text:
N(delta)-acetyl... vs N(alpha)-acetyl
Can anyone tell me what  the difference between these two compounds is:
N(alpha)-acetyl-ornithine
N(delta)-acetyl-ornithine
The only difference I can think of is the chiral properties of the nitrogen atom, i.e. nitrogen inversion.
If that is the case, does it make sense to differ between these to compounds, or will they be constantly interconverted?

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You did not link to N$^\delta$-acetylornithine but rather to a nonsymmetric version of N$^\alpha$-acetylornithine. I corrected your link (but it still needs to be peer reviewed).

Jan, "correcting" the link in a question like this changing the author's intent for the question.  The link as it currently exists does list "N(delta)-acetylornithine" as a MeSH descriptor synonym for some weird reason.  You should add the link to the right N(delta)-acetylornithine to your answer.  BTW, how did you find that molecule?  It did a pubchem structure search but it came up empty.

@CurtF. a search for acetylornithine did the trick.

Jan's edit has now been approved and as a result the accepted answer to the question (mine) now no longer makes any sense.  The question originally linked to a Pubchem page which listed N(delta)-acetylornithine as a synonym for a structure that was not actually N(delta)-acetylornithine.  That actually seems to have been the confusion experienced by the OP and it was what I tried to address in my question.  With the link altered to point to the correct structure for N(delta)-acetylornithine, none of this is clear.

Short answer: PubChem has an error.
Differences in structures as currently listed in PubChem
One other important difference is the stereochemistry of the sp3 carbon attached to the nitrogen atom, i.e., at carbon 2 of the 5-aminopetanoic acid chain.

N(alpha)-acetyl-ornithine has a computed IUPAC name in PubChem of 2-acetamido-5-aminopentanoic acid.  The stereochemistry is not listed, which leads me to assume that it is a racemate.
The molecule/PubChem page that you called N(delta)-acetyl-ornithine has a computed IUPAC name in PubChem of (2S)-2-acetamido-5-aminopentanoic acid.  The stereochemistry at the sp3 carbon bound to the nitrogen in this compound is clearly listed as S.

Inconsistencies between the structures and the names

Confusingly, both compounds are called L-ornithines on those pubchem pages.  But since carbon 2 is the α carbon, this is the position that should determine the L or D stereochemistry.  Thus, I'm not sure why N(α)-acetyl-ornithine isn't just called N-acetyl-DL-ornithine and what you called "N(δ)-acetyl-ornithine" isn't just called N(α)-acetyl-L-ornithine.
Secondly, there is the matter that the page you linked for "N(delta)-acetylornithine" lists a "main" name for the molecule of "N-acetylornithine" and "MESH descriptors" of "delta-N-acetylornithine".  This use of "delta" is wrong because it is inconsistent with the listed structure.  The α, β, γ, δ nomenclature exists to indicate which nitrogen atom is substituted.  In ornithine carbons 2 and 5 of the core pentanoic acid chain have nitrogen atoms, these correspond to N(α) and N(δ).  Thus, N(δ)-acetyl-L-ornithine should have the acetyl group substituting a different nitrogen atom than N(α)-acetyl-L-ornithine.

Thank you very much! Apparantly, I was confused by the inconsistency between the listed structure and the use of delta-N-acetylornithine as a MESH descriptor.

I should note that this answer addresses a previous version of the question, where the links to PubChem molecule pages were the ones provided by the OP, not by editors to the question.

The difference does not lie in stereochemistry. Rather, N$^\alpha$-acetylornithine and N$^\delta$-acetylornithine are regioisomers of one another.
The superscript Greek letter denotes which nitrogen is substituted with an acetyl group. The lettering is relative to the carboxyl group as shown in the image below:

Note that:

The carboxyl group’s carbon does not get the $\alpha$, because it is considered part as the functional group from with the labelling begins
The $\alpha$ carbon is asymmetric. Both L- and D-compounds (or (S) and (R) according to CIP nomenclature) are possible for both N$^\alpha$- and N$^\delta$-acetylornithine.

