Alkene Electrophilic Addition Reaction Stereoselectivity When Two New Asymmetric Centers Form I had a question about the stereoselectivity in electrophilic addition reactions (specific to alkenes) that form additional asymmetric centers in their products. I understand that when a reactant with one preexisting asymmetric center forms a product with an additional asymmetric center, the two possible product stereoisomers are a pair of diastereomers because the configuration for the preexisting asymmetric center will stay the same while the configuration of the  newly formed asymmetric center will vary. Since diastereomers differ in stability, their transition states differ as well, therefore the product with greater stability will come from a transition state with lower energy, which will require less activation energy to form. The product stereoisomer with greater stability therefore forms faster and becomes the predominant product.

However, the textbook also says that the electrophilic addition reactions that form products with two additional asymmetric centers from a reactant without any asymmetric center form four stereoisomers of which two pairs are enantiomers with each other.

It is also mentioned that this kind of reaction is not stereoselective. I don't understand how this is the case; the members that form enantiomers with each other might be energetically identical, but the stereoisomers that are diastereomers to each other must be energetically different, thus having different stability. If the logic from the first case applies the same here, wouldn't the reactions that form two new additional asymmetric centers also be stereoselective since they form diastereomers?

Stereoselective becomes a hand-wavy term as soon as you may end up with more than one type of stereoisomer. If you want to actually state what is going on, you would most likely use either of the terms enantioselective (selects for a specific enantiomer) or diastereoselective (selects for a specific diastereomer; this may or may not include both enantiomers of the diastereomer).

In the first case, there is only one selectivity that can be observed; you can only get diastereomers, no enantiomers. Therefore, there may be some reason in shortening the term to stereoselective. In the second case you should not resort to using stereoselective because it is not specific enough — and that is irrelevant of the question whether stereospecificity is observed or not.

The second reaction would likely indeed be diastereoselective to a certain extent while definitely not being enantioselective (because the product is symmetric). So calling it diastereoselective — and thus, by intrapolation, stereoselective — may be justified.

In practice, the selectivities of both reactions would be so low that no practicing organic chemist would call either ‘selective’.


Reason for triphenyl methane being less acidic than acetylene This question came through answering a question that talked about the acidic order of four compounds.
Upon researching the actual $\mathrm p K_\mathrm a$ values I noticed that even though there was a resonance destabilization of electrons (albeit partial) in triphenyl methane, the $\mathrm p K_\mathrm a$ values of acetylene were less than that of triphenyl methane? I was unable to find a solid reason for why this is the case.
Does anyone in the community have the reason?

As mentioned by @orthocresol, just comparing methane and acetylene shows a difference of $\approx 20$ $\mathrm pK_\mathrm a$ units and the amount of resonance required to overcome this difference is very high.
Comparing $\mathrm p K_\mathrm a$ of phenol($10.0$) and methanol($15.5$), we see that the presence of one resonating ring only produces a 5.5 $\mathrm p K_\mathrm a$ difference. Resonance in triphenyl system is effectively only one phenyl group at a time due to the steric effects due to the bulkiness of the phenyl groups.


What properties does a solvent with a high dielectric constant have? My professor commented that water has a higher dielectric constant than alcohol.

Okay, but on what basis?

He didn't elaborate; he asked us which one seemed "more electric" and most answered water. I guess from intuition.

But why exactly is water "more electric"? I think it has to do with the relatively high extent of auto-ionization in water, correct? Alcohols are generally less polar than water and thus cannot stabilize ions as effectively as water can. And I've even seem some computers submerged completely in oil. I'm guessing that only works because oil doesn't ionize much and thus isn't able to conduct a lot of electricity - or enough to short out any circuits.

Before I start, please note that I am not a chemist so be sure to verify what I say...

The electric conductivity of water depends on what is dissolved in it. For example, salt water will be far more conductive than distilled water. So if you're professor was talking about sea water, the answer could actually be different from purified water.

Due to the dissolved salt ions, sea water is a better conductor of electricity and so by definition it will be a poor dielectric, since according to this Wikipedia article: to be a good dielectric a substance must be both a good insulator and able to be polarized by an electric field.

Purified water on the other hand will be a better dielectric because it does not contain any significant amount of ions and is therefor not very conductive (i.e. is a better insulator than sea water).

I think (but don't quote me on an exam) that the reason alcohol is a poor dielectric is because it is not nearly as polar as water is, and according to the above, the second requirement for a dielectric is that it could be polarized by an electric field.

(This might also have to do with the size of the molecules, since water molecules are generally smaller than many kinds of alcohols, and can therefore re-orient themselves more easily in the presence of an electric field, but this is a guess.)

The Solvent Classifications section of this Wikipedia page also mentions that the polarity of a solution can be roughly gauged by the solution's dielectric constant. So this too seems to support the idea that polarity and dielectric properties are related.

If anyone has any improvements or corrections, please let me know in the comments and I'll update the answer accordingly.


Determining the stereochemistry of the product(s) when an alkyl shift in a carbocation generates a chiral carbon Consider this carbocation:

We know that this carbocation would immediately rearrange - by a methyl shift - to form the more stable carbocation.

Now, notice that the migration will produce a chiral center at carbon 1. Assuming that the leaving group was a chloride, will the configuration at 1 - after the methyl shift - be retained, inverted, or racemized?

Interestingly, assuming the nucleophile is $\ce{EtOH}$, in the final product - 2-d-3-ethoxy-3-methylpentane, carbon 2 is also a chiral center. I assume that it will be racemized, as in usual SN1 reactions, but I am not sure. Am I correct?

At a very detailed level, I was also taught that in SN1 reactions, 100% racemization does not occur. The inverted product is slightly more than the retained product. This is because - while the leaving group is still leaving - backside attack is preferred. So, does this phenomenon also have an effect on the molecule above?

To conclude, I wish to know the stereochemistry at both chiral carbon 1 and 2. Are they retained, inverted, or racemized? If racemized, then is the racemization 100%, or is the inverted product formed more than the retained one?

Since this problem is about the stereochemistry, I'd try to address it according to sophomore undergraduate Organic Chemistry text books used in the USA.

Briefly, any nucleophilic substitution reaction at primary carbon predominantly undergoes with SN2 mechanism while those reactions at tertiary carbon undergo SN1 mechanism exclusively. Meanwhile, reactions at secondary carbon would undergo either mechanism, based on reaction conditions, solvents, etc.

The "carbocation" given is a primary carbon even though it is a chiral carbon due to the presence of deuterium atom in place of hydrogen. Note that the stereo-effect caused by deuterium can be considered similar to that of hydrogen. Nonetheless, the bond energy difference (the $\ce {C–D}$ bond dissociation energy listed as $\pu{341.4 kJ/mol}$ while that of $\ce {C–H}$ as $\pu {338.4 kJ/mol}$) may cause the isotope-effects in kinetic of reaction mechanisms (that is a different avenue than what in discussion). Since $\ce{-CH2-Cl}$ and $\ce {-CHD-Cl}$ are similar and $\ce {C}$ atom in both cases is primary, the mechanism of the reaction should be SN2 exclusively. Therefore, the stereochemistry will be $\ce {100\%}$ inverted.

For reading for substitution reactions involving alkyl halides: article. Please also note that, one of example given on that website, the compound $\ce {Ph-CH(CH3)CH2-Cl}$ does not undergo methyl group migration to give benzylic carbocation during its nucleophilic substitution reaction.


What does "n" stand for in n-akyldiols, n-alkanes, n-dicholoroalkanes? I have read two conflicting answers upon a google search:

  " 'n' in this context stands for normal or the latin equivalent. That is why, like other Latin and Greek abbreviations, it is in italics. It refers to straight chain alkanes. If we weren't dealing with n-alkanes we would have branched chain isomers to consider and potentially very different answers. It is old nomenclature from the days when adjacent carbons in a molecule were designated alpha,beta,gamma, etc and when acidity was given in 'N' units of normality."

and

  " 'n' in science is like "x". It is just a reminder that a number needs to be there. You need to know that numbers can be replaced by x, x/y, n, or any variable and that groups can be replaced by their generic names."

The "n" stands for "normal", which in this context means "straight-chain".  It is not italisized, since it is not Latin, but English.  In contrast, the prefixes "i" or "a" for "iso" and "anteiso" are Latin and are properly italisized, but rarely are.

Your second choice, "n is like x", can be found in the subscripts of chemical formulae, where it does denote a variable amount.


Identifying a glass frosting chemical I have purchased an "unknown" white coloured powder from the market which can frost glass objects. Here are some properties of this powder

White in colour
When mixed with HCL or water, it results in the soluion becoming ice cold, the reaction is endothermic.
When kept for a while (10 minutes) the white powder settles at the bottom of the plastic vessel and the acid floats above it.
When stirred, the entire solution becomes milky
When glass is dipped in this solution for 90 seconds, it becomes frosted/translucent.

Since I am not sure what this chemical is, I cannot determine the safest method to dispose it. The person from whom I brought this told me to put it down the drain, but I wouldnt choose to do that before I know if it will not effect the water table near my locality.

Can someone point out what could this powder be, and the best way to dispose it off?

The most common glass-etching powders are solid ammonium bifluoride and solid potassium bifluoride. See for example the Material Safety Data Sheet for “EtchON Acid-Frost Glass Frosting Powder”, or this one. This is not a definite identification, but it fits your description, and those two are the most commonly available such chemicals.

Disposal is a bit tricky. Definitely don't put it down the drain! The first MSDS linked above says:

  Ecological Information:
  Noxious for the water inhabitants. Avoid disposal into rivers, sewer system and land penetration.

The second gives a hint at disposing of it:

  Disposal Instructions:
  All wastes must be handled in accordance with local, state and federal regulations. Material can be converted to a less hazardous material by weak reducing agents followed by neutralization.

So, contact your local waste management authorities, and ask them how to proceed. Do not attempt it neutralize it by yourself unless (i) you know what you are doing, and (ii) you have the appropriate hardware (fume hood, e.g.).


Colour intensity of transition metal complexes Even though they are both $\mathrm{d^7}$ complexes, the colour of $\ce{[Co(H2O)6]^2+}$ is pale whereas that of $\ce{[CoCl4]^2-}$ is more intense. Does the octahedral or tetrahedral geometry matter in determining the molar absorption coefficient of the two species?

