text:
Is this molecule possible? And if so, what is it called? (Cyclic groups)
I saw a little chemistry pun with a couple of hydrocarbons included, and it had me thinking if they were actually possible  and if so, what would the name given to it be.
The hydrocarbon I am curious about:

For those wondering what the pun is:

What do you think about their names?

What do you mean?

I mean do you have any idea about their names? Do you know IUPAC naming? These kinds of questions needs some opinion of OP, so that the answerer can clear his confusion.

Nice joke :) Both hydrocarbons are OK, however this exam would need to have quite hard on reactions ;)

I know about naming alkenes (E.g but-1,2-diene) and cyclic hydrocarbons (E.g cyclohexane) but I have never seen a hydrocarbon with either a double bonded alkyl group, so I'm mainly wondering what having an alkyl group but with double bonds does to the name. If I were to guess I would say the hydrocarbon would be 1-butylecyclohexane.

It's better to name it as butadiene derivative.

If you want to consider the diene as the main chain, you could call this 2-cyclohexyl-1,3-butadiene. This used to be the proper name for the compound, as double bonds used to take priority over any length of a carbon chain in IUPAC nomenclature. This compound exists, and you could theoretically purchase it at some place like this. In recent years, the proper priorities for IUPAC have changed. Now, the longer chain takes priority over double bonds, and thus this compound should be referred to using the cyclohexane as the main chain. This means that the compound should be called (buta-1,3-dien-2-yl)cyclohexane. For all practical purposes, either name should work, as they both describe the same molecule; however, the second name is consistent with IUPAC standards.

Please note that double bonds do not always have priority since the IUPAC recommendations have been changed. In the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the first criterion to be considered in choosing a preferred parent acyclic chain is the length of the chain; unsaturation is now the second criterion.

Would you then use methenyl as a substituent name? I don't think a double bonded carbon can be treated as a side chain when one of the sp2 carbons is part of the parent chain. Please correct me if I am wrong.

In general nomenclature, the name 2-cyclohexylbuta-1,3-diene may be used so that the unsaturated acyclic structure may be recognized (see Subsection P-44.1.2.2); however, the ring is preferred to the chain when constructing the preferred IUPAC name (buta-1,3-dien-2-yl)cyclohexane (see Subsection 52.2.8).

The after-exam molecule is simply referred to as:
2-cyclohexylbutadiene in Iron catalyzed
1,4-hydroboration of 1,3-dienes Journal of the American
Chemical Society 131(36): 12915–12917,
and even more simply:
cyclohexylbutadiene in Intermolecular Hydroamination of 1,3-Dienes Catalyzed by
Bis(phosphine)carbodicarbene−Rhodium Complexes J. Am. Chem. Soc., 2014, 136 (17), pp 6227–6230

2-cyclohexyl-1,3-butadiene. (after the exam)
[1,1,2-trimethylpropyl]cyclohexane (before the exam)
There are variants of these names, like locant numbers unnecessarily placed between prefix and infixes. There are trivial and substitutive names but these are more appropriate and clear.


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Molecular term symbols for excited state oxygen
What are the possible term symbols for excited state oxygen with configuration $(1\pi_u)^3$,$(1\pi^*_g)^3$ (all other occupied orbitals are closed shell)? Based on the possible values of S and $\lambda$ it seems that $^1\Delta_u$, $^3\Delta_u$, $^1\Sigma_u$ and $^3\Sigma_u$ are possible. However, I am not sure in this case how to determine the +/- labels for the $\Sigma$ term symbols. For ground state oxygen, triplet states pair with - and singlet states with + to give overall antisymmetric wavefunctions. However, in this case, two different orbitals are involved, so does this rule no longer apply?

What are the possible term symbols for excited state oxygen with configuration $(1\pi_u)^3(1\pi^*_g)^3$ (all other occupied orbitals are closed shell)? Based on the possible values of $S$ and $\lambda$ it seems that $^1\Delta_u$, $^3\Delta_u$, $^1\Sigma_u$ and $^3\Sigma_u$ are possible.

Term symbols for $(1\pi_u)^3(1\pi_g^*)^3$ - i.e. 1 hole + 1 hole - is equivalent to finding terms for $(1\pi_u)^1(1\pi_g^*)^1$, i.e. 1 electron + 1 electron. I will continue answering the question for the latter case as this is easier to handle.
The fastest way is to use group theory. $O_2$ molecule belongs to $D_\infty^h$ point group. We look at the character table for this group (e.g. here) and perform a direct product of irreducible representations corresponding to our electrons, i.e.:
$\Pi_u \otimes \Pi_g = \Sigma_u^+ \oplus \Sigma_u^- \oplus \Delta_u$
Note this only pertains to spatial symmetry, i.e. orbital angular momentum. It tells us nothing about possible spin states. However, since the two electrons are in different orbitals, all spin combinations are allowed in all cases. The resulting terms are:
TLDR: $^3\Sigma_u^+,$$^3\Sigma_u^-,$$^3\Delta_u,$$^1\Sigma_u^+,$$^1\Sigma_u^-,$$^1\Delta_u$
Your answer is thus correct, the missing piece being the +/- signs of the $\Sigma$ terms. Note that in answering we relied heavily on Pauli exclusion principle not playing a role - since the two electrons occupy different orbitals. Were that not the case, some of the terms would disappear (terms of e.g. $(1\pi_u)^1(2\pi_u)^1$ are $^3\Sigma_g^+,$$^3\Sigma_g^-,$$^3\Delta_g,$$^1\Sigma_g^+,$$^1\Sigma_g^-,$$^1\Delta_g$, while terms of $(1\pi_u)^2$ are a non-trivial subset, specifically $^3\Sigma_g^-,$$^1\Sigma_g^+,$$^1\Delta_g$).

However, I am not sure in this case how to determine the +/- labels for the $\Sigma$ term symbols.

The following derivations draw heavily from this pdf, read for further details.
We refer to definition of +/- signs as describing the symmetry with respect to reflection in plane that contains both atoms, let's use an arbitrary choice of xz-plane. Wavefunctions symmetric when reflected in this plane are "+" terms, anti-symmetric are "-" terms. However, imagining reflections is not practical. Therefore, we introduce operator notation. The "+" term corresponds to a wave function with eigenvalue of +1 (with respect to $\hat{\sigma}_{xz}$ operator); "-" term corresponds to eigenvalue of -1.
We note:
$$\hat{\sigma}_{xz} \pi_x = \pi_x$$
$$\hat{\sigma}_{xz} \pi_y = - \pi_y$$
$$\pi_1 = -N(\pi_x + i\pi_y)$$
$$\pi_{-1} = N(\pi_x - i\pi_y)$$
, where N is some boring normalization constant. Following up:
$$\hat{\sigma}_{xz} \pi_1 = - \pi_{-1}$$
$$\hat{\sigma}_{xz} \pi_{-1} = - \pi_1$$
We need combinations of configurations (Slater determinants = SD) that are eigenfunctions of $\hat{\sigma}_{xz}$.
E.g.:
$$^1\Sigma^+_u = N(|SD_1\rangle - |SD_2\rangle + |SD_3\rangle - |SD_4\rangle) =$$$$= N(|\pi_{u,1},\bar{\pi}_{g,-1}\rangle - |\bar{\pi}_{u,1}{\pi}_{g,-1}\rangle + |\pi_{u,-1},\bar{\pi}_{g,1}\rangle - |\bar{\pi}_{u,-1}{\pi}_{g,1}\rangle) $$
Let's apply the $\hat{\sigma}_{xz}$ operator:
$$\hat{\sigma}_{xz}{ }^1\Sigma^+_u = \hat{\sigma}_{xz} (N(|\pi_{u,1},\bar{\pi}_{g,-1}\rangle - |\bar{\pi}_{u,1}{\pi}_{g,-1}\rangle + |\pi_{u,-1},\bar{\pi}_{g,1}\rangle - |\bar{\pi}_{u,-1},{\pi}_{g,1}\rangle)) = $$
$$ = N(|-\pi_{u,-1},-\bar{\pi}_{g,1}\rangle - |-\bar{\pi}_{u,-1}{-\pi}_{g,1}\rangle + |-\pi_{u,1},-\bar{\pi}_{g,-1}\rangle - |-\bar{\pi}_{u,1},{-\pi}_{g,-1}\rangle) =$$
$$ = N(|\pi_{u,-1},\bar{\pi}_{g,1}\rangle - |\bar{\pi}_{u,-1}{\pi}_{g,1}\rangle + |\pi_{u,1},\bar{\pi}_{g,-1}\rangle - |\bar{\pi}_{u,1},{\pi}_{g,-1}\rangle) = $$
$$= N(|SD_3\rangle - |SD_4\rangle + |SD_1\rangle - |SD_2\rangle) = { }^1\Sigma^+_u$$
In going from first to second line we apply the operator, i.e. flip "1" subscript to "-1" and vice versa; in addition we multiply each orbital by "-1". Since we have two orbitals this multiplication disappears: ($-1*-1=1$), which is what we do in going from second to third line. In the last line we see that $\hat{\sigma}_{xz}|SD_1\rangle = |SD_3\rangle$;$\hat{\sigma}_{xz}|SD_2\rangle = |SD_4\rangle$ etc.
We could have used a WF in the form e.g.: $WF = |SD_1\rangle + |SD_3\rangle$. Such a WF is indeed an eigenfunction of $\hat{\sigma}_{xz}$. But it is not an eigenfunction of $\hat{S}^2$.
Similarly, a corresponding "-" term is:
$$^1\Sigma^-_u = N(|SD_1\rangle - |SD_2\rangle - |SD_3\rangle + |SD_4\rangle) =$$
$$\hat{\sigma}_{xz}^1\Sigma^-_u = N(|SD_3\rangle - |SD_4\rangle - |SD_1\rangle + |SD_2\rangle) =$$
$$ = -1 * N(- |SD_3\rangle + |SD_4\rangle + |SD_1\rangle - |SD_2\rangle) = -1 * { }^1\Sigma^-_u$$
The four SDs we use here are all eigenfunctions of $\hat{L}_z$ and $\hat{S}_z$ with $M_l = 0$ (i.e. one $\pi_1$ + one $\pi_{-1}$ orbital) and $M_s = 0$ (i.e. one alpha and one beta orbital). There are two more combinations of these SD and these are part of the $^3\Sigma^-_u$ and $^3\Sigma^+_u$ states.
Lastly, non-$\Sigma$ terms have both a "+" and "-" component that flips to each other upon reflection, e.g.: $\hat{\sigma}_{xz} \Delta^+ = \Delta^-$. But since $\Delta^+$ and $\Delta^-$ are different SDs (different configurations), we don't use the label in these cases - they are reserved for $\Sigma$ terms.
The take-away message here is that "+/-" label relates to some kind of symmetry. Similarly to "g/u" label, it only is relevant in some special molecules ("g/u" in molecules with inversion center; "+/-" in diatomic molecules). We use group theory where possible or a very technical procedure when not.