This is a good answer, especially since Jan edited the original question.  As a result of the edit, this answer now makes more sense than my own, but I'm not sure which version of the question (and answer) was more helpful to the OP.


text:
Minimising the formation of manganese dioxide in the titration of potassium permanganate
For a titration of potassium permanganate into a standard acidified solution of hydrated iron (II) ammonium sulfate, the solution seems to end up brown (due to the formation of manganese dioxide) during the titration unless the titration is carried out extremely slowly. How can I minimise this? Adding sulfuric acid to the solution seemed to help, but is there a physical way to prevent manganese dioxide forming when the titrant is added too rapidly?

Adding a large excess of acid helps greatly as the reduction to $\ce{MnO2}$ only takes place under neutral conditions. $$\ce{2H2O + MnO4- + 3e- -> MnO2 + 4OH-}$$ Under acidic conditions the $\ce{MnO4-}$ is reduced all the way to $\ce{Mn^{2+}}$.$$\ce{8H+
 + MnO4- + 5e- -> Mn^{2+} + 4H2O}$$ Also swirling or stirring the solution as you are titrating makes a big difference as it stops the area where the permanganate is being added from becoming temporarily depleted of acid.Even if you do get some $\ce{MnO2}$ forming then as long as you have enough acid in the flask a few swirls should be enough to reduce the $\ce{MnO2}$ down to $\ce{Mn^{2+}}$.

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sort: votes


text:
kWh of radioactive elements database?
I recently been studying different radioactive elements. The most difficult part is find the output of these elements in the form of kWh.
Is there a database or chart of these elements' output?
And most importantly does anyone know the kWh of $1~ kg$ of $\ce{^99Tc}$?

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I don't quite follow the question. kWh is a unit of energy. Are you asking what the total energy released if 1 kg of technetium-99m transitions to technetium-99 emitting mostly gamma rays? (I assume you mean Tc-99m because Tc-99 has an extremely long half-life, but correct me if I'm wrong)

According to Nuclides 2000, the isotopic power of Tc-99 is only 8.6E−06 W/g; the value for Tc-99m is about 4.4E+03 W/g.

I'm trying to calculate the energy out put of TC 99 like 1 kg uranium 235 puts out 50,000 kWh

Can you clarify if you are talking about Tc-99 or Tc-99m.

I am referring to TC-99m

$\ce{^{99m}Tc}$ is a nuclear isomer of $\ce{^{99}Tc}$, which is excited by 142.6 keV. Therefore, when one atom decays to $\ce{^{99}Tc}$, it releases 142.6 keV, mostly as gamma radiation, but also some internal conversion occurs. Therefore, for 1 kg of $\ce{^{99m}Tc}$ (10.1 mol):
$$
(142.6\ \mathrm{keV})(10.1\ \mathrm{mol})N_A = 8.673\times 10^{29}\ \mathrm{eV} = 139\ \mathrm{GJ} = 38.6\ \mathrm{MWh}
$$
In general, the total energy released by a nuclear reaction in a given mass is simply the decay energy (commonly known as the Q value) multiplied by the number of atoms. The Q value is simply the difference in mass between the starting material and final product converted to energy:
$$Q = (m_\mathrm{i}-m_\mathrm{f})c^2$$
Q values are easy to just look up, though. (Wolfram|Alpha is a good place)
Of course, depending on the application, all this energy is not accessible or useful. If one were trying to make an RTG, an enormous amount of shielding would be needed to turn the gamma rays into heat, since they penetrate matter very well, where things like Pu-238 work better because they release mostly alpha particles, which are easily absorbed, delivering their kinetic energy as heat.