At first, it is necessary to ask why some transition metal complexes are colourful at all — remember that alkaline metals, alkaline earth metals and earth metals usually form colourless solutions, e.g. sodium, magnesium, aluminium. The transition metals’ colour stems from the d-electrons they have and the shape of the complexes they create. A typical orbital scheme of a 3d-metal in an octahedral complex will look something like this (image originally taken from Prof. Klüfers web scritptum to his coordination chemistry course at the LMU Munich):

On the left are the metal orbitals, from top to bottom it is 4p, 4s and 3d. On the right are the ligand group orbitals, assuming six ligands each only donating electrons in a $\unicode[Times]{x3C3}$ fashion only. Any remaining metal electrons will be filled into the $\mathrm{t_{2g}}$ orbitals which correspond to $\ce{d}_{xz}$, $\ce{d}_{xy}$ and $\ce{d}_{yz}$ and into the $\mathrm{e_g}$ orbitals corresponding to $\ce{d}_{z^2}$ and $\ce{d}_{x^2-y^2}$

The energy difference marked as $10\,\mathrm{Dq}$ in the image would then be the energy of a photon required to excite an electron from $\mathrm{t_{2g}}$ to $\mathrm{e_g}$; this photon would thereby be absorbed, which is where we get our absorption coefficient from. It is usually in the visible range of light.

There are a few quantum mechanical principles that allow or disallow excitation from one energy level to another. One of these is the spin-allowance rule: Excitations are only allowed if the spin of the entire system remains unchanged. (So excitation cannot involve a spin exchange.) This is why $\ce{Mn^{II}}$ or high-spin $\ce{Fe^{III}}$ complexes are generally colourless: all d-electrons are of identical spin so there is no freedom to be excited.

A second rule is the Laporte-rule whose formal definition I never properly understood, but which states that d to d excitation is forbidden in centrosymmetric species such as octahedra. This rule would mean that all d-element complexes should be colourless. However, complexes are not static but the ligands vibrate back and forth, often in a non-centrosymmetric way. Whenever a photon of the correct energy hits a complex that is not ideally centrosymmetric, the excitation is allowed and the photon absorbed. Because this is a rather rare phenomenon, the colour of such a complex is weaker than it could be.

The $\ce{[Co(H2O)6]^2+}$ case is more complicated again, because in a perfectly centrosymmetric world it would be degenerate: Since it is $\ce{d^7}$, two spin-down electrons share three $\mathrm{t_{2g}}$ orbitals. It would be favourable to somehow make these orbitals dissimilar. This can happen by distorting the octahedral shape in a Jahn-Teller manner: The two water ligands on the $z$ axis are moved slightly further away, creating a slight stabilisation of all orbitals which contain a $z$ component, most notably $\ce{d}_{xz}$ and $\ce{d}_{yz}$. But since the final product is still centrosymmetric, this paragraph only explains the wavelength of absorption slightly better and can be disregarded.

Next question is why the chlorido complex of cobalt(II) is not octahedric but tetrahedric. This is due to a hypothetical $\ce{[CoCl6]^4-}$ being unbalanced in terms of large chloride ions around a small cobalt ion that only has a low charge to combat the $-6$ the chlorides have to offer. The tetrahedral complex $\ce{[CoCl4]-}$ has everything that the octahedral $\ce{[Co(H2O)6]^2+}$ has except the centre of symmetry. Because centrosymmetry is missing, the Laporte rule is no longer valid and every photon of the correct energy can excite electrons. More excitation equals a stronger extinction.

Also, the colour of both complexes is notably different because the orbital scheme is, too. $\mathrm{t_2}$ and $\ce{e}$ are now much closer together and their order is flipped. Combining that with chloride ions that usually generate low values of $10\,\mathrm{Dq}$, the energy difference is even smaller therefore the excitation wavelength longer therefore the colour we see blueshifted (complementary colour).


Can't nobelium form compounds like other lanthanides/actinides do? Over the past week I've been browsing through WebElements just to find out some interesting facts/properties about certain elements in the Periodic Table.

I've just come across the uncommon element nobelium and it looks like it doesn't form compounds with anything despite being an actinide. For example, actinium can form Actinium Trichloride and promethium can form Promethium(III) Fluoride.

Is there a reason why nobelium is so 'special'?

Wikipedia reports that nobelium does form compounds, but unlike other later actinides it could adopt a +2 oxidation state as well as +3, which complicates identifying specific compounds (e.g. nobelium reacts with chlorine but the specific product is not known).  The +2 state is stabilized by the full shell configuration $[\ce{Rn}]5f^{14}$.  Experiments in aqueous solution seem to show a preference for the +2 oxidation state, and $\ce{No}^{2+}$ has been coprecipitated into solids.  But we have yet to definitively identify a neat nobelium compound in either specific oxidation state.


What's the reddish powder that oozes out of a black rock? As an artist, some of the work I do involves the use of seashells, sea rocks/stones, etc.
I picked up a black rock from shore at one of the beaches here in Florida and it looked somewhat clumpy with other rocks and debris encrusted on it. As I do with everything I collect, I thoroughly washed all the rocks with fresh water and then placed them in a bucket filled with water and a good dose of bleach.

After drying the rocks, I noticed that the black one had small, reddish, powdery pimples all over it. It looked odd, so I broke the rock in half with a hammer,but there was nothing red inside. It actually looked like carbon, but the rock isn't brittle or powdery. I stuck the two halves back in a bucket, filled it half way with water, and them put more bleach in it. The water turned dark, almost muddy, and a reddish, almost gelatinous precipitate appeared and it was quite a bit of it. Any idea what the heck is in that rock that can cause that? I didn't notice any odor other than the bleach, albeit the rock did smell a  bit sulfury. Any idea what I have soaking in the bucket?

Is your rock magnetic?  It might not be, but what you are reporting sounds like iron oxide (i.e. rust), and so your rock could be an iron ore like taconite, or even magnetite.  Such an ore could look like carbon.  Even if not magnetic, it could still be an iron ore.


Why I can't use a single Slater determinant to describe the ground state of open-shell atoms and molecules? If we have a closed-shell atom or molecule (in the Born-Oppenheimer aproximation), and adopt the central field aproximation, we can write the ground state wave function of the system using a Slater determinant, constructed using the spin-orbitals that are the solutions of the equations

$$(h_{i} + V_{i}) \varphi_{i} = \varepsilon_{i} \varphi_{i}$$

with the lowests energies, where $h_{i}$ is the sum of the kinect energy and interaction with the nucleo(s) of the electrons, and $V$ is the potential. However, I read that if the system is open-shell, in general we can't write the ground state wave function using only a single Slater determinant, we have to take linear combinations of Slater determinants. Can someone explaim why this happpen, and indicate a reference? I would like to understand it.

You are looking for multi-reference cases.$^1$ Another term usually applied is "strong correlation". See What exactly is meant by 'multi-configurational' and 'multireference'?.

Instead of trying to answer myself on what those cases might be, let me quote Frank Jensen (Introduction to Computational Chemistry, 2nd ed., Wiley, 2007, p. 154):

  Consider again the ozone molecule with the two resonance structures shown in Figure 4.9 [which shows a) a zwitterion with a positive charge on the central oxygen b) a triplet configuration with single electrons on terminal oxygens]. Each type of resonance structure essentially translates into a different determinant. If more than one non-equivalent resonance structure is important, this means that the wave function cannot be described even qualitatively correctly at the RHF single-determinant level (benzene, for example, has two equivalent cyclo hexatriene resonance structures, and is adequately described by an RHF wave function). A UHF wave function allows some biradical characters, with the disadvantages discussed [earlier in the text]. Alternatively, a second restricted type [Configurational State Function (CSF)] (consisting of two determinants) with two singly occupied MOs may be included in the wave function. The simplest [multi-configuration SCF] for ozone contains two configurations (often denoted TCSCF), with the optimum MOs and configurational weights determined by the variational principle. The CSFs entering an MCSCF expansion are pure spin states, and MCSCF wave functions therefore do not suffer from the problem of spin contamination.

In some cases, a single determinant may give you an incomplete picture. This is more often the case for triplet or open-shell species, but open-shell singlet cases can also demand a multi-determinant treatment.

Note that multi-reference treatments may also be necessary to obtain accurate quantitative results if the physical ground and excited states are close in energy.

$^1$ The "reference" is due to the use of the Hartree-Fock determinant as the reference state for configuration interaction (CI) post-SCF calculations. There are systems where only bad results can be achieved when using a single reference in a truncated CI. (In full CI, you have every possible reference included.)


Will distillation do the job of purifying the water? I asked about my sophisticated water purification system for a generation ship on Worldbuilding Stack Exchange, and somebody mentioned this:

  If you're gonna go to all the trouble of distilling it, why not just
  have that be your only step?

And it is a good question. But, from my chemistry knowledge, this just won't work. Here are the problems I see with distillation being the only step:

Proteins in blood and urine are going to froth up from the heat. If these get into the water, the water is going to smell like urine and nobody is going to want to drink it.

This is the easiest issue to fix. Just make the apparatus tall enough and no urine froth will make it to the distallate. That or adding a trap to react with the urine froth and solidify it in between the 2 ends of the apparatus would work.

Any hydrochloric acid that is present from vomit is still going to be there because it changes the boiling point of water. You simply can't get rid of the acid via distillation alone, not even repeated distillation. You need either microbes that can break down hydrochloric acid or an acid base reaction to get rid of the acid.

Hydrochloric acid forms an azeotrope with water. In fact it forms several depending on the acid concentration. So at some point, strongly acidic water will come over. Worst part is that a mixture of hydrochloric acid and water is colorless and odorless, so the only way you could tell is by density. Start to see those lines that occur when you mix liquids of different densities and you know strongly acidic water is coming over. Also, checking for pH would help but that would mean either an acid resistant pH sensor in the container with the distillate or stopping the distillation periodically and using pH strips. Checking for a change in density is much easier.

If acid vapors escape the system, as I think some will, then everybody on the ship will get irritated lungs, and some might get "Acid Pneumonia" and die from both the irritation due to the acid and the infection that the hot acid predisposes them to. There would be some infectious agents on the generation ship to keep their immune system in their prime and to help prevent autoimmune disorders from spreading like wildfire. But if this is combined with acid vapors, it is basically a death sentence for whoever gets infected. This is not even considering the fact that acid in your eyes can make you go blind

No distillation system can 100% prevent vapors from escaping. Not even a vacuum distillation with a trap will do it.

So in my opinion, distillation is necessary but can't be the only step, otherwise WCS, you get water that smells like urine that nobody wants to drink and hot acid vapors spread all over the ship causing lung irritation and blindness and everybody dies from a combination of the acid vapors and dehydration related both to heat and infection very quickly and you just have an empty hull in space. BCS, everybody gets heartburn from drinking the acidic water.

So, is there a way to get rid of the hydrochloric acid when distilling the water? Or would the people distilling the water have to resort to an acid base reaction followed by a second distillation to get rid of the acid and any excess base?

The old school method of generating ultra-pure water by distillation was to add a small amount of potassium permanganate in a slightly alkaline solution. All the organics were oxidized by the permanganate ion in boiling water. A small of $\ce{MnO2}$ so formed during boiling would take care of the acid(s). This water was good enough for electrochemical experiments. HCl only forms an azeotrope at 20.2% composition and it boils around 110 Celsius.