For ground state oxygen, triplet states pair with - and singlet states with + to give overall antisymmetric wavefunctions. However, in this case, two different orbitals are involved, so does this rule no longer apply?

There is some confusion here. As mentioned above, for ground state ($(1\pi_u)^2$) the terms are indeed $^3\Sigma_g^-$ and $^1\Sigma_g^+$. But there is also the state $^1\Delta_g$ that has no "+/-" label, yet it is still anti-symmetric. The "+/-" label has nothing to do with anti-symmetry. Anti-symmetry is ensured by use of Slater determinants. The "rule" you mentioned is a result applicable to $\pi^2$ (and other doubly degenerate 2 electron) configurations, but is not related to anti-symmetry and gives no predictions for other configurations.

Is it possible to say something about the ordering of the energies? Hund's rules would suggest: $^3\Delta_u$ < $^3\Sigma_u$< $^1\Delta_u$ < $^1\Sigma_u$ (maximising S, then maximising $\Lambda$ but is this really applicable here since we have an excited state configuration?

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Why Fe(OH)2 and some precipitates do not dissolve in excess ammonia?
I understand that Cu(OH)2 can dissolve in excess ammonia because ligand exchange reaction occurs. But why ligand exchange reaction does not occur for Fe(OH)2?

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What are the products formed when CH3CH2OH reacts with NaH?
When $\ce{CH3CH2OH}$ reacts with $\ce{NaH}$, why are the products $\ce{CH3CH2ONa + H2}$? Why not $\ce{CH3CH3 + NaOH}$? Can you please explain in detail?

Think about the mechanism.

http://chemistry.stackexchange.com/questions/30982/why-does-a-hydride-like-nah-act-as-a-base-but-a-hydride-like-nabh4-act-as-a-nuc/30983?s=1|0.0732#30983

@bon Shouldn't Na combine better with OH-? (I sound dumb I guess, but I need a bit of a detailed explanation :))

There are a number of reasons why the reaction you suggest in much less favourable than the deprotonation. Firstly, hydride is a terrible nucleophile because its 1s orbital is too small to effectively overlap with anything except hydrogen.
Secondly, hydroxide is a poor leaving group for an $S_N2$ reaction and so you need a very good nucleophile in order to displace hydroxide from an alcohol group.
Thirdly, the reaction of sodium hydride as a base often produces a quantitative deprotonation because the product, $\ce{H2}$ is lost as a gas, which drives the reaction forward.

Consider firstly the ease of reaction: the hydrogen in $\ce{NaH}$ is basic $\ce{H-}$ and the hydrogen in ethanol is acidic $\ce{H+}$. The easiest reaction then, is a simple acid-base neutralization.
The reaction products you propose requires that a $\ce{C-O}$ bond is broken, or a situation wherein which the hydride attacks a carbon center which then ejects a hydroxide ion $\ce{OH-}$. In reality, however, you would have to eject $\ce{O^{2-}}$ which is much harder to remove.

Where does O2- come from?

@RamBharadwaj if you draw out the mechanism (assuming $S_N2$) it will look like $O^-$ is the leaving group, so after leaving it will be $O^{2-}$ because you would start out with ethoxide rather than ethanol.


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How do primary alcohols react with dihydropyrans
Can anyone also explain me the mechanism of this reaction?

I think it might be formation of acetal or hemiacteal reaction but I am not sure about it.

Why do you consider a the formation of  a hemiacetal under anhydrous conditions?

@KlausWarzecha sir actually I don't know. I am just having a guess because I think ester or ether will not formed from this reaction. Can you please explain me the mechanism and the product?

Please have a closer look at your starting materials and the reaction conditions.

The reaction is performed under protic, ahydrous conditions.
One of the starting materials is a primary alcohol.
The other starting material apparently looks like a cyclohexene where one carbon adjacent to the $\ce{C=C}$ double bond has been replaced by an oxygen atom. This is a dihydropyrane.

Now, let's try a mind game:

Supposed that your second starting material is cyclohexene, what would happen under anhydrous, protic conditions?

 The proton would probably add to the $\ce{C=C}$ double bond to form a cyclohexyl cation. Since the starting material is symmetric, there is only one secondary cation formed. Note that the situation would be different if you would start with methylcyclohexene! Then, a secondary or a tertiary cation could be formed. Remember inductive effects to judge which one is favoured.

How would the primary alcohol come into play here?

 It would probably add to the cyclohexyl cation and yield an ether.

In reality, your second starting material is a dihydropyrane, think cyclic vinyl ether. How will the lone pairs on the oxygen atom effect the protonation?

 If in doubt, draw! Your $\ce{C=C}$ double bond now has two markedly different carbon atoms. Protonation can result in two different cations. I suggest to draw some resonance structures to figure out which one is favoured. Remember the lone pairs ;-)

What happens in the presence of a primary alcohol now?

 Again, it will add to the cation.

How does the final structure look like?

 I'm certain that you have figured this out by drawing and once you have, you will easily identify the structure class :)

Note that the reaction in question is quite useful to reversibly protect alcohols in sequences of chemical transformations.

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How to rationalise with MO theory that CO is a two-electron donor through carbon?
A question I am looking at is as follows:

$\ce{CO}$ is isoelectronic with $\ce{N2}$. Sketch MO diagrams for $\ce{CO}$ and $\ce{N2}$. Point out key differences between the diagrams and use the diagram to explain why $\ce{CO}$ acts as a two-electron donor through carbon rather than through oxygen.

Understandably, the key difference between these molecules is that $\ce{CO}$ is heteronuclear, and thus will have differences in energy between the molecular orbital and the atoms.
But I can't explain why $\ce{CO}$ is a two-electron donor using MO theory, even though I can with Lewis?
Does anyone have any thoughts?

In the above MO diagram, the 5σ is the HOMO. But it is closest in energy to oxygen's 2p orbitals, so why is it centered on carbon?

I had to give this a good, long thought, after I initially thought this is a question which is hardly answerable with a molecular orbitals diagram by itself. Now that I have spent quite some time thinking about it, I must admit I enjoy this exercise a lot.
The key to explaining that carbon monoxide is a two electron donor through carbon is to start with with a very fundamental MO scheme. This is basically to get you dimensions right and not to mix things up later. It is also one of the things I would consider a prerequisite for these kinds of exercises.

The above diagram is for a symmetric diatomic molecule with no s-p mixing. This is true for the heavier main group elements; have a look at this explanation by Wildcat.
For dinitrogen we will have to adapt this scheme to include s-p mixing. Here it is important to know, that the interaction is still strong enough to raise the 3σ valence orbital above the 1π orbitals.

With pen and paper it is a bit too difficult to draw the MO, so it is probably easier to exaggerate the respective contributions to the orbitals. For example the 1σ orbital in the non mixing case is comprised of pure s-orbitals. When it is mixing, there has to be some s-p character which is difficult to draw. As a general guideline electron density concentrates along the bonding axis for bonding orbitals, while for anti-bonding orbitals in the lone-pair region. The π orbitals largely remain untouched by this.
From there we can use the fact, that carbon monoxide is isoelectronic with dinitrogen. We have to skew the energies of the atomic orbitals. The energy of the atomic orbitals is decreasing from left to right in the period; therefore carbon will have slightly elevated levels and oxygen's are lowered.
In this case it is easier to not tinker with the ordering of the MO as this is just a pen and paper exercise. However, one should be aware, that the levels will overall change in this molecule.

The atomic orbitals that are closer to the energy of the molecular orbital will have a larger coefficient. Therefore 1σ and 1π will be polarised towards oxygen. On the other hand 2σ and 3σ will be polarised towards carbon.
One can also argue this with Bent's rule (gold book), emphasis mine:

In a molecule, smaller bond angles are formed between electronegative ligands since the central atom, to which the ligands are attached, tends to direct bonding hybrid orbitals of greater p character towards its more electronegative substituents.

The rule itself (is an observation, which) applies to more complex molecules, but can be applied to this situation, too. Oxygen is more electronegative than carbon therefore the bonding hybrid orbitals will have more p-character directed towards oxygen. At carbon itself will be more s-character for the lone pair, which is what we observe for the HOMO, when we apply the polarisation, too.
From this deductive chain one can reasonably explain, that the HOMO of carbon monoxide must be of σ symmetry (two electron donor) and have the largest coefficient at the carbon atom and will therefore rather coordinate there than through the oxygen.

It seems reasonable but slightly awkward to apply Bent's rule here, considering that CO is a diatomic and thus, there are no "ligands" and "central atom".

Very detailed answer but are there references of having $(\sigma ^* 2\text s)^2$ as HOMO?

@Apurvium I don't understand your question. I don't use this numbering and I don't see in what way it would be applied here, i.e. I don't know what your numbering is referring to.

@Martin-マーチン Sigma 2s antibonding molecular orbital

@Apurvium It would be nice if you could write full sentences, that way we can hopefully understand each other. Anyway, I don't differentiate between binding and antibonding, especially not in numbering. It is a biased system and it doesn't make much sense when considering sp-mixing.

OK... My $(\sigma ^∗ 2\text s)^2$ refers to the $2\sigma$ in your last figure.
Also, one more confusion, I think the sp- mixing takes place at atomic orbital level before forming molecular orbitals?

@Martin-マーチン As you commented "... there are still sources out there stating that the HOMO of CO is antibonding in character". So, does the last figure accepted by most of the highly knowledgeable persons in chemistry?

@Apurvium I can only hope so, because it is qualitatively correct in the framework of MO theory. || Don't think about mixing of orbitals in a time manner, there is no before and no after. This is only a mathematical solution. The molecular orbitals do or don't exist just as the atomic orbitals do or don't exist. We can only observe electron density. So it's getting more cloudy. But from all the definitions of 'bonding' I know, the HOMO of CO fits these descriptions.

The electrons in the frontier orbital(s) play a special role for the chemical reactivity. In $\ce{CO}$, the HOMO is the $ 5 \sigma $ orbital (ref: your diagram), and has mainly $\ce{C}$ character. The two electrons in it act like a lone pair on the carbon.
The predominant $\ce{C}$ character of the HOMO accounts for the reactivity via carbon ($\sigma$ donor
Side note: $\ce{CO}$ is also a $\pi ^*$ acceptor due to the LUMO (which is also predominantly $\ce{C}$ in character.
To answer your comment:

So overall, how could I tell exclusively from the MO diagram that HOMO electrons are centered on C and not on O?

Don't think you can. The contribution to MOs is determind by the coefficients in the linear combination. You should see Martin's calculated values in the answer link he provided.

Well, there is a hint that the lower orbitals will generally be centred on the electronegative partner and the higher ones generally on the electropositive partner, but that’s a weak indication at best.

Agreed. It has its uses, say for something like HF where there are very few orbitals overlapping, but in CO it's really difficult (more precisely, I found it difficult) to use "intuition" because the C2s, C2p, O2p all interact significantly. Even the usual adage of "AOs close in energy to the MO contribute more" fails here because the HOMO is very close to O2p energy level.