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text:
H2SO4 with ethanoic acid reaction?
Earlier today I was preparing an ester (propyl ethanoate) and I noticed that instead of putting the propanol, ethanoic acid and $\ce{H2SO4}$ in the test tube and warming it (taking time and external energy) I could instead mix sulfuric acid with the ethanoic acid, which made the test tube go very hot, and then by adding the propanol and shaking it split into two layers. On pouring this into water I ended up with an immiscible layer which appeared to be an ester (it was sweet smelling and considering the reaction it would be likely to be an ester).
I cannot think of which reaction was causing the heating up initially. The test tube was clean so the only reactants were the sulfuric acid and ethanoic acid, and it appeared that a gas was being given off (as shaking it with a rubber cap on caused it to pop out of the tube, but no effervescence could easily be seen). Can anyone help? Furthermore, would the eventual product be propyl ethanoate or something else (it was suggested that it was dehydrogenating the ethanoic acid so the actual product would be something like propyl ethenoate, but from a quick look on the internet this would seem to require an iron catalyst (or another catalyst) and a high temperature, neither of which was present)?

Was the ethanoic acid completely anhydrous and was the sulphuric acid concentrated (usually is for esterification)? Concentrated sulphuric acid reacts very exothermically with water.

I believe so. It may have contained a little bit of water, which might explain it. Thanks.

My guess is sulphuric acid reacting with water to heat up the test tube which evaporated some of the ethanoic acid causing the bung to pop off.

Ah, that would make sense. Thank you for your help.

I would wait for someone else more experienced to confirm it as I'm not the most experienced but that seems likely.

This could be clarified as to whether the source container states "glacial" or not for the acetic / ethanol acid. Otherwise, could acetic (ethanoic) anhydride be formed, liberating water (see answers above) and the propanol reacts readily to give the ester?

I suspect that the ethanoic acid contains a residual amount of water and that the concentrated sulphuric acid is reacting with the water. Since this is a very exothermic reaction the test tube gets heated up which evaporates some of the ethanoic acid, causing the bung to pop off. If it's just slight evaporation this would also explain why you couldn't see any effervescence.

Good observation!
You have actually created the strongest acid ever. Stronger than Magic Acid, which is pretty darn strong (it's known as superacid, where superacid are acids that are stronger than 100% sulphuric acid.)
Combining Sulphuric Acid and Acetic Acid results into the formation of a polymeric acid.

This link has a great explanation for the reaction that takes place>> http://link.springer.com/article/10.1007%2FBF00908172#page-1

Magic acid is a superacid. Also, as far as I am aware, the strongest known acid is fluoroantimonic acid, a 2:1 mixture of $\ce{HF}$ and $\ce{SbF5}$.

You are not wrong, but sulphuric and acetic acid mixture is also not any weaker. However, the mixture doesn't seem to be as stable as fluoroantimonic acid mixture. Probably that's why it's not considered/used? Superacids aside, tell me what you think about the sulphuric acid and acetic acid reaction.

I think what happened is additon of conc h2so4 led to esterification of ethanoic acid and release of water. This water must have reacted with another h2so4 to produce heat without effervescence.


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Calculating pH in titration of monoprotic weak acid with strong base

$25.0\:\mathrm{ml}$ of $0.10\:\mathrm{M}$ $\ce{CH3COOH}$ is titrated with $0.10\:\mathrm{M}$ $\ce{NaOH}$. What is the $\mathrm{pH}$ after the addition of $30.0\:\mathrm{ml}$ $\ce{NaOH}$?

In broad strokes, as I understand it, the sodium hydroxide in this case would completely consume the acetic acid, leaving a solution of acetate ions, $\ce{Na+}$ and $\ce{OH-}$.
What makes sense to me then is to figure that the $\mathrm{pH}$ is now dominated by the concentration of hydroxide ions, so the $\mathrm{pH}$ would be given by
$\mathrm{pH} = 14-(-\log([\ce{OH-}])= 11.96$
where $[\ce{OH-}] = \frac{[(30.0 - 25.0) \times 10^{-4}]\:\mathrm{mol}}{(30.0+25.0) \times 10^{-3}\:\mathrm{l}}$
i.e., the remaining moles of hydroxide ($0.0005$) per liter of total volume ($55\:\mathrm{ml}$)
but is this correct? Do I need to take into account the conjugate base’s contribution to deprotonation of water? If so, how is that done?
I’d like to be clever and use the lack of $K_\mathrm{a}$ or related constants in the problem to indicate that I’m on the right track, but that would be giving too much credit to my professor’s design of these questions (and the prior question in this series does give $\mathrm{p}K_\mathrm{a}$ for acetic acid)