Today ultrapure water is produced by using quite expensive deionizers with UV lamps. UV destroys all the organics, and then an activated carbon filter absorbs them. Water produced from such deionizer has a specific resistance of $18.2  \rm{M\Omega.cm}$.


Reactions of tin (IV) iodide with potassium iodide and water In the reaction of tin (IV) iodide in acetone and water, the solution turns a urine yellow.

Is this simply due to the hydrolysis of tin (IV) iodide:

$$\ce{SnI4 + 2H2O -> SnO2 + 4HI}$$

why is the acetone needed though?

and for the reaction with $\ce{KI}$:

$$\ce{SnI4 + KI -> ?}$$
I understand it forms iodine, but what happens to $\ce{Sn}$ and $\ce{K}$ - do they react with the acetone for instance?

If iodide is oxidized to iodine (the cause of the yellow color), then something needs to be reduced in turn. Potassium is already in its most stable oxidation state (+I), so tin will be reduced, from $\ce{Sn}$(IV) to $\ce{Sn}$(II).

$$\ce{SnI4 + 2H2O -> SnO2 + 4HI}$$

$$\ce{SnO2 + 2HI -> I2 + SnO + H2O}$$

With $\ce{KI}$, a yellow-brown solution of $\ce{KI3}$ will be formed, which contains the triiodide anion $\ce{I3-}$.

$$\ce{I2 + KI -> KI3}$$

When acetone is added, the yellow color of $\ce{I2}$ will slowly disappear as iodoacetone and $\ce{HI}$ are formed in an acid-catalyzed reaction (source):

$$\ce{(CH3)2CO + I2 -> CH3C(O)CH2I + HI}$$


Sodium Chlorite vs. Chlorine Dioxide I ordered Sodium Chlorite in the mail and received Chlorine Dioxide. Can they be used interchangeably, or are they different? My background in chemistry is limited so laymen's terms would be appreciated. I appreciate any responses or feedback.

Sodium Chlorite is a white powder, which dissolves in water.

Chlorine Dioxide is a gas. This gas also dissolves in water to some degree and gives a yellow color.

Sodium Chlorite is used to make Chlorine Dioxide in the laboratory.

I strongly doubt, that someone has sent you Chlorine Dioxide, as it is to dangerous to store and to send.


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)

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.


Why does road salt reveal this bus's frame? Our city sprays brine on the roads in the winter to prevent people from sliding around. It works but it makes it impossible to keep vehicles clean. One thing I've noticed is that the brine sticks to the city buses outlining the frame underneath the body panels, like so:

Doing some research, these buses are made out of stainless steel to prevent rusting.

Is there a chemical reason for this happening?

No. the best two explanations don't need any chemistry at all.

If you look closely at the panelling on a bus like this you will find that there is physical  evidence on the outer skin about where the structural frame is. The skin immediately over the frame tends to stick out a little. These physical features are probably enough to ensure that more dirt sticks to those parts of the skin. You can see a similar effect above warm radiators when small bumps on the wall above tend to pick up a disproportionate amount of the dust in the moving air caused by the convection currents from the warm radiator, becoming darker as a result.

Another possible contributor would be the fact that the vibrations in the panel skin of the bus will also be higher at the centre of each panel farthest away from the frame. This might contribute something as well.


Equilibrium for the synthesis of methanol Liquid methanol is obtained with carbon monoxide gas and hydrogen gas in a reactor at $\pu{300 °C}$ and $\pu{250 atm}.$ $K_p = \pu{9.28E-3}.$ Find $K_c.$

With only these data, do I have to use the following equilibrium for the calculations
$$\ce{CO(g) + 2 H2(g) <=> CH3OH(l)},$$
or do I have to suppose first methanol is obtained as a gas and then liquified:
$$\ce{CO(g) + 2 H2(g) <=> CH3OH(g)}?$$

Note that under the given conditions, methanol cannot exist in the liquid state, however since you are given in the problem that methanol is obtained in liquid state, you will only consider the equilibrium (for problem solving purpose)
$$\ce{CO(g) + 2 H2(g) <=> CH3OH(l)}$$
And since $K_p$ is related to $K_c$ by the relation
$$K_p = K_c(RT)^{\Delta n_\mathrm{g}},$$ where $\Delta n_\mathrm{g}$ represents the change in the number of moles of gaseous species, so here $\Delta n_\mathrm{g} = -3,$ Therefore $K_\mathrm{c}=K_\mathrm{p}(RT)^3=9.28\times 10^{-3}\times (300 \times 0.0821)^3 \approx \pu{138.65 atm^3 L^3 mol^-3}$.


Why is a weak base more able to deprotonate a strong acid than a weak acid? Very basic question, but I'm rather confused.

Say I want to separate a carboxylic acid $(\ce{RCOOH},$ $\mathrm{p}K_\mathrm{a}~5)$ from a phenol $(\ce{ArOH},$ $\mathrm{p}K_\mathrm{a}~10)$ via acid-base extraction. Apparently I am able to use sodium bicarbonate $(\ce{NaHCO3}),$ a weak base, as a way to deprotonate the stronger acid, carboxylic acid. This allows the carboxylic acid to dissolve into the aqueous phase in an ether extraction, as shown in the far left of the attached diagram, while the phenol would remain in the organic phase (A at the bottom corresponds to the carboxylic acid, B is the phenol, C is the amine).

The question is:

Why does a weak base 'prefer' to deprotonate a stronger acid? Is it because, being a stronger acid, the carboxylic acid will have more anions in solution compared to the weaker acid, so the (relatively few) cations of the weak base will be more likely to find carboxylates than phenoxides?

If this is the case, could we still use a strong base, but just in a very very low concentration and have the same effect as using a weak base at a higher concentration?

The key is that water is the great equalizer and mediator. Let's deal with weak and strong base first: You are right that a sufficient quantity of a strong base would have the same effect. In this particular case, the hydrogencarbonate will separate into $\ce{CO2}$ and water on protonation and we can assume that most of the carbon dioxide will bubble out. So it will functionally not make a difference whether you started with $\ce{NaOH}$ (a strong base) or $\ce{NaHCO3}$. This is a part of the equalizer part: in an aqueous solution, you cannot have a stronger base than $\ce{OH-}$, because any stronger base will just deprotonate a readily available water molecule.

To some extent, even weak bases will do so in an equilibrium. To understand how a proton gets from an acid to a base, we have to understand the structure of water: A water-based solution like the one from the question is a huge network of water molecules engaged in hydrogen bonding with each other and mainly electrostatic bonding to ions. A when a proton "moves" through this network, it will basically jump from one water to the next, and that water will give up a proton to the next etc. (This is the reason for the huge mobility of protons in water as measured by conductivity experiments.) Thus, an acid and a base such as $\ce{RCOOH}$ and $\ce{HCO3-}$ do not have to come close to each other to effectively transfer a proton between them.

Thus the question "How does the base know to deprotonate the strong acid instead of the weak acid?" loses its basis. In an equilibrium, the stronger acids and bases will deprotonate and protonate more, respectively, than their weaker counterparts because that's what acid/base strength means.


Which elements can be diatomic? Which elements can be diatomic and why?

Motivation

Hydrogen, Nitrogen, Oxygen and the Halogens tend to be thermodynamically stable as a diatomic molecule at room temperature, and are usually labelled as diatomic nonmetals. (Although I am unsure about Astatine). These are the ones you learn in school and I am fine with their reasoning. I want to know about other elements in particular.

Looking at the rest of the non-metals: Carbon, Phosphorus and Sulfur have diatomic molecules in gaseous states. All with a common theme of being heated at very high temperatures.

This leaves us with diatomic Selenium which I could not find anything on.

Further notes

Diatomicity doesn't seem to be just a non-metal thing:

Lithium forms diatomic molecules and all diatomic alkali metals are claimed to be stable in gaseous states. Found from this question.
Silicon has been reported to have a diatomic molecule similar in structure to carbon.
Tungsten, Chromium, Uranium and Molybdenum are claimed to be diatomic with a sextuple bond.
Supposedly Noble gases can form dimer molecules but if these are considered diatomic is debatable.

This leaves us with diatomic Selenium which I could not find anything on.

Selenium forms a dimer with bond energy of 79 kcal/mol, according to table 8.6 (page 451) of Comprehensive Handbook of Chemical Bond Energies, which provides 5 references.

Tellenium dimer is 61 kcal/mol

Polonium dimer 45 kcal/mol


Test for a dipolar compound? Just surface level here, if I had two compounds and wanted to know which one was dipolar, could I just add them to water? If so, would the dipolar substance dissolve while the other compound is unaffected?

Thanks

In general substances whose molecules have a strong dipole tend to dissolve better in water than compounds which do not. If we consider for example diethyl ether and THF. The two are almost identical the only difference is that THF is held in by ring into one posiition. This makes THF have a dipole which is more permenent. The side groups on diethyl ether can move about thus allowing the molecule to adopt different conformations.

THF is totaly soluble in water, it is infinitely miscible with water. This means any combination of THF and water will mix perfectly. On the otherhand diethyl ether has a limited solubility in water. Only a very small amount of diethyl ether will dissolve in water.

If we look at the solvent accessable surface of THF where the atomic colours represent their partial charges we can see that THF has a strong permenent dipole. Here in these pictures the more blue an atom is the more negative and the more red the more positive it is.

While ether has less of a permennt dipole as the effects of the right hand side partial positive charge is canceled out by the left hand one.

Do be careful as more than dipole strength will determine the solubility of a compound. A better test for liquids would be to measure the relative dielectric constant at different frequencies. If we have a substance whose molecules have a dipole then at a low frequency the relative dielectric constant will be higher than at a high (often microwave) frequency. The reason is that at a low frequency the molecules have time to rotate to line up. At a high frequency this does not occur becuase the molecules have too little time to rotate. Below is shown a random array of dipoles and on the right some dipoles which have been aligned by an electric field.


Storing kinetic energy in bonds Let's assume a setup with a static linear molecule with three identical atoms connected by bonds and a single atom, identical to the other three, being shot at the molecule. Let's also assume that everything happens in 1D and can therefore be illustrated by this simple scheme:

(projectile) A ->    A-A-A (molecule)

Where the direction of the arrow is the direction of motion of the projectile and the dash signs in the molecule are the bonds between atoms.

[EDIT - must change next paragraph as the description does not fit what would happen: stand-by, no time to do it right now!]

Assuming a low-energy, non-relativistic, perfectly elastic collision, the projectile will stop dead in its track while the molecule will be set in motion at 1/3rd of the projectile's speed, as it has 3 times the projectile's mass (conservation of Momentum). Balancing also the equations for the Kinetic Energy easily reveals that 2/3rds of projectile's kinetic energy must have been locked away in internal vibrations of the molecule.

