text:
Where does the name ‘aliphatic’ come from?
Generally, carbons and the hydrogens bonded to them are classified as one of the following:

aromatic
olefinic
aliphatic
(something for triple bonds I don’t know)

The meaning of aromatic is rather clear to anybody: They belong to aromatic systems (even though the Hückel rule has nothing to do with a certain type of smell). Olefinic is also clear enough: Olefins were originally those gases which could generate an oily substance by reaction with e.g. chlorine ($\ce{C2H4 + Cl2 -> C2Cl2H4}$; the latter being 1,2-dichloroethane); the name olefin derives from French gaz oléfiant — oil-creating gas.
But I have no clue where the adjective aliphatic or the noun aliphat derive from, which are applied to saturated carbons. What was its original meaning and why do we use it?

Aliphatic = ali-"fat"-ic = fats = fatty acids = long, unbranched carbon chains.............. no? ): Bit of an Internet search says that it's derived from the greek ἄλειφαρ which means "oil" or "fat": http://www.perseus.tufts.edu/hopper/text?doc=Perseus:text:1999.04.0058:entry=a)/leifar

Olefins are hydrocarbons that contain at least a double bond. The term is used —sometimes misleadingly— interchangeably with alkenes.

Although the nomenclature is not followed widely, according to IUPAC, alkenes are acyclic hydrocarbons with one double bond between carbon centers. Olefins comprise a larger collection of cyclic and acyclic alkenes as well as di- and polyenes. — Alkene nomenclature – Wikipedia

The Goldbook page Wikipedia cites says the final words:

Olefins
Acyclic and cyclic hydrocarbons having one or more carbon–carbon double bonds, apart from the formal ones in aromatic compounds. The class olefins subsumes alkenes and cycloalkenes and the corresponding polyenes.

While Oxford dictionary seems to agree with the origin thingy:

Origin
Mid 19th century: from French oléfiant 'oil-forming' (with reference to oily ethylene dichloride).

Funny thing is, "aliphatic" also means "producing fats". Goldbook says:

aliphatic compounds
Acyclic or cyclic, saturated or unsaturated carbon compounds, excluding aromatic compounds.

Some definitions exclude alicyclic compounds from the list of aliphatic compounds, but IUPAC doesn't agree, so. Note that aliphatic compounds can be saturated or unsaturated.
So, back the main answer part,

Aliphatic; Etymology
From Ancient Greek ἄλειφαρ (áleiphar, “oil, ointment”). — Wiktionary

There are literally no articles on the Net speaking of the first discovered or documented aliphatic compounds, but I read in some textbook somewhere which mentioned that since these compounds were first referred to as what animal fats used in primitive or medieval lamps were consisted of, they're referred to as "aliphatic".
The only thing about a "first aliphatic" I could find is mentioning "Muconic acid is the first aliphatic compound formed as a result of aromatic ring splitting, which is further oxidized to fumaric acid and then participates in the cell metabolism as a standard endogenous substrate." by this paper.

M. P. Crosland's $\it{Historical; Studies; in; the; Language ; of ;Chemistry}$ has a footnote which reads: "The term 'aliphatic' was coined by A. W. Hofmann (according to Graebe, $\it{ Geschichte; der; organischen; Chemie}$, Berlin, 1920, p. 277n.). The class name 'Acyclic' is now often used instead of 'Aliphatic'.

@Todd indeed, the term's vanishing gradually. BTW, does it mention the reason Hofmann used this term?

@M.A.R. - No, it just gives the reference to the work by Greabe. Perhaps one of our German-speaking chem.SE regulars can dig up that reference and shed some light.


text:
Deciphering stereochemical notation in electrocylization reaction
I not agree with this solution, for me the red bond formed must be a dotted line in the drawing, and the hydrogen atom musn't be represent with a dot. Am I right or not?

Assuming that the "red dot" on I-2 is used in the conventional sense and indicates that the hydrogen is projecting above the plane of the drawing and on the same face of the molecule as the ester group in I-2, then you are correct, I-2 is not the correct product for the disrotatory motion shown by the arrows.  The disrotatory motion indicated by the arrows would place both the ester group and hydrogen below the plane of the screen.  The other possible disrotatory motion (move both arrowheads to the other end of the curved lines) would produce I-2.

@ron that exactly what i was thinking thank you :)

@ron I’m curious, how are arrows ‘defined’ to point above or below? What would the other arrow look like? Could you add that as an answer?

@Jan To me, the arrows do not indicate "above" or "below", but rather indicate that, in the present case, the tops of the two terminal p-orbitals are rotating inwards.  If the arrowheads were both moved to the other end of the curved lines, they would indicate the other possible disrotatory motion where the tops of both terminal p-orbitals are rotating outwards. The dotted lines indicate the pi bond axis, and usually the arrows are drawn closer to the dotted lines to better indicate what is really going on.

You are both correct and incorrect. (And this entirely without superimposition of quantum states!)
You are taking an achiral molecule and performing a reaction in achiral media to form a chiral product. Symmetry dictates that the reaction must be racemic, so there must be equal amounts of upward-pointing and downward-pointing carboxylates.
However, the drawing is also incomplete. It does not make sense to discuss the reaction without labelling the entire stereochemistry at the newly formed bond. It is important to realise that it generates only two of the four possible stereoisomers: the two syn ones ((S,R) and (R,S)). You should add an upwards-pointing hydrogen to the right-hand carbon of the new bond.

Judging by your comments, you are only interested in the arrow direction and resulting movement. Well, the arrows are not clear. They are 2D on a 2D plane and trying to signify a 3D movement. You could grab the curved bit of the arrow like a handle and:

either pull it upwards — then you have a rotation which would move the carboxylate below your screen’s plane and create the downward-pointing product ((R) on the α-carboxylate carbon),
or push it downwards — then you have the rotation which would move the carboxylate above the screen’s plane and create the upward-pointing product ((S) on the α-carboxylate carbon).

So the arrows’ directions are arbitrary, we do not know if the are shaped below or above the plane.

I hope I am correct with R and S. If anyone could supply me with the IUPAC numbering of atoms, that would be great; I don’t have access to ChemDraw again until Monday ^^'

I didn't said that the solution precise the final solution is racemic what i only ask is about the movement of the arrows and the product we have :)

@Shadock Well, racemic implies that both up and down are generated, so … ;)

Yes but to generate the other one the arrows have to go in the other way

For me if the blue arrows go in the other way i agree we have the product mention in the picture.

@Shadock There you go, an edit.

Well in fact I know what my teacher want to say when he draws arrows like this. Thank you for your answer. And draw in 3D on a 2D sheet is not very easy :P

@Shadock That's information clearly missing in the question ;)

Nothing is perfect ^^

Would it be correct to say that irrespective of absolute stereochemistry,  you wanted to establish that the disrotatory occurrence gave the functional group and hydrogen as syn rather than anti?

@Beerhunter Yeah, I didn’t notice the dot to mean an upwards-pointing hydrogen as ron pointed out in the question’s comments. To me it looked like the stereochemistry on that other atom was unestablished. So I tried pointing out that that thing needs stereochemistry, too; and we know that they are syn to each other.

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Is potassium bifluoride an ionic or a covalent compound?
The statement below is an excerpt from my textbook (Chemistry Part II, Textbook for Class XII by NCERT, ed. January 2014):

Because of the tendency of fluorine to form hydrogen bond, metal fluorides are solvated by $\ce{HF}$ giving species of the type $\ce{HF2-}$ like in $\ce{KHF2}$ $(\ce{KF.HF})$. Also, because of hydrogen bonding, $\ce{HF}$ dimerises to $\ce{H2F2}$ and exhibits dibasic nature and forms acid salt like $\ce{KHF2}$ on reacting with $\ce{KOH}$ whereas $\ce{KHCl2}$ does not exist because chlorine does not form hydrogen bond.

But since $\ce{K+}$ is a cation and  $\ce{HF2-}$ is an anion, it can easily form the ionic bond $\ce{^+K\bond{...}[HF2]-}$. So, is potassium bifluoride an ionic compound $\ce{K+[HF2]-}$ or a covalent compound $\ce{KF.HF}$?

Understand that the $\ce{KF}$ part of $\ce{KF.HF}$ is an ionic compound and only the $\ce{F-}$ somehow interacts with $\ce{HF}$ and the question answers itself.

I found this question very intriguing and while I support both answers by Ivan and Mithoron, I would like to add a little more insight.
When we consider compounds, we often tend to describe them as ionic or covalent. And often we can find both of these types in quite common molecules. We use these terms quite liberal, and when we talk about an ionic bond we usually mean a bond, that has predominantly ionic character, while a covalent bond has a predominantly covalent character.
Even in simple binary salts like sodium chloride we will not find the ideal ionic bond, and even methane the bond is not purely covalent. The truth - as always - lies somewhere in between.
What does that mean for potassium bifluoride? In this case it is especially difficult to classify the compound. This is less due to the potassium fluoride interaction, but to the interactions within the bifluoride subunit.
From the crystal structure itself one can clearly see, that there are not $\ce{KHF2}$ molecular units crystallising, but bifluoride ions in a lattice of potassium ions. The crystal structure itself is describes as follows.[1]

It may be regarded as an ammonium chloride arrangement of potassium atoms and
  fluorine dumb-bells.

This clearly speaks to the predominantly ionic nature of the potassium fluorine bond. To gain further insight into the topic I ran a couple of quantum chemical calculations.[2] Unfortunately I was unable to find a file with the crystal structure to run calculations on and I did not have the time to accurately reconstruct it from the parameters. Instead I would like to demonstrate on the example of $\ce{KF}$ how an (predominantly) ionic bond looks in the representation of the quantum theory in atoms in molecules (QTAIM).[3] The electron density at the bond critical point is very low with $\rho(\text{BCP})=0.053~\mathrm{e\, a.\!u.^{-3}}$ and the second derivative at this point is positive, $\nabla^2\rho(\text{BCP})=+0.288~\mathrm{e\, a.\!u.^{-3}}$.

These are criteria which we use to characterise a mostly ionic bond. The compound $\ce{KF}$ can be regarded as (one of) the most ionic compound,[4] so we can compare these values with values of other compounds.
Let's look at a covalent bond, before we look at the bifluoride ion. In this example I chose $\ce{HF}$ as it is related to it. Here we find a completely different picture. The electron density at the bond critical point is a lot larger, $\rho(\text{BCP})=0.366~\mathrm{e\, a.\!u.^{-3}}$ and the second derivative at this point is negative, $\nabla^2\rho(\text{BCP})=-2.858~\mathrm{e\, a.\!u.^{-3}}$. In the graphic one can see that there is charge accumulation along the bond path. All these criteria suggest, that this is a predominantly covalent bond.

When we do now look at the bifluoride ion, $\ce{[F-H-F]^-}$, we will see some intermediate bond, where it is hard to tell, which character is predominant. Note that due to symmetry reasons, we only need to look at one bond critical point. On the one hand we have a somewhat high electron density $\rho(\text{BCP})=0.177~\mathrm{e\, a.\!u.^{-3}}$ and the second derivative at this point is still significantly negative, $\nabla^2\rho(\text{BCP})=-0.416~\mathrm{e\, a.\!u.^{-3}}$. On the other hand we also see that the charge accumulation is not constant between the bonding partners, it is separated by an area where the Laplacian becomes positive, which indicates some ionic character.