Not only that, but there are still sources out there stating that the HOMO of CO is antibonding in character. If we can't even get that right by intuition, it is not hard to imagine that we will fail at predicting more complicated features.

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Copper sulphate solution dissolves iron nail?
I tried an experiment of the Galvanic series and put three iron nails (slightly rusty, but brushed) into a copper sulphate solution and waited a few days until the nails get coated with copper.
Strangely something entirely different happened. The nails got almost dissolved, yet quite a bunch of residue added up.
I was not expecting this to happen. Could someone tell me what I did wrong?
Thank you

Displacement reaction occurred, but the copper would not automatically stick to the iron. I reckon that electroplating is needed to make them stick.

I got the instruction from that German page: http://netexperimente.de/chemie/118.html - but now did some research in the English parts of chemistry-pages. So far I couldn't find any information anything else would be required to make the reaction work (for example: https://schoollabs.wordpress.com/2013/02/23/iron-nails-copper-sulphate/ )

Well, the video in the page is curious... it worked in the video?

@bon - which of my writings would you like to have formatted?

None in particular. I'm just putting it there for future reference.

@KennyLau: The video and the page use a different time frame than I did. They mention "short time" and in the video 2h. I used days as I had been curious what would happen. The more the result was unexpected.


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is it possible for covalent compounds to produce colour during flame test?
I have heard it is the metal ions that are the reason for flame colour during a flame test. so in covalent bond, there are no metal ions does that mean covalent compounds don't produce flame colour. if not what is the reason, if yes why doesn't glucose produce colour. Also is it possible to identify whether a substance is ionic, or covalent with just the flame test. I have searched the web but there are no clear explanations.

When a compound is put in a flame, pretty much all bonds are broken, so it makes no difference whether they were ionic or covalent.

Glucose does not produce a distinct color because it is a hydrocarbon, so in combustion looks pretty much like other hydrocarbons. So, somewhere in your question is a conflict between compounds, metals, and how colors are made in the flame.

You have wrong assumption - there's no real connection between degree of covalence and  flame colour. Also most compounds of metallic elements is in fact primarily covalent.

The colors produced in flame tests are the result of electrons becoming excited and returning to their ground states, releasing photons of characteristic colors. Different colors from flame tests result from different energy gaps between the electron shells in different atoms. Since flame tests determine properties of individual atoms, they cannot be used to determine if a compound is ionic or covalent. In fact, bonds are normally broken when a compound is placed into a flame, so it would be nearly impossible to use a flame test to find anything about the bonding nature of a compound. Glucose does not produce a color when it is burned because it just undergoes combustion - more or less the same reaction that is creating the flame in the first place.
On the other hand, there are other ways to determine if a compound is ionic or covalent. Ionic compounds normally have ridiculously high melting points, and if you put an ionic compound into distilled water, it will conduct electricity. There are lots of tools to determine if a compound is covalent or ionic; however, a flame test is not one of them.


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How to calculate the new concentration of a solution after adding more solute and convert it to ppm?
My question is the following:

$2~\mathrm{g}$ table salt is added to $0.5~\mathrm{m^3}$ water whose salt concentration is $10~\mathrm{mg\,L^{-1}}$. Compute the salt concentration of the mixture in ppm.

I am fairly new to chemistry actually this is my first time taking a chemistry course of this level so I'm having quite a hard time. Therefore, if anyone could recommend some websites where I could receive help or even videos, such as khanacademy that would be greatly appreciated!
As for the following question all I know is that $2~\mathrm{g} = 2000~\mathrm{mg}$
and ppm is calculated by:
$$\text{ppm} = \frac{\text{grams of solute}}{\text{grams of solution}}\times  1000000 $$
However, im confused as to how much grams of solute and grams of solution there will be in the mixture after adding the $2~\mathrm{g}$ of salt?

You should start by figuring out how much solute there is in the original solution - this you can find out by multiplying the concentration by volume. It's safe to assume that the volume will not change substantially.

Let's think about it :
You have the formula of the mass concentration $C_m$:
$C_m = \frac{m}{V}$,
where $m$ is the mass in grams of solute and $V$ the volume in liters of the solution.
A first step for you would be to calculate the mass of solute already in the solution : your solution has a volume of $0.5~\mathrm m^3$, which is $V=500~\mathrm L$, with a massic concentration of $C_m=10~\mathrm{mg\cdot L^{-1}}$. Therefore, there is already a mass
$m=C_m~\times~V = 10\cdot10^{-3} \times 500 = 5~\mathrm g$
of salt in your solution.
To this mass, you add $2~\mathrm g$ of salt, so the mass of solute in your solution is now $m=5~\mathrm g + 2~\mathrm g = 7~\mathrm g$.
So, at end you have a mass $m=7~\mathrm g$ of salt.
I let you compute the mass of solution, knowing that $1~\mathrm L$ of water weights $1~\mathrm{kg}$.

Thank you so much for your help! Also do you happen to have any suggestions for some resources that could help me better understand this topic. As for the mass of the solution I get 500 000 grams? does this make sense? As my final answer I get 14 ppm.

For the mass of the solution, you indeed have $500~L = 500~kg=500\cdot 10^{3}~g$, and the value in ppm seems right. As for the ressources I can't really help you. I suggest you do as many exercices as you can from the ressources available in you course. KhanAcademy also looks like a good ressource for you. Also, please consider accepting my answer if it helped you.


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Naming of 2-methylpropane
I can't understand why 2-methylpropane has/needs the 2 in front of it?
It appears to me that if you were to add the methyl group at the first or last carbon on the chain then you get plain old butane and so defining 2-methylpropane is unnecessary when it could just be methylpropane.
Have I missed something or is this just a convention? Apologies if it's a really simple question; it's my first ever time studying organic chemistry.

Related (concerning 2-methylpropanoic acid): IUPAC rules on omission of locants in carboxylic acid

You are correct. It is unnecessary to include the 2 in order to provide an unambiguous name. This compound is also often called isobutane which is an older name that is still in common usage.

Are there any relevant IUPAC rules on this?

@matt_black According Subsection to P-61.2.1 of the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the preferred IUPAC name (PIN) is "2-methylpropane". The name "isobutane" is no longer recommended.

Yeah. You're right. If we add methyl group(-CH3) to first carbon(or last carbon) on the chain of 2-methylpropane(isobutane) you would get butane.
The below figure is butane.
The below figure is isobutane or 2-methylpropane.


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Are there any other elements that get less dense in their solid state?
I was learning about how water, because of its hydrogen bonds, actually gets less dense as it goes into its solid state - I was just wondering, what other elements do this? Are they similar to water?
P.S. I want to make sure that I understand why ice is less dense than water in its liquid state, so please correct me if I am wrong.
In water, the hydrogen bonds have the ability to move around and bounce with other hydrogen bonds. They also have the ability to break and reform. Their ability to move around and bounce around allow them to be more 'closely knit'. But, as they get cooler, they, the molecules, loose their kinetic energy and thus stop moving around and become more framed and rigid. They loose the ability to come closely together. And thus, ice is less dense.
What I don't understand also, is why making water cold makes the hydrogen bonds spread out. Why don't they just freeze in the position they are in, instead of spreading out?

Elements do not form H-bonds.  E.g. silicon, gallium, germanium, bismuth, and plutonium have higher liquid than solid density at standard pressure.

And antimony as well. It really isn't that unusual, but many people seem to think water is unique. It isn't.

Not an element, but "type metal", an alloy of lead, antimony, and tin, uses that property of expanding on freezing to ensure that the type fills the mold without leaving voids.

Without checking it I would say you could look for metals and elements that do not form close-packings. Usually metals prefer known structures in their solid state like the Cu, Mg and W-type. But there are some like Ga, In, Mn, Po, ... that do not form close or dense packings but have a considerably lower density. I can imagine this changes when they melt and may perhaps lead to a higher density in their molten state. But that is just my thoughts without further verification by literature or so.

We can add lithium to the list, at high pressure. The fcc phase tends away from a close-packed structure towards a more open structure, which may contract as it melts, because of the separation of electronic charge from the ionic cores.

@JonCuster Are the density measures for liquid and solid plutonium and antimony false in Wikipedia?

Those are not at the same temperature - the solid is given at room temperature and the liquid at the melting temperature. And in the case of Pu, there are 6 allotropes between room temperature and the melt to further complicate things, each with a density change at the solid-solid phase transition. The solid phase at melt is bcc, which is not a close packed crystal structure.

Your terminology is a little off: water isn't an element.
There isn't really a simple single explanation of why some solids are less dense than their associated liquids. The general explanation (which doesn't really explain much) is that the solid has a structure and, sometimes, that structure takes us more space than the average structure that occurs in the liquid.
In water, ice has a definite structure where each oxygen sits at the approximate centre of a tetrahedron with two bonds to hydrogen and two hydrogen bonds to the hydrogens of other water molecules. In liquid water the structure is far more fluid and averages out to be slightly more dense that the (more ordered) solid. Exactly why this is true in this case requires a lot more than could fit into an answer here. If water was frozen extremely quickly you might be able to freeze the structure in the liquid; normal cooling allows enough time for the bonding to become more organised.
But the phenomenon isn't unique to water, though the specifics will be very different for other substances. Among elements Gallium, for example, also expands on freezing (don't store it in glass bottles: see this other answer here).
Other examples of elements are germanium and silicon. Silica is also less dense as a solid than as a liquid. In all cases the substances form a relatively ordered crystal structure which just happens to be less dense than the (less ordered) liquid phases.

If we cool running water it will even reach below zero temperatures but remains liquid. What will the density be in this case?


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Mate & Carbonated water reaction
I know that when adding carbonated water with Yerba Mate leaves there is a reaction that removes the carbonation. (Empirical results)
I wonder what actually causes such reaction but looking in Scholar even for similar reaction such as soda + mate wasn't very productive.
What is the reaction that undergoes there? Is mate just a catalyst ?

I don't know much about chemistry, but I am trying to come up with an okay question.  At least try to indicate how to improve it rather than just down-voting it.  :(

The carbonate is supersaturated in solution and unstable. That is why if you just leave a bottle of carbonated water open it will go flat. I'm guessing that the Yerba Mate plant goo has particulates in it which act as nucleating sites for bubbles to form. With a lot of bubbles forming you'll lose the fizz quickly.

Adding like anything to carbonated water accelerates loss of CO2 http://chemistry.stackexchange.com/questions/42457/what-does-nutella-add-to-the-cokementos-reaction?s=1|1.0737 http://chemistry.stackexchange.com/questions/51167/are-the-ingredients-of-pepsi-in-different-countries-different-enough-to-affect?s=2|1.0200 http://chemistry.stackexchange.com/questions/36977/how-does-the-kreosan-coca-cola-rocket-work?s=4|0.1250


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Can we make a salt using metal oxide?
Get a metal like sodium and then you burn it to make sodium oxide because burning something is adding oxygen. Then you add the sodium oxide to an acid like hydrochloric acid. That should make a salt (sodium chloride) and water. Evaporate or filter the water away and you are just left with the salt itself.
Is this possible to do or am I just completely wrong?

