starting point.

The excellent MultiWFN is able to produce both functions and luckily, a reference for it has explanations for both as well:
Electrostatic potential
According to the reference above,

$$V_{tot}(r)
    = V_{nu}(r) + V_{ele}(r)
      = \sum_A \frac{Z_A}{|r - R_A|} - \int \frac{\rho(r')}{|r - r'|} dr'$$
  This function measures the electrostatic interaction between a unit point charge placed at $r$ and the system of interest.
  A positive (negative) value implies that current position is dominated
  by nuclear (electronic) charges.
  Molecular ESP has been widely used for prediction of nucleophilic and electrophilic sites for a long time.
  It is also valuable in studying hydrogen bonds, halogen bonds, molecular recognitions, and the inter-molecular interaction of aromatics.

Average local ionisation energy
Again, according to the reference above,

$$\bar{I}(r)
    = \frac{\sum_i \rho_i(r) |\epsilon_i|}{\rho(r)}$$
  where $\rho_i(r)$ and $\epsilon_i$ are the electron density function and orbital energy of the $i$th molecular orbital, respectively.
  This function has many uses, for example, reproducing atomic shell structure, measuring electronegativity, quantifying local polarizability and hardness, and predicting sites for electrophilic or radical attack.

Thus, while the electrostatic potential represents the hypothetical electrostatic interaction energy between the molecule and a positive unit point charge at $r$, the average local ionisation energy is the orbital density-weighted absolute orbital energy average.
Finally, resonance/delocalization/electronegativity are built in the method you use to produce the electron density $\rho(r)$ (edit: a nice discussion about it can be found here).


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Comparison of organic and inorganic acid strengths
So I know how different factors affect acid/base strength, but I am not sure when they are all in the same question, which affects acid/base strength more.
So, for example this question:
How do I know that $\ce{HBr}$ is the strongest acid...because I also have the inductive electron withdrawal from Fluorine in answer choice B.

pKa value is great indicator of acid base strength, lower the pKa value stronger is tha acid and conjugate base of that acid is week base due to high pKa value.
HBr has pKa value (-9) strongest acid after HI (pKa -10).


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What is the pKa of methylamine
I completed a practice (not for marks) assignment for an organic chemistry course, and one of the questions asked for the pKa of methylamine.
I looked on PubChem, and it gave a result of $10.6$. Marking myself, I found that the answer key states the $\mathrm{p}K_\mathrm{a}$ of methylamine is $33$. Especially on a log scale, that is not a trivial difference.
I talked to my professor about it; he was confident the true $\mathrm{p}K_\mathrm{a}$ is in the low thirties and stated that I need to be skeptical of material online. His point in part was that there was no way methylamine was more acidic than water.
I am not personally convinced on either number yet.
What is the $\mathrm{p}K_\mathrm{a}$ of methylamine, and what evidence supports that number?

The value 10.6 is the $\mathrm pK_\mathrm a$ of the conjugate acid of methylamine.

Both sources are correct. However, they are referring to two different $\mathrm{p}K_\mathrm{a}$ values.
$$\ce{H3C-NH3+ <=>[$K_\mathrm{a1}$] H3C-NH2 + H+ <=>[$K_\mathrm{a2}$] H3C-NH- + 2 H+}$$
The second value is the one that your marking scheme refers to when it asks for the $\mathrm{p}K_\mathrm{a}$ of $\ce{H3C-NH2}$. However, it is of little practical relevance. Most of the time, the question is not how acidic methyl amine is but how basic. And for that, the $\mathrm{p}K_\mathrm{a1}$ value of the conjugate acid, methyl ammonium, is used. $\ce{H3C-NH3+}$ has a $\mathrm{p}K_\mathrm{a1}$ value of approximately $10$, and it is the value you found online.
Formally correct usage would only name $\mathrm{p}K_\mathrm{a2}$ as the value of methyl amine. Colloquial chemists’ usage mainly uses $\mathrm{p}K_\mathrm{a1}$.

Nice answer. I know what you mean for the equation with Ka,2 but that is a really weird way to write the reaction. I really think you should have written the reactions as two separate equations. $$\ce{H3C-NH3+ <=>[$K_\mathrm{a1}$] H3C-NH2 + H+ }$$  $$\ce{H3C-NH2 <=>[$K_\mathrm{a2}$] H3C-NH- + H+}$$

@MaxW That was actually my first thought as well, and I would have done it your way, but mathematically it cancels out so I suppose it is fine.

Also @Jan http://i.imgur.com/ZtWjUOY.jpg the Ka's above the arrows display a bit weird on mobile. (+1 anyway)

Of course it isn't Jan's fault that the interpretation of  the formatting is different on mobile phones. That absolutely isn't his problem. // Stupid !@#$%^&* mobile phones...

@orthocresol They don’t look perfect on PC either. Especially the equilibrium arrows are way to short. I think I should file a bug report to mhchem ;)

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Most likely intermolecular force in aqueous ethanol

The most likely interaction to be found in an aqueous solution of ethanol is
A. London dispersion forces
B. Hydrogen bonding
C. Dipole-dipole interactions
D. Ion-dipole interactions

I am split between B and C. I actually thought it was both of them but the question requires me to choose only one.

I really think the question is poorly worded. The phrase "most likely" is a probability consideration. The only answer that is really eliminated is (d) since few of the water molecules will be ionized. HINT - The only way to answer the question is to think of "most likely" to mean "most significant," that is the strongest type of bonding from the remaining three. So what do you think?

(A) is out as London dispersion (instantaneous dipole-induced dipole) forces are weak, always present, and do not contribute much to the main intermolecular forces of attraction.
(D) is out as auto-ionisation of water $$\ce{2H2O -> H3O+ + OH-}$$ is negligible.
That leaves (B) and (C).

Looking at their structures, hydrogen bonding will be the more dominant force as the Hs highlighted in the structures can be exposed to hydrogen bonding.

Some comments. (1) It isn't just about the right answer. This about the chemistry too. (2) Your answer doesn't really address the poor wording of the question. (3) In the last paragraph, hydrogen bonding involves more than just hydrogen atoms. In this case hydrogen bonding will be hydrogen atoms from one molecule to the oxygen atom on the other.

All  the lower alcohols are soluble in water due to hydrogen bonding(the compounds which can form hydrogen bond with the solvent molecules are soluble in such solvents).since ethanol is an alcohol thus it is bonded with water molecule  by hydrogen bond.see the bond between an alcohol and water in the picture


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Rank the following protons in order of acidity
I have an acid and base organic chem quiz tomorrow and I need help determining how acidic protons are. I understand the concept of atoms, resonance, induction, and orbital when considering the acidity of protons. However, when I'm given these molecules, it overwhelms me so I don't understand how to break them down. I believe that the first step is that I have to find the conjugate base for each one and then just compare? Below is photo of my worksheet. I have already filled in the correct answers. Thanks in advance for your help.

Authors managed to ignore amide H atom...

@Mithoron I think it's quite common (and deliberate) to ignore the amide hydrogen. I would guess that the overall topic is CH-acidity here.

Maybe you could try answering by the inductive effect on the carbon containing the acidic hydrogen. The electron cloud of the carbon c is more depleted than d which is more depleted  than b which is more depleted than a. However, I am not sure.

@Martin-マーチン Being common and potentially deliberate doesn’t make it better, though …

@Jan if I were the author, and I actually have asked these questions quite often, I would not do it any other way. The hetero atom is too obvious to count. If I were having a test about CH acidity, I would deduce points for giving that answer. You don't know the intend of the question, so you should not judge if it is better or who worse.

@Martin-マーチン In my opinion, that leads to people outright forgetting the heteroatom, and then wondering why they need two equivalents of LDA in a Myers alkylation (as the first example that hit my head). Each such question should be accomagnied by a note along the lines of ignore the amide proton.