So in this case it is quite hard to tell if we are talking about a covalent bond with high ionic character or an ionic bond with high covalent character. But to be perfectly honest here, this is the case with many hydrogen bonds.
I expect that these findings are comparable to the alkali metal bifluorides,  which are all known up to at least caesium.[5]

Notes and References

Richard M. Bozorth, J. Am. Chem. Soc. 1923, 45 (9), pp 2128–2132.
Gaussian 09 Rev. D1; DF-BP86/def2-TZVPP; AIMAll 13.11.04;
The depicted QTAIM plots are the Laplacian distribution within the molecule. Solid blue lines indicate charge depletion $∇^2ρ>0$, dashed red lines indicate charge accumulation $∇^2ρ<0$, red spheres are bond critical points, solid black lines are bond paths, dark red lines are zero flux surfaces (these separate the atoms).
Why is KF the most ionic compound? (Chemistry.se)
R. Kruh, Keiichiro Fuwa, T. E. McEver, J. Am. Chem. Soc. 1956, 78 (17), pp 4256–4258.

What does it take to get lithium bifluoride? Article says this is unstable in ambient conditions.

Yes, $\ce{KHF2}$ is an ionic compound and a covalent compound. There is nothing exceptional about it; in fact, most compounds that we call ionic (not $\ce{NaCl}$, though) have covalent bonds in them. Consider $\ce{MgSO4}$ with ionic bonds between $\ce{Mg^2+}$ and $\ce{SO4^2-}$, and covalent $\ce{S-O}$ bonds within $\ce{SO4^2-}$. Consider $\ce{Na2CO3}$ with similar $\ce{C-O}$ bonds. These are the ways of the world.

$\ce{KHF2}$ is an ionic compound with cation $\ce{K+}$ and anion $\ce{HF2-}$, and writing $\ce{KF.HF}$ doesn't suggest covalent bond between K and F, but say that $\ce{HF2-}$ anion is adduct of F- anion and HF molecule.
The bifluoride anion is quite interesting as its bond is not only a rare symmetric hydrogen bond, but also the strongest hydrogen bond known. All hydrogen bonds have some covalent character, but here it can be estimated as $50\%$, reflecting its mesomeric structures:
$$\ce{^-F\bond{...}H-F <-> F-H\bond{...}F-}$$
These also mean that it is a so-called 3-center 4-electron bond.

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What can one learn from a plot of calculated molecular orbitals?
What can one learn from the plots of the various iso-probability surfaces/orbitals of a given molecule/system? Can any predictions be made about bonding or kinetics by looking at these plots? Can an absorption spectrum be extracted from it? etc.
(I am aware this is a broad question, and a quick bullet list would suffice)
For example:

Energy diagram of the molecular orbitals explored for the D6h
  structure (left) and the D3h distorted geometry (right) (source)

@Jan this is a general question, and the other one is more particular. I don't think there's much overlap. If you insists, I'd redirect the other one here.


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What chemical process occurs when dead leaves make a mark on concrete?
In autumn, leaves that fall on concrete make marks that remind me of fossils.

Is this caused by a chemical reaction that occurs between the concrete and the leaf? Or would a leaf leave a stain on any surface as it decays? What circumstances can intensify the process?

Tannins in the leaf are leached out by water. Getting leaf wet then rolling a car tire on top would extract more fluid from the rehydrated leaf. https://en.wikipedia.org/wiki/Tannin

If you check shortly after a rain you'll see wet, brown puddles.  I suspect that the reason that they leave such a strong mark on concrete is that it can act as a base and tannins are mildly acidic, and their products are colored and relatively insoluble.  Check out blackwater aquariums for more extreme examples of this coloration, though not involving concrete.

The shape of the leaf is consistent with a type of maple tree leaf, and the brown coloring is characteristic of the water soluble tannins that would be extracted by water. For comparison, here are two photographs, one of tanic acid in water and one of a dried mixture of tannins. The photos were taken from this Wikipedia article.  Although the tannic acid in the photograph below was extracted from different parts of a redwood tree, it is a ubiquitous tannin found in most hardwood tree leaves and the color is characteristic of many such tannins.


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Electron configuration and the 2p orbital
I looked through the questions already asked about this topic and they were very specific and over my head. I'm having trouble with the basics.
When drawing orbital diagrams I know electrons are placed in the 2p boxes singly first then paired because of they have parallel rotations and the Pauli exclusion principle states that an orbital can only house 2 electrons and those electrons must be of opposite spin. My question is why do electrons behave this way in the 2p orbital but not the 1s or 2s? And why does the p subshell contain three 2p orbitals?
As far as electron configurations and orbital diagrams are concerned I've gotten to Ne and the 2p configuration confused me so I though I'd ask before moving on.

they have parallel rotations . Electrons do not rotate. They are not  little classical balls orbiting the nucleus. why do electrons behave this way in the 2p orbital but not the 1s or 2s. What do you mean by this? The Pauli exclusion principle applies for all electrons.

I mean why do the three 2p orbitals fill singly first then pair up instead one orbital filling up at a time.

Because the 3 p-orbitals are degenerate but putting two electrons into the same orbital (with opposite spin of course) has an energy cost associated with the repulsion between the electrons.

Cross posted to Physics SE. Bad form...

Thats not allowed? What wrong with different answers from different people?

Crossposted from http://physics.stackexchange.com/q/218512/2451


text:
What electrodes to use to minimise corrosion during water electrolysis?
I did the $\ce{H2O}$ electrolysis experiment a few times, but every time a huge amount of the oxygen reacted with its corresponding electrode. What material should the electrode be made of in order to maximise the amount of oxygen I get?

a platinum electrode would be inert in such an electrolysis experiment.

Graphite electrodes also work quite well, they can be found in pencils and if they by any chance do erode, they can be replaced cheaply.

It depends on the electrolyte as well as the electrode material. For example, $\ce{NaCl}$ releases atomic chlorine at the positive electrode, quickly attacking most materials except metals of the $\ce{Pt}$ group, as @MaxW mentions.
$\ce{NaOH}$ or $\ce{KOH}$ are a bit gentler on the electrodes. Run the apparatus at a low over-voltage to avoid damaging them. Coated nickel electrodes are suggested for use with alkaline solutions.
Phosphoric acid is also suggested as an electrolyte.


text:
How to minimise solution concentration changes due to evaporation?
Lets say I have a precise solution, and would like to store it long term (up to two years) with as little as possible change in concentration. The volume is 10-100ml. What container should I choose? What temperature? Is there any other factors? I would prefer not to use sealed glass container. Is there a way to approximate a solvent loss?  What if I stored some water in exactly the same conditions together with my sample, would I be able to approximate water loss in my sample from the water loss in this reference container, or does the difference in the concentration of the solute affect the rate of evaporation too much?

A sealed glass ampule would be the most certain way of ensuring your solution stayed the same concentration, removing the possibility of solvent loss. Even in that case, concentration is likely to change as the reagent in the solution would likely degrade to some extent (unless it was a very simple solution such as NaCl in water).  Since a glass container isn't what you want to use, I'd suggest storing the solution in a tightly capped vial, sealing the cap with Parafilm,and then storing this vial in the freezer. This should minimise solvent loss and change in concentration due to degradation.

What if I stored some water in exactly the same conditions together with my sample, would I be able to approximate water loss in my sample from the water loss in this reference container, or does the difference in the concentration of the solute affect the rate of evaporation too much?

That would be a gross approximation, not suitable for a standard solution. Your solution and pure solvent do have different vapor pressures according to Raoult's law.

What temperature?

It depends on solvent, solute, and their interaction. I'd check two main things: i) solubility at the storage temeprature, which must be as close as possible to the solubility at the working temperature; ii) vapor pressure of the solvent, that gives you a rough indication of the storage temperature. You could take a look at the temperature dependence of the solvent vapor pressure and solute solubility - or, if seriously in doubt, even try simulating the thermodynamics of your solution using a software.

What container should I choose?

Volumetric flasks are usually the best choice. They minimize evaporation compared to other glassware (bottles, beaker, etc.) and, most important, they have the most accurately placed graduation mark. Use the appropriate rubber cap and close accurately with laboratory film. PTFE band between cap and flask cone might also help preventing losses.
At times, a slight overpressure above the liquid can be desirable. In this case, a gas valve (or a quality glass faucet) will be placed on top of the flask and nitrogen will be flushed inside. The flask can be stored capped by the sealed valve. Also consider the opportunity of degassing your solution by placing sieves on the bottom of the flask or by bubbling nitrogen through the liquid. Finally, consider the need for a dehydrating agent in the case of a moisture-sensitive solution.
But, most of all, don't keep standard solutions for two years but replace them every ~6 months.


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Point Groups and MO Theory
I am a little confused about how point groups give information about MO of polyatomic molecules.
Is it true that all molecules of a given point group such as, $\mathrm{C_{2v}}$, will have the same MO diagram, the only thing that will change is the number of electrons?

You also have to consider the atoms being used to form the MO's and the orbitals available. For instance if one atom is highly electronegative the diagram will change (different orbitals may interact than in the case where both atoms are close in energy). Equally an atom further down the periodic table has more orbitals available for bonding, also changing the diagram.

Not in the slightest.  Point groups provide a simple tool to divide orbitals into isolated interacting sets. That's all.

Is it true that all molecules of a given point group such as, $C_{2v}$, will have the same MO diagram, the only thing that will change is the number of electrons?

No. There are so many different possibilities of how molecules with $C_{2v}$ may look, that they cannot possibly all have the same energy level diagram. Take water and acetone for mild examples.
What you can do, however, is estimate how different molecular orbitals will look based on the atomic orbitals you put into the estimation. Let’s take formaldehyde as it is a nice and simple example.
We have two hydrogen atoms, one carbon and one oxygen. Carbon and oxygen both lie on the $C_2$ axis so we can consider them separately, but we should use group orbitals for the hydrogens (I am defining the $yz$ plane to be the plane of the molecule). Luckily, they only have s orbitals, so it doesn’t get too complicated. Check out how the s-orbitals transform with the different symmetry operations to find their irreducible representations: $\mathrm{A_1 + B_2}$. Carbon and oxygen each have an s orbital and three p orbitals which will transform in the same way: the s orbital as $\mathrm{A_1}$ (naturally), the p orbitals transform as $\mathrm{A_1 + B_1 + B_2}$ (for $\mathrm{p}_z, \mathrm{p}_x$ and $\mathrm{p}_y$, respectively).
Thus, we have five atomic (or group) orbitals of the $\mathrm{A_1}$ type which will combine to form most of the σ bonds, three of the $\mathrm{B_2}$ which will also combine to σ bonds and two of the $\mathrm{B_1}$ type which will combine to form π bonds. Had we chosen water, we would only have had six orbitals altogether transforming as $\mathrm{3 A_1 + B_1 + 2 B_2}$ — a clear distinction showing that the diagrams cannot always be the same.
We now have to estimate the energy levels the orbitals initially have. It is a good guess to put the highest-energy atomic orbital at the respective atom’s ionisation energy; then you would just have to esimate a lower level for the s orbitals of carbon and oxygen. With these levels in hand, you can educatedly guess the different energy levels of the resulting molecular orbitals and educatedly guess what the orbital would look like: The resulting MO must transform in the same way as the AOs used to build it. So any $\mathrm{A_1}$ orbital will be symmetric to any symmetry operation while $\mathrm{B_2}$ will always have a nodal plane in the $\ce{C=O}$ bond axis, etc. I’m sure you can now come up with a well-educated guess of the formaldehyde orbitals; you can check out what they actually look like on the homepage of Professor Zipse. (You can ignore the accompanying German text. The higher-level antibonding MOs LUMO+2 and LUMO+3 that are required by theory are not shown; one must be $\mathrm{B_2}$, the other $\mathrm{A_1}$.)