The strong base contributes so much more (in magnitude) that the conjugate base (weak) can be neglected for a reasonable approximation. Thus, for the scope of the question, I believe you did it correctly -- many questions are designed for such approximations.

Your calculation is right.  Regarding your question to take into account the conjugate base's contribution to deprotonation of water, I'd say:
If you calculate the equilibrium constant of the neutralization reaction $$\ce{CH3COOH + OH- <=> CH3COO- + H2O}$$
$$K= \frac{K_a}{K_w}=\frac{10^{-4.75}}{10^{-14}}= 10^{9.25}$$
Now, if you write  of reaction of acetate ion with water:$$\ce{CH3COO- + H2O <=> CH3COOH + OH-}$$
It's the opposite of the neutralization reaction. Its  equilibrium constant  is  $10^{-9.25}$.
You can see clearly that the neutralization reaction is the "predominant reaction". So, the pH is determined by this reaction as if it's the only one happening in the aqueous solution. So, you can neglect the reaction of of acetate ion with water.

The K value is actually $10^{9.25}$.

@ Ed V  Many thanks. I'll correct it

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text:
Why is the reaction of calcium oxide with sulfur dioxide to form calcium sulfite not a combination redox reaction?
I understand that a combination reaction is one where two substances combine to form a third substance, like with
$$\ce{2Na + Cl2 -> 2NaCl}.$$
I do not understand why $$\ce{CaO + SO2 -> CaSO3}$$ is not a combination redox reaction. After all, it is creating a third substance, so why is it not a combination redox reaction?

To determine whether a reaction is a redox reaction, you first have to find the oxidation number of each element in the reactants and products. You do this by either counting the number of electrons in each bond and assigning them to the most electronegative atom in the bond, or by memorizing some rules and applying them. When you do that for this reaction, you should get the following:
\begin{align}
  \ce{   &&CaO + SO2 &-> CaSO3}\\
  \ce{Ca:&& +2 &-> +2}\\
  \ce{O: && -2, -2 &-> -2}\\
  \ce{S: && +4 &-> +4}
\end{align}
Since the oxidation number hasn't changed for any element, no reduction or oxidation has taken place, and this is not a redox reaction.

This question can be solved the easiest if we break it down to its fragments. The requirements of a combination redox reaction are that the reaction must be a redox reaction and a combination reaction at the same time.
The question of whether this is a combination reaction or not is pretty easy: we get one compound from the original two, so yes, it is a combination reaction.
However, this is not a redox reaction, because no particles' oxidation number changes, so no oxidation/reduction actually happens.
So in conclusion this is a combination, but not a redox reaction, so it doesn't qualify as a combination redox reaction.

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text:
What is the Kovat's Index used for?
As mentioned in the title, why do we use the Kovat's Index in gas phase chromatography? It seems a little arbitrary to compare everything to the order that n-alkanes (such as n-pentane, n-hexane, and n-heptane) emerge from the column, and I am curious to know reasons why we use such an index.

Why do we use the Kováts index?