Now, it's easy for us to calculate these numbers by considering the projectile and the molecule as just two separate systems and conveniently attaching a total mass to each. But in reality the universe doesn't group atoms into molecules, does it? Each bond in the setup doesn't "know" about the existence of the other bond and each atom could be said to be unaware of all but the atoms bonded or colliding with it.

In this context, not "knowing" total masses, how does the universe "calculates" that in this case the molecule must store 2/3rds of the projectile kinetic energy into internal vibrations, no more, no less?

The universe doesn't calculate anything. Energy and it's conservation and everything that follows from it is a human-made concept, a physical model. It describes the things that happen in nature quite well (in the right circumstances), but it is only a model. In how far this lends some kind of "reality" to the concept/model I couldn't say. It's more of a philosophical question. Physical laws only describe how nature works, but not why it works the way it does.

Edit:

Concerning the physical content of the collision you described: I think of it in the following way: The projectile hits the first atom of the molecule, let's call it A1, and transfers its momentum to it. Now, this atom moves in the direction along the bond axis directly towards the next atom, A2. While it does so, it has to work against the spring force of the bond between A1 and A2. Thus, kinetic energy of movement is converted to kinetic energy of vibration. A2 will now be accelerated in the direction of the next atom, A3, and it has to work against the spring force of the bond between A2 and A3 and(!) the restoring force brought about by the bond between A1 and A2. Again kinetic energy is converted to vibrational energy. But now you have a system of coupled oscillators. This process will be repeated for every atom in the molecule until every bond vibrates. So, it is not true that

  Each bond in the setup doesn't "know" about the existence of the other
  bond and each atom could be said to be unaware of all but the atoms
  bonded or colliding with it.

They are coupled through the spring forces of the bonds and thus each atom "feels" the mass of every other atom in the molecule as well as the bond strength of every bond in the molecule.


How is a reaction with more products than reactants spontaneous? If I have a reaction that has $\Delta G^\circ = 0,$ where there are more products than reactants (for example, $\ce{A <=> 2B + 2C}),$ then if I change the concentration of all the reactants and products (i.e. to 20 mM), then shouldn't the reaction favor the reactants, since you have more total product than reactants (in my example, you would have 40 mM of both $\ce{B}$ and $\ce{C}$ and then just 20 mM of $\ce{A}),$ and therefore produce a positive $\Delta G$ value? I get that if you use the equation $\Delta G = \Delta G^\circ + RT\ln Q,$ you would get a negative $\Delta G$ for my example, showing that it is a spontaneous reaction.

However, can anyone explain this problem from a conceptual perspective, since it would seem that if you have more products than reactants with $Q,$ that would drive the reaction to favor the reactants, according to Le Chatelier's principle?

Note that whether the forward or reverse reaction is favorable depends entirely on the relative magnitudes of Q and K (the reaction quotient versus the equilibrium constant):

$$\begin{align} Q&>K \rightarrow \text{reactants favored}  \\ K&>Q \rightarrow \text{products favored}  \end{align}$$

Next, consider that when you declare a value for $\Delta G^\circ$ you can translate that into a value for K, using the following equation:

$$\Delta G^\circ = -RT\log K$$

For instance, if $\Delta G^\circ = 0$ then $K=1$.

For the example here, this means that

$$\begin{align} Q&>1 \rightarrow \text{reactants favored}  \\ 1&>Q \rightarrow \text{products favored}  \end{align}$$

Next, consider the standard state of all substances. For solutions this is typically $\pu{c^\circ = 1 M}$. Standard states are important because the reaction quotients and equilibrium constants, when described in terms of concentrations, contain ratios of the concentrations in solution to the standard concentrations, for instance

$$Q= \frac{\prod_{\text{products i}} (c_i/c^\circ)^{\nu_i}}{\prod_{\text{reactants i}} (c_i/c^\circ)^{\nu_i}}$$

If all substances are present at the standard concentration, $Q=1$. In the present example $K=1$ so the equilibrium state consists of all substances being at standard concentrations ($c_i=\pu{1 M}$).

Finally, consider the following way to relate Q, K and $\Delta G$:

$$\Delta G = RT\log \left( \frac{Q}{K}\right)$$

If $K=1$ as in the present example then

$$\Delta G = RT\log Q $$

This just reinforces what was stated earlier for the special example presented here: when $Q<1$ the free energy difference is negative and products will be favored, and vice-versa when $Q>1$.

Inserting the definition of the reaction quotient

$$\begin{align} \Delta G &= RT\log \left( \frac{\prod_{\text{products i}} (c_i/c^\circ)^{\nu_i}}{\prod_{\text{reactants i}} (c_i/c^\circ)^{\nu_i}}\right) \\ &= RT\log \left( \frac{c_B^2c_C^2}{c_Ac^{\circ\,3}}\right) \end{align}$$

If we assume all substances are present at the same concentration c, then

$$  \Delta G= 3RT\log \left( \frac{c}{c^{\circ}}\right) $$

Clearly if $c<\pu{1 M}$ then $\Delta G<0$ and products are favored.


How do potentiometric cells do not proceed through a spontaneous reaction like in a galvanic cell? It seems to me that the cell setup of a galvanic cell is pretty much the same as in a potentiometric cell. However, galvanic cells just proceed the spontaneous direction of the net redox reaction of the anodic and the cathodic species in the cell, while in a potentiometric cell, it seems that the equilibrium activity ratios of the measured solution are not to be affected.

I've read that in a galvanic cell, current flows as to follow the natural tendency of electron flow--from the anode to the cathode. But in a potentiometric cell, it's said that the current of electrons flowing is kept to a minimum. How does a potentiometric cell does that? How does it avoid for reduction at the measured cell or oxidation at the reference electrode to happen?

P.S. I know that the question looks uncohesive, but it's because I'm really confused how these concepts occur. Please try to clear up the misconceptions that might have been apparent from my question.

First, I wouldn't classify cells as potentiometric or galvanic.

If you take any electrochemical cell with non-zero total cell potential and connect a wire between the two electrodes it will discharge. This will happen spontaneously in the galvanic direction.

In potentiometric measurements of electrochemical cells, what you do is a potential measurement which ideally happens at zero current(definition of a voltmeter), and practically at picoampere or lower current. Therefore you are not allowing the electrons to flow from the outside circuit and hence your cell doesn't discharge.


Calculate vapor pressure from volume For a chemistry lab, I'm required to calculate the vapor pressure of some organic solvents... methanol, ethanol, and propanol, in addition to an "unknown" sample of some solvent. In the lab, we were instructed to heat these solvents in a sealed syringe along with 5 mL of air. While they were being heated, I recorded the volume and temperature of the air and solvent in the syringe. Now, from that data, I need to calculate the vapor pressure of these substances at the temperatures at which I recorded the data. How can I do this?

You should treat everything as ideal gas. Then you calculate the volume change from the air itself, using ideal gas law and pressure being constant (atmospheric, maybe). Calculate also the number of moles of air in the syringe, it will come handy soon.

The remaining change in volume is then attributed to the solvent, again using ideal gas law, you put in the given volume, temperature, pressure and obtain number of moles, which turned into vapor.
Having this, you know number of moles of solvent and number of moles of air, so again assuming everything ideal (Raoult's law), you know the partial pressures (i.e. vapor pressure).

For added fancines, you can take the decrease of the liquid volume into account.

Side note: While using ideal gas and ideal mixture of ideal gases approximation, all quantities are nicely additive, so don't be afraid.


Purifying washing soda I bought a box of A&H Washing Soda, the packaging of which implied that it was pure washing soda (sodium carbonate).

On dissolving it in osmotically-purified water, however, I found that the water took on a brown tinge, which would seem to indicate a contamination or adulteration of some type. Also, the washing soda tended to form hard chips or clumps, which I have not seen in previous experiments doing the same thing.

I suspect there may be some kind of organic compound being used to adulterate the washing soda.

How can I determine if there is an organic adulterant and remove it?

If there is an organic contaminant in the $\ce{Na2CO3}$, you could try to rinse the sodium carbonate with an organic solvent in which $\ce{Na2CO3}$ is insoluble.  Based on this paper, acetone and methanol would qualify.  Since those solvents, especially methanol, are toxic, I'm not recommending either one of those.  Based on the data for methanol and acetone, I'm guessing absolute ethanol would also be incapable of dissolving sodium carbonate, so it would probably the best (relatively) non-toxic solvent for extraction.  If you can't get absolute ethanol due to tax or beverage laws, you could probably used denatured ethyl alcohol instead.  It usually has small percentages of methanol, acetone, and/or isopropyl alcohol added to make it non-potable.

After you wash the sodium carbonate in this solvent, you should air dry it for several days, and then dry it in an explosion-proof drying oven at 100 - 200 °C.  Be very careful when combining flammable solvents with heated environments!

The problem could also be with the water you used, although you did say you used osmotically purified water.  The reason I think the water may be the culprit is that if the washing soda contained a brown contaminant, I would think it render the powder visibly brown.  If no brownness is evident until adding the soda to the water, it makes me think the high pH from the washing soda is precipitating minerals from the water, in particular iron.


Reaction between phenylamine and halogens I understand that phenylamine is considerably more reactive than benzene, meaning that it is able to react with bromine water to give 2,3,6-tribromophenylamine. But I was then wondering why I haven't seen anywhere mention reactions with other halogens. So will phenylamine react with other halogens to produce a similar compound, I see no reason why it wouldn't happen? Do text books just use bromine water as it is convenient?

Will  $\ce{C_6H_5NH_2 + 3Cl_2 -> C_6H_2NH_2 Cl_3 + 3HCl}$ happen?

Chlorination of aniline with chlorine is rapid and gives 2,4,6-trichloroaniline according to the references cited in this paper here. The earliest reference cited is 1845 by AW Hofmann so it has been around a long time. More modern approaches use N-Chlorosuccinimide as the reaction is more controllable


Cahn-Ingold Prelog rules concerning tie breaking on second level comparing the cahn-ingold prelog priority of the above groups, I should first compare the atoms attached to the first carbon.
So we get:
(A): {O,C,H}
(B): {O,C,H}
(C): {O,C,H}
(D): {O,C,H}
therefore we have to consider the second level, and this is where I get confused:
Do I compare it this way:
(A):{C,H,H,H}
(B):{O,H,H,H}
(C):{F,C,H,H}
(D):{C,H,H,H}
I did this by rearranging all the "second level atoms" in descending order of atomic numbers.
since A and D are still in a tie, I rearrange all the "third level atoms" in descending order of atomic numbers:
(A):{H,H,H}
(D):{F,H,H}
so the priority is C>B>D>A
Is that correct?

look, all i can say is that $ D > C > A $

why ?