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why do we denature proteins before mass spectroscopy?
I'm learning about mass spectroscopy in particular the use of it in analyzing protein structure. they said we have to use trypsinolysis to cut the protein to segments, but why does the protein have to be denatured first before trypsinolysis?

This should be on [biology.se].

I'm voting to close this question as off-topic because it would be more appropriate on biology.SE.

I disagree that this belongs on biology.  Enzymes are chemical entities that do chemistry!

I put it as a chemistry question because I'm learning about this on my chemistry course in biomedicine- so thought its more appropriate here :)

Trypsin is an enzyme, and has to bind to a target protein in a specific way in order to be active. Without first denaturing your protein, most of the protein will be conformationally/sterically inaccessible to trypsin, and you will only get partial digestion (or maybe no digestion at all).


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Geometry optimization of a small molecule inside of a larger one
I'm very new to this and trying to optimize the structure of a larger molecule that "wraps" around a smaller molecule. I'm using ORCA. I'm having two issues with the task:
a) I've attempted to use the following:
! RI BP86 def2-SVP def2-SVP/J D3BJ TIGHTSCF Opt Grid3 FinalGrid5
but have not had any success.
b) How can I reliably export a two-molecule starting configuration in .xyz format starting from ChemDraw? I've tried exporting in .cml, loading in JMOL/Avogadro/Molekel and printing in xyz, but it invariably destroys the bonds.

Regarding a), could you elaborate more on what the problem is? Regarding b), what do you mean by "destroys the bonds"? Btw, you shouldn't even care about bonds when importing geometry in Avogadro; just check that atoms are located at the right positions.

b) *.xyz does not contain bonds information. Whenever you export to *.xyz, you loose it.

Bond is an irrelevant concept in computational chemistry. The more relevant term is bond length. Visibility of bond in molecular visualizer is just for better understanding. And also are you trying to do some sort of constrained optimization?

I don't think you've provided nearly enough detail about your problems, particular "a".
But let me comment about "exporting from ChemDraw." Keep in mind that ChemDraw is a 2D tool. Consequently, the lengths between atoms are designed for depiction on paper, not necessarily for accurate representation of the 3D geometry necessary for calculations.
Avogadro can import 2D geometries from ChemDraw and attempt to assign reasonable 3D geometries. In your case, this seems especially troublesome since you want to have one molecule inside another.
I'd either:

Sketch the two molecules separately in ChemDraw and then use Avogadro to "insert" the smaller one into the larger after geometry optimization.
Use Avogadro to create the full 3D geometries of each molecule and then, select the smaller molecule and use the manipulate tool to move it into position.

As several people have commented, if you want to preserve the appearance of bonds, XYZ is not the right format. It only stores coordinates, not bonding information. Avogadro and other tools will attempt to re-perceive bonds and bond orders, but this is not a perfect task, particularly in the presence of multiple nested molecules.
Instead, I'd keep a file in CML format with the bond orders intact. This is more-or-less the "native" file format for Avogadro.


text:
Can a hydride shift be concerted?
I see that in most textbooks Alkanide shifts are shown as concerted mechanisms, but Hydride shifts are not. Why is this done? Shouldn't the rearrangement be concerted when the carbocation formed is primary and not concerted when the carbocation formed is secondary, regardless of whether or not it is an alkanide or hydride shift? Thank you.

It depends upon what hydride shift is occuring. For instance a 1,5-hydride shift is a type of sigma tropic rearrangement, which is a class of pericyclic reaction and very much concerted in its mechanism.


text:
amino acid pka of carboxylic acid
Why is the pka of the carboxylic acid part of alpha amino acids (pka around 2.3 for most amino acids) quite a bit less than the pka of a normal carboxylic acid, say acetic acid (pka =4.76)?

Look at mesomer forms of the conjugate base. The more you find the more the acid is strong.

The amino group in $\alpha$-amino acids is electron-withdrawing via the inductive effect. Mesomeric (resonance) effects cannot occur here because there is no conjugation of the nitrogen lone pair with the carboxyl group. There are two equivalent ways of looking at how the EWG affects the acidity:

It stabilises the conjugate base via withdrawal of electron density from the negatively charged oxygens;
It polarises the $\ce{O-H}$ bond further than in the unsubstituted carboxylic acid, making the proton more prone to dissociation.

Therefore, the $\mathrm{p}K_\mathrm{a}$ of glycine is $2.34$, significantly lower than the unsubstituted acid (acetic acid, $\mathrm{p}K_\mathrm{a} = 4.76$). Similar effects are seen in 2-chloroacetic acid ($\mathrm{p}K_\mathrm{a} = 2.86$).

As for the comparison between glycine and 2-chloroacetic acid, one thing to bear in mind is that glycine exists as a zwitterion at neutral pH. Therefore, the relevant acid-base equilibria are:

For $\mathrm{p}K_\mathrm{a1}$, which you are interested in, note that the substituent is no longer merely a $\ce{-NH2}$ group but is now a $\ce{-NH3+}$ group. Despite nitrogen and chlorine having extremely similar Pauling electronegativities, the positively charged nitrogen is now more electron-withdrawing than a simple uncharged chloro substituent.
One could draw some parallels with aromatic chemistry: aniline is rather unreactive (compared to chlorobenzene) in electrophilic substitution under acidic conditions, where it is protonated. Of course, the reactivity of aromatic rings is a kinetic effect whereas $\mathrm{p}K_\mathrm{a}$ is a thermodynamic effect, but the underlying rationale is the same: $\ce{-NH3+}$ is more electron-withdrawing than $\ce{-Cl}$.

Cheers for this reply, upvoted, but I'm not going to accept yet as I'm not fully convinced for the following reason: chlorine has the same electronegativity as nitrogen so why is there still a non-negligible difference in the pka of 2-chloroacetic acid and glycine. Furthermore I'd expect the former compound to be even more acidic because nitrogen pulls electron density from 3 bonds (2 hydrogens and the rest of the molecule) whereas chlorine is only pulling electron density from the molecule itself, the only moiety it is attached to.

Furthermore, fluoroacetic acid (fluoro electronegativity > nitrogen) is still observed to be less acidic (pka 2.6) than glycine.

Unless the substituents are extremely similar (e.g. F, Cl, Br, I), it is difficult to simply compare substituents just on the basis of Pauling electronegativity, and a direct relationship may or may not exist. Nevertheless, it is a good question and I will expand the answer a little bit because there is a second aspect of glycine that I neglected earlier.


text:
Predictions about hydrolysis reactions
Predict whether these $1.0~\mathrm{M}$ solutions are acidic, basic, or neutral. Lets start with $\ce{CH3NH3+NO3-}$
How do I approach a problem such as this?
I certainly do not know all the chemical compounds that exist thus what is my method?
I am guessing that $\ce{NO3-}$ does not hydrolyze and $\ce{CH3NH3+}$ does.
Thus, $\ce{CH3NH3+ + H2O -> CH3NH2 + H3O}$ = ACID
(b) $\ce{NaI~(aq)}$
$\ce{Na}$ does not hydrolyze and neither does iodine, but why?
(c) $\ce{NaNO2}$
$\ce{Na}$ again does not hydrolyze but from the first example neither does $\ce{NO2}$
Does this have something to do with either the solutions have a hydrogen atom or $\ce{OH-}$ molecule?

authors: سجاد رحيم ياسر


text:
Soil batteries?
In an attempt to demonstrate the basic principles of common batteries, my teacher assigned me to demonstrate a simple experiment to the kids at the elementary school. A tank filled with saltwater, copper ring used as cathode, cola tin as anode and a motor is powered etc. Surprisingly one little guy there asked if we could use unprocessed soil with natural salts and follow the same technique on a large scale.
I am not keen on disappointing future (maybe) scientists so I answered: yes, but using the soil as an electrolyte is, as it seems inefficient. My teacher said I should search and come with a whole answer so here I am. What about on other worlds? What about on mars? Perhaps using supercritical $\ce{CO2}$ could be the thing. Can a soil battery be useful at all? I like the concept.

Great question. Galvanic corrosion in soils in a big deal - see this slideshow for a decent overview - and the setup of the "battery" is essentially with soil (usually hydrated to some extent) taking the place of the aqueous salt bath electrolyte.

The slideshow was very useful for my overall search. Thank you for commenting

Unfortunately, the little guy would be a bit disappointed down the line. The energy of the cell comes from the oxidation of the aluminum (the soda pop can) and that Al anode is eventually consumed. Aluminum is produced at relatively high cost, using electrolysis, so wet soil (or seawater) is not the issue. Same deal with the standard Zn and Cu electrodes stuck into a lemon, potato, watermelon, or other fruit/vegetable: the oxidation of the anode metal is the ultimate energy source and the electrolyte medium just plays a necessary supporting role. (Of course, some hydrogen gas is produced at the cathode, but that does not help much.)  Still, this makes a nice demo for the kids and the corrosion tie-in is great!

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Dissociation percentages of N2O4 at different pressures using Gibbs' free energy

\begin{align}
\ce{N2O4(g)&<=>2NO2(g)} &
\Delta_\mathrm{r}G^\circ &= \pu{4.7 kJ mol^-1}
\end{align}
Knowing that the standard values for pressure and temperature are $\pu{1 bar}$ and $\pu{298 K}$ respectively, find:
a) the dissociation grade (noted as $\alpha$) of $\ce{N2O4}$ at equilibrium, in standard conditions
b) the dissociation grade of $\ce{N2O4}$ at equilibrium and at $\pu{298 K}$ / $\pu{10 bar}$

First, we find $K$ from $\Delta G$
Then we write the partial pressures as the molar fraction of the compound times the total pressure of the system (at equilibrium)
After that, we write the relation for $K_p$ using partial pressures
and I'm stuck...
\begin{align}
  \Delta_\mathrm{r}G^\circ &= -RT \ln K &
  \Rightarrow K&=0.15 \mathrm{~(unitless)}
  \\
  P_{\ce{N2O4}} &= \frac{1-\alpha}{1+\alpha}\cdot P &
  P_{\ce{NO2}}  &= \frac{2\alpha}{1+\alpha}\cdot P &
  K_p = \frac{(P_{\ce{NO2}})^2}{P_{\ce{N2O4}}}
  \\
  \Rightarrow K_p &= \frac{4\alpha^2}{1-\alpha^2}\cdot P
\\
\end{align}
My question is: how can I convert between $K$ (which is unitless) and $K_p$ (which is not) so that I can find the dissociation grade at $\pu{1 bar}$ and $\pu{10 bar}$?

See my answer to this recent question: http://chemistry.stackexchange.com/a/40586/16683 The short answer is that the dimensionless $K$ is calculated using $p_i/p^\circ$ instead of $p_i$, where $p^\circ = 1~\mathrm{bar}$.

Thanks for the clarification! I looked it up first, but I somehow managed to miss that post.