There's instructional value in including this heteroatom imo. Not saying it's better or worse but it's also useful to know an approximate pKa for amide NH and amine NH compared to the various CH protons there (the 1,3-dicarbonyl and the carbonyl). In any case the question does strongly imply that only the labelled hydrogens are to be labelled, so there is no fault in the question.

the c hydrogen is more acidic because it is stacked between two electron withdrawing carbonly groups, and after the loss of hydrogen the carbanion is more stabilized, where as the same thing is not observed in case of d hydrogen the carbanion is in conjugation but only with one of carbonly group followed by b and c hydrogen.
now in case of second example, e is more acidic because it is allylic as well as next to the carbonyl group which means the anion after loss of hydrogen is more stabilized as come to f and g.


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Lewis structure
Which of He, F, P or N can not be the central atom in a lewis structure? Its a multiple choice question with only one answer. I have seen somewhere that He, F and H cannot serve as central atoms so I don't know how to go about this.

See http://chemistry.stackexchange.com/questions/7800/why-fluorine-doesnt-act-as-the-central-atom-in-inter-halogen-compounds for some insight.

"Central" seems a poor constraint given the atoms. Diatomic molecules don't have a "central" atom. I think the question would be  as Which of He, F, P or N can not be an atom in a polyatomic Lewis structure?

@MaxW Isn't a diatomic molecule already polyatomic?

@Martin - マーチン - Yes but I was trying to reword the question to avoid the He monoatomic molecule. I can draw a Lewis structure for a helium atom/molecule. Since it is the only atom in the molecule it is also the "central" atom. I also avoid the use of "central" since a diatomic molecule doesn't have a "central" atom. $\ce{N2}$ and $\ce{F2}$ of course have valid Lewis dot structures.

@Max I agree, the question is very ambiguous and the concept of "central atom" is also not very well defined. There is enough to go wrong here. I am not even sure if the original author of the question understands the very restricted nature of his choice of Lewis structure.

On another note: The question could use a more descriptive title.

@ Martin - マーチン - I agree that the question author wrote the question very poorly. // Also you certainly couldn't buy a bottle of some He compound. But we could debate if He forms compounds under extreme conditions. So if any He compound exists under any conditions then the whole question is shot to hell.

lekarane - Out of curiosity, what class was this for? what book are you using?

I know this is late after all the discussion, but I'll point out that the questions asks about a "lewis structure." That doesn't eliminate an atom, or ions.

The way I interpret the term "central atom" is that in a binary compound or ion $\ce{EX_$n$^{$z\pm$}}$ (with $\ce{E} \neq \ce{X}$), $\ce{E}$ is the central atom. This interpretation is obviously up for debate, especially with regard to diatomic molecules (where $\ce{E} = \ce{X}$), since the term is not well-defined.
What I want to point out is that fluorine and hydrogen can both act as central atoms, even if you neglect the case of $\ce{F2}$, which (according to "my definition") does not contain a central atom per se.
In liquid hydrogen fluoride1 the following autoionisation equilibrium ($K \approx 2 \times 10^{-12}$ at $273~\mathrm{K}$) is established2,3
$$\ce{3 HF <=> H2F+ + HF2-}$$
$\ce{H2F+}$ (the fluoronium ion) has fluorine as the central atom, is isoelectronic with $\ce{H2O}$, and is therefore bent.
$\ce{HF2-}$ (the bifluoride ion) has hydrogen as the central atom, with a 3-centre-4-electron bond, and is linear. (Some discussion can be found here.)

The only answer I would accept is therefore $\ce{He}$.

1 Please do not make liquid hydrogen fluoride at home. Unless you absolutely have to do so, please do not make it in the lab either.
2 Housecroft & Sharpe, Inorganic Chemistry, 4th ed., p 277
3 Weller et al., Inorganic Chemistry, 6th ed., p 147
n.b. for our German friends, I am sure Holleman-Wiberg describes it somewhere too but I don't have access to the book now.

But a He atom is a molecule. Since it is the only atom in the molecule it is also at the center of the molecule. // The question stinks. The author of the question didn't anticipate such fine reasoning about this question.

@MaxW IUPAC defines "molecule" as an electrically neutral entity consisting of more than one atom.

So PV = nRT doesn't work for He? // I know I am splitting hairs, but is this a question from Chem 101 or some Advanced Inorganic Chemistry class?

@MaxW I don't know where it says that the ideal gas law only applies to molecules. No assumption is made of the type of particles in an ideal gas, because it doesn't affect the model. To be honest, I do not really know what the answer is intended to be, nor am I interested in guessing the intent of the question author. I merely wanted to point out the existence of counter-examples to the statement that "F and H cannot be central atoms".

We're having a violent agreement. I think your answer is a black swan that makes the question even murkier than I supposed.

@MaxW Such is life, when you get poorly-worded, ambiguous questions in high school... :)

Interestingly there is quite a lot now about helium containing ions, but in my first skimming, I could only find examples where helium would act as a terminal compound. That is if you neglect $\ce{He_{(1;2)}@C_{60;70}}$.

@Martin - マーチン - Well the question doesn't eliminate ions and the article you pointed to talks about $\ce{{He_n}^+} $ clusters. So for $\ce{{He_3}^+} $ there must be a central He.

Ho-Wi: Lehrb. d. Anorg. Chem. (102. Auflage), Kap. XII 2., p. 449. (german edition). "Das Ionenproduct des in reinem flüssigen Fluorwasserstoff vorliegenden  Autoprotolysegleichgewichts $$\ce{ 3HF <--> H2F+ + HF2- }$$ beträgt ca. $10^{-10.7}$ bei $0~^\circ\mathrm{C}$ [...]." Roughly translated: The ion product of the auto-ionisation equilibrium in pure liquid hydrogen fluoride  (...) is about $10^{-10.7}$ at $0~^\circ\mathrm{C}$ [...]."

This answer is true until $\ce{HeF2}$ is found? ;)

I'm not sure what the official definition for a central atom is, but as I understand, it is a device used to figure out how to draw a Lewis structure.  Whichever is the most electropositive atom that can form enough bonds is, in general, a central atom.
In that case, $\ce{F}$ would be a "central atom" in the molecule $\ce{F2}$; $\ce{P}$ would be in something like $\ce{H3PO4}$; and $\ce{N}$ would be in $\ce{NH3}$.
However, since $\ce{He}$ is a noble gas in the first row, it never participates in covalent bonding and is thus never a "central atom." A
If the question is a multiple-choice question that comes from that kind of a trick-question place, then it might be possible.  If it is some kind of super-advanced chemistry, I am sorry that I am not qualified to give you a better answer.


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Rounding of atomic weight isotopes
When we look at atomic mass, for example, $\ce{^134Ba}$ (barium) has a mass of $133.90 \,\mathrm{Da}$. Why is this rounded to $134 \,\mathrm{Da}$? This is shown in my book.

More to the point, that 134 is the sum of the number of protons plus neutrons in the nucleus, so it must be an integer.

The way my teacher first explained this is due to the definition of atomic mass. According to Google:

The mass of an atom of a chemical element expressed in atomic mass units. It is approximately equivalent to the number of protons and neutrons in the atom (the mass number) or to the average number allowing for the relative abundances of different isotopes.

Now, the key words are

...the average number allowing for the relative abundances of different isotopes.

What does that mean? We go back to the definition of an isotope. An isotope of an element is a species that has the same number of protons in the nucleus, yet a different number of neutrons. For example, $\ce{C^12}$ and $\ce{C^14}$ are isotopes of each other. Sidenote, isotopes are not to be confused with allotropes, which are different forms of the same element. For example, graphite and diamond are allotropes of carbon, since both, either, or neither might be made up of $\ce{C^12}$ ($6p$, $6n$)  or $\ce{C^14}$ ($6p$, $8n$). Now how do we represent this with the periodic table? Simply, we take the weighted average. Now, I do not have enough information or data to calculate the relative abundances of each isotope of carbon, and the periodic table does not either. We have an average. The atomic mass of $\ce{C}$ is given as $12.011$. That means a few different things. Firstly, there are more isotopes that are closer to having a sum of $p+n=12$, and less that have $p+n=14$ or $15$. Secondly, this means that although this number is not exact, it represents the most common $p+n$. Averaging all of those different $p+n$'s up, from each isotope, multiplying they by their relative abundances, and averaging should give our periodic table value of $12.011$.
Now it is important to remember that this number is not rounded. You might be asking, "Then why, in our calculations, do we sometimes round this number?" Well it depends on the problem, because for many problems, you do not round, but for the ones that you do, it is because it is more convenient to assume that the number that is given rounded is the most common isotope. Was that sentence confusing? Essentially, it is easy and generally correct to say that $12.011$ for carbon can be rounded to $12$, and $12$, meaning $\{p,n\}\in\{6,6\}$. We can do this by assuming that carbon in its most common form has $6p$, and $6n$.
In orthocresol's comment above, he is also correct. He has a perfectly valid point regarding common word usage in speaking. However, I do not think that this is the main reason.
Regarding Jon Custer's comment above, yes, that would be true, but only if we had a single isotope, which we do not.
Now, make sure you can differentiate between atomic mass and atomic weight! :)
Hope this helps!