Why do you even think it could be wrong?

You can't filter anything out of a solution. Other than that, it's a feasible plan.

Your description is almost correct - note that when sodium metal burns in air, it forms primarily sodium peroxide in addition to sodium oxide. Sodium peroxide will react with hydrochloric acid to give sodium chloride and hydrogen peroxide. Heating will decompose that to water plus oxygen. You should still be able to get the sodium chloride product that you expected.

Fizzle Nizzle's idea is nearly correct. First, all metals can be oxidized and form oxides. Second, all oxides can react with acids like $\ce{HCl}$. And a salt (like a metallic chloride) is always obtained in solution. But ! But it is not always possible to save the obtained chloride out of this solution. By evaporating the solution, the salt is often hydrolyzed, and by heating it, its solution sometimes may undergo the reverse reaction of its formation. This is particularly true for metals oxidized to the oxidation number +III or higher.
Aluminium is a good example of this behavior. Aluminium can be oxidized into aluminium oxide $\ce{Al2O3}$ (and into hydroxide $\ce{Al(OH)3}$ if water is present). These oxides or hydroxides can react with $\ce{HCl}$ solutions to get a solution of $\ce{Al(III)}$ ions, the formula of which can be written as $\ce{Al^{3+}}$ or $\ce{[Al^{3+}(H2O)_n]}$. But its evaporation will not produce $\ce{AlCl3·nH2O}$, whatever the value of $n$. Its evaporation will have the ion $\ce{Al^{3+}}$ get hydrolyzed (= react with water) and produce successfully $\ce{AlCl2(OH)}$, $\ce{AlCl(OH)2}$, then $\ce{Al(OH)3}$ each time with more $\ce{HCl}$ relieved in the vapor, in equations like : $$\ce{Al^{3+}+ +3 Cl^- + H2O -> AlCl2(OH)(s) + HCl(g)}$$ $$\ce{AlCl2OH + H2O -> AlCl(OH)2(s) + HCl(g)}$$ $$\ce{AlCl(OH)2 + H2O -> Al(OH)3(s) + HClg)}$$ If the non hydrolyzed compound $\ce{AlCl3}$ is to be obtained, this must be done by another way, and not by neutralization of an hydroxide by an acid. It may be done by direct syntheses, and for example by the reaction $$\ce{2Al + 3 Cl2 -> 2 AlCl3}$$ or by indirect ways avoiding the presence of water like : $$\ce{Al2O3 + 3 C + 3 Cl2 -> 2 AlCl3 + 3 CO}$$

You are correct. Metal oxides will react with acids to form water and the corresponding salt. It's the same with metal hydroxides, they also react with an acid to form water and the salt.


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How do i make an molecular orbital diagram step by step?
My professor explained me, but i'm still unable to do a correct molecular orbital, for a diatomic molecule

http://chemistry.stackexchange.com/questions/50279/how-should-i-construct-the-orbitals-of-dilithium-using-only-pen-and-paper

http://chemistry.stackexchange.com/questions/15205/energetic-placement-of-atomic-orbitals-in-the-hcl-molecular-orbital-diagram


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Difference between polarizability and dipole moment?
In my head, any molecule that is very polarisable, has a dipole moment. Is this true? I am very confused and can't seem to find anything that helps my thought, and I can't seem to distinguish between the two without both intertwining.

No, it's rather the other way round, typical examples of highly polarisable molecules don't have dipole moment.

@Mithoron - that's actually quite misleading. There are many molecules with both large polarizability and large dipole moment.

First, the quick definitions. A dipole moment ($\mu$) implies a permanent separation of charge across a distance. For example, HF has a positive charge near the hydrogen atom and a negative charge near the fluorine:

Instead, polarizability $\alpha$ indicates the degree to which electrons (and thus charge) re-arrange in an applied electric field. Indeed, when a polarizable molecule experiences any type of electrostatics, there is an induced dipole moment:
$ \mu_{induced} = \alpha \vec{E} $
Now this induced dipole moment (from the polarizability) is completely distinct from the permanent dipole moment. For example, benzene is non-polar (no net dipole moment) but highly polarizable. In the HF case, I could find some large applied field $\vec{E}$ that pushes electrons from the F to H.
Here are my mental analogies:

A dipole moment is like a hill or tilted plane - water (charge) moves "downhill" and it's always there.
Polarizability is like hitting the water with a cannonball. I can drive water from one side of a bathtub to the other!

The two properties can be related, but it's possible to find highly polarizable molecules with no dipole moment - particularly if they are symmetric. Here's a plot with ~80,000 molecules from PQR - note that all combinations occur.

I'll mention that it's a bit harder to find molecules with >40 Debye dipole moment and large polarizability - mostly because they'd be huge.

If it is permanent dipole moment, they should call it permanent dipole moment. When talking about intermolecular forces, induced dipole and permanent dipole are distinguished and give me the impress that dipole moment includes both.

A dipole moment is a mathematical calculation relating to the unequal distribution of charge within a compound. In other words, the higher a compound's dipole moment, the more polar the compound is. On the other hand, polarizability is the tendency for a compound to form a dipole when confronted with an external electric field. In other words, a compound's polarizability is its tendency to become polarized while a compound's dipole moment is a measurement of how polarized a compound is in its ground state.


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What are some of the more easily manipulated residues on an enzyme?
I am trying to attach a cross-linker to an enzyme and I am looking for candidates for reaction sites. I have several amino acids that are present on the surface of the enzyme (hydrophilic) but I am hoping someone can give some tips for amino acids that are easier to attach linkers to.
I originally was planning on attaching a linker to lysine because it is among the more reactive residues but my enzyme lacks a sufficient number of lysine groups.
Could any one give me some general advice on which amino acids would be better to investigate first? I know the specific residue I choose will depend mostly on the structure of the enzyme, but if you could give me a suggestion of which amino acids are typically easier to manipulate (ie. Lysine > glycine) I would greatly appreciate it.

I'm assuming the OP concerns chemical means of modifying residues. Multiple protein residues other than Lysine may be labeled with an adduct. Here are some examples:

Histidine residues may be labeled using an epoxide attached to a ligand. (Chen et al., 2003)

Cysteine residues may be labeled with a nitrile derivative where the sulhydryl acts as the nucleophile. (Obella et al., 2007)

Arginine residues may labeled using a derivative of the naturally occurring methylglyoxal. (Oya et al., 1999)

Carboxylate functionality such as in Aspartate and Glutamate residues as well as the C-terminus may be labeled via the formation of an amide linker using a carbodiimide such as N-ethyl-3-N',N'-dimethylaminopropylcarbodiimide (EDC). (Gilles et al., 1999)

The above list of residues is not comprehensive, nor are the labeling agents. I recommend checking out the reviews by Basle et al. (2010) and Chen and Wu (2016).
References
Basle, E, Joubert, N, Pucheault, M (2010) "Protein Chemical Modification on Endogenous Amino Acids" Chemistry & Biology 17:213-227. DOI: 10.1016/j.chembiol.2010.02.008
Chen, X and Wu, Y-W (2016) "Selective chemical labeling of proteins" Org. Biomol. Chem. 14:5417-5439. DOI: 10.1039/C6OB00126B
Gilles, MA, Hudson, AQ, Borders, CL (1990) "Stability of water-soluble carbodiimides in aqueous solution" Anal. Biochem. 184:244-248. DOI: 10.1016/0003-2697(90)90675-Y
Gong Chen, Alexander Heim, Doris Riether, Dominic Yee, Yelena Milgrom, Mary Ann Gawinowicz, and Dalibor Sames (2003) "Reactivity of Functional Groups on the Protein Surface: Development of Epoxide Probes for Protein Labeling" J. Am. Chem. Soc. 125:8130–8133. DOI: 10.1021/ja034287m
Oballa, RM, Truchon, JF, Bayly, CI, Chauret, N, Day, S, Crane, S, et al. (2007) "A generally applicable method for assessing the electrophilicity and reactivity of diverse nitrile-containing compounds" Bioorg. Med. Chem. Lett. 17:998–1002. DOI: 10.1016/j.bmcl.2006.11.044
Oya, T, Hattori, N, Mizuno, Y, Miyata, S, Maeda, S, Osawa, T, Uchida, K (1999) "Methylglyoxal modification of protein. Chemical and immunochemical characterization of methylglyoxal-arginine adducts" J. Biol. Chem. 274:18492-18502. DOI: 10.1074/jbc.274.26.18492

If you try to modify one of the 20 naturally occurring amino acid, you have to make sure there is at least one on the surface, and that you don't have too many. In some instances, active site cysteines, serines, or histidines are more reactive than the same residues elsewhere on the surface (e.g. when they have a different $\mathrm{p}K_\mathrm{a}$ because of their surrounding).

@KarstenTheis Indeed these are great points. I think the OP noted their concern about the lack of sufficient number of lysine groups. I inferred from this that their intent wasn't about targeting one specific residue, but a residue type that is sufficiently abundant for their experiments. I suspect the suggestion of using the EDC reagent may be reasonable in beginning to answer their question.


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Strong Acid/Strong Base Titration
Bromothymol Blue is an indicator that turns yellow in acid, blue in base.
If I were to titrate NaOH with HCl, what color should I look for at the equivalence point? Should I titrate until the solution turns from blue to yellow or will there be an intermediate colour of green?
I know the salt produced (NaCl) has a pH of 7.

http://antoine.frostburg.edu/chem/senese/101/acidbase/indicators.shtml
This is a link to a webpage that talks all about pH indicators and the different pH ranges if you're looking for more info.
When it comes to your question, the color should go from yellow to green because in your breaker is where you add the indicator fluid and HCl acid and you titrate that with the NaOH base. If you titrate further than your green color you have now made your solution more basic.


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Electrolysis of dilute and concentrated sodium chloride and the Nernst equation
Interested to know if there is a critical concentration at which $\ce{Cl-}$ will oxidise over $\ce{H2O}$ and if this can be determined using the Nernst equation.
That is, obviously if $[\ce{Cl-}]>3.0\ \mathrm{mol/l}$ then it will oxidise, but if $[\ce{Cl-}]<1.0\ \mathrm{mol/l}$ it will not, since
$$\begin{alignat}{2}
\ce{2H2O(l) \;&<=> 4H+(aq) + O2(g) +4e-}\quad&&E^\circ=-1.23\ \mathrm V\\[6pt]
\ce{2Cl- (aq) \;&<=> Cl2(g) + 2e-}\quad&&E^\circ=-1.36\ \mathrm V
\end{alignat}$$
However, in attempting to use the Nernst equation to determine the exact concentration at which $\ce{Cl-}$ will oxidise over $\ce{H2O}$, I get an obviously incorrect number,
$$\begin{align}
1.23\ \mathrm V&=E^\circ-\frac{2.569\times10^{-2}\ \mathrm V}{2}\ln\left( {\frac{\left[\ce{Cl- (aq)}\right]^2}{\left[\ce{Cl2(g)}\right]}}\right)\\
\frac{0.13\times2}{2.569\times10^{-2}}&=\ln\left[\ce{Cl- (aq)}\right]^2\\
\ln\left[\ce{Cl- (aq)}\right]&=\frac{0.13}{2.569\times10^{-2}}\\
\left[\ce{Cl- (aq)}\right]&=\mathrm e^\frac{0.13}{2.569\times10^{-2}}\\
&=157.643\gg3
\end{align}$$
In this above relationship, I have set the $E$ value to that of $\ce{H2O}$'s reduction potential to determine the concentration of chlorine ions required to get the same voltage. But obviously my number is too large!