Don't worry about it, there are a ton of questions here about equilibrium constants and Gibbs free energies which are difficult to search.


text:
Can atomic partial charges be measured in molecules experimentally?
Can we measure atomic partial charges in molecules experimentally? The charge of isolated ions can be measured, but when atoms are part of a molecule, the case is much more difficult. We do not really know where one atom starts and where another one stops (the atomic radii are diffuse). So I am inclined to say that partial charges must be based on theoretical concepts, such as the Mulliken population analysis (or a range of other methods), or be derived (estimated) from measurables such as dipole moment.
However, is it possible to probe molecules at the atomic level, using STM, TEM, or some other electron probing technique, and get the partial charges that way, perhaps based on electron-electron repulsions? Still, it seems to me, that the border between the atoms need to be properly defined in order to get the partial charges.
According to the Wikipedia page "Partial charge", several experimental techniques can be used to estimate the partial charge, e.g. XPS, NMR, EPR, UV/vis, with more. But do these techniques measure something which is a direct result of the effect of partial charge, or is some other physical observable measured, from which the partial charges are "guessed"? For example, using XPS to get the kinetic energy of the photoelectron seems like a "direct" measurements (although I know that what the detector sees is not really the kinetic energy, but something with which the kinetic energy correlates: calibration is needed). However, the idea of partial charge seems to be somewhat vague. Numerical values of partial charges depend on the definition of where one atom starts and where another ends, which is an abstract concept: this border cannot be observed or measured.
Another issue that comes to mind, is that the partial charge is not necessarily uniformly distributed around each atom, but will depend on bonds, lone pairs, interactions with other molecules, etc.

Partial charge is a notational fiction anyway.  If you're working at the level of detail that your question implies, wouldn't it be cleaner to work directly with real observable phenomena (such as the dipole moment)?

Let me put it this way. There are no atoms in molecules. There is just a continuous cloud of electron density, and as you correctly pointed out, we have no way to pinpoint where one atom ends and another starts. Or rather, we have many ways to do that, all slightly different and all inherently arbitrary. (My personal favorite is the concept of Bader charges, but that's a matter of taste.)
So yes, the very idea of partial atomic charges is unavoidably vague.

"There are not atoms in molecules." Some times we talk about "perturbed" atoms. "Most of" the atoms do not change very much when bonded, like the nuclei and core electrons; mostly the valence electrons take part in bonding. So surely it is safe to say that atoms do exist in molecules? But you agree, then, that partial charges cannot be measured experimentally? I am not sure how you answered the question.

I answered the question with the words you see above; if they don't translate to a clear "yes" or "no" - well, sorry about that, but such is the nature of the matter. Bader charges, though, are measured experimentally. The ambiguity is not in the measurement; it is in the agreement of what to measure, i.e. in the definition of atoms.

Perhaps my question has become more philosophical than I intended. I get your point, thank you for clarifying.

However, is it possible to probe molecules at the atomic level, using STM, TEM, or some other electron probing technique, and get the partial charges that way, perhaps based on electron-electron repulsions? Still, it seems to me, that the border between the atoms need to be properly defined in order to get the partial charges.

It is possible (though not easy) to map electronic density in a molecule (and somewhat easier in crystalline solids). Having electronic density map it is possible to define interatomic borders as a surface with density lower than in areas the surface divides. This allows to attribute well-defined value of 'atomic charge' to an atom.
Another possible option is to fit the attributed charges (without references to real electronic density) so they produced electric field matching the real electric field. The 'real' field may be calculated from experimental data or ab-initio model. This procedure is implemented in some QC packages, but it is very sensitive to the definition of 'similarity' of the fields.

What I read is that "we are able to obtain estimates of partial charges based on experimental data", but not that we can actually measure them. Would you agree with that?

But then I guess that is the way all things are "measured".

@AndersMB 'We are able to obtain electron density data from the experiment. There are several procedures that we can use to extract some values from the data. If you accept them as atomic charges is up to you, but they are close enough in some cases.'


text:
Meaning of steady state (kinetics)
In kinetics, we use the steady state approximation to relate the concentration of intermediates to the concentration of reactants. For example, given the mechanism
$$\ce{A->[k_1]B}$$
$$\ce{B->[k_2]C}$$
using the steady state approximation for B gives $k_1[A]=k_2[B]$, so $[B]=\frac{k_1}{k_2}[A]$. Since B is at steady state, $[B]$ should be a constant through time. But the equation says that $[B]$ is a function of $[A]$ (and $[A]$ changes with time), so how can $[B]$ be constant? Am I misunderstanding the meaning of steady state here?

Good question.  Let's examine your system in more detail.
$$ \frac{dA}{dt} = - k_1 A $$
$$ \frac{dB}{dt} = k_1 A - k_2 B $$
$$ \frac{dC}{dt} = k_2 B $$

The third equation is redundant because $C$ can always be found from the conservation equation $C = A_0 - A - B$.
This system can be solved exactly without any further assumptions.  So thank you for picking a system that has this nice feature.  An exact solution will let us compare to any approximate solutions obtained via simplifying assumptions.  Assuming that the initial conditions are $A(0) = A_0$ and $B(0) = 0$, then the exact solution is

$$ A(t) = A_0 e^{-k_1 t} $$
$$ B(t) = \frac{k_1}{k_2-k_1} A_0 \left (e^{-k_1 t} - e^{-k_2 t} \right )$$

What about the quasi-steady state approximation (QSSA)?  The QSSA is an assumption that $B$ does not really vary with $t$ as far as $C$ is concerned.  That is, predicting the rate of formation of $C$ can be done by assuming therate of formation of $B$ is zero.  So if $k_1 \gg k_2$, then $A$ is more or less "instantly" and quantitively converted to $B$, after which $B$ converts to $C$ via the usual first order kinetics with rate constant $k_2$. So this case, $k_1 \gg k_2$, is already a clue that the QSSA is invalid for this mechanism.
Why is the QSSA invalid for this case of $k_1 \gg k_2$?  Usually, the QSSA is invoked on catalytic reaction schemes.  In these schemes, it is assumed that the concentration of catalyst is very small relative to the concentration of substrate.  Thus there is an upper bound on the possible concentration of the intermediate, a bimolecular complex of reactant and catalyst: the total amount of catalyst added limits how much bimolecular complex there can be at any time.    This assumption that substrate is in excess relative to catalyst must be satisfied for the QSSA to make sense.
What about the other case?If $k_2 \gg k_1$, then at time scales where $k_2 t \gg 1$, then $B \approx \frac{k_1}{k_2}A_0 e^{-k_1 t}$.  The maximum concentration of $B$  in this case is the ratio of rate constants, a ratio that we've already assumed to be much less than one.  And since the maximum concentration of $C$ is not similarly bounded, this is a case where the QSSA makes sense.  This implies that the conservation equation can be written as $A_0 = A + B + C \approx A + C$.  Proceeding with the QSSA assumption to find the rate of formation of $C$:

$$\frac{dB}{dt} \approx 0$$
$$B = \frac{k_1}{k_2}A$$
$$A_0 = A + B + C \approx A + C$$
$$\frac{dC}{dt} = +k_2 B \approx k_1 A = k_1 (A_0 - C) $$
That last equation is now a single-variable first-order linear ODE and could be solved for $C(t)$ if desired.  But the main thing is that with the QSSA, $B$ was eliminated from the expression for $C$

I can't think of a physical reason why a system where $k_1 = k_2$ exactly couldn't exist.  But someone else will have to come along and explain how to get the math to work out.
If $k_1 = k_2 = k$, then the system of equations is :
$$ \frac{dA}{dt} = - kA$$
$$ \frac{dB}{dt} = k(A - B)$$
The solution is:
$$A = A_0 e^{-k t}$$
$$B = A_0\; k t\; e^{-k t}$$
In this case, there is still no generally valid way to apply the QSSA.  The derivative of $B$ is $B'(t) = k A_0 e^{-k t} \left(1 - kt \right)$ and is equal to zero only when at the precise time where $kt=1$.  It isn't approximately equal to zero in any other time range, except as $t \rightarrow \infty$.  So, the QSSA isn't valid to the case when $k_1 $ is exactly equal to $k_2$, and it isn't valid when $k_1 \neq k_2$ either.
Bottom line: using the QSSA in this problem is equivalent to assuming $k_2 \gg k_1$.  You might not have thought so without going through this whole analysis.  The key fact is that the QSSA requires that variation in the concentration of the intermediate be small relative to the variation in concentration of the products or substrates that you care about.  In mechanisms involving catalysts, this happens any time that catalyst levels are small relative to substrate(s).  However, in other mechanisms, this happens only with arbitrary restrictions on the kinetic parameters that may not be obvious a priori.

It seems to me that the conditions in (5) or the condition in (4) give a "quasi-steady sate." The OP's problem was expecting an "absolute" steady state.

I'm not 100% sure but I don't think so.  In 5, B is varying exponentially.  That isn't steady state!

Ah, yes... (5) further depends on $k_1 \ll 1$ and $k_2 \ll 1$ so that the exponential can be approximated with an expansion of $e$.

@CurtF. So you're saying that we can't use steady state approximation in this case? However, many problems I've seen that involve steady state approx do not involve a catalyst. Ex. $$\ce{2NO<=>N2O2}$$  $$\ce{N2O2+H2->N2O + H2O}$$ Also, this wikipedia article link says the condition for steady state approx is $k_1<<k_2$, so it seems like steady state approx is applicable here...

In the previous comment, I meant $$\ce{N2O2 + H2 -> N2O + H2O}$$. Sorry, I didn't know SE wouldn't let me edit my comment after 5 min. :(

I saw the wikipedia discussion after I posted my answer.  I can see what they are saying, and I guess if $k_2 \gg k_1$ then you could say that the QSSA approximation applies.  Tomorrow I'll try to edit my answer to make this clear. I'm still a bit uncomfortable calling it the QSSA because it seems very different than the examples I am used to.

OK, edited.  Hope the improved version makes more sense.


text:
Determining water of crystallisation in a hydrate from the mass percentages
How would I find the amount of water in a hydrate if all I was given was the mass percentage of one element in the compound? For example, a question asks:

$\ce{Zn(ClO3)2. $x$H2O}$ contains $21.5\%$ zinc. Find the value of $x$.

After thinking for some time I now know that I need to find the mass of $\ce{Zn(ClO3)2}$, and then find the mass and amount of water, and then divide by the amount of $\ce{Zn(ClO3)2}$. But I can't seem to find the mass of $\ce{Zn(ClO3)2}$.

In this case you only have one variable to solve for ($x$) and you have one piece of information (Zn being 21.5% by mass) so you have enough information to solve the question.
The molecular mass of the hydrate can be expressed in terms of $x$:
$$\begin{align}
M(\ce{Zn(ClO3)2.xH2O}) &= M(\ce{Zn}) + 2\cdot M(\ce{Cl}) + (6+x)\cdot M(\ce{O}) + 2x\cdot M(\ce{H}) \\
&= \pu{65.38 g mol-1} + 2(\pu{35.45 g mol-1}) + (6+x)(\pu{16.00 g mol-1}) + \,\\
&\qquad 2x(\pu{1.01 g mol-1}) \\
&= (232.28 + 18.02x)~\pu{g mol-1}
\end{align}$$
Now
$$\begin{align}
\frac{M(\ce{Zn})}{M(\ce{Zn(ClO3)2.xH2O})} = \frac{\pu{65.38 g mol-1}}{(232.28 + 18.02x)~\pu{g mol-1}} &= 0.215 \\
65.38 &= 0.215(232.28 + 18.02x) \\
x &= 3.99 \approx 4
\end{align}$$


text:
Is dextrose an isomer of glucose?
Is dextrose an optical isomer of glucose?
If a microorganism metabolises glucose, would it also be capable of metabolising dextrose? Do both 'fit' into the same enzyme that binds to glucose?

There are two enantiomeric forms of glucose, D- and L, with D-glucose being the naturally occurring form. Dextrose is just another common name for D-glucose, so if a microorganism can metabolise D-glucose then it can by definition metabolise dextrose as they are just two words for the same thing.