When a specific isotope like $\ce{^134Ba}$ is given, then there is no rounding or averaging necessary. The atomic mass is simply taken as the mass number, i.e. the the sum of the protons and neutrons in the nucleus. Thus the atomic mass of $\ce{^134Ba}$ is simply $\pu{134u}$.


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How to calculate the number of prochiral carbon atoms in a complex organic molecule?

Calculate the number of prochiral carbon in the following compound.

The Wikipedia page on prochirality states:

In stereochemistry, prochiral molecules are those that can be
  converted from achiral to chiral in a single step. An achiral species
  which can be converted to a chiral in two steps is called
  proprochiral.

My approach:
I first considered the sulfur atom to be prochiral as it satisfies the above condition, but it was pointed out, that this is incorrect as the S atom is already chiral.
My doubt:
But should even the double bond containing $\ce{sp^2}$ carbon atoms be considered prochiral?
According to what Martin postend in the comments I conclude that ("all" carbon atoms-4) in the compound are prochiral. I subtracted 4 chiral carbon atoms in the compounds. Tell me if there is still any mistake!
However, I am feeling that a carbon containing two hydrogen cannot be considered prochiral. The Wikipedia article states

If two identical substituents are attached to a sp3-hybridized atom, the descriptors pro-R and pro-S are used to distinguish between the two.

But H atoms are not considered substituents in organic chemistry, see substituent .

The S atom is already chiral.

Well, sulfur is not carbon. You should consider any carbon that could be a chiral centre with one conversion. Hint: Almost all of them are.

@orthocresol Thanks.I get you.But what about the double bonded carbons?

@Martin-マーチン You are correct.S should not be prochiral here.I guess you are referring to the double bonded carbons?

No, I am referring to all carbons. There is not a bit of symmetry in that compound.

@Martin-マーチン So say even a carbon containing two H should be considered prochiral?

Yes. It's in the very same article you linked: "If two identical substituents are attached to a sp3-hybridized atom, the descriptors pro-R and pro-S are used to distinguish between the two."

I consider the concept of prochirality as a property of an achiral compound (or centre), that can be turned into a chiral compound (or centre) in one step (proprochiral: in two steps) by means of substituting or addition. Hence all carbon centres with three different ligands (substituents or hydrogen) can be considered prochiral.
Prochirality is unfortunately a grown concept, which was used in different ways throughout the years and can be confusing at times. I find the definition in the IUPAC gold book quite enlightening and much better than the Wikipedia page that spawned off of it.

prochirality
Also contains definition of: proprochirality
This term is used in different, sometimes contradictory ways; four are listed below.

The geometric property of an achiral object (or spatial arrangement of points or atoms) which is capable of becoming chiral in a single desymmetrization step. An achiral molecular entity, or a part of it considered on its own, is thus called prochiral if it can be made chiral by the replacement of an existing atom (or achiral group) by a different one. An achiral object which is capable of becoming chiral in two desymmetrization steps is sometimes described as proprochiral. For example the proprochiral $\ce{CH3CO2H}$ becomes prochiral as $\ce{CH2DCO2H}$ and chiral as $\ce{CHDTCO2H}$.
The term prochirality also applies to an achiral molecule or entity which contains a trigonal system and which can be made chiral by the addition to the trigonal system of a new atom or achiral group. For example addition of hydrogen to one of the enantiotopic faces of the prochiral ketone $\ce{CH3CH2COCH3}$ gives one of the enantiomers of the chiral alcohol $\ce{CH3CH2CHOHCH3}$; the addition of $\ce{CN−}$ to one of the diastereotopic faces of the chiral aldehyde shown below converts it into one of the diastereoisomers of the cyanohydrin. The two faces of the trigonal system may be described as Re and Si.

The term prochiral also applies to a tetrahedral atom of an achiral or chiral molecule which is bonded to two stereoheterotopic groups. For example, the prochiral molecule $\ce{CH3CH2OH}$ can be converted into the chiral molecule $\ce{CH3CHDOH}$ by the isotopic replacement of one of the two enantiotopic hydrogen atoms of the methylene group. The carbon atom of the methylene group is called prochiral. The prochiral molecule $\ce{HO2CCH2CHOHCH2CO2H}$ can be converted into a chiral product by esterification of one of the two enantiotopic $\ce{–CH2CO2H}$ groups. The carbon atom of the $\ce{CHOH}$ group is called prochiral. The chiral molecule $\ce{CH3CHOHCH2CH3}$ can be converted into one of the diastereoisomers of $\ce{CH3CHOHCHDCH3}$ by the isotopic replacement of one of the two diastereotopic hydrogen atoms of the methylene group. The carbon atom of the methylene group is called prochiral. The stereoheterotopic groups in these cases may be described as pro-R or pro-S. Reference to the two stereoheterotopic groups themselves as prochiral, although common, is strongly discouraged.
  See: chirality centre
The term prochirality is also applied to the enantiotopic faces of a trigonal system.

The first definition applies especially to tetrahedral carbon atoms.
You can also work off of another definition which might be a bit easier to comprehend.

prochirality centre
  An atom of a molecule which becomes a chirality centre by replacing one of the two stereoheterotopic ligands attached to it by a different ligand, e.g. C-1 of ethanol; C-3 of butan-2-ol.

stereoheterotopic
    Either enantiotopic or diastereotopic. (In contrast the term constitutionally heterotopic has been used to describe identical groups located in constitutionally non-equivalent positions).

As a rule of thumb you can always substitute a $\ce{-CH2\bond{-}}$ group with a $\ce{-CHD-}$ group and ask your self if it is a chiral centre now.
I personally find it easier to identify stereoheterotopic ligands as this is also important for spectroscopy (esp. NMR).
This is somewhat also mirrored in the third definition.
The second (and fourth) definition applies to doubly bonded carbon atoms especially. If there are three different ligands, then you can assign a Re- and Si-face of that centre. Hence I would also consider these carbons prochiral.
In your case that means that all carbons (except for the ones that are already chiral) are prochiral.

Marked in pink are all chiral elements (the chirality of the nitrogen depends on the inversion barrier).
Marked in dark blue are all carbon centres who contain diastereotopic protons and hence can be considered prochiral according to the first definition. This is also evident from the $\ce{^1H-NMR}$ estimation of ChemDraw.
Marked in green are all carbon centres where you can add a nucleophile to create a stereo centre and are hence prochiral according to definition two.
Marked in orange are all carbons who are part of an aromatic system. Strictly speaking they are also trigonal carbons and technically you can assign a Re/Si to them. Accordingly they fall within the second definition and can be considered prochiral centres.

@glorfindel Thanks for finding all those broken images. I have removed the 'source' statement, as it is completely unnecessary, because the image was already part of the blockquote. (Also I personally dislike using  for these statements; it makes it appear less important, when they are most of the time not.)

@Martin-マーチン thank you for the improvement. The bot does scan the neighbour area of the image for links to the same domain (and if it finds one, it doesn't add attribution), but not as far as necessary to detect this link. I think I can manage to improve it by letting it consider the entire blockquote + neighbour area. In any case, this is one of the benefits of having them submitted as suggested edits, it encourages reviewers to improve the edit beyond what is possible through automated means.


text:
What is the percentage of d-tartaric acid in a mixture of d and l isomers?