Lets compare A and C
$ OCH$3 and $ H $ are common to both. one has CH3 and the other has CH2F. of course, C > A.

compare A and D in the same way and find that D > A

compare D and C now.
there is clear proof that D > C because it has F in two places while C has H in those two exact positions.
thus it is seen that D > C > A

now see B and C
in B : after O we have CH3 and after C we have { H , H , F }
in C : after O we have H and after C we have { H , H , O }

looking at second layer after oxygens : B > C
looking at second layer after carbons : C > B.
thus even i am confused when it comes to B
but definitely D > C > A


Dynamic equilibrium - effect of adding inert gas Well consider the reaction $A + B ⇌ C$
Adding inert gas to a container where this reaction is taking place will increase the pressure of the system. In accordance with Le Chatelier's Principle, the equilibrium should shift in the direction which opposes this change in pressure, i.e. in the direction where fewer number of molecules are created. In this specific reaction, it happens to be in the forward reaction. However, I have been taught that adding an inert gas at constant volume has no effect on the equilibrium. Why is this?

At constant temperature, only changes to the concentration (technically activity) of a reactant or product, shift the equilibrium.

Adding inert gas does not change the concentration of any reactant or product, so it does not affect equilibrium.


Problems with a lithium battery I have a lithium battery that is separated from any device and is producing a whitish-gray smoke every time the wires move. Sometimes it makes hissing noises.

Is there a way to solve these problems?

You probably want to drain its charge and put it into recycling.  It's probably a fire hazard at this point, and the damage that a fire from it could cause will cost more than a new one.


How to predict the effusion/diffusion for a mixture of gases? Suppose a a container contains a gaseous mixture of methane and hydrogen. The mixture effuses out through a hole in the container. Now hydrogen is lighter gas so it must effuse faster and all of hydrogen should come out of the flask earlier than methane (before all methane effuses out) and a stage should come when there is only methane left in the container.
Is this how it happens?

Sort of, but not quite.

If our hole is large (say, as if pierced by a toothpick), then the gases will go out in pretty much the same composition that is found inside, so their ratio in the mixture will stay nearly the same till the very end.

If the hole is small, about the size of free path of a molecule in a gas, then it's a different story. BTW, these conditions are called Knudsen flow. Methane weighs 16, and helium weighs 4 (I know you've said "hydrogen", but let's assume it was helium, just to make the calculations easier). So helium particles are 4 times lighter than those of methane, hence they fly 2 times faster. So there would be two times as many of them in the outgoing stream - that is, if the mixture inside was 1:1. See, it's not like all helium flows outside, while methane molecules stand politely in the line, saying "After you".

As the story progresses (which is not going to be fast with just one tiny hole), the mixture inside gets enriched with methane. So does the outgoing flow, but still it remains richer in helium relative to the mixture inside, and hence continues to shift its composition in the same direction. Yet still, pure methane is a limit which is never actually attained.

That being said, pressing gas mixtures through porous membranes is a valid way to separate the components by molecular mass regardless of their chemical properties, and as such, was used for isotope enrichment.


Will using vinegar harm car paint? After having parked my car in a newly constructed airport parking garage, I came back to find something had dripped through a crack in the ceiling of the structure and left considerably thick film on my car.  On a glass window surface, I tested soap and water, mineral spirits, two types of latex paint remover (i.e. goof off), and household chlorine bleach.  None were effective.  But plain white 5% vinegar dissolved the film if I let it soak or scrubbed on it with a cloth.

My question: is an acid of the strength of vinegar a risk to automotive "paint" (the top layer is probably "clear coat").

That is a valid question, because acid rain has been known to damage paints, particularly high-gloss, i.e. clear coat. Best would be to test the effect of vinegar on a hidden area, and double-check under bright light from different angles. Your idea to test on glass is good, because it shows how long a soak is needed.

If it appears unaffected, rinse afterwards with a weak solution of baking soda or ammonia, and then rinse again with plain water to remove any remaining acid.
If damage is slight, can car wax effect a good repair?
If vinegar does damage in the test spot, you might look at various car polishes or even cutting compounds, and then wax (or reapply a gloss coat, which may mean refinishing a large section of bodywork to get it to match).


Why is it so difficult to create mock NMR spectra of macromolecules? From what I understand, if you have a NMR spectra, you can determine the structure of a macromolecule but it is very difficult to determine what the NMR spectra of an expected structure would look like? Why is this the case?

Macromolecular NMR spectra are difficult to compute because macromolecular electronic structures are computationally expensive to solve with accuracy. Obtaining precise chemical shifts and splitting requires accurate treatment of electron density and geometry, especially for simulation of NOESY, which I understand is important in macromolecular structural elucidation. This requires good methodology and large basis sets.
An interesting case study by Saielli and Bagno demonstrates the use of DFT (with cc-PVTZ basis) in predicting NMR spectra for two candidate structures and the correlating them with experimental data, however the mean error in chemical shifts is nontrivial1 (0.2 ppm in the best match). The same people do $^{1}\ce{H}$ and $\ce{^{13}C}$ NMR predictions of a number of complex molecules2 (with intriguing discrepancies in shifts) however this is a far cry from predicting, say, a large protein $^{1}\ce{H}$ NMR spectrum.
I have seen EPR spectra of transition metal clusters simulated as abstract spin systems (for instance by EasySpin) however I presume that this is very qualitative.

(1) Saielli, G.; Bagno, A.; Can two molecules have the same NMR spectrum? Hexacyclinol revisited.; Org. Lett.; (2009) 11, pp. 1409-1412
(2) Bagno, A.; Rastrelli, F.; Saielli, G.; Toward the complete prediction of the $\ce{^1H}$ and $\ce{^{13}C}$ NMR spectra of complex organic molecules by DFT methods. Application to natural substances.; Chem. Eur. J.; (2006) 12, pp. 5514-5525


Gallium-induced structural failure of aluminium I have seen youtube videos of people treating some aluminium object with liquid gallium and after letting it sit for a while, they are able to break up the aluminium into tiny little pieces with their bare hands.

Can someone tell me what is exactly going? I haven't been able to find a satisfactory explanation as of yet.

Does the gallium somehow incorporate itself into the metal lattice, disrupting the bonds? or is it forming an amalgam of sorts and actually dissolving the aluminium?

Is it a combination of these two, or are my ideas off completely? Also, what would be the exact mechanics of a process like this?

EDIT: How exactly does gallium "unzip" the grain boundaries?
 Also, besides Ga or Hg, what other metals would produce a similar effect? What mechanism/property allows them to behave in such a manner?

As you state, it forms an amalgam, just as mercury does. @Jon Custer's explanation is correct, and furthermore, the alloy does not form a protective oxide, so the aluminum oxidizes to $\ce{Al2O3}$ and quickly crumbles. This happens even faster with an Hg-Ga alloy!

BTW, gallium should be safe enough, though expensive, but I would not like to work with mercury without a hood, and without careful containment.

Thanks for the question; this encouraged me to view that amazing demo on Vimeo!


Elastic scattering of x rays in 3D: the principle for XRD analysis I have some trouble visualising scattering of x-rays in 3D, since all textbooks and examples present it in 2D. Under I will share some reflections (no pun intended) about the elastic scattering of x rays, and the diffraction effects observed, which is the basis of XRD analysis.

Here you see a figure of two crystal planes, separated by the distance d. Two parallel and in-phase x-rays hit an atom in the planes. The scattering cones are drawn. The darker area represents those scattering directions going into the crystal itself, and thus not detectable and uninteresting for XRD analysis. The light area are those scattered rays exiting the crystal.

Thinking in three dimensions is different from thinking in one dimension. To achieve constructive interference, the rays must travel in the same direction and also be superimposed on each other. Even though the scattered rays ' angle is, say, 45 degrees, they may not travel in the same direction.

However, if Bragg's law is valid for 45 degrees, all rays of 45 degrees scattering angle will interfere constructively. So we would get a diffraction cone of increased intensity. However, we cannot detect the whole diffraction cone, partly because some of the rays travel into the crystal, and partly because the detector surface area only covers parts of the diffraction cone; we would measure a curved segment of the diffraction cone. This is called the Debye-Scherrer ring, yes?

Though we get constructive interference in all directions of 45 degrees scattering angle, the scattering angle itself is random. This means there are scattered rays which do not undergo constructive interference in all directions, regardless of angle (though smaller angles are more probable). So, in an ocean of random scattering, we have diffraction cones of much greater intensity, since Bragg's law is fulfilled.

Is this a correct assessment of the XRD situation?

In 3D you will see spots not cones from a single crystal.  If you Google for "XRD spot pattern" images, you will probably find yourself enlightened about how many different shapes, and forms diffraction patterns can take.

Debye-Scherrer rings are one special case which occurs when you have a diffraction pattern from a very fine powder.  In such a powder you are summing the diffraction patterns from thousands to millions of individual crystallites.  In this case, you can imagine a large number of spot patterns rotated randomly around the beam axis so each spot projects into a cone.  Each individual crystallite, however will produce a spot pattern, and there is never anything but a spot pattern from a single crystallite.  (Sometimes spots can be smeared when there is strain in the crystal, or when the crystal is too small to produce high quality diffraction, but these are really cases of imperfect crystals.)

You can see diffractions from rays transmitted through the crystal.  However, X-rays tend to be absorbed within a few microns to millimeters in most materials so that limits the size of crystal you can use for transmission X-ray diffraction.

Finally, I noticed you said "Two parallel and in-phase x-rays..."  I just want to note that diffraction occurs with a single photon.  A single photon can scatter, and produce constructive/destructive interference with ITSELF.  You need to do this with many photons to see a full interference pattern, however, and this is usually the case in experimental situations.  If this is new to you, then take a look at Young's double slit experiment.

Hopefully this clarifies things a bit.


Is it possible for a carbon atom with a negative formal charge to have ONE bond and ONE lone pair? I was completing some problems from my textbook and found that the answer of this problem seemed incorrect to me.

From my limited understanding, carbon atoms with a negative formal charge must have 5 electrons (3 bonds and 1 lone pair), but this answer didn't match up with this rule.

I would really appreciate it if someone could tell me what I'm missing here.

The hydrogen atoms are not shown, so this is really $\ce{CH3CH2CH2CH2-}$ where the rightmost carbon atom has a lone pair and a formal negative charge.

The species is the n-butyl anion:


More ions form pairs when concentration of acid increases, does that decrease the speed of reaction? Electrical conductivity decreases as an acid becomes more concentrated because more ions form pairs. However, does that affect the speed of reaction of the acid as there is a smaller concentration of Hydrogen ions?

In general, the conductivity of any liquid increases with the concentrations of its ions, and decreases with their dimensions. In pure sulfuric acid, the only ions existing are $\ce{H3SO4+}$ and $\ce{HSO4-}$. They are big and their concentration is low. So the conductivity of pure sulfuric acid is low. But when you add some water, the $\ce{H2O}$ molecules are totally transformed into $\ce{H3O+}$ and $\ce{HSO4-}$ ions which are numerous and sometimes small (for $\ce{H3O+}$). The conductivity must increase.