As for the question of whether the enzyme would recognise both optical isomers (enantiomers), this would probably not be the case. Enzymes are 'chiral', as are the two enantiomers of glucose. The active sites of enzymes/proteins are often specific for a given isomer, hence why many drugs are only biologically active in one enantiomeric form, with the other enantiomer being either inactive, less active, or in certain cases toxic.
In particular, the first step in glucose metabolism is phosphorylation of the C-6 hydroxyl group by the enzyme hexokinase. However, hexokinase is incapable of phosphorylating L-glucose.


text:
Confusion about the graph plotted from the Arrhenius equation
According to Arrhenius equation $k = A\mathrm e^{(E/RT)}$, where $k$ is the rate constant, $E$ is the activation energy, $R$ is the universal gas constant and $T$ is the temperature.
Taking log both sides,
$$\log k = \log A - \frac{E}{2.303RT}$$
Why is the graph plotted between $\log k$ and $1/T$ linear? We know $E =$ threshold energy—average energy of the molecules. Now average energy is given by $\sqrt{ (8RT/\pi M)}$, it is itself a function of temperature. So the slope is variable and hence should have been some other curve.

Activation energy can vary but it isn't average energy, but but certain value depending on mechanism of reaction. BTW A varies with temp. but this effect is minor for your calculation

Arrhenius equation
$k = A \exp{\dfrac{-E}{RT}}$
where:

k is the rate constant
    E is the activation energy
    R is the universal gas constant
    T is the temperature

The line

Why is the graph plotted between $\log k$ and $1/T$ linear?

Taking log both sides,
$\text{log}(k) = \text{log}(A)-\dfrac{E}{2.303RT}$
A straight line for $y$ in terms of $x$ is:
$y = a + bx$
so:

$y$ is $\text{log}(k)$
  $a$ is $\text{log}(A)$
  $b$ is $-\dfrac{E}{2.303R}$
  $x$ is $\dfrac{1}{T}$

so plotting $\text{log}(k)$ versus $\dfrac{1}{T}$ gives a straight line.
E "confusion"

We know $E =$ threshold energy—average energy of the molecules. Now average energy is given by $\sqrt{ (8RT/\pi M)}$, it is itself a function of temperature. So the slope is variable and hence should have been some other curve.

As you note there is some temperature dependency. But from the Wikipedia article:

"Given the small temperature range kinetic studies occur in, it is reasonable to approximate the activation energy as being independent of the temperature."

It is sort of circular reasoning but look at the Arrhenius equation in log form.

$\text{log}(k) = \text{log}(A)-(\dfrac{E}{2.303R})\dfrac{1}{T}$

then

$\dfrac{d(\text{log}(k))}{d(1/T)} = -\dfrac{E}{2.303R} $= constant

if $\text{log}(k)$ and $\dfrac{1}{T}$ are not linear over the temperature range of interest because of a temperature dependency, then the reaction isn't of the "Arrhenius equation" type. Such plots are in fact used to verify confirmatory to the Arrhenius equation over the temperature range of interest.

I think, OP is asking ‘isn’t $E = f(T)$?’, and ‘Shouldn’t that change the graph?’

This doesn't provide answer, you probably didn't understood what he's asking. Please be more careful with answering.

Jan & @Mithoron - Ok I think I fixed the problem with my answer. Thanks for pointing out the problem. Based on the wording of the question I should have known that the OP was smart enough to know what a "line" was. ;-)

The equation you gave for E is the average energy of the molecules, not the activation energy of the reaction.


text:
Why can copper have multiple coordination numbers?
Is there a way to explain why the coordination number of for example $\ce{Cu^2+}$ can be either 4 or 6?

It usually is 6.

Yes, but it can also be 4, as fas as I know. But I am not really interested in the number itself, but more in why it is exactly that number.

There's a short and a long answer to the question: the short one is that there are things we can't really guess by looking at the periodic table. You can probably guess what it is, but as I only have my trusty periodic table to use, I can't give you.

And what is the long answer?

Possible duplicate of Are coordination numbers of elements fixed?

I disagree with the duplicate. This question clearly asks for the why and unlike the other one not for the if. Also this is specifically targeted at copper, while the other question is quite general. Duplicate questions should be essentially the same question in different wording. (cc @Jan)

@Martin-マーチン Alright then; I’ll answer ;p

http://chemistry.stackexchange.com/questions/24873/why-is-copper-ii-coordination-number-so-big

the coordination number of for example $\ce{Cu^2+}$ can be either 4 or 6.

Allow me to preface this answer by debunking that myth.
Balamurugan and Palaniandavar published a set of interesting sterically strained trigonal planar copper(I) complexes, among them 1. Under cyclic voltammetric conditions, they observe two distinctly different oxidation potentials, one of which they attribute to the oxidation to the corresponding trigonal copper(II) complex. Unfortunately, they do not report isolation.[1]
Tetracoordinated copper(II) species are not as common as hexacoordinated ones, but they are also rather well-known. For example, Kitajima et al. synthesised the structurally interesting complex 2, a formal copper(II) and boron(III) binuclear complex. Naturally and as expected for tetracoordinated copper(II) complexes, its structure is tetragonal around both the copper and the boron centres.[2]
Slightly more interesting is the second complex Kitajima et al. were able to crystallise: it is the formal addition of a DMSO solvent molecule to complex 2 to give complex 3. Here, copper(II) is pentacoordinated and equally interestingly the coordination sphere represents a square pyramidal structure, not the usual trigonal bipyramid.[2]
I’ll skip hexacoordinated complexes, because they are damn common and just note that everybody who witnessed complex chemistry at any educational institution has come in contact with $\ce{[Cu(H2O)6]^2+}$ and $\ce{[Cu(NH3)4(H2O)2]^2+}$.
Finally, Martínez-Bulit et al. were able to synthesise an interesting heptacoordinated copper(II) complex 4. Unfortunately, the complex did not crystallise like the corresponding cobalt(II) and zinc(II) counterparts so we are left with analogy to the other syntheses, MS and IR data for establishing it’s structure. In figure 2, you can see a depiction of the coordination sphere.[3]
A variant of heptacoordinated, albeit only characterised for copper(I) to the best of my knowledge, is $\ce{[CuCp(TMS-C#C-TMS)]}$, complex 5.[4] Cyclopentadienyl — even though it is a $\unicode[Times]{x3b7}^5$ complex — should be characterised as pentacoordinating because there are five coordinating atoms. Likewise, acetylene is dicoordinating for a total of 7. It should be possible to extend the concept of sandwhich complexes to $\ce{[CuCp2]}$ 6; however, that complex is currently unknown outside of calculations.[5]

Figure 1: Structures of the complexes mentioned above.

Figure 2: Three-dimensional representation of 4’s ligand sphere.

Now why does copper(II) assume these structures? Well, you can argue things like complex stability, Jahn-Teller distortion, steric strain, electronic effects and much more. In the end, it all boils down to: if the MO scheme of a complex provides sufficient stabilisation with respect to the individual components, the complex is viable to be formed. If removing a ligand will substantially destabilise the complex, it is likely to remain inert. Unfortunately, I was unable to find a linear copper(II) complex (or any linear copper complex for that matter); however, I would assume it to exist if a ligand provided a large cavity with one coordination site and only a single anion was present to coordinate the other. Or maybe not even anions but neutral donors; the anion being $\ce{F- . SbF5}$, a very non-coordinating ion. I’m also sure that the cuprocene sandwhich complex will be synthesised given sufficient time. That would extend copper(II)’s coordination count range from 2 to 10 — always given beneficial circumstances for a certian coordination number.
So tl;dr: There is no simple explanation for the choice of exactly that coordination number.
Note that in aquaeous solution the ligand sphere of the seemingly so stable complex $\ce{[Cu(H2O)6]^2+}$ is in constant fluctuation with dissociating and reassociating ligands. Otherwise, there would be no rapid exchange mechanism that would allow the formation of $\ce{[Cu(NH3)4(H2O)2]^2+}$ upon addition of concentrated ammonia. Compare iron(II): while the hexaammin complex $\ce{[Fe(NH3)6]^2+}$ is undoubtedly stable and characterised,[6] you won’t be able to get it by reaction $\ce{[Fe(H2O)6]^2+}$ with ammonia. Instead, $\ce{Fe(OH)3}$ will precipitate, demonstrating that ligand exchange in iron(III) complexes is slow with respect to deprotonation, while in copper(II) complexes it is reasonably fast.

References:
[1]: R. Balamurugan, M. Palaniandavar, Inorg. Chem. 2001, 40, 2246. DOI: 10.1021/ic0003372.
[2]: N. Kitajima, K. Fujisawa, Y. Moro-oka, J. Am. Chem. Soc. 1990, 112, 3210. DOI: 10.1021/ja00164a052.
[3]: P. Martínez-Bulit, A. Garza-Ortíz, E. Mijangos, L. Barrón-Sosa, F. Sánchez-Bartéz, I. Gracia-Mora, A. Flores-Parra, R. Contreras, J. Reedijk, N. Barba-Behrens, J. Inorg. Biochem. 2015, 142, 1. DOI: 10.1016/j.jinorgbio.2014.09.007.
[4]: D. W. Macomber, M. D. Rausch, J. Am. Chem. Soc. 1983, 105, 5325. DOI: 10.1021/ja00354a023.
[5]: D. W. Clack, K. D. Warren, Inorg. Chim. Acta 1978, 30, 251. DOI: 10.1016/S0020-1693(00)89045-3.
[6]: G. L. Schimek, J. W. Kolis, Chem. Mater. 1997, 9, 2776. DOI: 10.1021/cm970095c.

$\ce{[Cu(H2O)4].SO4.H2O}$ or blue vitriol is also a common copper complex with CN of 4.

@Apurvium That’s copper(II) sulphate pentahydrate, correct? If so, it is hexacoordinated: the sulphates bridge two copper centres to complete the distorted octahedron.

One tetrahedral complex with CN of 4 is very common: $\ce{[CuCl4]^2-}$ https://pubs.acs.org/doi/10.1021/jp109723v


text:
Calculating the age by radioactive dating
The following was an example done in class, so I know the answer and (maybe?) the formula, but I still can't derive it for myself, and I would appreciate some help.

Given a sample of biotite with $\ce{K2O} = 9~\%,~~ \ce{^{40}Ar} = 2.242~\mathrm{\mu g/g}$, find the age.

The constants given were $x_e= 0.581 \cdot 10^{-10}\ \mathrm{y^{-1}}$, $x=5.543 \cdot 10^{-10}\ \mathrm{y^{-1}}$ where $x$ is the decay constant, isotopic abundance of $\ce{^{40}K} = 0.01167~\%$, and relative atomic masses $M_\mathrm{r}(\ce{K}) = 39.098, M_\mathrm{r}(\ce{O}) = 15.999, M_\mathrm{r} (\ce{^{40}Ar}) =39.96$.
The answer shown was $2308 \cdot 10^6~\mathrm{y}$.
The formula I'm used to for this type of problem is $t=\frac{1}{x} \ln(\frac{\ce{^{40}Ar}^*}{\ce{^{40}K}} (x/x_e +1))$ but I'm not sure how to apply that given just a percentage of $\ce{K}$ instead of a mass.  I thought of just finding the weight percentage of $\ce{K}$ from the $9~\%$ (its $80.46~\%$ of the $\ce{K_2O}$, or $7.2~\%$ of the total), but that didn't work.  I'm not sure if I need to consider atmospheric $\ce{Ar}$ either.
I tried $^{40}K = .09 * .83022 * .0001167 = 8.719 *10^{-6}$ where .833022 is the weight percentage and it didn't give the correct answer.
I also tried $^{40}K = .09 * .6666667 * .0001167 = 7.002 *10^{-6}$ using the molar percentage of K, but also didn't get the correct answer.