The Kováts index is used to normalize GC data. This Wikipedia page(though not very long), sums up why you would want to do so.
Retention times of the same compound on even two different versions of identical instrument with the same temperature and pressure program using the same column from the same manufacturer might not be the same! Retention time is dependent on so many things:

The identity of the analyte
The identity of the stationary phase
The length of the column (did you know that columns sometimes get trimmed as they age to remove deposits?)
The diameter of the column
The thickness of the stationary phase film
The flow rate of the carrier gas
The temperature or temperature program of the oven

With so many variables, gas chromatography (indeed all chromatography) can seem like a forbidding black art in which one has no hope of being able to compare results from one instrument to the next or one lab to the next.
Fortunately, we're smarter than that. We can always compare our analyte's retention time to some retention times of well behaved standards. If we know the elution order (will not change) of a series of well-behaved standards on several instruments, and we know when our analyte elutes in relation to those standards on one instrument, we have a good starting point for predicting the retention time of that analyte on another instrument.
Ervin Kováts took this idea a step farther and created a quantitative index relating the retention time of an unknown analyte to the retention times of known standards in a way that produced system-independent constants describing the elution behavior of compounds. The system-independent part is important, given all of the factors listed above that can influence retention time. Every analyte should have (nearly) the same Kováts index value on any GC system under any set of parameters. Thus, if we determine the Kováts index of analyte $\ce{A}$ on one GC, we can know something about how $\ce{A}$ will behave on other instruments or with a different temperature program, etc.

Why the linear alkanes?

The choice of the n-alkanes was probably not arbitrary, but I cannot say for certain. Of the two links from the Wikipedia article that might contain a rationale, one is dead and the other is in German behind a paywall. However, any well-understood series of standards would have worked. My guess would be that Kováts chose the linear alkanes for the following reasons:

They form a homologous series increasing one carbon at a time.
A great many of them are (were - Kováts developed this index in the 1950s) available commercially in high purity (high enough purity that they can be easily purified further). At Sigma Aldrich, you can buy all of the n-alkanes at least up to eicosane $\ce{C20H42}$.
The first 20 linear alkanes cover a wide range of boiling points (and thus retention times): Methane $\text{BP}=-161.5\ ^\circ \text{C}$, pentane (the first liquid at room temperature) $\text{BP}=35.9\ ^\circ \text{C}$, decane $\text{BP}=173.8\ ^\circ \text{C}$, and eicosane $\text{BP}=343.1\ ^\circ \text{C}$.
There is a relatively reliable linear relationship between $\log(t_r)$ and the number of carbon atoms in the chain.

How do you determine and use the Kováts index?

Lets say we have an analyte $\ce{A}$ that has a retention time of 193 seconds on some system. First, we identify the linear alkanes that have retention times on either side of the analyte. Lets say that heptane has a retention time of 170 seconds and octane has a retention time of 200 seconds in this system.
The formula for the index looks like this (for non-isothermal programs):
$$I=100\times \left[n+\left(N-n\right)\dfrac{t_{r,a}-t_{r,n}}{t_{r,N}-t_{r,n}} \right]$$

$I$ is the Kováts index
$n$ is the number of carbon atoms in the alkane with the lower retention time
$N$ is the number of carbon atoms in the alkane with the higher retention time
$a$ is the analyte
$t_r$ is retention time, so $t_{r,a}$ is the retention time of the analyte

For our example:
$$I=100\times \left[7+\left(8-7\right)\dfrac{193-170}{200-170} \right]$$
$$I=100\times \left[7+\dfrac{23}{30}\right]=777$$
Now, we want to analyze $\ce{A}$ on a different instrument, where the retention times of heptane and octane are 217 and 289 seconds, respectively. We can plug values back into the equation for the index to predict the retention time of $\ce{A}$ on this system:
$$I=100\times \left[n+\left(N-n\right)\dfrac{t_{r,a}-t_{r,n}}{t_{r,N}-t_{r,n}} \right]$$
$$777=100\times \left[7+\left(8-7\right)\dfrac{t_{r,a}-217}{289-217} \right]$$
$$7.77=\left[7+\dfrac{t_{r,a}-217}{72} \right]$$
$$0.77=\dfrac{t_{r,a}-217}{72}$$
$$55.2={t_{r,a}-217}$$
$$t_{r,a}=272$$

Interesting, so the Kováts index works by extrapolating all elution behaviour to the n-alkane series, and by doing so it essentially means "this compound elutes as if it were an n-alkane with 7.77 carbon atoms". Don't know why I didn't realize that before.