How can an insoluble compound be a strong electrolyte Here's a quote from Petrruci General Chemistry (pg. 160):

Silver chloride, $\ce{AgCl}$ is an insoluble ionic compound. When $\ce{AgCl}$ dissolved in water, it is 100% dissociated into $\ce{Ag}^+\text{ and } \ce{Cl}^-$ ions; there are no $\ce{AgCl}$ pairs.

I'm confused:

1) If $\ce{AgCl}$ is insoluble, how is it dissolved in water?

2) If  $\ce{AgCl}$ is a strong electrolyte but insoluble, does it mean a  $\ce{AgCl}$ molecule is ionized, but isn't separated - i.e. the silver and chloride remains close (they're not solvated by water molecules) but each becomes and ion?
I'm confused, can someone explain what's going on?

It sounds contradictory, but soluble and insoluble are relative terms. Silver chloride's $\mathrm{K_{sp}}$ is $1.77\times 10^{-10}$, so one can generally think of it as insoluble, but actually about a milligram will dissolve in a litre of water.

Much like the terms strong acid and weak acid, strong electrolyte and weak electrolyte refer to the dissociation of a substance in a solvent, though they include all electrolytes, not just acids. In the case of silver chloride, though little of it dissolves, what does is present only as $\ce{Ag+}$ and $\ce{Cl-}$, not solvated  $\ce{AgCl}$. In water, all salts are strong electrolytes, but in other solvents, things can be different. For example, while perchloric acid is a strong electrolyte in water, it doesn't dissociate completely in acetic so it's a weak electrolyte.

Source: (1) Electrochemical Dictionary; Bard, A. J.; Inzelt, G.; Scholz, F., Eds.; Springer Berlin Heidelberg: Berlin, Heidelberg, 2008.


Differences between Liquid and Solid Infrared Spectrometry In general, when should I use liquid or solid IR spectrometry to analyse my compound? What should I expect to see? Would I see higher or lower frequency, broader or narrower transitions? My specific case is a tricobalt cluster, Co3(CBr)(CO)9. But I'm also curious for all compounds in general.

Some basic comments can be found in Spectrometric Identification of Organic Compounds. Although it is primarily concerned with interpretation, it has some notes on the types of instrumentation and samples that would typically be encountered in an (in)organic chemistry lab. I'll only reproduce parts that mention the quality of the obtained spectrum.

  Vapor-phase spectra resemble those obtained at high dilution in a nonpolar solvent: Concentration-dependent peaks are shifted to higher frequency compared with those obtained from concentrated solutions, thin films, or the solid state (see Aldrich, 1989).

  Thick samples of neat liquids usually absorb too strongly to produce a satisfactory spectrum. (...) For example, thick samples of carbon tetrachloride absorb strongly near 800 cm-1; compensation for this band (using a reference beam with pure solvent) is ineffective since strong absorption prevents any radiation from reaching the detector.

  Solids are usually examined as a mull, as a pressed disk, or as a deposited glassy film. The suspended particles in a mull must be less than 2 microns (in diameter) to avoid excessive scattering of radiation. The quality of the spectrum (when using a pellet) depends on the intimacy of mixing and the reduction of the suspended particles to 2 microns or less.

  In general, a dilute solution in a nonpolar solvent furnishes the best (i.e., least distorted) spectrum. Nonpolar compounds give essentially the same spectra in the condensed phase (i.e., neat liquid, a mull, a KBr disk, or a thin film) as they give in nonpolar solvents. Polar compounds, however, often show hydrogen-bonding effects in the condensed phase. Unfortunately, polar compounds are frequently insoluble in nonpolar solvents, and the spectrum must be obtained either in a condensed phase or in a polar solvent; the latter introduces the possibility of solute–solvent hydrogen bonding.

One of the reasons you don't want very concentrated samples (for structural determination) is that you increase the amount of intermolecular interaction present in the sample, which can change peak positions and widths. A prototypical example is water:

For the vapor (gas phase), the water molecules are almost non-interacting, leading to well-defined and narrow peaks. The structure here may still be from some small number of environments where a molecule is interacting with a handful of others, but it may also be rotational fine structure (see rovibrational spectroscopy). In the (condensed) liquid phase, these individually-identifiable peaks coalesce into one broad peak due to the many hydrogen bonding environments induced by collsion with other molecules. The ~400 cm-1 redshift of the peak center originates from the nature of a single hydrogen bond:

where the molecule on the right (the h-bond acceptor) donates electron density into the $\ce{O-H}$ bond of the molecule on the left (the h-bond donor), specifically an antibonding orbital, which weakens the bond. Weaker bonds are lower in energy, and wavenumbers (cm-1) are units of energy. Finally, for the solid phase, most common polymorphs of ice have regular structure with a finite number of hydrogen bonding environments, which leads to slightly more distinct peak structure in the solid over the liquid. These peaks are further redshifted due to the decreased $\ce{O-O}$ distance in ice over liquid water (2.76 versus 2.82 Å), leading to even greater charge transfer between water molecules. Broadening and shifting can be also seen for glycine peptides of different lengths prepared as both a solid mull and in continuous films due to the terminal $\ce{-NH3 / -NH2+}$ and backbone $\ce{-NH}$ groups.

I don't have any references for it right now, but you can also obtain peak broadening due to solute–solvent chemical exchange; I have only heard of this for hydrogen.

To summarize,

increasing the number of distinct chemical environments for a particular vibrational oscillator is going to broaden the peak associated with that oscillator, and
intermolecular interactions that directly alter the electron density of the vibrational oscillator are going to shift its peak position; whether it will redshift or blueshift roughly corresponds to a decrease or increase in bond order.

References

Silverstein, Robert M.; Webster, Francis X.; Kiemle, David J. Infrared Spectrometry. Spectrometric Identification of Organic Compounds, 7th ed.; John Wiley & Sons, Inc.: New Jersey, 2005; pp. 79-80.
http://www1.lsbu.ac.uk/water/water_hydrogen_bonding.html
Blout, Elkan R.; Linsley, Seymour G. Infrared Spectra and the Structure of Glycine and Leucine Peptides. J. Am. Chem. Soc. 1952, 74, 1946-1951.


Infinite Increase in Entropy when Energy added to Absolute Zero My textbook states the following:

  If a system were at absolute zero, an additional small amount of heat energy would lead to an infinite increase in entropy. Such a state is impossible. Absolute zero can never be achieved.

It also provides the equation:

$$ \Delta S_\text{surroundings} = \frac{-\Delta H_\text{system}}{T}$$ Where $T$ is given in kelvin.

From the statement, will the entropy of the system or surroundings increase; from what I can deduce, I would say the entropy increase would be within the system as the surroundings is losing energy, but I am not sure.

The second part of my question is, why would this lead to an infinite increase in entropy? Please provide a comprehensible mathematical explanation and also, preferably, an analogy.

Textbook: Pearson Baccalaureate: Higher Level Chemistry, 2nd Edition. By Catrin Brown and Mike Ford Pages: 254-255

The textbook is referring to the entropy change of the system.  While the textbook is correct that absolute zero can never be attained, its statement that the entropy change is infinite is wrong. The authors' rationale for thinking it is infinite likely stems from a misinterpretation of the definition of entropy change:
$dS = \frac{\text{đ}q_{rev}}{T}$
where $\text{đ}q_{rev}$ is the reversible heat flow into the system.
Let's start with a system at $0 K$, and then warm it to a temperature $T = T'$.  Then:
$\Delta S =\int_{0}^{T'} dS= \int_{0}^{T'}\frac{\text{đ}q_{rev}}{T}$
Let's consider what happens when we flow the first, infinitesimal amount of heat, $\text{đ}q$, into the system.  And to make the calculation easy, let's assume the heat flow is reversible (it doesn't have to be; if it weren't, we'd just need to find a reversible process that gets us to the same final state).  As soon as we do this, the temperature is no longer zero! And once the temperature rises above zero, the singularity disappears and we no longer have a concern about infinite entropy change.
But, you may ask, what about the mathematics right at the very beginning, when the temperature is indeed zero.  Here, to properly calculate the entropy change, we need to use a limit, recognizing that both $T$ and $\text{đ}q$ are approaching zero.  The easiest way to understand this is to consider the Debye $T^3$ law, which models the constant-volume heat capacity of solids as they approach absolute zero:
$C_v = C_v(T) = k T^3$, where $k$ is a constant, and where I've written "$C_v(T)$" as an explicit indicator that $C_v$ is temperature-dependent.
Since, for a constant-volume reversible process, $\text{đ}q_{rev} = C_v(T) dT$, we have:
$\Delta S= \int_{0}^{T'}\frac{\text{đ}q_{rev}}{T} = \int_{0}^{T'}\frac{C_v(T)}{T} dT = \int_{0}^{T'}\frac{k T^3}{T} dT=\int_{0}^{T'}k T^2 dT$
I.e., in evaluating what happens to $dS = \frac{\text{đ}q_{rev}}{T}$ as $T \rightarrow  0$, it is necessary to consider what is happening to both the numerator and the denominator. The textbook authors' mistake was in not understanding this.  Clearly, once we find a reasonable functional form for $\text{đ}q$ in the limit as $T \rightarrow  0$, the singularity disappears.
Think of it this way (borrowing from one of my comments): At constant volume, $dS=\frac{C_v}{T} dT$. As you approach absolute zero, it's not just $T$ that's going to $0$; $C_v$ is going to $0$ as well. And since $C_v$ is going to $0$ faster than $T$ is going to $0$, $\frac{C_v}{T}$ goes to $0$ rather than infinity (for the same reason that $\frac{x^3}{x}$ goes to $0$ rather than infinity as $x$ goes to $0$). I.e., the mistake would be in assuming that $C_v$  is a constant; it's not.
Here's another way we can understand that the textbook's statement is wrong:  Thermodynamics allows us, in principle, to determine absolute entropies for any substance.  The absolute entropy is given by integrating the entropy change from absolute zero to whatever temperature the substance is at:
$\text {Absolute entropy at } T' \equiv S(T') =\Delta S_{0 \rightarrow T'}= \int_{0}^{T'}\frac{\text{đ}q_{rev}}{T}$
If the entropy change between $0 K$ and any higher temperature were infinite, the absolute entropies of all real substances would likewise also be infinite! [Unless the substance were at absolute zero, which is unatainable.]
Here's a link showing how standard molar entropies might be calculated (I don't know if the procedure described here is what is used to determine the official CODATA values):
https://www2.stetson.edu/~wgrubbs/datadriven/entropyaluminumoxide/entropyal2o3wtg.html.
Essentially, it mentions using the Debye extrapolation up to ~$15 K$, and then using a differential scanning calorimeter to determine the reversible heat flow between $15 K$ and $298.15 K$


Is Acetanilide toxic to fish? I understand that acetanilide can hydrolyze to aniline and assume that will happen if that gets added to an aquarium, thus causing toxicity?