?!? "Given a sample of biotite with $K_2O$=9%, $^{40}Ar$=2.242 ug/g, find the age." // $K_2O$=9% that means 0.09 g  $\ce{K2O}$  per gram of  sample // $^{40}Ar$=2.242 ug/g means 2.242 ug $^{40}Ar$ per gram of sample.

I hope I didn’t mess up your formula while improving the MathJax ^^'

@Jan, thanks for fixing that!  I'm new to using MathJax and I always for get how to do fractions.


text:
High pressure modification to the van der Waals equation
I get that a gas behaves non-ideally under high pressure and low temperature conditions. At high pressures, the density of a gas increases and attraction forces operate among molecules due to short intermolecular distance. This allows us to somehow modify the ideal gas equation. If a molecule is approaching the wall of container in order to collide, the neighbouring gas molecules soften their impact.
$$p_\text{ideal}=p_\text{real}+\frac{an^2}{V^2}$$
How was the term $\frac{an^2}{V^2}$ derived?
My thoughts about that term, $\frac{n^2}{V^2}$ may be the extra pressure thats why the $V$ lies on the numerator (inverse relationship)

The most rigorous way to derive corrections to the real gas equation is the virial expansion. At second order this gives Van der Waals law, but this derivation is completely general and can be pushed to higher order.
My answer assumes a basic knowledge of classical mechanics, statistical physics and analysis.
Hamiltonian of the system
We consider a system of identical particles described by the following Hamiltonian:
$$H=\sum_i\frac{\vec{p}_i^2}{2m}+\sum_{i\lt j}U(|\vec{r}_i-\vec{r}_j|)$$
If you don't know what a Hamiltonian is, you can consider it as the energy of the system.
Canonical partition function for the perfect gas
For a perfect gas, you consider that the particles are non-interacting, so that
$$U(|\vec{r}_i-\vec{r}_j|)=0,\quad\forall i,j.$$
In this particular case, life is easy! In fact you can easily compute the Canonical partition function (this is a definition)
$$Z_\mathrm C=\frac1{N!h^{3N}}\int\mathrm d^N\vec{p}\,\mathrm d^N\vec{r}\,\mathrm e^{-\beta H}$$
which, in the case of a non-interacting gas, is simply
$$Z_\mathrm{PG}=\frac1{N!h^{3N}}\int\mathrm d^N\vec{p}\,\mathrm d^N\vec{r}\,\mathrm e^{-\beta\sum_i\frac{\vec{p}_i^2}{2m}}=\frac{1}{N!}\left(\frac{2 \pi mk_\mathrm BT}{h^2}\right)^{\frac{3N}{2}}$$
In order to compute this integral, you can simply use the definition of Gaussian integrals. This is the partition function of a perfect gas and you can now derive all the properties of this system.
Canonical partition function for an interacting gas
If you want to consider an interacting gas, your partition function becomes
$$Z_\mathrm I=\frac1{N!h^{3N}}\int\mathrm d^N\vec p\,\mathrm d^N\vec r\,\mathrm e^{-\beta\left(\sum_i \frac{\vec{p}_i^2}{2m}+\sum_{i\lt j} U(|\vec{r}_i-\vec{r}_j|)\right)}$$
The integration over momenta can be easily computed as before (using Gaussian integrals). In contrast, the integral
$$Q_N=\int\mathrm d^N\vec r\,\mathrm e^{-\beta\sum_{i\lt j}U(|\vec{r}_i-\vec{r}_j|)}$$
is particulary difficult (even for simple forms of the interacting potential $U$). We can write temporarily the partition function of the system as
$$Z_\mathrm I=Z_\mathrm{PG}Q_N=\frac1{N!}\left(\frac{2\pi mk_\mathrm BT}{h^2}\right)^{\frac{3N}{2}}Q_N$$
Grand Canonical partition function for the interacting gas
Now, as you sais, the critical parameter is the density $\rho$: at low density you recover the perfect gas behavior. Therefore, it seems a good idea to develop our formalism in powers of $\rho$. It is easier to work in the Grand Canonical ensemble (you don't have to worry much about this), where the partition function is given by (this is a definition)
$$Z_\mathrm{GC}=\sum_N^\infty\mathrm e^{\alpha N}Z_\mathrm C$$
This is just a generalization of the Canonical partition function $Z_\mathrm C$. With this definition we have, for the interacting gas,
$$Z_\mathrm{II}=\sum_N^\infty\mathrm e^{\alpha N}\frac1{N!}\left(\frac{2\pi mk_\mathrm BT}{h^2}\right)^{\frac{3N}2}Q_N=\sum_N^\infty\frac{z^N}{N!}Q_N,$$
where we defined for simplicity
$$z=\mathrm e^{\alpha}\left(\frac{2\pi mk_\mathrm BT}{h^2}\right)^{\frac32}$$
Relation between the Grand Canonical partition function and the equation of state
From $Z_\mathrm I$ in the Canonical ensemble we can now compute the ratio $p/(k_\mathrm BT)$. The Grand Canonical potential is defined by
$$\phi=-k_\mathrm BT\ln(Z_\mathrm{II})$$
and this Grand Canonical potential is linked to volume and pressure by
$$\phi=-pV,$$
so that you finally have a link between the Gran Canonical partition function and the pressure:
$$\frac p{k_\mathrm BT}=\frac{1}{V}\ln(Z_\mathrm{II}).$$
Since $Z_\mathrm{II}$ is a development in powers of $z$, the same is valid for $\ln(Z_\mathrm{II})/V$. This gives
$$\frac1V\ln(Z_\mathrm{II})=z+\frac{a_2}{2!}z^2+\frac{a_3}{3!}z^3+\cdots$$
Density
Using an easy argument that I will not show here (unless explicitly required), you can then write the density as
$$\rho=\frac{\langle N\rangle}V=z\frac{\partial}{\partial z}\left(\frac1V\ln(Z_\mathrm{II})\right),$$
where $\langle N\rangle$ is the average number of particles in your gas.
Therefore, for the density we can write
$$\rho=z+a_2z^2+\frac{a_3}{2!}z^3+\cdots$$
Developement of the state equation in powers of $\rho$
Since we said that $\rho$ is our critical parameter we can write
$$\frac p{k_\mathrm BT}=\rho+b_2\rho^2+b_3\rho^3+\cdots.$$
In fact the perfect gas law was the first-order expansion with respect to the density. For a diluite gas, $\rho \ll 1$ and therefore we simply get
$$\frac p{k_\mathrm BT}=\rho.$$
Virial coefficients
It is now easy to link the coefficients $a$ to the coefficients $b$, by equating the powers of $z$. In our case you find:
$$b_2=-\frac{a_2}2,\qquad b_3=-a_2^2-\frac{a_3}3.$$
It now remains to link these coefficients to $Q_N$. By definition of $Z_I$ we can write
$$\ln(Z_\mathrm I)=\ln\left(1+Q_1z+Q_2\frac{z^2}{2!}+Q_3\frac{z^3}{3!}+\cdots\right)$$
By Taylor expansion of the logarithm we get
$$\ln(Z_\mathrm I)=Q_1+Q_2\frac{z^2}{2!}+Q_3\frac{z^3}{3!}-\frac{1}{2}\left(Q_1z+Q_2\frac{z^2}{2}\right)^2+Q_1^3\frac{z^3}{3}+\cdots$$
For $a_1$ and $a_2$ we have
$$a_1=\frac{Q_1}V\qquad a_2=\frac1V\left(\frac{Q_2}2-\frac{Q_1^2}2\right)$$
which gives
$$b_2=\frac1{2V}\left(Q_1^2-Q_2\right).$$
Computation of $Q_i$
Now, if we compute $Q_1$ and $Q_2$ explicitly (which is much easier than computing $Q_N$ for a general $N$) we get
$$Q_1=\int\mathrm d^3\vec r=V$$
$$Q_2=\int\mathrm d^3\vec{r}_1\,\mathrm d^3\vec{r}_2\,\mathrm e^{-\beta U(|\vec{r}_1-\vec{r}_2|)}=Va+V^2$$
Van der Waals law
With the expression for the virial coefficients and the analytical expressions of $Q_1$ and $Q_2$ we finally write
$$b_2=-\frac a2$$
Therefore we finally get Van der Waals law
$$\frac p{k_\mathrm BT}=\rho-\frac a2\rho^2.$$
At the second order in $\rho$ we get Van der Waals law, but this procedure is completely general and can be used to compute higher order corrections to the perfect gas law. Computing the coefficients $Q_N$ is quite cumbersome, but can be performed using the cluster expansion.

There are dozens of "real gas equations" which try to rationalize one factor or another. Wall effects are never one of the problems considered. Such real gas equations are for the "bulk" gas not what is happening at the surface.
Wall effects are generally ignored entirely...
(1) In general the notion is that the wall collisions are "on average" an adiabatic process.
(2) There is surface adsorption/desorption which is a whole different problem. In general the container's surface absorption is assumed not to be able to change the concentration of the bulk of the gas.

The pressure is exerted on the wall, and therefore, I think that the wall really plays an important role. A molecule moves in an attractive potential well generated by all other molecules in the gas. To collide on the wall, it has to escape this potential well halfway.

@higgsss - If you're trying to model the couple of atomic layers on the surface of the wall then quantum well are important. In general the minute amount of material adsorbed on the wall is insignificant to the total amount of gas.

This doesn't have to do with the detail of the wall. In the bulk of the gas, each molecule is surrounded by other molecules from all directions, whereas very close to the wall, there are no molecules beyond the wall.  Hence the attractive potential is always weakened near the wall.

@higgsss  - The "volume" of gas near the wall is in general insignificant. The only situation I can think of where the wall effect would be significant is gas moving through a long narrow tube.

The volume of gas near the wall is indeed negligible compared to that of the entire container. But only the gas molecules near the wall are able to collide on the wall, thereby exerting pressure on it.

It is not "derived" as you may expect. It is an empirical formula based on the guess that you made - the quantity/volume unit is related to pressure. It is squared because you expect 2 molecules to attract each other. Then you stick in an empirical constant to make the whole stuff work.

There is certainly a statistical mechanics derivation of the van der Waals equation of state, although it seems that OP wants to know the kinetic aspect of this.

@higgsss is right, there is a statistical mechanics derivation of Van der Waals equations of state (see my answer)! For the Van der Waals correction, intuitive arguments can be sufficient, but for higher order correction this is no longer true.

authors: จักรพงษ์ วรรณปกาสิต


text:
Why is CH4 a gas at room temperature but CH3Cl a liquid?
I cannot figure out a question on my review sheet. The full question is:

Methane ($\ce{CH4}$) is a gas at room temperature but chloroform ($\ce{CH3Cl}$) is a liquid. Explain why changing just one $\ce{H}$ atom for a $\ce{Cl}$ atom changes the compound's state at room temp.

I thought about it, and I think it does have to do with $\ce{CH4}$ is non polar, so it does not tend to stick to each other (gas state), while $\ce{CH3Cl}$ is polar, so it does stick, such as like $\ce{H2O}$, which is liquid and is cohesive, so $\ce{CH3Cl}$ would be as well. Is anyone able to tell me if I am on the right path?

see http://chemistry.stackexchange.com/questions/6859/how-to-determine-whether-a-compound-is-gas-or-liquid-or-solid

The question itself if just wrong: chloromethane is not chloroform and chloromethane is not a liquid but a gas at room temperature. That's a pretty bad epic fail for a chemistry question.

Yes, I think that is the right answer expected for this question.