Specific rotation for pure tartaric acid is $-12 ^\circ$. A mixture of d-l isomers is prepared which has a specific rotation of $+6^\circ$. What is the percentage of d-tartaric acid in a mixture of $d$ and $l$ isomers?

The pure tartaric acid must have been $l$-tartaric acid.
My attempt:
$$\frac{l-d}{l+d}=\frac{+6}{-12}=\frac{1}{-2}$$ So $d=3l$. Which implies $$\frac{d}{d+l}=\frac{3}{3+1}=0.75 $$
However the answer given in my textbook is $66.67 \%$. Most probably I didn't do any calculation error. So where did I go wrong?

Your calculation is correct.
You can check your result since you know that pure (+)-tartaric acid has a specific rotation of $+12^\circ$ and that pure (−)-tartaric acid has a specific rotation of $-12^\circ$. Therefore, a mixture of $75\ \%$ (+)-tartaric acid and $25\ \%$ (−)-tartaric acid has a specific rotation of
$$0.75\times(+12^\circ)+0.25\times(-12^\circ)=6^\circ$$
A similar exercise can be found in Hart, Hadad, Craine, Hart Organic Chemistry – A Short Course: Google Books
The corresponding solution can be found here: Google Books

And that results in $50~%~\mathrm{e.e.}$


text:
Is a sequence of two Wagner-Meerwein rearrangements possible?
We've just studied about the Wagner-Meerwein rearrangement in carbocations.
Our teacher told us that it occurs in order to stabilize the carbocation intermediate.
Knowing the carbocation stability order,

$\mathrm{Tertiary > Secondary > Primary} > \ce{CH3+}$

I decided to test the "limits" of this particular rearrangement reaction.

I drew a primary carbocation (A):

Upon subjecting it to a Wagner-Meerwein hydrogen-shift (in my head), I obtained the secondary carbocation (B):

Which was all fine and dandy; well within what I've been taught.
But seeing the potential for additional stability another Wagner-Meerwein rearrangement could provide, I conducted a Wagner-Meerwein alky(methyl)-shift and ended up with the following tertiary carbocation (C):

I started with a primary carbocation, changed it to a secondary carbocation and finally ended up with a tertiary carbocation.

Now I showed this to my teacher, but he said that in this particular case the carbocation (B) won't convert to carbocation (C); i.e- carbocation (A) isn't going to undergo two Wagner-Meerwein rearrangements (1,2-anionotropic shifts), because in effect it'd be a 1,3-shift and not the 1,2-shift as the Wagner-Meerwein rearrangement calls for.
A few days later, I took a different sheet with carbocations (B) drawn on it and carried out a Wagner Meerwein alkyl(methyl)-shift to obtain carbocation (C) just as before [except that this time around, my starting molecule was B not A]. I showed this to my teacher again, but the rearrangement I carried out in this case doesn't seem to perturb him at all. His argument is that, since my starting intermediate is carbocation (B), to reach carbocation (C), it undergoes only one Wagner-Meerwein rearrangement which is perfectly fine, unlike my first case which involved two rearrangements.

Now my question is, if he's right, then how would carbocation (B) "know" whether it already underwent a Wagner-Meerwein rearrangement [from carbocation (A)] before considering rearranging itself to become the more stable, carbocation (C)? How does it "know" if it's the starting intermediate or not.

Carbocations are reactive intermediates. They usually get snapped up by some other nucleophile around it. Essentially they do not have so much time to migrate so many hydrogens/alkyl groups to find what is the most stable form. However, if you do have the correct conditions, multiple alkyl shifts are indeed possible. A famous example being the synthesis of adamantane (the one with the AlCl3 step goes via a series of cationic rearrangements).

possible duplicate http://chemistry.stackexchange.com/questions/27835/can-hydride-shift-and-methyl-shift-happen-one-after-another-in-a-carbocation/

AHHH. Why do people keep proposing primary carbocations?! This is just not plausible under normal reaction conditions.

There is no reason why a sequence of two Wagner-Meerwein rearrangements wouldn’t happen — assuming low concentrations and thus the absence of directly capturing nucleophiles. In fact, a professor of mine once said:

The only secondary carbenium ion is the prop-2-yl ion; all others will rearrange to form tertiary carbenium ions.

(He did explicitly exclude carbenium ions stabilised by other means from that discussion; of course an oxocarbenium ion is well-known as is an allyl cation.)
A very well-known example of this is the biosynthesis of lanosterol, in which the enzyme lanosterol synthase takes squalene-epoxide, orders it in a boat-chair-boat conformation, catalyses a domino sequence of carbocations being captured by double bonds, followed by a sequence of four Wagner-Meerwein rearrangements and finally deprotonation to generate the backbone ring-structure of chloresterols. See the following scheme.

If nature can do it, it will happen in the lab at least to a minute extent.


text:
Can course salt melt into water
What can cause course salt to melt? If you put it on a flat surface, can it melt on its own into water? What causes this reaction to make the salt melt?

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. This appears to be a homework question, please share your thoughts and attempts towards the solution. Please also clarify to me what course salt is?

Salts can't "melt" at room temperature, much less into a water solution. What might happen for some salts is that they're able to draw humidity from the environment with such avidity that they become a liquid solution. If a compund can do this we call it deliquescent.
Source: https://en.wikipedia.org/wiki/Hygroscopy#Deliquescence


text:
Silica turning transparent during column chromatography with chloroform
Initially, my column had been loaded with n-hexane and ethyl acetate, however I ended up having to use a more polar solvent to elute my product. It was my first time running chloroform through a column packed with silica and I noticed that as the chloroform made its way down the column, the silica turned from white to transparent.
My initial hypothesis was that chloroform and silica must have the same refractive index. I looked up the values, and found them both to be ca. 1.45 (see references), so I am inclined to believe this explanation.
However, if someone here could weigh in, and either confirm/deny this. (Or, alternatively, venture a new hypothesis).
I am guessing no chemical transformation takes place when silica looses its opalescence, because if that were the case then the whole procedure becomes completely useless. However, I could be wrong.
References:
i. http://refractiveindex.info/?shelf=main&book=SiO2&page=Malitson
ii. https://en.wikipedia.org/wiki/Chloroform_(data_page)

Sounds right to me.

I even observe this using MC:MeOH as solvent system..

I have done a lot of rock and mineral collecting. If you wet the rock with water you can often see much more detail of the rock since less light is scattered. The rock looks a lot like you have polished it.

You are correct, sir!  (or madame)  The silicic acid becomes translucent (not transparent) in chloroform because their refractive indexes are similar.  This is a chance to check for air bubbles in the column, which reduces efficiency.  Also, any material still on your column will probably not have the same RI, and will be visible as a white or colored band, probably at the top of the column.

It would be more appropriate if you posted this as a comment. This doesn't add much to the discussion. Cheers!

He asked if it was the refractive index responsible, and I answered him.


text:
Why do the halogenations of benzene and toluene lead to different products?
The reaction of  benzene with $\ce{Cl2}$ in presence of sunlight breaks the double bonds and 6 chlorine atoms attach to the ring.
However, the same reaction done with toluene does not lead to removal of the double bonds. Instead, the $\ce{H}$ atom of $\ce{CH3}$ is substituted.
Can somebody please explain why does this happen. If more amount of $\ce{Cl2}$ is taken in sunlight and the reaction is carried out for long time, will the double bonds break?

the reactions are discussed here for instance... http://www.chemguide.co.uk/organicprops/arenes/halogenation.html

@MaxW I went through the link but I wanted to know will the double bonds be broken once all H atoms from CH3 are substituted?

Sorry, I was just trying to point out a source where the reaction conditions  were discussed in more detail. I agree that the source i referenced doesn't have an answer.