@NicolauSakerNeto - That's how I interpret it, although I admit to being not very familiar with the concept before I  started answering this question. I am now envisioning a very interesting inquiry-based experiment for an undergraduate analytical or instrumental lab course to test the validity of Kovát's index over a range of GC temperature programs and columns.

Good answer. Just one question...Are you sure kovats index independent of stationary phase? I think they are not, that's why we have different McReynolds indexes in different phases. Also, according to this site https://goldbook.iupac.org/html/R/R05360.html they are temperature-dependent..

accoording to this website https://chem.libretexts.org/Core/Analytical_Chemistry/Chromedia/01Gas_Chromotography_(GC)/Gas_Chromotography%3A_Basic_Theory/25Qualitative_analysis%3A_identification kovats index is stationary phase-dependent


text:
What can I use instead of AOT in the following protocol?
We want to nanoencapsulate our pesticide using the following protocol; but we don’t have AOT surfactant. So, what can I use instead of AOT in the following protocol?

Sodium alginate solution in water was emulsified into dioctyl sodium sulfo-succinate (AOT) solution in methylene chloride by sonication for 1 min over ice bath. The primary emulsion was further emulsified into 150 mL of aqueous PVA solution by sonication for 1 min over ice bath to form a secondary water-in-oil-in-water emulsion. The emulsion was stirred using a magnetic stirrer, and 50 mL of aqueous calcium chloride solution was added gradually to the above emulsion. The emulsion was stirred further at ambient conditions for 18 h and then under vacuum for 1 h to evaporate methylene chloride.


text:
Only-C chain molecules
Are (cyclic) $\ce{C}$-chain molecules as in the picture likely to occur or do they exist at all?
 and so on...
EDIT:
I'm considering only double bonds between $\ce{C}$ atoms. This rules out most common allotropes.

Diamond, graphite... are carbon only

And buckeyballs and carbon nanotubes

I don think those in particular will exist due to high angle strain. But larger ones nay. However I reckon they would be more stable as alternating triple and single bonds rather than all double bonds.

Regarding $\ce{C2}$ with quadruple bonds, see this.

Can't seem to find anything that exactly matches, but the closest might be "cyclic allenes" - this paper| may be relevant.

I guess these will help : [https://en.wikipedia.org/wiki/Allotropes_of_carbon] and [https://en.wikipedia.org/wiki/Diatomic_carbon]

The second and third would never exist due to angle strain.

The term you are looking for is "allotropes", and carbon has a whole bunch of them, but the three most readily spotted in nature are graphite, diamond and fullerene, however, none of them are simple chains such as those postulated in the answer.
Graphite exists in large honeycomb flakes held together on two axes by covalent bonds, and on the third by weaker Van der Waals forces, which is why they come off readily under mechanical shear, leading to their popularity in pencil lead.
Diamond are huge, multiply interlocked complexes of crystalline carbon, although often with impurities locked into their structure, perverting the crystal structure and yielding interesting colors as a byproduct.
Fullerene is a molecular allotrope, consisting (usually) of 80 carbon atoms arranged in a buckyball. Initially thought to be a lab-grown species, it has since been identified as a natural result of certain combustion processes.
Now, none of these are chains like what you describe; while some of the compounds you sketch out are likely possible and present, they are also likely to be highly reactive, perhaps even to the point of decomposing spontaneously.
While amorphous carbon chains may very well exist, without a crystalline structure to shield their electrons, their reactivity would likely be such that they would be unable to exist for very long at all.
Simplistically speaking, the long stable chains of carbon found in hydrocarbons aren't entirely characteristic of carbon, as such, but rather, characteristic of hydrogenated carbon.