Three factors to consider.
(1) Most importantly there four intermolecular forces which effect BP
Ionic bonds  > Ion-dipole interactions > H bonding >  dipole-dipole interactions > Van der Waals dispersion forces.
(2) Increasing the number of carbon atoms (e.g. n-alkanes) increases the BP
(3) Branching on carbon chain lowers the BP.

The answer being given though is a bit of hand-waving that oversimplifies the situation. Dipole moment alone can't explain the overall trends. Van der Waals dispersion forces must be considered also.
For instance with increasing substitution $\ce{CH2Cl2}$, and $\ce{CHCl3}$ both have lower dipole moments than $\ce{CH3Cl}$ but higher BP's. $\ce{CCl4}$ has no dipole moment like methane, but has the highest BP of all.
                                          Bond Lengths
           Dipole     B.P.       Mol.
           Moment     (°C)       Wt.       C-H     C-X
 CH4         0        −161.49    16.04     108.7    ---
 CH3Cl       1.9       −23.8     50.49     111      178.3
 CH2Cl2      1.6        39.6     84.93     106.8    177.2
 CHCl3       1.15       61.15    119.37    107.3    176.7
 CCl4        0          76.72    153.81    -----    176.6

Also look at the halomethanes as a series. $\ce{CH3F}$ has a smaller dipole moment than $\ce{CH3Cl}$ because the bond length is shorter.
           Dipole   B.P      Mol.      Electro-  C-X     C-H
           Moment   (°C)     Wt.       Negat.    Length  Length
 CH3F       1.85    −78.4    34.03     4.0       138.5   109.5
 CH3Cl      1.87    −23.8    50.49     3.0       178.4
 CH3Br      1.81      4.0    94.94     2.8       192.9
 CH3I       1.62      42     141.94    2.5       213.9

The trend for molecular weight looks good, but the deuterated form of methane, $\ce{CD4}$, has the same (or at least nearly the same) boiling point as $\ce{CH4}$.

The answer to predict boiling points of the four substances would indeed be expected to be given as:

$\ce{CH3Cl}$ is a dipole. The dipole-dipole interactions are much stronger than the van der Waals interactions present in methane, so it’s boiling point is much higher.
When comparing $\ce{CH4}$ to $\ce{CCl4}$, the latter has a much higher boiling point due to its larger number of electrons meaning stronger van der Waals forces.

That said, the teacher that gave you the above information should be removed from chemistry classes immediately. Here’s why:

$\ce{CH3Cl}$ is not chloroform but methyl chloride or chloromethane. It is a gas at room temperature with a boiling point $\vartheta_\mathrm{b} = - 23.8~\mathrm{^\circ C}$.

Chloroform, which is indeed a liquid at room temperature ($\vartheta_\mathrm{b} = 61.2~\mathrm{^\circ C}$) is $\ce{CHCl3}$ or trichloromethane. You need to substitute three hydrogens with chlorine atoms to create chloroform.


text:
Why is vulcanization called "vulcanization"?
I know it might sound a dumb question but whenever I come across a note on importance of vulcanization I get stuck as to the meaning of the term and why it was adopted.

From Wikipedia:

Although the curing of rubber has been carried out since prehistoric times, the modern process of vulcanization, named after Vulcan, the Roman god of fire, was not developed until the 19th century, mainly by Charles Goodyear.

And from the Online Etymology Dictionary:

vulcanize (v.) 1827, "to put into flames," from Vulcan (q.v.), name of the Roman god of fire, + -ize. As a treatment for rubber, first recorded 1846. Related: Vulcanized; vulcanizing.

Who coined the word? Thomas Hancock, a British scientist and inventor:

On 21 November 1843, Hancock took out a patent for the vulcanisation of rubber using sulphur, 8 weeks before Charles Goodyear in the US (30 January 1844). He mentioned in his "Personal Narrative" that his friend William Brockendon invented the word vulcanisation from the God Vulcan of Roman mythology. Hancock did not credit himself with discovering the reaction of sulphur with rubber; he instead said that in 1842 Brockendon had showed him some American rubber samples which had been treated with sulphur.

According to Thomas Hancock's 1857 Personal narrative of the origin and progress of the caoutchouc or india-rubber manufacture in England

In preparing to carry out the new process on a
  large scale, it appeared desirable to give the mate-
  rial a more definite name than " the change,"
  which I had adopted for the purposes of my speci-
  fication; and, whilst discussing the subject amongst
  my friends, Mr. Brockedon proposed the term
  " Vulcanization," and, as no better suggested
  itself, I determined to adopt it, and it is now
  known by that name both in Europe and America.
  It owes its derivation to the Vulcan of mythology,
  as in some degree representing the employment of
  sulphur and heat, with which that mythological personage was supposed to be familiar. I shall,
  therefore, of course use the term in the remainder
  of my narrative.


text:
What is the preferred IUPAC name of 4-propionyl biphenyl?
Is the IUPAC name of the following compound 4-propionyl biphenyl?

The compound that is given in the question is a ketone. According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the preferred names for ketones derived from alkanes are constructed using the suffix ‘one’.
Furthermore, the selection of a preferred parent structure is based on the seniority of classes, which gives priority first to characteristic groups expressed as suffixes and then to parent hydrides when different classes are present.
Therefore, the preferred parent structure for the compound given in the question is ‘propan-1-one’.
The parent structure is substituted with a ring assembly R at the first carbon atom. Thus, the resulting name is ‘1-(R)propan-1-one’, where the name R of the ring assembly is determined as follows.

Generally, according to the current IUPAC Recommendations, ring assemblies of two identical cyclic systems with a single bond junction can be named by one of two methods:

(1) by placing the prefix ‘bi’ (…) before the name of the corresponding parent hydride enclosed in parentheses, if necessary. Parentheses are used to avoid confusion with von Baeyer names;
(2) by placing the prefix ‘bi’ (…) before the name of the corresponding substituent group (…), enclosed in parentheses, if necessary.

For the unsubstituted parent structure of the given ring assembly, the traditional name ‘biphenyl’ is retained.
With regard to numbering of locants, the current version of the IUPAC Recommendations reads as follows:

Each cyclic system is numbered in the traditional way, one with unprimed locants, the other with primed locants. Lowest possible locants must be used to denote the positions of attachment. These locants must be cited in preferred IUPAC names; they can be omitted in general nomenclature when no ambiguity would result.

Therefore, the preferred IUPAC name (PIN) of the unsubstituted structure of the given ring assembly is ‘1,1'-biphenyl’ and the locants ‘1,1'’ are compulsory. (The name ‘biphenyl’, without locants, may be used in general nomenclature.)

Concerning substituent prefixes derived from such ring assemblies, the current IUPAC recommendations read:

Names of substituent groups derived from ring assemblies are written in two ways:
(1) the suffix ‘yl’ is added to the name biphenyl;
(2) names denoted by locants are placed in brackets; the suffixes ‘yl’ and ‘ylidene’ are added with elision of the final letter ‘e’ in the name of the parent hydride.

Low locants are first assigned to ring junctions (here: ‘1,1'-’), then to free valences (here: ‘4-yl’).
Therefore, the complete name of the compound given in the question is 1-([1,1'-biphenyl]-4-yl)propan-1-one.


text:
At what temperature do complex hydrocarbons become ashes?
When heating a metal tube with polypropelene in it, i found that i still had brittle black foam and flakes in it even though it had been in a blue flame at around 500'C

What happens to complex hydrocarbons/plastics when they turn into black foam at high temperatures?
At what temperature do tars and heavy hydrocarbons completely lose all their Carbon and only ashes are left?

In short, there is no such temperature. You can't remove carbon by heating alone. Or rather, you can, but first you will vaporize your metal tube and everything else, so that in the end there would be no ashes left. No, the presence of oxygen is the key.

What happens depends whether it is being heated in the presence or absence of oxygen.  For example, in the manufacture of coke (mostly carbon), coal is heated without oxygen and from Coke Production For Blast Furnace Ironmaking:

The coal-to-coke transformation takes place as follows: The heat is transferred from the heated brick walls into the coal charge. From about 375°C to 475°C, the coal decomposes to form plastic layers near each wall. At about 475°C to 600°C, there is a marked evolution of tar, and aromatic hydrocarbon compounds, followed by resolidification of the plastic mass into semi-coke. At 600°C to 1100°C, the coke stabilization phase begins.

Without oxygen, the carbon will remain upto its boiling point of 3825 °C.
With oxygen, carbon as coke with start combustion at its autoignition temperature of about 700°C

I wanted to add on to DavePhD's fine answer with a simple point.  "Ashes" in general refers to the solids left over after oxidative combustion of hydrocarbons or carbohydrates.  The primary components of "ashes" are inorganic salts.  Sodium and potassium oxides, phosphates, sulfates, and carbonates, along with lower amounts of calcium and magnesium salts of these same anions, would be the primary constituent.


text:
Le Châtelier’s Principle and Addition of Water
With regards to gas volume, Le Châtelier’s principle predicts that an increase in the volume of the container causes the system to favor the side of the reaction that produces more moles of gas. However, would this principle apply equally well to an aqueous system, at equilibrium, that has now had water added to it? That would be analogous to expanding the volume of a gaseous system, wouldn’t it? Would adding water to an aqueous system at equilibrium affect the position of equilibrium at all?

It is not exactly analogous to expanding the volume, but adding water affects the concentrations of both the reactants and products, which in turn affects the position of equilibrium.

related: http://chemistry.stackexchange.com/questions/38417/using-le-chateliers-principle-to-explain-why-percent-dissociation-of-acids-base

Yes adding more water will affect the position of the equilibrium simply because you are adding more volume to the solution by adding more water. So adding volume is inversely proportional to pressure where the system needs to reduce the amount of water. Then the equilibrium will shift to the side with increased amount of pressure.

We are talking about liquids here, not gases, so volume is not inversely proportional to pressure. In fact, pressure is not really relevant at all.

It will not be liquid as a whole because we add water to a system which is already aqueous. So this results in an aqueous system again. If it is either gas or liquid it definitely has no link to the equilibrium constant for sure. And as the Boyle's law states, pV = nRT, it is clear that pressure is inversely proportional to temperature. The volume will impact on the pressure and the pressure will decide which way should the system shift, in relevence to the coefficients of the products and reactants.

pV=nRT only applies to a gas, not a liquid.

Yeah true. That's why I stated that liquid or solid could not be considered when calculating the equilibrium constant which implies the meaning that Boyle's law has nothing to do with liquid or solid.

But the entire question is only about solutions so any discussion of gas laws is irrelevant.

Yes, I was adding the law for the "That would be analogous to expanding the volume of a gaseous system, wouldn’t it?" part of the question.

But in this case the osmotic pressure works analogously to gas pressure, so if it's for the purpose of linking the two scenarios up, I think mentioning pressure isn't that bad a thing.


text:
Solubility of cholesterol and phospholipids in acetone
I'm having difficulty with this specific part of an assignment. I know that it is to do with polarities of the different parts of cholesterol and that it is because cholesterol is amphiphilic, but I don't know what it is about acetone that means it can dissolve amphiphilic substances but not phospholipids.

In fact, you seem to have gotten it the wrong way round. Phospholipids are a "standard" example of amphiphilic compounds. That's exactly why they form the cell membrane. And are you actually sure cholesterol is amphiphilic?

It has polar and non-polar parts, the OH group is polar and the hydrocarbon chain and rings are not, so it is surely?

It is probably a little bit amphiphilic, but I would say, perhaps not as much as a phospholipid. The -OH group is relatively small and there are four fat hydrophobic rings.


text:
Why does a mixture get warm in an exothermic reaction?
The production of ethanol by fermentation is exothermic. My textbook says that the yeast stops working if the mixture gets too warm. Why would the mixture get warm if the reaction is exothermic?