I have a solution that is $3\%$ hydrogen peroxide and $0.25\%$ acetanilide that I'd like to add to my fish and shrim aquarium to kill some algae.

When I checked the SDS of acetanilide on Sigma-Aldrich website, I saw that it is harmful to aquatic life.
So, there is no exception whether it is an alga or fish to kill. I suggest you not use acetanilide in the aquarium.


How do chemists find a procedure to produce a given substance? How do they find out exactly what reactions they can use to produce some substance they want to produce? Is there some reaction database that helps them find a way to make the target substance out of available chemicals?

For example, let's say they want to produce ethanol. There are multiple ways to do this: they could use the fact that yeast can transform sugars into ethanol and carbon dioxyde and use distillation to produce concentrated ethanol, or use ethylene hydration to get the same thing.

The question is: how could they find those ways? How do they know what will be produced by a given reaction, or inversely, what reaction will yield a given substance? Some reactions are well known and were just observed and studied, but what about the obscure ones?

Thanks.

Chemists use many sources of information to do this.

Part of being trained as a chemists is learning, in general terms, what kinds of reactions are known and what their best uses are. That training can go 5-10 years beyond a bachelors degree and is very intensive.

Beyond what you can keep in your head, one of the most important information sources is the "primary literature", consisting of:

Patents
Journal publications

As an example, see this patent about ethanol manufacture.

Alternatively, see this very recent journal publication dealing a new catalytic reaction.

Having all of this information out there is a start, but it's a vast collection spread out over many publishers and 100 years or more. To find what you need, it's essential to have a good indexing system. This allows you to search large databases of reactions and molecules for references that exactly match, or which are close to what you need.

This last part - being able to reason by analogy - is critical to being able to get the most out of the primary literature. Maybe I don't find an exact match for the exact reaction I want to do. But I find something pretty close. Here's where my training and reasoning ability come in to either decide to adapt what I've found or keep looking for something better.

An example of one of these search systems is Chemical Abstracts Service, but there are others.

As a professional chemist, you're expected to keep up with all of the main journals in your field so you stay up to date. You do this by reading articles, talking to people, giving talks, and going to conferences.

There's another category of information consisting of 'reviews'. These are articles that summarize a large area of chemistry - for example, recent large-scale methods for the production of ethanol.

The discovery/invention of new reactions is a very active area of current chemical research.


Cell notation for the lead-acid battery Reactions for the lead acid battery are:

$$
\begin{array}{}
\text{Oxidation}&\ce {Pb(s) + HSO4^-(l) &-> PbSO4(s) + H+(l) + 2e-}\\
\text{Reduction}&\ce{PbO2 + HSO4^-(l) + 3H+(l) + 2e- &-> PbSO4(s) + 2H2O}\\
\text{Total reaction}&\ce{Pb(s) + PbO2(s) +2HSO4^- +2H+ &-> PbSO4 + 2H2O}\\
\end{array}
$$

What will be the cell notation for this battery? My attempt:$$\ce{Pb(s), Pb^2+(s)| HSO4^-(l)| PbO2(s), Pb^2+(s),Pb(s)}$$

The things that I have in mind,

In both sides the electrode material is $\ce{Pb(s)}$
Separate every element that is in the same phase with a comma
Separate every element with different phases with a single bar
Separate the two half reactions with a single bar (as I find no salt bridge here)

I am having a tough time learning cell notation. Anyways this was a challenging one for me and I tried my best shot. I am really not sure whether everything is right or not. So please identify my mistakes and show me the right way to do it.

There are a couple of things wrong here. First off, your final reaction is unbalanced. Once you've fixed the balancing, read the other mistakes:

The ions do not exist in the liquid state! They are solvated/hydrated by the solvent. Since the solvent is water here, we'll say that the ions are in the aqueous (aq) phase instead.
While there is certainly no salt bridge here, there is still an electrolyte - an aqueous solution of sulphuric acid. Hence, you must mention it in your cell notation, between the anode and the cathode.
The $\ce{PbSO4}$ formed at the anode is in solid state. Hence, writing it as $\ce{Pb^2+(s)}$ is incorrect, as it is not dissociated into the ions $\ce{Pb^2+}$ and $\ce{SO4^2-}$.

With all these corrections, your final, correct cell representation should be:

   $$\small{\ce{Pb(s), HSO4^-(aq) | PbSO4(s), H+ | H2SO4 ($\pu{x~M}$) | PbO2(s), HSO4-(aq), H+(aq) | PbSO4(s)}}$$

Some websites (like KhanAcademy) and texts (NCERT 12), cite the lead-acid battery reaction as this instead:

$$
\begin{array}{}
\text{Oxidation}&\ce {Pb(s) + SO4^2-(aq) &-> PbSO4(s) + 2e-}\\
\text{Reduction}&\ce{PbO2 + SO4^2-(aq) + 4H+(aq) + 2e- &-> PbSO4(s) + 2H2O(l)}\\
\text{Total reaction}&\ce{Pb(s) + PbO2(s) +2SO4^2(aq)- + 4H+(aq) &-> 2PbSO4(s) + 2H2O(l)}\\
\end{array}
$$

assuming the bisulphite ion to be further ionized. The cell notation in this case would then be:

   $$\small{\ce{Pb(s), SO4^2-(aq) | PbSO4(s) | H2SO4 ($\pu{x~M}$) | PbO2(s), SO4^2-(aq), H+(aq) | PbSO4(s)}}$$


Role of amalgamated zinc in Clemmensen reduction What is the role of amalgamated zinc in the Clemmensen reduction mechanism?

As the illustration below would show, you can use either amalgamated zinc  (in a Clemmensen reduction) or hydrazine (in a Wolf-Kishner reduction). This reaction as drawn hides some of the secondary products which should be zinc peroxide and mercury oxide*

Metallic zinc is “activated” by mercury by “dissolving” it into individual atoms of zinc that are distributed evenly in the alloy. You're basically creating a zinc solution.

The reaction of zinc with HCl releases hydrogen gas.
However, when this reaction takes place with zinc amalgam, the individual hydrogen molecules that are formed by the reaction of zinc with HCl remain in a reactive state (sometimes called “nascent hydrogen”) and reacts with the ketone to initiate the reduction reaction.

So the mercury in zinc amalgam serves to “trap” the active hydrogen as it is formed, and allows it to attack the carbonyl compound rather than be released as a gas.

*Disclaimer: I have never actually performed this reaction and I could be dead wrong.


Deuterium depleted water through fractional distilation Could a packed fractional still of the same type used to distill ethanol to 95% be used to produce deuterium depleted drinking water? The still has a 1000 watt electric heating element and has a fractioning column that is 120cm long which is packed with stainless steel wool. If it is feasible what kind of performance might I expect?

Relying on the difference in vapor pressures of $\ce{HDO}$ and $\ce{H2O}$ is equivalent on relying on the equilibration of

$\ce{HDO(g) + H2O(l) <=> H2O(g) + HDO(l)}$

The fractionation factor $\alpha_{\ce{D}(l)/\ce{D}(g)}$ for this reaction is 1.015 at STP.

Deuterium ($\ce{^2H}$ or $\ce{D}$) is separated from protium ($\ce{^1H}$) commercially, usually via the Girdler sulfide process, which instead of relying on the above reaction, relies on

$\ce{ HDS + H2O <=> H2S + HDO}$

The fractionation factor $\alpha$ for this process is about 2.2.

Steam distillation for deuterium enrichment is certainly possible, but is not nearly as favorable as the industrial process. Given the very low concentration of deuterium in natural waters, the Rayleigh fractionation equation would be applicable:

$$R_{\ce{D}/\ce{H}} \approx R_{\ce{D}/\ce{H}_{init}} \times f^{\alpha-1}$$

In this equation:

$R_{\ce{D}/\ce{H}}$ is the ratio of deuterium to hydrogen in the liquid water
$R_{\ce{D}/\ce{H}_{init}}$ is the initial ratio of deuterium to hydrogen, or about 0.000156.
$f$ is the fraction of the liquid water remaining after distillation.
$\alpha$ is the fractionation factor mentioned above.  However the value of 1.0015 is for the condensation of vapor phase water into liquid; performing the reverse reaction we should use the inverse, i.e. for distillation of existing liquid into vapor, $\alpha = 0.985$

With this equation and with mass balances (the total amount of deuterium and of protium doesn't change in the distillation process) you can find the amount of deuterium in the distillate.

If you distill off 100 grams of an initial 1000 grams of water, then $f=0.9$, $f^{\alpha-1} \approx 1.000156$, meaning the remaining 900 grams of liquid water will have a deuterium-to-hydrogen ratio of 0.0001562 instead of 0.000156.

The mass balance for deuterium would be:

$1000 \times 0.000156 = 100 R_{\ce{D}/{\ce{H}_{distillate}}} + 900 \times 0.0001562$

If I did all my math right, that means your distillate would have a deuterium-to-hydrogen ratio of 0.0001542, or about 1.2% less than the original water.

Hopefully this exercise convinces you that you would have to do a lot of distillation and burn through a ton of water to generate reasonable amounts of deuterium depleted water through simple steam distillation.  There's a reason they don't do it that way commercially.


How to derive the relation between gibbs energy and equilibrium constant? I want to understand the derivation between gibbs energy and equillibrium constant $$\Delta G=\Delta G^o+RT\ln Q?$$
I have seen a similar post on CSE  Derivation of relationship between equilibrium constant and Gibbs free energy change   which seems to be incomplete and still confusing so I am again asking this question.

The derivation that was written in the post was as follows:

  Using the fundamental equations for the state function (and its natural variables):_
  $$dG=-SdT+VdP$$
  $$V=(\dfrac {\partial G}{\partial P})_T$$
  $$\bar G(T,P_2)=\bar G(T,P_1)+\int_{P_1}^{P_2}\bar V dp$$
  Here $\bar x$ represents molar $x$, i.e. $x$ per mole
  $$\bar V=\frac {RT}{P}$$
  $$\bar G(T,P_2)=\bar G(T,P_1)+RT \ln\frac {P_2}{P_1}$$
  Defining standard state as $P=1\text{bar}$ and $\bar G=\mu$
  $$\mu(T,P)=\mu^o (T)+RT\ln \frac{P}{P_o}$$
  consider the general gaseous reaction $aA+bB\rightarrow cC+dD$
  $$\Delta G=(c\mu_C+d\mu_D-a\mu_A-b\mu_B)$$
  for "unit progress" in reaction. Using $\mu_i=\mu^o_i+RT\ln \frac{P_i}{1bar}$
  $$\Delta G=(c\mu^o_C+d\mu^o_D-a\mu^o_A-b\mu^o_B)+RT\ln \frac{P_C^cP_D^d}{P_A^aP_B^b}$$
  $$\Delta G=\Delta G^o+RT\ln Q$$

I know the relation between Gibbs free energy,Enthalpy,Entropy and Temperature as $$G = H - TS$$
how is the relation $dG=-SdT+VdP$ derived from the above formula?
Why do we take a path where Entropy and volume are constant  in the equation $dG=-SdT+VdP$ as in a chemical reaction both volume and entropy can change?