The problem with your approach is that the Nernst equation is defined: $$E=E^\circ - \dfrac{RT}{nF}\ln Q$$ and the reaction quotient $Q$ can only be defined for complete reactions, not just half reactions. You need to define a reduction half reaction for your system - $\left(\ce{2H+ + 2e- -> H2}\right)$ might be convenient.  However, then you have additional species to worry about in $Q$.

You actually get the same result when you consider the reduction of water with each of the above-mentioned oxidation equations. Is taking a theoretical approach here invalid as it does not consider practical phenomena such as overvoltage, which would increase the 1.23 V value to make the required chlorine concentration smaller?


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Molecular Orbitals of Hydrogen Fluoride (HF)
This source states that the three s-orbitals of hydrogen and fluorine interact to form three new molecular orbitals, while other sources say that the 2s orbital is non-bonding.
Which one is more correct? Also, if they indeed form three new molecular orbitals, how do they look like?

Bearing in mind the energy and symmetry conditions, in the case of $\ce{HF}$ one can construct MOs using the $1s$ AO on H and the $2s$ and $2p$ AOs on F.

In general, the contribution to MOs is determined by the coefficients in the linear combination.
Here, one observes that the $1s$ electrons are almost completely localized on the $\ce{F}$ atom. Also, $1 \pi$ electrons are completely localised on the $\ce{F}$ atom because the $2p_x$ and $2p_y$ orbitals on F have a zero net overlap with the $1s$ orbital on $\ce{H}$.
Electrons in MOs localized on a single atom are referred to as nonbonding electrons.
Also, I would note that the $3 \sigma$  MO has less bonding character and the $4 \sigma^*$ MO has less anti-bonding character.
Note that the total bond order is approximately one because the $3 \sigma$ MO is largely localized on the F atom, the $3 \sigma$ MO is not totally bonding, and the $1\pi$ MOs are completely localized on the F atom.
On account of fluorine being a more electronegative atom, in the $2 \sigma$ bonding orbital the electron density is much greater on the more electronegative fluorine than on the hydrogen. However, in the anti-bonding $4 \sigma^*$ orbital, this polarity is reversed.
Caveat: The preceding paragraph may support your intuition and may be right in a few simple cases, but I wouldn't rely on it too heavily.
Given below is a diagram showing $2 \sigma$ , $3 \sigma$ and $1 \pi$ MOs in HF


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Cleavage of benzyl group from alkane?
Is there a method to selectively cleave the C-C bond between a benzyl group (-CH2Ph) and a linear or cyclic alkane? Can this be done under standard conditions with somewhat widely available reagents? Can the benzyl group be forced out if it is under significant steric clash when it is attached?

Are you (formally) trying to replace $\ce{-CH2Ph}$ with $\ce{H}$?

The following is just a mind game:

Oxidize at the benzylic position to get an arylketone using oxone
Convert the ketone to an ester by Baeyer-Villiger reaction
Cleave the ester
Run a Barton-McCombie deoxygenation on the resulting alkanol

Please note that I havent't done it and I'm not planning to. I'd rather try to avoid the benzyl group in the starting material if possible.


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How to determine number of proton environments in these compounds?

Is my drawing correct? Thanks in advance!

When you say "peaks", you count a doublet as one, right? You should call it "signal" then. In that case, you are right, there are five different chemical environments for protons in the ortho and meta isomers, and three in para.

Ok! lemme edit my question @Karl

@Karl is it also true that the indane-1 3-dione derivative has 5 signals instead of 3? as there is no vertical line of symmetry

Correct, the left and right side are not chemically equivalent. But the signals will again be complex and overlapping. I don't think anyone can predict any of the spectra by hand. Those are tough cases.

In a word, no, your drawings are not correct. And incorrect enough for me to have a rant about NMR teaching at the introductory level.
Some background knowledge before tackling your molecules.

'Environments' is a vague term, often misused and misunderstood, unless you specify what type of environment: chemical or magnetic. As an argument about chemical environments alone, you need only look at the symmetry of the molecule to make a judgement. This can be understood and determined by someone who has never heard of NMR. It might also be an appropriate assessment if you do not intend to make any interpretation of the splittings of peaks from coupling. If you want to look at it from an NMR perspective (and you ARE asking as an NMR question), then you have to consider magnetic environments also. In all four molecules you have drawn, every single proton is in a separate environment.
Equivalence of the proton environments can be assessed on whether a particular nucleus is identical to another site, based on a substitution test. If you substitute, in turn, one of the 1H nuclei in the molecule below, you end up with identical molecules, therefore the protons are chemically equivalent. They will have the same chemical shift.

An additional element of equivalence needs to be considered also, and this relates to magnetic equivalence. Nuclei that are chemically equivalent may not couple to other nuclei in an identical manner, and these nuclei are therefore magnetically non-equivalent. Although they hgave the same chemical shift, their coupling patterns are very complicated, and usually involve coupling to each other.

In the case above, the two protons shown in red are in different environments; the same chemical environment, but different magnetic environments. Without understanding this key concept, you would not be able to rationalise the appearance of the spectrum, which actually looks like:

Some standard couplings that you should commit to memory for aromatic rings:

3JH-H ortho couplings are typically 7-8 Hz
4JH-H meta couplings are typically 2-3Hz
5JH-H para couplings are typically <1Hz, and are not commonly observed.

Splittings in a spectrum can be reported as either

observed splitting pattern so a doublet of doublets with similar couplings is likely to be reported as a triplet
expected splittings based on molecular structure. So a reported doublet of doublets with similar couplings may look like a triplet.

Although both methods are used in the literature, the strong preference is for the latter, as it provides structural information, and a means for justifying the appearance of the spectrum. It is much more difficult to start from the appearance of a spectrum and work backwards to determine/justify structure. A triplet coupling implies coupling to 2 identical nuclei

...And you're back in the room....
So, coming to your molecules......

the 1,2 disubstituted and 1,3 disubstituted case

You can work through predicting what you would expect based on known typical couplings. Do this for each proton. A couple of workings are shown, and the theoretical expectation of these peaks is shown.

Now, unfortunately, this particular group of substituents are both strongly electron withdrawing, and what you would probably get is compounded complications though strong coupling as the chemical shift differences are not sufficient enough to allow simple first order analysis ie a dog's breakfast, as shown below.

Having spent time determining the number of chemical and magnetic environments is pointless because the spectrum looks so complicated that it looks nothing like you would expect, although this is hardly presumed knowledge for NMR101.
A more suitable choice of substituents, -NH2 and -COOH or -CH3 and -COOH, would have provided a much better example.

1,4 disubstituted and the indanedione thingy

There are two chemical environments directly on the ring, but they are magnetically non-equivalent. So again, time spent trying to predict what the splitting might look like leads to frustration because we get a second order AA'BB' system that, although some might try to convince themselves looks like a doublet, is in fact a much more complicated spectrum (like in the case for furan above). Both give textbook splitting patterns that indicate a 1,4-disubstituted or symmetrical 1,2-disubstituted rings, but nothing like what you might expect if you trying to determine expected splittings for the first time.
And so the correct answer for how many proton 'environments' in these compounds is:

.End Rant.

You'll see no triplets, because there are no two chemically equivalent protons in ortho and meta.
I guess you might be right in the case of the para-isomer, because the 5J coupling through the ring might be too weak to see (it is not zero, 3/5 are not magnetically equivalent, neither are 2/6).
In the meta isomer you will see some 4J coupling between position 2 and 4/6, and 5 will definitely show a dd signal.
ortho: 3+6 will each be dd (3J+4J), and 4+5 will each be ddd (2x3J+4J). Plus you'll see a lot of second order coupling, because the CS difference is so small. That means the dd will not look like a regular dd.
("dd" means doublet of doublets, i.e. when a proton has two different neighbours. dd can look like a triplet, when the two coupling constants are (nearly) identical.)

How about low resolution nmr? Sorry I'm still a high school student I don't quite understand about 4J coupling and dd signal...

At 60 MHz or so, second order coupling will likely melt all peaks into one multiplet. The normal rules for doublets, triplets etc. only work if the signals of each coupling group are well separated in the spectrum. 0.5 ppm are only 30 Hz at 60 MHz. That is not a lot more than the regular 3J coupling in the aromatic ring.

Can you tell me how many proton environments there are for each isomer please? @Karl

That you can work out yourself, im sure? Hint: Only the para isomer is symmetric.

So is it 5 for ortho meta isomers and 3 for para isomer?:) @Karl

I also counted the -COOH:p @Karl

oops. you're right.

So is it 4 for ortho meta isomers and 2 for para isomer? @Karl


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Does reaction quotient Q take into account pure solids and liquids?
I understand that the equilibrium constant does not take into account the concentrations/ pressures of pure solids and liquids. How about the reaction quotient?
In the following reversible reaction, am I right to say that since water is a pure liquid, it is not reflected in the reaction quotient, hence removing water (without changing the temperature) will not shift the equation left/right after applying Le Chatelier's Principle?
$$\ce{H3PO4 (aq) + H2O <=> H3O+ (aq) + H2PO4- (aq)}$$

You're wrong, it will.

@Mithoron: Thank you so much for your reply. Could you help me understand why? Do we actually have to include pure solids/liquids in the reaction quotient Q? Or do do we not use reaction quotient in applying Le Chatelier's Principle? Grateful for your help!

They shouldn't appear in the reaction quotient, indeed. However, by removing water, you concentrate the solution, which leads to a shift of the equilibrium.

It isn't that solids don't appear in equilibrium constants and/or reaction quotients. They are simply assumed to have an activity equal to the value of 1! Hence people don't show them, but they should. I think it is important to understand and remember this assumption.
Also, equilibrium constants and reaction quotients are fundamentally a ratio of activities! The use of concentration and/or pressures in equilibrium constants/ reaction quotients is the results of approximations in themselves and this simplicity permeates chemistry and should be stopped or at least respected.
for the reaction you show, the equilibrium constant is
$$K_\mathrm{eq} = \frac{a_\ce{H2PO4-}a_\ce{H3O+}}{a_\ce{H3PO4}a_\ce{H2O}}$$
See the Chemistry LibreTexts for more details about solid and liquid activities and equilibrium constants.


text:
Does water evaporated from sea water contains any salts in it?
I want to know whether water evaporated from sea water is liable for drinking or not?