@MaxW It's ok. If you have any explanation, do tell

Both halogenations can be, according to procedures on SciFinder, performed at ambient temperature, simple sunlight and in reasonable reaction times.
The first step in both transformations is the formation of chlorine radicals by photoinduced homolytic cleavage of the $\ce{Cl-Cl}$ bond (equation $(1)$).
$$\ce{Cl-Cl ->[$h \nu$] 2 Cl^.}\tag{1}$$
These chlorine radicals want to react with something to lose their radical configuration. In the case of toluene, the kinetically favoured attack is that generating the stabilised benzyl radical as shown in equation $(2)$.
$$\ce{Ph-CH3 + Cl^. -> Ph-CH2^. + HCl}\tag{2}$$
This benzyl radical can then react with another chlorine molecule forming benzyl chloride (equation$(3)$), and so on in two further steps to form trichloromethylbenzene (equations not shown but analogous to $(2)$ and $(3)$).
$$\ce{Ph-CH2^. + Cl2 -> Ph-CH2-Cl + Cl^.}\tag{3}$$
Once we have reached the fully chlorinated methyl group, we can again imagine a chlorine radical looking for a reaction partner. Again, we should ask us which the favoured attack is — the one leading to the most stable products. And rather than attacking the aromatic ring, the benzylic carbon can again be attacked as shown in equation $(4)$:
$$\ce{Ph-CCl3 + Cl^. -> Ph-CCl2^. + Cl2}\tag{4}$$
Thus, we are ‘destroying’ our product because this attack is still better (leads to more stable products) than attacking the aromatic system.

Why does benzene react in the way it does? Well, equation $(1)$ is identical. However, there is no benzylic group; there are only two possibilities how a radical can react with benzene:

abstract a hydrogen radical forming a phenyl ($\mathrm{sp^2}$-hybridised) radical
attack the aromatic ring generating a 6-chlorocyclohexa-1,3-diene-5-yl radical

The second reaction is clearly favoured over the first, since while it does break the aromatic system, it still generates a p-orbital centred pentadienyl-type radical which is much more stabilised than a radical in an $\mathrm{sp^2}$ orbital.
However, toluene or its reaction products will hardly ever go this route since a benzyl radical is still much more stabilised than a pentadienyl-type one. The minute quantities of toluene products that do chlorinate this way will be fully chlorinated, but their yields will be very insignificant.

I very well understand the reaction you mentioned for toluene. But, in case of benzene, if the reaction intermediates have quite less stability  , why do they react after all? (Though I understood why the aromaticity is lost if the reaction takes place). And yes, thanks for the answer.

@AmritanshSinghal Because the chlorine radical is a very reactive species, so it wants to react. It cannot react to form the more stable radicals, so it just chooses the least instable one. And once a single double bond of benzene is broken, it will continue to chlorinate all of that molecule’s double bonds since the stabilising aromaticity is gone.


text:
dow process, oxidizing toluene to phenol
Can  I bubble Oxygen thru Toluene to produce Phenol? Or do I need a catalyst?(Dow Process)
Before anyone says: "You can't"; google Wiki.

Can you just quote the relevant sections of Wikipedia instead of being defensive and telling us to go to Wikipedia before anybody even said anything?

I would like to note that according to the so-named Wikipedia article, the Dow Process has nothing to do with this question. The Dow Process either refers to the electrolytic production of bromine from brine or the hydrolysis of chlorobenzene with aqueous sodium hydroxide to form phenol. An analogous process for phenol production from alkyl benzenes is the cumene or Hock process where isopropylbenzene is reacted with oxygen to form phenol and acetone.

Orthocresol I am Notr":"O the defensive"> I just got some scathing answers from Yahoo Answers, like":"It can't nbe done."

Ortho: I am Not:"on the defensive"> Look for the worst and yopu'll surely find it.

Ortho, I am Not:"on the defensive". Seek the worst and youll surely find it. On a Positive note; I just got a lot of scathing comments on YahooAnswers. like:  "It can't". T  he Wiki article is under Phenol. Or:  https:/wikipedia.org/phenol .  And its possible Wiki is wrong. They tend to be right on very technical subjects and with a ref. ref.

Here it is, Verbatum:  "Because of phenol's commercial importance, many methods have been developed for its production.,
oxidation of toluene, as developed by Dow Chemical:
C6H5CH3 + 2 O2 → C6H5OH + CO2 + H2O"                                         C6H5OH is Phenol, is it not?

Norris, it says: "production", Dow process, to C6H5OH, which is phenol, I believe. Perhaps article is oversimplified, misses a step, as I feared. Please describe this step and  google  Wiki, or , edit it. It says exactly as I quote. I am not running back-and-forth to prove it.Im not lying.

Toluene oxidation is a two-step process.
According to Ullmann’s Encyclopedia of Industrial Chemistry (sixth edition), the first step is the oxidation of toluene with atmospheric oxygen to benzoic acid, which is carried out in the liquid phase at temperatures of about 100–150 °C and an absolute pressure of about 3 bar. A cobalt naphthenate is used as a soluble catalyst at concentrations of 0.1–0.3 %.
In the second step, the oxidation of benzoic acid with atmospheric oxygen and steam uses molten benzoic acid as the reactant and solvent at a temperature of about 230–240 °C and atmospheric pressure. Copper(II) benzoate is used as a soluble catalyst. Magnesium salts may be added to act as a promoter.
In this reaction, copper(II)benzoate decomposes to copper(I) benzoate and benzoylsalicylic acid (2-(benzoyloxy)benzoic acid). The copper(I) benzoate is regenerated to copper(II)benzoate with atmospheric oxygen. The benzoylsalicylic acid is hydrolyzed with steam to benzoic acid and salicylic acid (2-hydroxybenzoic acid). The salicylic acid is decarboxylated rapidly to phenol and carbon dioxide.


text:
Acid catalysed bond hydrolysis in 3-bromobut-1-ene
I came across this doubt while solving a problem. I can't understand to which path (1 or 2) I should give priority, as both sound equally good to me.

Without looking up data, I would firstly assume that the $\ce{-CHBrCH3}$ group acts like any other alkyl group, and that path I would be preferred (by Markovnikov's rule). Bromine is electron-withdrawing via the inductive effect, but only very weakly.
I did go and look up data for this particular substrate. The reference is J. Org. Chem. 1936, 1 (4), 393–404. On page 399 the authors describe the addition of $\ce{HBr}$ to the above substrate.

In the absence of air and peroxides, however, 3-bromo-1-butene gives mostly (60 per cent.) 2,3-dibromobutane.

2,3-Dibromobutane is the product arising from path I, which confirms my assumption.

I don't know whether Markovnik rule can be applied here, but another way to arrive at why 1 is preferred is due to the higher stability of the secondary carbocation than the primary due to inductive effect of alkyl groups and hyperconjugation

That is precisely the rationale for Markovnikov's rule, so you are saying the same thing I did, albeit in a more detailed fashion. I was rather lazy to type all that out because I assumed whoever was reading it would understand why the rule applies.

Might be. I had commented because the markovnikov rule I've read talks about the addition of negative part of the addendum while in this case there is no mention of the negative part of the addendum. Though I think OH- can be regarded as the negative part and then we can say that markovnikov rule is applicable.

But if HBr along with peroxides is supplied then what would be the pathway? Actually this doubt is a part of the problem where HBr + ROOR was added as reagents.

@DIP If that was part of your question you should have asked it at the start, then I would have known to answer it. ): The same article that I cited describes the addition with peroxides. The selectivity is reversed, as is expected from a radical addition, which proceeds with anti-Markovnikov regioselectivity. The product distribution is 80% 1,3-dibromobutane and 20% 2,3-dibromobutane. [...]

[...] However, the matter is complicated by the fact that 3-bromobut-1-ene and 1-bromobut-2-ene are in equilibrium under peroxide + HBr conditions (the product distribution is the same for both alkenes). The authors write that it is impossible to conclude which alkene the addition products arise from.