I have edited your question, but in future, please use appropriate titles and tags: 1) if your question is about why the mixture gets warm, indicate so - the title you chose says that you are actually asking about why the fermentation stops. 2) if your question is about the exothermicity of the reaction, then it has nothing to do with organic chemistry, biochemistry, or alcohols. Instead, the thermodynamics tag would be more useful.

In thermodynamics, we like to divide any situation into two parts, the system and the surroundings. In general the system is the part of the universe that is being studied while the surroundings is everything else that interacts with the system. Between the boundary of the system and the surroundings, energy and heat can be transferred (in an open system).
Now in fermentation, the yeast and glucose molecules are said to be the system while the surround water is the surroundings. Since the reaction is exothermic, this means that energy is moving from the system to the surroundings. Therefore the water should gain more heat energy which will increase its temperature.
However I doubt that the heat generated from the reaction is enough for the enzymes in the yeast to stop working. When I did this experiment we used a Bunsen burner to heat up the solution to approximately 35 degrees. So really what is going to cause the solution to heat up to above 45 degrees (this approximately where yeast begin to stop working) is that you keep the Bunsen burner on and keep heating it.

No worries, I am happy that I was able to help you.

An exothermic reaction is a reaction whereby energy is released, typically in the form of heat. As fermentation is an exothermic reaction, it releases heat to the surroundings, i.e. the mixture warms up.
Hope that helps!

Since it is an exothermic process it releases heat which leads to the heat up of the mixture, second it stops working because in yeast fermentation requires enzymes to work. These enzymes have a optimal working temperature when the temperature rises above it they start to work slower (or above a critical temperature they denaturate and stop working for good, but I seriously doubt this will happen with heat coming only from the fermentation).
I don't think it would stop working. It will be more of an equibilirium, because a slower reaction causes less heat which in return (provided the system is not isolated) will cool down which again speeds up the reaction.

Nice first answer +1 although it's also limited by yeast cells itself not only enzymes

@daveid So then the mixture heats up from the heat it produced during the reaction?  That's odd because the enthalpy of mixtures reduces during exothermic reactions. Also I'm assuming for the mixture to heat up, it must be in a closed system? Otherwise heat will be lost to the surroundings and not absorbed by the mixture.

This is actually an open system as energy and matter is freely able to pass between the boundary of the system and the surroundings. However it is the water that absorbs the energy from the reaction. And then it loses some of this energy to the surrounding air.

We have something like this.
Reactants  ---> Products + Heat
Now by Le Châtelier's principle, we can say that if we provide heat to the reactant, (that is, raise the temperature) we are more likely to not get the product.

This reaction is enzymatic so this principle doesn't have much to do here.

@Mithoron Thanks for the knowledge. Should I delete the answer?

Not necessarily, it probably has slight influence. Also might be a useful info for OP as he seems more interested about thermodynamics than mechanisms


text:
Most energy-dense commercial battery available
What's the most energy-dense and currently commercially available chemical battery technology? Would it be the Lithium-Ion Polymer or has it been surpassed already?
There's is a chemical non-rechargeable battery which is more energy-dense than LiPo, even if it's less power-dense?
I'm trying to have a general sense of the current state-of-the-art technology, as the Wikipedia articles and tables seems to be very dated.
Edit: I meant "dense" as in respect to mass. I believe that "specific energy" would be the right terminology.

This has a list of common battery types with energy densities.
Lithium ion primary batteries appear to have the highest specific energy at around 300 Wh/kg. Lead-acid batteries have a specific energy around 10% that of lithium-ion batteries, but are extremely reliable (depth of discharge, self-discharge, etc.), and do not have nearly the same heating/fire issues (though they can explode if you mess with them), which is one of the reasons they have endured so long in automotive applications.
More exotic chemistries have higher specific energies, but I am unsure as to whether they are available to the average consumer.

Lithium-air batteries are NOT production-ready.

Not neccesarily truth, I don't have exhausting knowledge in the area

What's the most energy-dense and currently commercially available chemical battery technology

Assuming we talk about energy per mass unit.
In case one doesn't care about reversibility, it is air-metal batteries (say, alumunium-air). In case one does, it is probably Lithium-sulfide batteries.
Both are inferiour to common internal combustion engine, though.
Assuming we are talking about power per mass unit
Lead batteries are close to top, AFAIK. Otherwise they would be pushed out from cars already.

Do you know of any commercially available air-metal battery?

http://www.microbattery.com/zinc-air-battery-specs || zinc-air batteries are definitely available, see the link. They are faaar inferrior to potential of aluminium-air batteries, but still beat any rechargeable Li batteries. Aluminium-air batteries were announced to be produced this spring and are said to be used in various prototypes, but no other relevant traces in the web were found.

This would be a decent answer if it contained some actual numbers not vague hand waving opinions and it referenced whether the tech was a prototype or a commercial product.

@matt_black I converted it to community wiki, feel free to fill in the details. The numbers can be pulled from Wikipedia.


text:
What is the geometrical structure of Pb3O4?
I'm confused by the structure of Pb3O4. There are some sites claiming that it is a six membered ring of alternating Lead and Oxygen atoms with another Oxygen atom in the center bonding with 2 out of the 3 Lead atoms.
There are other sources which claim that it is actually two 4 membered rings of alternating lead and oxygen atoms with a lead atom in common between the 2 rings.
Moreover, isn't Lead a metal? Then How can it combine with oxygen in such "bonds" whereas ionic bonds are non-directional?

The Wikipedia article probably has the latest information. https://en.wikipedia.org/wiki/Lead%28II,IV%29_oxide // "Ionic" and "covalent" bonds are abstractions that chemists invented not atoms. Think of black and white paints. There is no "pure white" and "pure black."

Maybe below 3d structure of Pb3O4 will help?
3d structure of Pb3O4
Also in the below link of a pdf, just before the discussion section the paragraph clearly states that there are both ionic and covalent bonds present in Pb3O4.
REsearch Paper on Pb3O4 structure
Thus we can conclude that it is Tetragonal, tP28 .(as per the link provided by Maxw in the comment to your question)
Also metals can form covalent bonds. Refer to below question and its answers:
Metals forming covalent bonds


text:
How to calculate the amount of citric acid to neutralise the pH within wastewater?
We have an effluent tank, with a consent to discharge at between 5.5 - 11 pH, the tank holds 100000 liters of dirty water.
When the pH reaches 10.5 we add citric acid to bring the pH down. How many liters of citric acid with a pH of 2.2 should we add to the system to bring the pH down to around 7. We always end up putting too much in then add caustic to bring it back up.

Gotta start with my favorite formula:
$$Q = CV$$
$$C_\textrm{eff}V_\textrm{eff}+C_\textrm{acid}V_\textrm{acid} = C_\textrm{out}V_\textrm{out}$$
$$(10^{-10.5}~\mathrm{mol/L})(10^5~\mathrm{L})+(10^{-2.2}~\mathrm{mol/L})V_\textrm{acid} = (10^{-7}~\mathrm{mol/L})(10^5+V_\textrm{acid})$$
$$V_\textrm{acid} = 1.51~\textrm{L}$$

Well, we normally add somewhere in the region of 200 litres of acid, the pH then drops below our bottom end (5.5). So are you saying to add 1.58 litres of acid in total?

Yes, it appears that the amount of acid required is extremely small compared to the total volume (I was honestly expecting something on the order of 100-300L, so it was surprising to get ~1.5L)

Does it make any difference that this is all being diluted with water? The tank gets filled from the drains within the factory which normally runs at about 6.4pH, the pH of the effluent tank only gets so high when we carry out internal cleaning with the caustic ph of about 14.4. 1.5litres seems incredibly small

What do the calculations say will happen to the pH when you add 328 litres of citric acid?

Using my same formulas, I get a pH of 4.7 with 328L of acid

If you are going to be on here for another 5 minutes, I will go and check on the monitoring station what the pH dropped to and report back. Many thanks for this, greatly appreciated.

OK, so I had a pH of 10.7, which I subsequently added 328 litres of citric acid to, the pH then dropped to 4.78! The maths do not lie. Thank you very much that is exceptional. Can you explain why the rate of drop in pH is so different?

pH is a logarithmic scale, so the amount of acid/base you add to the "solution" can have drastically different effects

Well the "math" does lie in this case. How much citric acid to add depends on what else is in the dirty water. So getting a pH of 10.5 with NaOH is different than getting a pH of 10.5 with NH3. It really seems that you should do a quick test on a sample of the water. You wouldn't need super accuracy. It would seem that you could just count drops of citric acid solution. So amount of citric acid depends on what other "stuff" is in the dirty water, and how much.

@MaxW absolutely. these calculations assume there are no bases to neutralize the added acid. thank you for the added info

Citric acid is not really a strong acid, treating it as such seems a bit wrong. What about the equilibrium?

@Martin-マーチン The highest pKa of citric acid is about 6.4 (per Wik). For solutions with pH above about 7.5 to 8, citric acid will behave essentially as a (triprotic!) strong acid. Beyond that, at the target pH of 7, with a second-highest pka of ~4.8, it will still behave roughly as a diprotic strong acid, though with some buffering coming into play from that 6.4 pKa.


text:
Discrepancy in MO diagrams for Period 2 homonuclear diatomics
I've recently noticed that different sources on the internet and in books have different ideas of whcih atomic orbital influences which molecular orbital. In the diagram, where the lines connect the AO to the MO to show their influence, I noticed that in some of the MO diagrams, the $\mathrm{2p}$ orbital does not connect to the $\sigma_1$ orbital (the lowest in energy) and the $\mathrm{2s}$ orbital does not connect to the $\sigma_4$ orbital (the highest in energy). On the other hand, I also see the case where the opposite is true and they DO connect (this is the case for my own chemistry lecturer's notes).
I asked her if she could explain this difference, and she said that the $\sigma_1$ should have some $\mathrm{p}$ character and that the $\sigma_4$ should have some $\mathrm{s}$ character, so they do connect. She didn't however say that the other way was wrong. As a result I'm not sure whether the other way is wrong, or whether it's just like a difference in notation.
I hope someone can help clarify for me. I've attached two photos of the different cases.
Thank you.
$\hspace{15 mm}$
(source)
versus
$\hspace{30 mm}$

Welcome Daire to Chemistry Stack Exchange. I am pretty sure that both ways of drawing the MO diagrams are correct. Generally you would see the MO diagram as being drawn without showing the s and p orbital mixing (so your first image).

As molecular orbital diagrams are qualitative ways to understand bonding, there are different ways in which they may be drawn.
In the simplest possible cases, molecular oxygen (O=O), for instance, it is okay to assume that s-orbitals mix only with s-orbitals (of similar energy), and that the p-orbitals only mix with p-orbitals. This works fine for molecular oxygen, and explains many of its properties: double bond between O and O, paramagnetic etc.
In other cases, molecular nitrogen being the common example, this is not sufficient, and we have to account for orbitals mixing with orbitals of other kinds. This is called s-p mixing.
Firstly, you construct a MO diagram where orbitals of the same kind mix (the same as you would have done for oxygen), you then allow these MO's to interact with other MO's of the same symmetry to produce a new set of mixed orbitals. When the MO's interact in this way, they produce new MO's of higher and lower energy, which accounts for the different order of the MO's between diagrams where s-p mixing has been taken into account.
In general, any homonuclear diatomic molecules below oxygen (lithium, nitrogen, boron etc.) need s-p mixing to be considered.
For more information about s-p mixing (and other kinds of orbital mixing) do a search here on stack exchange. There are plenty of other similar questions with answers to suit.