Presumably a very large chain of linear acetylenic carbon or carbyne could close on itself and form a cyclic molecule made purely with carbon. I think this allotrope would be stable with respect to spontaneous decomposition in carefully controlled conditions, but would be extremely reactive

If you Google "carbyne ring" you get several relevant hits including this rather speculative paper: http://evgeny-belenkov.narod.ru/CR07_52_0343_en.pdf


text:
Why is europium used in Euro bank notes?
Europium is apparently used as an anti-counterfeiting measure in Euro bank notes because of its fluorescence under UV light.
Is there any reason why it is specifically $\ce{Eu}$ and not any of the other lanthanides that fluoresce as well?
Is it because $\ce{Eu}$ is relatively rare and expensive so it's harder to fake? Is it because it has a very characteristic fluorescence? Is it because they just thought it's funny to use $\ce{Eu}$ in Euros?

I want to think its because of the name but I wouldn't think its the best reason.

@Alizter It might well be. See my answer for more detail.

I specifically know of one case where rare earth infrared fluorescence (near IR) was used in a European currency prior to Euros.

As already noted by the OP, many rare earths fluoresce under ultraviolet light and there seems to be no particular reason why Europium is a better choice than any other element, purely based on its fluorescent properties. (See update at bottom)
The European Central Bank keeps the exact nature of the compounds used a secret. However, it was discovered in 2002 by Frank Suijver and Andries Meijerink [1] that europium, complexed with two $\beta$ diketone molecules provides the red light but the other colours were more difficult to identify.
However, the reason for the choice of Europium is unclear. According to Hugh Aldersey-Williams[2]:

It [the European Central Bank] wilfully misunderstands my request to know who fought for Europium and drearily asks for my 'understanding that, for security reasons, we cannot comment on the chemical components of the euro banknote security features.'

Apparently, in the original paper by Suijver and Meijerink they say that they were visited several years beforehand by European Bank officials who were investigating potential luminescent materials so it may be that they inadvertently gave the bank the idea for using Europium in its banknotes.
Europium is one of the more expensive lanthanides and so this is probably a factor in its usage to prevent against forgeries, as the OP suggests.
I suspect there is also simply an element of irony in the fact that Europium is used in Euro bank notes and it may be that it was just some bank official tasked with finding a suitable compound to use who thought that it would be funny to use Europium.
Anyway, the bottom line appears to be that nobody knows the reason because the European Central Bank refuses to tell anyone anything about the compounds they use.
UPDATE: In response to @Nicolau Saker Neto's comment I did some more research into the fluorescence of Europium. Europium(III) (red) is regularly used alongside Europium(II) (blue) and Terbium(III) (green) in trichromatic lighting systems. It may therefore be the case that Europium was chosen because it is already well studied and in common use and that they created a $\beta$ diketone complex in order to fine tune the exact wavelength of the emissions and make it unique to their banknotes.
References:
[1] http://dspace.library.uu.nl/handle/1874/25831 (Paper in Dutch which I can't read. However, Williams refers to some of the content in his book and I have alluded to this above. If anyone can translate the original paper and add anything to the answer that would be very helpful.)
[2]Hugh Aldersey-Williams; 'Periodic Tales: The Curious Lives of the Elements'; ISBN 978-0-141-04145-2 http://www.amazon.co.uk/Periodic-Tales-Curious-Lives-Elements/dp/0141041455

I once attended a criminal forensics seminar which discussed mixing small amounts of lanthanides in gunpowder, and I seem to recall that europium (red) and terbium (green) were found to be particularly useful, I think due to strong fluorescence. I've forgotten the details, though.

Interesting. I will do some more research in a bit.

Well, knowing exactly which lanthanides are in Euro bank notes is not that hard - just stick one into an ICP-MS instrument, found in just about every university and research lab in the world.

@Michael Apparently when they were investigating the emissions of the banknotes they were warned that 'any further investigation into what causes the luminescence of euro notes would constitute a violation of the law' so what you are suggesting would probably also be illegal.

@bon, only illegal in the EU, I presume.