Primary carbenium ions are very unstable. Typically, they will not form unless the positive charge is very well stabilised — so a benzyl cation ($\ce{Ph-CH2+}$) will form but an ethyl cation ($\ce{H3C-CH2+}$) will not form under ambient conditions.
Secondary carbenium ions are much more stabilised by double hyperconjugation, but they are still subject to rapid Wagner-Meerwein rearrangements to form more stable tertiary carbenium ions. However, if there is a sufficient possibility of stabilisation inherent to the molecule, forming them is no problem. Stabilisation could arise from:

a geminal heteroatom donating an electron pair to delocalise the cation onto said heteroatom
a vicinal group that can donate an electron pair in a similar manner, forming a three-membered ring
a neighbouring double bond that can resonate with the carbocation generating an allyl-type cation

In your case, the vicinal bromine is able to form a bromonium ion as in your path I. Thus, the secondary carbenium ion is reasonably stabilised and will be the major product.
Note that protonation of the double bond is reversible. Protonation leading to the primary carbocation will — if no immediate Wagner-Meerwein migrating hydride is observed — quickly eliminate back to the double bond, further favouring path I.

This observation is a classic example of Markovnikov’s rule. It states:

When adding $\ce{H-X}$ to a double bond, the hydrogen will add to that side of the double bond which already has a higher number of hydrogens.

Markovnikov’s rule is not strict (e.g. it fails completely for borane addition) but its basis is in the formation of the more stable carbenium ion as first addition step of the mechanism.


text:
Difference between some terms relating to liquids
So my chemistry teacher, when she talks about liquid substances she talks about them in reference to 3 words. She refers to liquid substance with one of these 3 words-
1)Aquesous form
2)Solution form
3)Molten form
Can anyone please tell me the difference between these terms?

http://chemistry.stackexchange.com/questions/32008/molten-vs-liquid

well what about solution form??

Take table salt, sodium chloride, for example.
If you take table salt and dissolve it in water, you are making a salt solution.
Any solution that uses water as the solvent is an aqueous solution. If, for example, you dissolved salt in ethyl alcohol, that would still be a solution but it would not be aqueous.
Now imagine you took salt and heated it really, really strongly and melted it. Just like ice melts to become water, table salt can melt too, just that it's at a much higher temperature. That's a molten liquid. In this case, there is no solvent added.

so I will explain you.
will aqueous is the term used for those solutions in which water is used as a solvent.
solution consists of two components that is solvent and solute and it is a type of mixture.
molten state is used for those when a solid substance is heated and it is melted and when cooled again it changes into solid state.


text:
Difference between subshell and sub-subshell?
I consider the s p d f orbitals as a subshell, but I have heard about the term sub-subshell. what is the difference between these two?

I haven’t really heard it, but I can imagine it meaning the $\mathrm{t_{2g}}$ and $\mathrm{e_g}$ sub-subshells of the d-subshell, for example.

There seems to be some usage in relativistic chemistry, Google throws up some stuff but I hardly understand them. For example in this question by ron, I believe the p(1/2) and p(3/2) can be called sub-subshells sometimes, since they are not quite orbitals but they arise from the splitting of a subshell (which is degenerate in the non-relativistic picture).

As you pointed out the the s, p, d, and f denoted subshells of atoms.
The term sub-subshells is very uncommon. I did find a couple of uses.
Materials Science
By G. K. Narula, K. S. Narula, V. K. Gupta
pdf file about Modern Atomic Mechanical Theory, page 24
What the authors are noting is that the 2p subshell will have three orbitals which can support two electrons each. Each of these three orbitals is being referred to as a sub-subshell.
As I said it seems a very uncommon term, and I'd suggest you avoid using it.

so u mean that instead of sub-subshell I use the word orbitals?

and what if orbital is used for s p d f and then what is a subshell

Let's just agree that  the s, p, d, and f denote subshells of atoms and not peel this back that far.

agree with that

I can't think of a good word to replace the concept of sub-subshell in the way that the authors cited used the term. In general in chemistry if the orbitals in a subshell weren't degenerate then we'd use nomenclature to explain the specific electronic configuration.  If the orbitals are degenerate then we'd say "the 2p orbitals," or "the p orbitals."

you are right .

I would use just orbital instead of subsubshells. For example: A subshell is a combination of orbitals. However $\mathrm{s}$ is an orbital and a subshell.

A subshell/orbital for example 2p is further divided into Px,Py&Pz.These Px,Py,&Pz are called sub-subshells/suborbitals.It is also same in case of s,p,d,f orbitals.But for s-orbitals there is no suborbitals and s is suborbital itself.
Hope this help.


text:
What is the difference between λ and Λ with regard to molar conductivity?
I came across these equations (boxed in red) in a book:

$$
\begin{align}
&\therefore & t_\mathrm{c} &= \frac{u_\mathrm{c}}{u_\mathrm{c} + u_\mathrm{a}}\\
&\text{and} & t_\mathrm{a} &= \frac{u_\mathrm{a}}{u_\mathrm{c} + u_\mathrm{a}}
\end{align}
$$
  At infinite dilution,
  $$\text{Also,} \quad t_\mathrm{c}^\infty = \frac{u_\mathrm{c}^\infty}{u_\mathrm{c}^\infty + u_\mathrm{a}^\infty}$$
  Since, $\lambda_\mathrm{c}^\infty = u_\mathrm{c}^\infty\times F$ and $\lambda_\mathrm{a}^\infty = u_\mathrm{a}^\infty\times F$
  $$
\bbox[3px,border:1px solid red]{
\begin{align}
&\therefore & t_\mathrm{c}^\infty \times \Lambda_\mathrm{elecrolyte}^\infty &= \lambda_\mathrm{c}^\infty\\
&\text{and} & t_\mathrm{a}^\infty \times \Lambda_\mathrm{elecrolyte}^\infty &= \lambda_\mathrm{a}^\infty
\end{align}
}
$$

Where $t$ refers to the transport number of an ion, and the subscripts $\mathrm{a}$ and $\mathrm{c}$ refer to an anion and cation, respectively.
From what I've learnt, the uppercase lambda $\Lambda$ refers to molar conductivity. But then what does the lowercase lambda $\lambda$ stand for?
Different sections within the same book give different definitions for $\lambda$, one being 'ionic conductance' and another being 'ionic conductivity', and I'm pretty sure 'conductance' and 'conductivity' are not the same thing.
I did try looking it up online, but Wikipedia wasn't much help.

Look in the Wikipedia article Ion transport number.
Lower case lambda, $\lambda$, is the molar ionic conductivity
Upper case lambda, $\Lambda$, is the molar conductivity.
$\lambda = t \dfrac{\Lambda}{\nu}$
Where $\nu$ is the number of ions for the cation or anion in the molecule. So $\ce{Na3PO4}$ would have 3 cations per molecule, but only 1 anion per molecule.


text:
Heating CuO and Urea
My little brother had to do this experiment at school, but couldn't finish it because of a lack of time. My chemistry knowledge isn't as well as it used to be, so I couldn't be much of an help.
There is a test tube (Tube A) filled with $\ce{CuO}$ and urea. With a glass straw, this is connected to a second test tube (Tube B), which is filled with lime water (which is if I recall correctly $\ce{Ca(OH)2}$?)
A quick sketch:

I really wanted to recreate this at home but I sadly don't have the tools for it.
The question here is: what happens when tube A is heated? (What happens to the material in tube A? What happens to tube B? What is left behind in A? etc).
My guess would be that $\ce{CO2}$ would flow from tube A to B, but I am not sure even about that.

Heating a hydrocarbon with $\ce{CuO}$ generally evolves $\ce{CO2}$ and $\ce{H2O}$. Urea is no exception for this rule.
The $\ce{CO2}$ evolved will turn lime water milky(this is a test for $\ce{CO2}$ gas).

Urea is not a hydrocarbon (not that it matters much, though).