Related: Vaporized sodium hydroxide?

Water evaporated from sea water is water vapour. After condensation it is just pure water and no dissolved salts. It should be okay for drinking !

Distilled water is pure water, right? Evaporation + condensation = distillation. Thus your condensed water must be pure, salt-free and good for drinking purposes. I am assuming, of course, that the pieces of equipment one uses to evaporate and condense the H2O are exempt from impurities, so that during its transit through the condenser the water does not get contaminated by such undesired substances, whatever they may be.
As a side note, distilled water is tasteless and will not quench one's thirst as mineral water would.

You'd have to desalinate it to get rid of the salts.
https://en.wikipedia.org/wiki/Desalination


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Latent Heat and Kinetic Energy
The kinetic theory says that temperature is the measure of the average kinetic energy. That would mean then: The temperature rises with the rise in kinetic energy. We know that when matter changes state 'latent heat' helps in 'breaking the bonds' at the transition phase and that this 'extra heat' is not noticed on a thermometer. But energy can neither be made nor destroyed. Taking that into consideration latent heat has to convert into some form of energy. Since temperature is not rising, it's not kinetic energy it's getting converted to. What energy does it get converted to?
Note: I already know why the temperature does not change.
What I want to know is what the latent heat energy gets converted to.

Possible duplicate of During a phase change in matter, why doesn't the temperature change?

@Mithoron I recommend not closing as duplicate. This question is asking about the form the absorbed energy takes, while the linked question asks why the temperature doesn't change.  The answer is more or less the same, but the question is distinct.

Mainly potential energy

Latent heat isn't heat; it is energy released or absorbed during a constant temperature phase change.
The kinetic theory of gases has little to say about phase changes. Phase changes, involve liquids or solids in general and most often don't involve gases.
It's an error of comprehension (or articulation) to claim a phase change releases latent heat. If I slowly accelerate a tank of gas, its kinetic energy increases with no increase in temperature. Temperature is a measure of kinetic energy, not the only measure of kinetic energy. (We can notice this from the fact that most solids at their melting point do not increase in temperature when heat energy is added to them.)
In a more general sense, latent energy is the energy difference between the two phases. So where it goes to, or comes from, is the conversion of one phase to the other. The assumption here is that one phase is lower in potential energy compared to the other. You may know that entropy is defined in terms of heat and temperature. In classical thermodynamics energy and entropy are two fundamental concepts, they are on equal footing. But in statistical thermodynamics, energy is fundamental, while entropy can be derived from the ensemble of possible states that a system can be in. Whoops! Too much!
In the case of ice melting, the intermolecular bonds (mostly hydrogen bonding) are 'broken' (which means there are a lot fewer of them and they are (on average) weaker). The energy that was in the form of potential energy of the bonds, now becomes "kinetic" energy as chemists understand the term in this context.
In this context (as contrasted with the kinetic energy of a bullet or a ballistic electron), the kinetic energy includes all of the energy of translation, rotation and vibration: the rotation of a molecule doesn't change its speed, nor do the movements of the two hydrogen atoms with respect to the oxygen atom - their bond to the central O atom can be looked at as a (coil) spring and they can stretch and bend that spring. Faster and greater spring movement occurs during melting. (The rotation of the H atoms also occurs but the energy required to spin a H atom about its bond is very small as bond energies go, so is generally neglected with little loss of accuracy; this wouldn't be true for bonds between larger groups, where rotation around the bond could require quite a bit of energy). So, in the isothermal melting of ice, the place where the energy is going is in these intramolecular movements. (intra = inside). And in the isothermal freezing of water, the place where the energy is coming from is these same intramolecular motions.
There is no one unique definition of kinetic energy (K.E.) and another for potential energy (P.E.). In a group of atoms or molecules, the 'potential' energy is 'internal' and kinetic is 'external' and yet if one focuses in on a single molecule, some of its potential energy is due to its movement (or movements of its parts), so what is and what is not kinetic energy or p potential energy may change with the context. In fact, the basic laws of energy conservation may be K.E. + P.E = constant or K.E. = P.E. (which is often expressed as K.E. - P.E. = 0. These two approaches are embodied in Hamiltonian and LaGrangian Mechanics which are the two related but complimentary ways modern physics is expressed. The LaGrangian way is L = T - V (where T is K.E. and V is P.E.) and the Hamiltonian way is H = T + V. These are used in both classical Physics and in quantum Physics.

This helps me. I was looking for a description of the "coil" nature of the intramolecular bond angles in water as an explanation for how its temperature doesn't rise or fall when freezing or melting. I posted my question on the physics site, but have had limited success there.

Is it the case that this "ensemble of energy states a system can be in" is described by the Shrodinger equation?

Experimentally heat is absorbed to melt a solid or to vaporise a liquid and as the temperature remains constant at the melting or boiling temperature, the entropy must be increasing. If the entropy change from one phase to the other is $\Delta S_{1,2} = S_2 - S_1 $ and the heat absorbed (i.e. enthalpy or latent* heat) is positive $\Delta H_{1,2}$ in transforming phase 1 to phase 2, the entropy change is positive,  $\Delta S_{1,2} = \Delta H_{1,2}/T$.
Any solid or liquid is held together by intermolecular potential energy,  in general called van-der-Waals potentials (ion-dipole,  dipole - dipole etc).  Starting from a temperature just below the melting point as energy is added the molecules absorb this both as kinetic and potential energy (vibration and rotation quantum numbers increase) both internally (normal mode vibrations & rotations)  and also as motion in the intermolecular potential.  The potential interaction between molecules is unchanged as heat is added, but the molecules start to gain enough energy to overcome the well formed by the potential and that holds them in place.
There is now an increase in entropy as there are more energy levels that the molecules can occupy and so more ways of occupying them, i.e the  number of possible configurations $\Omega$ (or complexions) increases. Entropy is proportional to the number of these as $S=k_B\ln(\Omega)$.
Another form of the increase in entropy is that of position, which is strictly fixed in a solid, but exchange of a molecule’s position occurs all the time in a liquid. Positional exchange largely accounts for the entropy increase in atomic solid-liquid transitions, $\approx 8~ \pu{J K^{-1}mol^{-1}}$ e.g. solid - liquid argon.
Eventually there is enough energy that molecules start to escape from the intermolecular potential and so the phase changes.
Generally the process of initiating a phase change, say, freezing relies on nucleation, then growth of the nucleus. If this does not occur supercooling or superheating is observed. In supercooling, which is common, water for example, can still be liquid many degrees below normal freezing point. Part of the effect of overcoming the intermolecular potential is to form a nucleation/seed site, say a small region of vapour inside a liquid or a region of crystal inside a liquid. Next this region will either grow and form some bigger region or shrink to nothing in size.
The nucleus/seed will not grow spontaneously due to the relatively large surface energy which is relatively larger when a nuclei is small, however, a fluctuation in the structure due to some random accumulation of thermal energy may increase its size and so it now becomes stable. The competition is between the negative free energy gain from the volume of the new phase 2 being lower than that of phase 1, and the loss in free energy due to the positive surface energy. As the radius of the nucleus/seed increases the volume term wins out and growth is now spontaneous.
Finally, the average translational kinetic energy of a molecule is $3k_BT/2$ irrespective of whether the molecule is in the gas, liquid or solid phase. In the liquid or the solid the motion is merely restricted to a narrower range around the minimum of the intermolecular potential, than it is in the gas phase. Thus the notion that it is the kinetic energy that molecules acquire is the reason for molecules to go into the vapour phase is incorrect.

The term ‘latent heat’ of vaporisation or fusion is rarely used nowadays  and instead enthalpy of vaporisation or fusion are used. ‘Fusion’ is an old -fashioned word for melting.


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Detection of NaF levels in aqueous solution also containing NaCl
Is it possible to determine the concentration of sodium fluoride, more specifically fluoride ions, in a solution that also contains sodium chloride?
I have looked into using an ion-specific electrode, but this seems very expensive.

Depending on the equipment you have available, you could buy some $\ce{CaCl2}$ de-icer, dissolve it in water, and add it to a known volume of your solution.  Highly insoluble $\ce{CaF2}$ will precipitate out; keep adding the $\ce{CaCl2}$ solution until no more precipitation is observed.
You can then filter the precipitated fluoride from solution, rinse and dry it, and weigh it.  The original fluoride concentration can then be back-calculated based upon the mass of obtained $\ce{CaF2}$, the percent by weight of fluoride in $\ce{CaF2}$, and the original solution volume.


text:
IUPAC name for citric acid
I wonder what the name is of the following structure is:

I think it's something like 3-hyrodxypentanoic diacid
However, I wasn't able to find how to name a side chain if it contains a $\ce{COOH}$ group.
Maybe I'm completely wrong, I hope someone can tell me how to name this structure.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the order of seniority of classes in decreasing order of seniority is as follows.

(…)
  7. acids
  (…)
  17. hydroxy compounds (includes alcohols and phenols)
  (…)

Thus, the carboxylic acid group has priority to be cited as a suffix.
Systematic names of carboxylic acids are generally formed using the suffix ‘oic acid’ or ‘carboxylic acid’ or the prefix ‘carboxy’.
The relevant rule in this case reads as follows:

P-65.1.2.2.1 If an unbranched chain is linked to more than two carboxy groups, all
  carboxy groups are named from the parent hydride by substitutive use of the suffix
  ‘carboxylic acid’, preceded by the appropriate numerical prefix ‘tri’, ‘tetra’ etc. and
  appropriate locants.

Note that the suffix ‘carboxylic acid’ expresses the substituent group $\ce{-COOH}$. Thus, the name of unsubstituted tricarboxylic acid corresponding to the compound given in the question is ‘propane-1,2,3-tricarboxylic acid’.

Therefore, the preferred IUPAC name (PIN) of the complete compound given in the question is ‘2-hydroxypropane-1,2,3-tricarboxylic acid’.

However, according to Subsection P-65.1.1.2.3, the name ‘citric acid’ is also retained for general nomenclature.


text:
Formal charges can mislead when writing arrow-pushing mechanisms
We know opposing charges attract and like charges repel. By this logic then, in the two examples on the left, the curved arrows would be show the formation of an oxygen-oxygen bond. However, these mechanisms are incorrect, and instead the mechanisms on the right take place.

Can anyone explain this with some more examples?

Because they depict different reactions? I'm slightly confused as to what you are asking about. Can you clarify what is confusing you?

@bon I am simply asking that why 1st reaction is wrong i.e why does oxygen atom does not attack on oxygen atom .

It would cause hypercoordination and oxygen isn't prone to this.

All those can be explain using with quantum mechanics. If you are able to follow it I can explain it to you in few days (my timetable is full). And there is no "attraction" it is a misunderstanding to say that. Electrons go where there are the less possible density of charge. Also I've never seen a positive oxygen which want to stay like this. And due to its electronegativity the density of electrons will be more around it than around the carbon that's with the oxygen on the alcool will react with the carbon.

@Mithoron what is hypercoordination ?