Firstly, if you know the relation $G= H-TS$, you can reach at the differential form of it just by taking differential at both sides, i.e.$$dG= dH-SdT-TdS$$ Now, recall the definition of enthalpy as $H= U +PV$, So, we can write, $dH= dU + PdV + VdP $, and also recall the heat supplied to the system as $dQ= TdS $ and by the first law of Thermodynamics, $dQ= dU+ PdV$ .Thus combining we have $$dG= dU+PdV+VdP-SdT-dQ = VdP -SdT$$ Thus you derive the required relationship.

Now coming to your next question, it's not about taking path where Entropy and volume remains constant. A state in a thermodynamic system can be defined in terms of three parametrs which are interrelated i.e Pressure, Temperature and Volume. In fact, you can only independently vary any two out of these three parameters and third one will automatically get adjusted. Here the independent variables are Temperature and Pressure. By changing the Pressure and Temperature independently, the change in the Gibbs' Free Energy is calculated. The coefficients are volume and Entropy only but that doesn't mean they are kept constant. So, the interpretation of the equation $dG= VdP- SdT$ needs to be done correctly.


Polluting agents generated by road traffic in a city I want to measure the air quality according to road traffic in a city.
What are the polluting agents (generated by fuel combustion) strictly related and generated by road vehicles that is worth measuring?

Among $\ce{CO}$, $\ce{CO2}$, $\ce{NO_$x$}$, PM2.5, PM10, $\ce{NH3}$, $\ce{O3}$, $\ce{SO2}$ which are the most significative? Should I add or remove any agent?

You might also want consider the metals which appear in road dust, I recall having seen some work on neutron activation anaylsis of dusts collected near roads. The rubber tyres are likely to contain some metals. Cars will be likely to release metals in other ways. I would consider the dust from the operation of disk brakes, the wear on tyres, fuel additives and also corrosion of bodywork and engine parts. I know that zinc and cadmium plated parts can release both ionic and particulate forms of these metals.

One paper on the subject was by Siddique et. al. im Radiochimica Acta, 2013, volume 101, pages 273-277.

It is interesting that lead levels in the air have fallen since leaded petrol was banned in the UK, but some workers have reported finding platinum group metals in the air near to roads.

Using ICPMS Simitchiev et. al, in TALANTA, 2008, volume 77, pages 889-896 found Pt, Rh and Pd in road dust.


Identifying a Positive or Neutral Ligand Inside Coordination Sphere? There are some coordination compounds like $\ce{Na2[Fe(CN)5NO]}$, where ligands like $\ce{NO}$ are used which exist as both $\ce{NO+}$ (nitrosonium ion) and $\ce{NO^.}$ (neutral nitrosyl). How to identify which ligand is being used in such complexes?

A similar question was also on Quora but none of the answers were satisfactory.

Please don't answer with respect to Raman/infrared using stretching frequencies etc. since all that is out of scope at the level I'm currently studying.

Books have been written on the topic of whether a certain nitrosyl ligand is the nitrosyl cation, the nitrosyl radical or — you forgot this one — the nitrosyl anion $\ce{NO-}$. This ligand is termed a non-innocent ligand, meaning we cannot just look at it and tell what it is.

You asked for an answer that does not require IR or Raman spectroscopy. Well sadly, that means one has to resort to something even more exotic such as Mössbauer spectroscopy and ESR techniques. If you get lucky, a crystal structure may also help. Because the only chances we have to somewhat determine what we are dealing with are answers to the questions:

what is the $\ce{NO}$ bond length?
what is the $\ce{M-N-O}$ angle?
what is the spectroscopic oxidation state of the central metal (not the formal one)?
what is the count of unpaired spins?

One of the most obvious data points is the bond angle. A nitrosyl cation will typically only coordinate in a practically linear matter with an angle of $\approx 180^\circ$ while the nitrosyl anion will typically coordinate in a ‘bent’ manner with an angle of $\approx 120^\circ$. However, we also have the nitrosyl radical which has been shown to adopt both bond angles so we just lost our key piece of evidence.

Even using ESR to determine the spin state (i.e. number of unpaired electrons) can be inconclusive. Taking the related example of the binding of oxygen to haemoglobin, the current consensus accepts a structure that does not immediately conform to the observed spin of $S = 0$. The answer here is antiferromagnetic coupling that keeps to spins antialigned even though they could also not couple and observe Hund’s rule.

Thus, without a plethora of experiments you cannot determine in which state non-innocent ligands such as nitrosyl, dioxygen (which can be $\ce{O2, O2^.-}$ or $\ce{O2^2-}$ etc. bond to the metal. You may want to look at the related question on the electronic structure of $\ce{[Fe(H2O)5(NO)]^2+}$ for which I went through the trouble of actually looking up literature.

Because the oxidation state of iron and nitrosyl are so unclear, a curly bracket notation has been suggested. In this case, we would be dealing with $\ce{\{FeNO\}^6}$ (if I did the maths correctly).

Thankfully, you can always name these compounds by IUPAC nomenclature. Because IUPAC nomenclature desires to allow naming even when the actual electronic structure is unknown, it has defined nitrosyl to always be a neutral three-electron donor for nomenclature purposes. Thus, we would be able to call the complex you are interested in disodium pentacyanidonitrosyliron(III).


How is titanium metal obtained from natural resources? How does titanium occur in nature?

Is it available as a native metal or only in the form of ores?

Which precesses are typically used to convert the natural resources to the solid metal?

Any hints on where to find further information are welcome. Thank you very much in advance.

Titanium is found as an ore, according to the Titanium Occurrences Wikipedia page, specifically, the 2 most economical ores are:

Ores of titanium

Rutile $\ce{TiO2}$ and Ilmenite $\ce{Fe^{2+}TiO3}$

Extraction of titanium metal from ore

According to the chemguide page Titanium,

  Titanium can't be extracted by reducing the ore using carbon as a cheap reducing agent. The problem is that titanium forms a carbide, TiC, if it is heated with carbon, so you don't get the pure metal that you need. The presence of the carbide makes the metal very brittle.

Rather, the ore is chlorinated, then reduced as shown in the schematic from the webpage Titanium Exposed (the chemical processes are included):

Chemguide states that either $Mg$ or $Na$ can be used in the reduction stage, which adds to the cost of production as they need to be extracted themselves.

Asides from the links used in this answer, another good resource is Chemistry of Titanium (Professor John Lancashire - University of the West Indies)


Alternative python modules to RDKit to convert SMILES to structure Python module RDKit is great, but it cannot be installed by pip, and therefore it is difficult to use it sometimes in a virtual environment.

I am wondering that if there are some alternative python modules that can convert smiles to chemical structures?

I am not aware of any other Python tool not mentioned in the question @Mithoron linked to. Openbabel is available via pip by the way.

That RDKit is not available on PyPi should not hinder you from using it in a virtual environment. RDKit is available via the conda package manager which is easy to use and combinable with pip.

If you are on Linux, the probably best solution for managing different versions of Python (also Anaconda Python) and virtual environments is pyenv. I highly recommend to check it out.


What is nowadays called silazane? Example from the Recommendations 2013. Left: Alleged disilazane ; Right: My version

Clearer example without obvious flaws in alternative structure. Left: Alleged disilazane ; Right: My version

I am confused with the contemporary IUPAC nomenclature of organic compounds. According to Recommendations 93 (available at acdlabs.com), silazanes are heterogeneous hydrides where both terminal atoms are silicon atoms (R-5.1.4.2). The same can be concluded from the general rule on heterogeneous hydrides (R-2.2.3.2). However, in the introductory chapter of Recommendations 2013 (which is freely available at RSC Publishing site) on page 82 the compound given in the image is given two names, with the least priority (non-PIN) variant being 1,1′-(2-ethenyl-1,2,3-trimethyldisilazane-1,3-diyl)bis(ethan-1-one). I am confused by this name, as, by rule R-2.2.3.2, terminal atoms should be lowest in the Hantzsch–Widman order, which preserves unambiguity of the structure. From this rule, I expect disilazane to be something like the structure on the second image (although I acknowledge that the tetravalent nitrogen must be positively charged, which changes the name automatically).

TL;DR So the question is:

  Do Recommendations 2013 change something about naming heterogeneous hydrides? And if so, what is now called silazanes?

It seems this name should not be in the blue book. It is an error.

The following instruction is what Corrections to Nomenclature of Organic Chemistry. IUPAC Recommendations and Preferred Names 2013 has given for your problem:

  Page 82, P-15.3.2.4.1, example 4 on this page.
  Delete 1,1′-(2-ethenyl-1,2,3-trimethyldisilazane-1,3-diyl)bis(ethan-1-one).

As Igors Mihailovs (OP) pointed out in comment section to the question, the Recommendations 2013 are not error-free. You may see how nany errors have been corrected in this web page. For example, page 75 has 4 errors in that page alone!


Why would rinsing the flask with acid in this experiment affect its accuracy? The following question is #21 from this test form (PDF link):

The apparatus in Figure 4 was set up to measure the time taken for $\pu{20.0 cm3}$ of sodium thiosulfate solution to react with $\pu{5.0 cm3}$ of hydrochloric acid in a $\pu{100 cm3}$ conical flask at $\pu{20 °C}$. The timer was started when the sodium thiosulfate solution was added to the acid in the flask. The timer was stopped when it was no longer possible to see the cross on the paper.

What is likely to decrease the accuracy of the experiment? [1 mark]
A) Rinsing the flask with acid before each new experiment.
B) Stirring the solution throughout each experiment.
C) Using the same piece of paper for each experiment.
D) Using different measuring cylinders to measure the volumes of acid and sodium thiosulfate

The answer given as correct is (A). Why is rinsing the conical flask with acid likely to decrease the accuracy of the experiment? I would imagine that stirring the solution, option (B), introduces error since the rate of stirring is difficult to control.

Probably because the use of hydrochloric acid (which is a reactant involved in this reaction!) could result in some residual aqueous HCl in the flask, which may add to the HCl added later.
Anything you use to clean the flask should be inert/not involved in that reaction. If you really want to use HCl, then you should later wash the flask with a lot of water, but that's not said in the answer.

Also note that the volumes involved here are not so large (only 5 ml of HCl are used), so a "small" residue might have a large effect on the result.

As for the stirring, the colour of the reaction is caused by the sulfur precipitate. But I think that, if you stir in each experiment, the times might probably decrease, but you would still observe a trend with temperatures.

That answer would be wrong if it involved stirring in some experiments and not in others.