Cupric oxide is easily reduced at low temperatures to copper (~$\pu{250^\circ C}$ by $\ce{H2}$ or $\ce{CO}$), and even without a reducing agent, it disproportionates to cuprous oxide and $\ce{O2}$ at ~$\pu{800 ^\circ C}$ (Cotton and Wilkinson: Advanced Inorganic Chemistry). (In aqueous solution, cupric salts are more stable than cuprous.)
Depending on the ratio of $\ce{CuO}$ to urea and temperature, you could get:

$\ce{3 CuO + CO(NH2)2 -> 3 Cu + CO2 + 2 H2O + N2}$, or
$\ce{6 CuO + CO(NH2)2 -> 3 Cu2O + CO2 + 2 H2O + N2}$

The $\ce{CO2}$ passes into the limewater (dilute solution of $\ce{Ca(OH)2}$) and precipitates $\ce{CaCO3}$. The $\ce{N2}$ just bubbles through. The residue in tube (1) could be finely divided copper (black perhaps with a reddish tinge), or yellow (or red) $\ce{Cu2O}$.
Reaction mixture (2) might be more complex. If reaction mixture (2) is heated slowly, the low temperature reaction (forming $\ce{Cu}$) would occur first, and might go completely, leaving a mixture of $\ce{Cu}$ and $\ce{CuO}$ instead of $\ce{Cu2O}$. It would be interesting to control the temperature and follow the progress of the reaction by the amount of gas produced and see what product remains in test tube (1).


text:
Significant figures in conversion factors?

Example: I have $\pu{64 cm^{3}}$ of milk. How much is that in gallons?

When I use the conversion factors $\frac{\pu{1 mL}}{\pu{1 cm^3}}$
$\frac{\pu{1L}}{\pu{1000 mL}}$ $\frac{\pu{1.0567 qt}}{\pu{1 L}}$ $\frac{\pu{1 gal}}{\pu{4 qt}}$.
I know the final answer should only have two significant figures, but how many significant figures should the intermediate conversion factors have?
I have heard that:

conversion factors should have one more sig fig than the least precise measurement. (So three in this case)
But I also heard that we shouldn't round anything until the end.
(So use all 5, and round after conversions are complete)

The reason why I got confused is because of my textbook:

check your conversion factor. $1 \text{cm}^3$ is a milliliter which is much much less than a quart.

With a modern calculator, I wouldn't round anything until the end. When doing calculations with a slide rule or log tables, then things were a bit different.

This is case dependent. You need to realize what kind of mathematical object are you deal with.

@MaxW, so is my textbook wrong in rounding in this example?

@ user1420303 In the attached textbook example, they rounded an exact conversion factor. But when I did the math on my own and didnt round untill the end, I came up with a slightly different answer. Does that mean we are both right?

eromod - Your textbook is wrong to imply that there is any particular rounding method that will always yield the "right" result. Rounding has always caused problems. The overall point is just that you have to realize that rounding will cause problems and be careful. There isn't any magic procedure.

@Max, how would you do it?

As I said, if I had a calculator, I wouldn't round anything until the end. If I had to do all the calculations by hand, then I'd be inclined to round intermediate values. I'm not going to multiple two 8-digit numbers by hand and then round to 2 digits.

So to conclude: round conversion factors only if you dont have a calculator because its tedious.

Use significant figures as much as you can in intermediate conversion factors,and then round off the final answer to two significant figures,using more significant figures in intermediate conversion factors will lead to a accurate answer.
I let you conclude.

that was my initial impression too, but have a look at my textbook example I attached.

you should  use as much as sig.fig.,i have seen many examples and solved many exersises regarding this topic,and have never seen rounding off before the final answer unless significant figures are too much and cannot be taken under arithmatic operations.

I'm not sure where your conversion factor came from as it is incorrect but regarding the correct factor:
Because the inch is defined using the metre, and the gallon defined using the inch, there is an exact conversion between them—1 US gallon is exactly 3.785411784 L.
Thus, like any exact quantity or conversion factor, we treat the gallon to litre conversion factor as having infinite significant figures.
Because we have defined it exactly, there is no implicit uncertainty in the number.
There are other conversion factors that are not exact and do actually have uncertainties, but for most practical purposes, these have far more significant figures than a normal measured quantity.
See this question for more details.

conversion factors were wrong, I updated them. But what is the conclusion though? How many sig figs should I use? That link says I should look it up in the CRC Handbook. That what you recommend too?

As my answer says, the conversion factor is exact.

So if my measured quantity is $\ 64 cm^{3}$ , how many significant figures should I use in the qt/L conversion? All of them because its an exact conversion? Or just three sig figs (one more that the two I have to work with)

@eromod The conversion factor is only exact if you use that exact conversion factor. For example, an inch is defined as 2.54 cm. If you use that number, it has infinite precision. However, if you say "ahh, 2.5 is close enough" then the conversion factor only has 2 significant digits. Likewise here. If you use the exact conversion factor, there's infinite precision. If you round it, it has the rounded amount of precision.

@ R.M. But in my textbook, they rounded the conversion factor...I added a screenshot


text:
How does toluene favor ortho/para in terms of kinetic and thermodynamic control?
If nitration of toluene is carried out at a very low temperature, presumably the reaction would be carried out under kinetic control. During these conditions that favor kinetic products, is more ortho or para formed when compared to normal laboratory conditions?

Related: How does toluene react at higher temperatures and why?

@Loong Thank you very much for finding that. Would it be possible to extrapolate that "para" is the kinetic product?

Is nitration an equillibrium reaction like F-C-alkylation?


text:
Comparing the experimental and calculated UV/vis spectra for ethene
I am trying to compute UV/Vis absorption spectrum for Ethene using Gaussian 09.
I created the ethene molecule in GaussView 5, and cleaned it. Then I optimized the molecule and checked if I got the minimum energy conformation with the following:
# freq=noraman cphf=noread b3lyp/6-31g(d) geom=connectivity formcheck
 integral=grid=ultrafine scf=maxcycle=1000

Next I performed the energy calculation as follows:
# td b3lyp/6-31g(d) geom=connectivity

As a result I got a single peak with the maximum at 134 nm (see picture below) which seems to be way off from what I googled around on the Internet, which is like 173 nm or 180 nm.

Can anyone please help me figure out if I have done it correctly and whether I am way off the experimental data?

In short, there are two obvious problems with the setup OP uses for TD-DFT calculations:

B3LYP functional is not a good choice for TD-DFT.
6-31G(d) basis is usually too small.

At M06-2X/Def2-TZVP level I get a maximum at ~160 nm, which, taking the accuracy of the TD-DFT approximations into account, is close enough to the experimental value.

Thank you @Wildcat. I tried to redo the calculation according to your suggestion. I just took the molecule optimized at b3lyp/6-31g(d) and calculated the energy as follows: # td M062X/Def2TZVP geom=connectivity. I got the maximum 158.53 nm, so close to what you kindly reported. I wonder however if it's OK what I did in terms of optimizing the geometry at a different level than the one used to do the TD calculation for UV/Vis spectra? Looks like it works here. Do you think I do not need to reoptimize all of my molecules then?

B3LYP/6-31G* might be a bad choice for a lot of calculations but it is not to be doomed a priori. We calculate a lot with it and reach acceptable accuracy. It always depends on the molecule that you are investigating. M06-2X for example does not give acceptable excitation energies for my complexes. Not even with triple zeta bs while B3LYP/6-31G* is way better for us.

B3LYP/6-31G* might occasionally spit out some good numbers for excitation energies, right, but I would not trust this level of theory anyway since even when it works fine, it is very likely just the result of a fortunate cancellation of errors.

@RicardoMoreno, B3LYP/6-31G* geometry of ethene should be good enough, so that there is no need for reoptimization.

@pH13-YetanotherPhilipp That is because M06-2X is not meant to be used for excitation energies, only ground-state properties such as thermochemistry and non-covalent interactions.

@pentavalentcarbon, why do you say that "M06-2X is not meant to be used for excitation energies"? It is one of the best hybrids according to benchmarks (10.1002/qua.24438).

For smaller molecules like ethene, the webbook by NIST is a good source of experimental data.  Search results for ethene in particular here yield both a figure of a vacuum UV spectrum, indicating an absorption maximum centred around 160-165 nm with $\log \epsilon$ of about 4.2, as well with a primary literature reference (Platt, J.R.; Price, W.C., J. Chem. Phys., 1949, 17, 466).
Of course, other databases like Scifinder/Chemical Abstracts by the American Chemical Society may indidicate more recent data.  A specialized compilation, more suitable for ethene in particular, is the MPI-Mainz UV/VIS Spectral Atlas of Gaseous Molecules of Atmospheric Interest here, with an entry about ethene.
Side note: NIST database equally offers access to other thermophysical data (like Vibrational and/or electronic energy levels) potentially of interest for you, too.
Addendum:  With the searchable Computational Chemistry Comparison and Benchmark DataBase, NIST equally hosts numerous results of calculations of energies and structures for small molecules.  This that may serve as a valuable reference, comparing the methods with each other, as well against experimentally determined data.