In a bond, the more electronegative element will have a greater share of the electrons, and a partial negative charge to reflect this greater electron density.  The less electronegative element will have a partial positive charge to reflect the lack of electron density.
Because in chemical reactions, electrons will flow from areas of high electron density to areas of low electron density. Knowing where the partial charges are is an important first step in determining where the molecule will react.
Formal charge is NOT the same as electron density.

Arrow pushing is a tool that helps you think about the possible reactions available for a given species or set of reactants: where electrons are likely to move in order to form new bonds and break old ones. But you still have to look at the species that you get after doing the arrow pushing and consider whether they will be favorable or not. Sometimes you can get a reaction between a (formally) negatively charged atom and a positively charged atom - for example between methoxide and tert-butyl cation. Do the arrow pushing and see how that works out.
The reactions you have labeled as incorrect can be seen (via arrow pushing) to produce species that have too many bonds or electrons around a single atom (that was what was meant by the term "hypercoordination" - you could also say hypervalent - for example, having 10 valence electrons around oxygen), whereas the ones labeled as correct lead to stable or at least normally bonded species.


text:
Vapor pressure of immiscible liquids
I have read that vapour pressure of a mixture of two immiscible liquids is
$$P=P^0_A + P^0_B$$ where A and B are immiscible liquids.
If they are immiscible, they would form separate layers. Thus since the more dense liquid would be below the less dense liquid, it would have no interface with the gas phase. Given this lack of an solid-gas interface, should the pressure due to the lower layer be zero?

Will the equation be more plausible if you actually achieve a mixture by stirring the two layer system strong enough to achieve a dispersion, in which the former layers are no longer observable?

What if their densities are the same? What then is the driving force to form horizontal (rather than vertical) separation? Something else to ponder - why would a layer of B over A impact the equilibrium vapor pressure of A? (Even being 'immiscible' there is always some concentration of A in B - you can't beat the entropic contribution to the free energy).

They don't necessarily have to form separate layers, they could just be in separate containers such as the arrangement shown here: https://chemistry.stackexchange.com/questions/112854/will-evaporation-from-two-immiscible-liquids-blow-up-a-balloon

This is not just some vapor pressure. This is the equilibrium vapor pressure. Thermodynamics is all about equilibrium, you know. And equilibrium, roughly speaking, is what takes place in a closed container after a billion years.
Immiscible as they are, the liquids still have some solubility in each other (maybe extremely low, but anyway). Over the course of centuries, molecules of the lower liquid will slowly creep one by one through the upper layer and vaporize. Eventually they will saturate the gas phase and create the same vapor pressure as if the upper liquid were not there.
Stirring the whole mixture, as Klaus suggested, is quite likely to speed things up a bit.


text:
The distinction between graphene supercapacitors and reduced graphite oxide supercapacitors
Looking at this study, Direct laser writing of micro-supercapacitors on hydrated graphite oxide films, a consumer laser cutter is used to synthesize reduced graphite oxide supercapacitors from hydrated graphite oxide film.
In a similar study at UCLA (scalable fabrication of high-power graphene micro-supercapacitors), graphite oxide film is used again but "LightScribed" by a LightScribe DVD player to form graphene supercapacitors.
What is the distinction between reduced graphite oxide and graphene. And why are the products of laser treated graphite oxide different in these papers?


text:
Calculating time taken for a set percentage of a reaction to occur based on half-life

The half-life of a first order homogenous gaseous reaction: $\ce{SO2Cl2 -> SO2 + Cl2}$ is 8.0 minutes. How long will it take for the concentration of $\ce{SO2Cl2}$ to be reduced to 1% of the initial value?

I solved it this way:
Let initial mass be 100, so that the final mass is 1.
$$100 \cdot (\frac{1}{2})^n=1$$
I solve this for n, and multiply it by 8.0 to get the time taken.
Answer is spot on, and the method is identical to the one used in half life questions in Physics. Yet my Prof insists that I solve for the rate constant first, and then find half-life from that, and that my answer is wrong because n, the value for number of half-life periods, cannot be a decimal value.
The questions I found online all explicitly ask for the rate constant, which the question I got doesn't ask for.
Is my method right/wrong? Can you guys provide proof?

See if you can derive your equation algebraically by working with the rate constant approach.

I know that approach, I'm concerned specifically with this method. I usually stick with methods I know are impossible for me to screw up, so I use this method. I just want to know what change is there to half life in chem that makes my method wrong.

Ryan showed that the rate constant is given by $k=\frac{ln(2)}{t_{1/2}}$.  So the concentration at time t is given by
$$C=C_0\exp(-kt)=C_0e^{-ln(2)\frac{t}{t_{1/2}}}$$But,
$e^{-ln(2)}=\frac{1}{2}$.  Therefore, $$C=C_0\left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}}$$Substituting $n=\frac{t}{t_{1/2}}$ gives
$$C=C_0\left(\frac{1}{2}\right)^n$$

Holy smokes, why in the world did my attempts never get here... Thanks a ton, you probably saved 5% of my grade.

This is what I was trying to get you to do in my Comment on your question.

n can very easily be a non-integer. What would be the answer for n if you were asked how long does it take the concentration to drop by 25%?

@Beerhunter Who said it was an integer?  It would be a non-integer.  So what!  Are you actually asking me how to determine the value of t in this specific problem when $C = 0.75C_0$ using this equation?

N can be a non integer, it was my prof who was insisting otherwise. Thanks a ton, folks.

My comment was around what Arv's prof stated

Maybe Arv's prof thinks that whenever the symbol n is used, it has to be an integer.  I wonder if he also thinks that in the ideal gas law PV = nRT, n has to be an integer.


text:
Why did my iron nails not dissolve in hydrochloric acid?
I was trying to dissolve iron metal in HCl to make iron chloride, but ran into some problems. I'll give a brief background of my experiment. I took iron nails and put them in a container filled with muriatic acid (chemicals for my pool). The acid contains an equivalent of concentrated HCl since it is composed of HCl by 37.5%. The other 62.5% is listed simply as "other ingredients".
The faint yellow acid began to turn a more solid green yellow. Gas immediately began to bubble on the nails' surface. I left it for two days to dissolve and when I checked back the iron nails were not dissolved. They look shiny like the rust was cleaned off. The solution is black/dark green and there is a white/green precipitate at the bottom. The nails have small green crystals growing on them
I did this experiment once previously (with the same type of nail and HCl) with just a single nail inside of a hollow plastic pen tube and it seemed to have worked very well; it formed a dark brown solution at the bottom and the rest of the nail looked very rusted to a point where I could not remove it because the rust had stuck it to the inside of the pen.
Is there any explanation for the two different results and the iron nails not dissolving in my other trial? Also is there any guess as to what my solution and precipitate contains? Let me know if any more information is needed.

Do you have a picture of the vessel where the nails did not dissolve?  And were/are you able to check the acidity of your dark green solution?  We (and you) might get some clues from those observations.

@OscarLanzi I just added an image of the solution

Btw, nobody still speculated about nature of white precipitate on the photo.

It seems like there is not enough acid on the nails.

Perhaps a depleted layer is forming next to the nails. To avoid that issue, use a magnetic stirrer or provide a source of vibration.
There might not be enough $\ce{HCl}$ for the large amount of iron in the nails. Check the total amount of $\ce{HCl}$, given the volume and that ~2/3 is "other ingredients" (likely water). Weigh the nails. Calculate the number of moles of  $\ce{HCl}$ needed to react with the number of moles of  $\ce{Fe}$. Is the product  $\ce{FeCl2}$ or  $\ce{FeCl3}$?
Have you removed all the grease (from the manufacturing process) from the nails? A little soap or detergent and rubbing may be needed.
Could there be some difference in the nails you used? If some are galvanized (zinc coated, to prevent rust), then that would affect the reaction.

Let us know how you do: we learn more from our mistakes than from successes.

It could be you ran out of acid.  37.5% $\ce {HCl}$ seems like a lot, but solid iron can consume 23 times its volume in 37.5% $\ce {HCl}$.  The volume occupied by two moles of $\ce {HCl}$ in the "concentrated" solution is that much greater than the volume occupied by one mole of really concentrated, solid metal.
Check this by seeing if your solution is still acidic, if you still have it.


text:
Why can't equations of state be replaced by machine learning models?
The Peng-Robinson equation of state, for example, has no underlying physical meaning, and is just a model that was fit to data. Computer scientists have developed much, much better models for data than messing around with polynomials and exponentials, like neural networks. Why can't standards organizations take a bunch of data about the chemical, fit a neural network or whatever to it, and publish that as the "EOS", instead of coming up with a meaningless equation that almost certainly wouldn't do as well?

If you have a hammer, they say, every problem looks like a nail. All these equations of state were proposed long before the current progress in machine learning. Also, they can be written on paper, which was an advantage in those days.

Furthermore, having an actual equation in front of you lets one do thought experiments that help you see what the equation is saying. Plopping down a USB drive and saying there is a neural network model on it is not revealing to a human at all.

There is no qualitative difference between neural networks and "messing around with polynomials and exponentials". It is only a matter of taste and availability of data what kind of data model you put on your data points.

It is true that a high order polynomial can fit any training set. But that is not a strength - an unfalsifiable model overfits. In particular, a polynomial of order n is only likely to be predictive if the true function is n times differentiable. Since chemical space is discrete, and for many purposes some molecules are special cases, polynomial models are a poor choice in cheminformatics.
Neural nets can work, with an explicit regularizer or with dropout. It's true that the resulting model is a black box - after building it, there is the challenge of understanding the predictions. However, running the model to make predictions is cheaper than doing a lot of experiments. After gaining some understanding of the model you can do more targeted experiments to check it against reality.
With some problems, we have had good results with a fingerprint-based SVM.

For n data points there exists an n-1 order polynomial that perfectly fits the data.
Therefore there is no basis for "a neural network or whatever" being better.
Furthermore, it simply isn't true that the Peng-Robinson equation "has no underlying physical meaning".
The Peng-Robinson equation (like Van der Waals) recognizes that atoms/molecules occupy space and that attractive forces exist between them.
As Peng and Robinson say in their article:

Since two-constant equations have their inherent limitations,
  and the equation obtained in this study is no exception,
  the justification for the new equation is the compromise
  of its simplicity and accuracy.

That being said, neural networks have been used to find equations of states.  See for example Equation of state and artificial neural network to predict the thermodynamic properties of pure and mixture of liquid alkali metals

Generally, ANN is powerful and successful method for complex non-linear systems due to unique advantages such as high speed, simplicity and large capacity which reduce engineering attempt. In recent years, ANN modeling has been successfully used for prediction of thermophysical properties of pure and mixture fluids [24], [25], [26] and [27].

And for n data points there exist entire families of n-th order (much less (n+m)-th order) polynomials that fit the data too.

Beware the old saying "with four parameters, I can fit an elephant"

I'm currently working with a R410a Equation of State and I found ML useful in the following cases:

Getting a formula from a data set predicted by a simulator
software. (Software companies knows this and are getting harder and
harder to download large samples)

Multilinear regressions when you have a table with data but you don't have a formula. (refrigerants , water, air, etc).

Classification when you have to use different equations depending on
the zone (gas, liquid, mixture, crystallization form, etc).

You will hear a lot about overfitting of neural networks, this is not a problem of neural networks, is a problem of low data density. When you have only a few of inputs in your model, there is not so hard to have a dataset big enough to easily avoid overfitting.
Adventages:

If you only have data and more than one input variables is the easy way.
You can perform several trainings changing the input parameters, so you don't have to iterate later.

Disadventages:

Equations of state HAVE (a vague) underlying physical meaning.
It's harder to generalize for different gases and compositions. So, it's also hard to be seen in a chemical simulation software.
You can't write the trained model in a book.
It's harder to check if the model has a good behavior.
When there is a small zone with a big nonlinearity, your neural network will tend to ignore that zone. For example, it's a personal example, if you train a model to predict the Methane Number of natural gas without inerts. You will need to add A LOT of data of methane numbers near 100 to reach good predictions in that zone.


text:
How to separate butane from air?
I work in a manufacturing company specializing in polyethelene foam and I have been struggling to separate butane from air in our warehouse during curing period where all the butane gas used to make our product is just released in the air. I have been thinking of ways how to separate and recycle the butane? Can someone help me?

Wait... butane in the air, just like that? Well, hopefully nobody is smoking in the warehouse. As for the separation, I'd probably rely on freezing out the butane (it has boiling point about $0^\circ$C).

Are you sure there is enough butane to make recycling worthwhile?

The noble gases tag doesn't seem to be right for butane. And I'd echo matt_black...quantity available to capture could be very small.

Butane has pretty high  boiling point (around $\ce{0 °C}$). This means, that just cooling air to $\ce{-30 °C}$ should condense a lot (if not all) of butane from it.
However, as I guess, most of butane used in foam production remains in the foam, otherwise the foam would collapse.  So, it is very possible that there is not much to recycle in air itself. The gas, of course, will be released eventually, but only when foam itself is destroyed.


text:
How to determine the magnetic character of heteroleptic complexes?
I was asked to find out the magnetic behavior (and essentially electronic configuration) of heteroleptic complexes using Valence Bond Theory and Crystal Field Theory.
But I am confused about how to do it. Especially in the ones which have both strong field and weak field ligands together; how am I to decide whether pairing of electrons will take place or not ?
For a more concrete example consider the complex $\ce{[Cr(NH3)_4Cl_2]^{+}}$

I think my first question is how many valence (d) electrons do you need to consider - what is the formal oxidation state on the Cr ion?

Are there electron counts where it doesn't matter whether the ligands are strong field or weak field?

@Geoff Cr is in +3 oxidation state so there are 3 electrons in d.

Based on the available information, we would start by figuring out the 3D geometry of the complex. $\ce{Cr^3+}$ should be octahedral, with two possible "combinations", cis or trans. Let's start with the trans version. The symmetry of this system is $D_{4h}$ so we call this "tetragonal".
With a tetragonal geometry, the MO's would be arranged in such a way that there will be two degenerate orbitals at the lowest energy, and three higher-energy orbitals. The MO diagram typically looks something like this:

Chromium(III) has an [Ar] $3d^3$ electronic configuration so we need to fill up three levels. Now the question is whether or not the splitting caused by the ligands are large enough to cause pairing.
We know for a fact that the trans version of the example is a green solution, so it absorbs red light (700 nm or ~14,300 $cm^{-1}$). Pairing energy is typically in the range of 20,000 - 30,000 $cm^{-1}$ so the 'gap' between yz and xy is probably about only ~14,000 $cm^{-1}$. As such, I would say the trans isomer is magnetic and not paired, because it is energetically more favorable to promote the 3rd electron to the higher energy level than to force it to pair.
Similarly, we know for a fact that the cis version is violet. This means it absorbs yellow light, about 570 nm or 17,000 $cm^{-1}$. This is still lower than the pairing energy, so I'd say that The tetraamminedichlorochromium(III) is magnetic.


text:
How to calculate the amount of substance of barium sulfate that is precipitated?

Determine the amount (in mol) of barium sulfate that will be precipitated when $200.0~\mathrm{cm^3}$ of $0.450~\mathrm{mol\,dm^{–3}}$ barium nitrate solution is added to an excess of sodium sulfate solution, given that the equation for the reaction is:
  $$\ce{Ba(NO3)2 (aq) + Na2SO4 (aq) -> BaSO4 (s) + 2NaNO3 (aq)}$$

I have just tried doing it and I got an answer of $0.09~\mathrm{mol}$, but I am not sure if this is correct, because I have only just been introduced to $\mathrm{dm^{-3}}$ so didn't really know what I was doing. I converted $200~\mathrm{cm^3}$ to $0.2~\mathrm{dm^3}$ (is this right?), and then multiplied this value by the concentration, $0.450~\mathrm{mol\,dm^{-3}}$, and this gave me $0.09~\mathrm{mol}$. Since the stoichiometric ratio between $\ce{BaNO3}$ and $\ce{BaSO4}$ is 1:1, I just suspected that the amount of substance of $\ce{BaSO4}$ would also be $0.09~\mathrm{mol}$.
If anyone could tell me if I have done this correctly, it would be greatly appreciated.

Yep! That's basically correct :)

Yes, your answer is correct.
However, since you are still learning I recommend, that you take the time to write out all the equations, with proper units and every step. This should help you to gain a better understanding of the chemistry that is actually going on.
One of the key words of the exercise is excess of sodium sulfate solution, which indicates that the reaction will be complete, i.e. entirely on the product side.
Hence, you can (and you correctly did) determine from the reaction equation,
$$\ce{Ba(NO3)2 (aq) + Na2SO4 (aq) -> BaSO4 (s) + 2NaNO3 (aq)},$$
that the stoichiometric ratio is one, $n(\ce{Ba(NO3)2})/n(\ce{BaSO4}) = 1$. Therefore you can also write $$n(\ce{Ba(NO3)2}) = n(\ce{BaSO4}).$$
Now you simply have to calculate the amount of substance of barium nitrate. You have been given the volume of your solution $V(\ce{Ba(NO3)2}) = 0.200~\mathrm{cm^3}$ and the concentration $c(\ce{Ba(NO3)2}) = 0.450~\mathrm{mol\,dm^{-3}}$. Using the relation
$$c = \frac{n}{V}$$
you can calculate the amount of substance of barium nitrate.
You have also correctly converted cubic centimetres to cubic decimetres. In any case, you should get familiar with metric prefixes since they will come up in every science class you will take.


text:
Difference between Excess property and property changes of mixing
I have some confusion in my mind about the difference between an excess property and a property change of mixing. I know that the Excess property is defined as the difference between the actual value and those of an ideal mixture:
$$\ M^E(T,P,x_i)=M^{actual}-M^{id}(T,P,x_i)$$
On the other hand the property changes of a mixture is:
$$\ M^{mix}=M{actual}-\sum n_i*M_i$$
So the latter seems to me that the whole change of a mixing process is with respect to the pure, ideal gas, species.
Excess and mixing are the same for enthalpy and volume but are different for entropy and Gibbs free energy.
Now come the problem: property changes of mixing and excess properties are related each other by the residuals properties (deviations from ideal gas)?
Otherwise I do not understand why they are different.
Thanks a lot

The difference is subtle, however they are different as you suspect.
You can look at the property of mixing as the change in property resulting from mixing the pure components (with everything at the same temperature and pressure) together to form the mixture.
The excess property is deviation between the actual property and the property in an ideal mixture at the same temperature and pressure.
They are the same for volume and enthalpy because the volume of mixing and the enthalpy of mixing in an ideal solution are both 0. This is not the case for G or S.
In other words the difference arises because $M^{id}=\Sigma x_i M_i$ is not true for all properties. It is only true for enthalpy and volume. So the second term in your expression for the excess property above, cannot in general be written as $M^{id}=\Sigma x_i M_i$ (if it could, the excess property and mixture property would be the same). It should instead be $M^{id}=\Sigma x_i M_i + M_{mix}^{id}$.
The extra term for mixing the components would be be given by $$
\frac{G^{id}_{mix}}{RT} = \Sigma x_iln(x_i)
$$
and
$$
\frac{S^{id}_{mix}}{R} = -\Sigma x_iln(x_i)
$$
for Gibbs free energy and entropy, respectively.


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How obtain mass-weighted coordinates used to plot intrinsic reaction coordinates
I have obtained the energies from an IRC calculation. Usually these are plotted in papers vs the mass weighted coordinates; often something like this is labeled on the x axis: $Å \times \sqrt{\mbox{amu}}$.
How do I obtain these coordinates?
Possible solution
With some reading, I came up with a possible solution. Apparently the IRC step size (used in Molcas calculations) are in units of Bohr in normalized mass-weighted coordinates. So all I need to do is to do the following

Set the TS data point to 0
Multiply the step number (TS is "step 0") with the step size
Multiply the above with the Borh-to-Ångström conversion factor (0.529177249)
The direction to reactant should be negative
The direction to product should be positive

Could you include a paper where they do that as an example?

See Fig. 2 in West et al: http://pubs.acs.org/doi/10.1021/acs.jpca.5b03402

You seem to have overlooked the contribution of the correction step (during the iterations of the IRC method) to the mass-weighted coordinates.


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Electroplating copper onto steel
I'm rather unfamiliar with chemistry but I want to electroplate steel with copper.
From Wikipedia:

The cations associate with the anions in the solution. These cations are reduced at the cathode to deposit in the metallic, zero valence state. For example, in an acid solution, copper is oxidized at the anode to $\ce{Cu^2+}$ by losing two electrons. The $\ce{Cu^2+}$ associates with the anion $\ce{SO4^{2−}}$ in the solution to form copper sulfate. At the cathode, the $\ce{Cu^2+}$ is reduced to metallic copper by gaining two electrons. The result is the effective transfer of copper from the anode source to a plate covering the cathode.

I have also found a YouTube video where a guy uses $\ce{HCl}$ instead of $\ce{H2SO4}$.
What are the requirements for the reaction to work? Does it have to be an acid solution? Can I use vinegar?

You should be able to use copper acetate, which can be made from warm vinegar and hydrogen peroxide on copper (be very careful with this, warm peroxide can form dangerous compounds with some other organic chemicals). I suspect it would require higher voltage, however. I would have to confirm that before writing it as an answer. See: http://www.instructables.com/id/High-Quality-Copper-Plating/

You can attempt to use a copper item as the anode in a vinegar solution; the $\ce{Cu}$ is oxidised to $\ce{Cu^2+}$ and the acidity of the vinegar prevents the precipitation of $\ce{Cu(OH)2}$. The copper acetate formed in solution will migrate to the cathode, where you can place your item to be plated.
However, as my own experiment has shown, too high a current will result in a black deposit, not pink as the copper crystals are plated to quickly to form large crystals and thus scatter light. So use a low current!