Dear @Buttonwood, thank you so much for your guidance and the awesome links! By the way, do you think that the # td M062X/Def2TZVP level as suggested by Wildcat (and it seems to be working fine for ethene) will be good enough for UV/Vis computation for larger molecules, let's say 20-25 heavy atoms like carbon, and only a few oxygens or nitrogens around?

@Ricardo Moreno You are wellcome.  Yet: My practice in computational chemistry is too limited to provide you sound advice about the method / basis set to be deployed other than too look out how similar questions were solved by others in the field.  Drawing a structure, (pre)optimising here and then with a force field & PM3 as a starting point / comparision e. g. for powder diffraction differs on what your question adresses and (in terms of computational cost) is at least a league below b3lyp and other hybrid functionals.

@RicardoMoreno Take a tour d'horizon in CCBDB just discovered.  (Ethene is present, too.)


text:
Do all electrons in an element move at the same speed and have the same momentum angular vs non-angular momentum?
Do electrons' speed, energy, angular momentum increase when they are in the p, d, and f rather than in the s orbital? Do all electrons in a specific orbital, say for example, the f orbital all have the same speed, energy, angular momentum? Do electrons in the f orbital have higher angular momentum as a result of their distance from the radius or because of their higher speed, energy, or something else?

Electrons are not tiny little moons in orbit about the nucleus. One could say the wave function 'is' an electron. There is an angular momentum quantum number associated with each wave function, however.

In simple words, we can say almost nothing about properties of an individual electron in a many electron system


text:
Mathematical form of four hybrid orbitals

The mathematical form of the four $\ce{sp^3}$ hybrid orbitals are given below.
  \begin{align}
\ce{
\tag{a} sp^3 &= 1/2s + 1/2p_x + 1/2p_y + 1/2p_z\\
\tag{b} sp^3 &= 1/2s + 1/2p_x - 1/2p_y - 1/2p_z\\
\tag{c} sp^3 &= 1/2s - 1/2p_x + 1/2p_y - 1/2p_z\\
\tag{d} sp^3 &= 1/2s - 1/2p_x - 1/2p_y - 1/2p_z\\
}\end{align}
  For each hybrid orbital the number in front of the $\ce{s}$ and three $\ce{p}$ functions, called a coefficient, describes the contribution and relative ratio of each canonical orbital to the hybrid wave function. Add up the coefficients and prove to that these orbitals are $\ce{s^1p^3}$.

Yes, I know that the grammar is terrible on that last sentence, that is how it is written. So, I know that for an $\ce{sp^3}$ orbital it is one part $\ce{s}$ orbital and three parts $\ce{p}$ orbital. However, I have no idea what this question wants me to do. Adding up the coefficients gives me $2$, $0$, $0$, and $-1$, but I have no idea if that is how it wants me to add it up (you could ignore the sign and get $2$ every time, and show that $1/2$ is a quarter of $2$, thus proving that each component contributes a quarter to the hybridized orbital, but this seems too simple and incorrect).

The sign on the $\text{p}_z$ orbital in (d) is wrong. It should be +, not -.

Well, that may be true, but that's how it was given to me. There could be a typo though.

Could you provide a source for the question, or was it given to you by your teacher? I would like to remind everyone, that the notation is a contraction: $$\ce{sp^{n} = s^{\frac{1}{n+1}}p^{\frac{n}{n+1}}},$$ omitting the common factor of $\frac{1}{n+1}$.

This was given to me as a practice quiz in my organic chemistry class.

@Martin-マーチン so does that mean each of the four orbital mathemtical forms has naming as s^1/4p^3/4 and when we combine all four we get the sp^3 hybridised carbon ?

If a molecular orbital $\psi$ (in this case, a hybrid orbital) is constructed from an orthonormal basis set of atomic orbitals $\{\phi_n\}$ via a linear combination
$$\psi = \sum_n c_n \phi_n$$
then loosely speaking, the "amount" of $\phi_n$ in $\psi$ is given not by $c_n$, but rather by $\left|c_n\right|^2$. (See, for example, Griffiths, Introduction to Quantum Mechanics, 2nd ed., Section 3.4.)
Therefore, summing the four values of $c_n$ for each AO has no physical meaning.
Instead let's take wavefunction (b) for example. The coefficient in front of the s orbital is $1/2$, so the "amount" of s-orbital character is simply $\left|1/2\right|^2 = 1/4$. The coefficient in front of the $\mathrm{p}_y$ orbital is $-1/2$, so there is $\left|-1/2\right|^2 = 1/4$ $\mathrm{p}_y$-character.
Armed with this knowledge, it's incredibly easy to see that each orbital has exactly $1/4$ s-character and a total of $3/4$ p-character ($1/4$ from each p orbital), because none of the $\pm$ signs matter (the minus signs are removed by the absolute value).

Thanks for the answer... Unfortunately quantum mechanics isn't a prerequisite for the class that I'm a part of for which this question was used, that's next year, I got this question in sophomore organic chemistry class. Your answer makes sense, but there should be another way to explain this? I dunno, it could just be a bad question. Thanks again though!

@vin2 I think this is the only "correct" way of explaining it. You could apply other heuristics to argue that the contribution is 1:3, such as what you described as 1/2 being a quarter of 2, but that would be an incorrect way of reaching the correct answer.

@vin2 The simplest (and ultimately correct) way is to think that individual unhybridized orbitals (s,p,d...) form an ortonormal basis set in multi-dimensional vector space by definition. School teaches to work with such a basis (cartesian coordinates) in 2d and 3d spaces, it should be pretty easy to extrapolate. This approach is surprisingly common in applied mathematics.


text:
Release of sulfur dioxide from odored fuel gases
Often, fuel gases such as butane and natural gas (methane) are mixed with some foul-smelling impurities such as thiols (mercaptans) and hydrogen sulfide. Presumably, anything containing sulfur would burn to sulfur dioxide. The odor of sulfur dioxide is not easily noticed while burning such fuels, but I presume that the large amount of fuel gases burnt in the world would cause a significant amount of $\ce{SO2}$ emissions.
Why is this not usually brought up? Are the amounts actually so small that they are completely insignificant?

insignificant compared to release of SO2 due to burning of coal around the world,

The odorants might be partly responsible for the buildup of a white crust (corrosion?) around pilot-light nozzles, though. But as @Khan states, the amount of sulfur added is minuscule.

According to this North American Energy standards board publication , the total concentration of sulfur in natural gas, both natural and from oderant additives, must not exceed $\pu{70 ppbv}$. Concentrations are actually more commonly about a tenth of this. For example, from the same source, the Union Gas system typically contains about $\pu{5 ppbv}$ sulfur.  Furthermore, the sulfur content of coal is many orders of magnitude higher than that of oderant-enriched natural gas at about $\pu{0.5 to 5\%}$ sulfur, and the worldwide combustion of coal is much greater than that of natural gas (although much of the sulfur from coal combustion is now scrubbed out at the source of combustion). Other fossil fuels like diesel also contain significant amounts of sulfur (regulatory sulfur content of diesel fuel in the US and EU was 10 ppm as of 2009).
The global atmospheric background concentration of $\ce{SO2}$, according to
Owen, Lewis A.; Pickering, Kevin T (1997). An Introduction to Global Environmental Issues. Taylor & Francis. pp. 33–. ISBN 978-0-203-97400-1, is around $\pu{1 ppbv}$, meaning the sulfur content of the Union Gas product, for example, is only five times that that found throughout the Earth's atmosphere.
Summary - TL/DR:
Based on the background $\ce{SO2}$ concentration and the overwhelming sulfur emision source from the combustion of coal and other fossil fuels, the sulfur originating from natural gas oderant additives represents a negligible contribution to the global atmospheric $\ce{SO2}$ concentration.