Europium-based compounds are widely used in other places to provide a distinctive red fluorescence (old CRT monitors & TVs, plus most fluorescent tubes, plus glow-sticks, plus glow in the dark paint). Perhaps it was just the easiest choice.


text:
Why super glue does not stick inside their own bottle?
We usually use super glue to stick any two objects but I was wondering like why don't they stick inside their own bottle ?

Related: What is the chemical basis for superglues?

It does, if moisture enters the bottle http://www.reddit.com/r/mildlyinteresting/comments/38d2hl/my_superglue_hardened_in_the_shape_of_the_bottle/

From explainthatstuff.com:

Why doesn't glue stick to the tube?
Adhesives are designed to work when they leave the tube—and not
  before. Different adhesives achieve this in different ways. Some are
  dissolved in chemicals called solvents that keep them stable and
  non-sticky in the tube.
When you squeeze them out, the solvents
  quickly evaporate in the air or get absorbed by the surfaces you're
  sticking to, freeing the adhesives themselves to do their job.
Plastic modeling glue works like this. It contains molecules of
  polystyrene in
  an acetone solvent. When you squeeze the tube, the glue spurts out and
  you can usually smell the very strong acetone as it evaporates.
Once it's gone, the polystyrene molecules lock together to make strong
  chemical bonds.
  Glue doesn't smell when it's dry because all the
  solvent has vanished into the air.
Some glues (such as synthetic,
  epoxy resins) have to be mixed together before they work. They come in
  two different tubes, one containing the synthetic resin and the other
  containing a chemical that makes the resin harden. The two chemicals
  are useless by themselves but, mixed together, form a tough, permanent
  adhesive.

From sciencefocus.com:

How super glue works

The main ingredient in super glue is a chemical called cyanoacrylate, an acrylic resin that forms a bond (cures) almost
  instantly;
The only trigger it requires are the hydroxide ions found in water;
Traces of water are found on almost everything and in the surrounding air, which causes the glue to become sticky;
It works best in moist conditions, although it is not advised to start gluing on a soaking wet surface.

Therefore the lack of water in the tube, or the chance of solvents evaporation, doesn't allow the glue to stick to the tube.
In other words: that strong smell that comes out when we open the tube's cap is due to the solvents evaporation, so once they're gone the glue sticks. If a superglue tube is left open for a long time (exposed to air, allowing the water to come in contact with the glue), it's impossible (or almost) to squeeze it out since it got stuck to the tube.

@bon thanks, altered as mentioned in Wikipedia: "cyanoacrylate is an acrylic resin that rapidly polymerises in the presence of water (specifically hydroxide ions)"

If you leave the cap off your superglue, it certainly will stick to the bottle.  But, as noted in the answer, it needs water to react...

@JonCuster traces of water are found in the air, so once the cap is off, the water is in contact with the cyanoacrylate, which makes the glue stick to the tube!


text:
What does boiling point of a metal depend on, specifically manganese?
Boiling point, as I understand, generally depends on molar mass and the intermolecular interactions in the liquid phase.

What are the other influencing factors?
What intermolecular forces are present in liquid metals?
Why does the boiling point of manganese deviate?

Here are the graphs for the melting point and boiling point of the transition metals, showing how manganese has anomalous behaviour. I've been looking for an explanation myself for a while now.

Well, Mn is clearly different given that the room temperature crystal structure is cbcc_A12, rather than standard bcc for Cr and Fe (although Fe would not be bcc except for the magnetic term in the free energy). For Cr and Fe, the cbcc phase has more than 10kJ/mol excess free energy over the stable bcc at room temperature.  Mn than transitions to cub_A13, than fcc_A1, than finally bcc_A2 before melting.  Clearly Mn has difficulty figuring out what it wants to do, bonding wise. However, cbcc_A12 Cr would melt around 1000K, and cbcc_A12 Fe not much higher.

I think the dip in melting point curve for Manganese is because while the others release 2+ electrons into the pool and get the Argon configuration, Manganese releases 2 to get half-filled stability...then as the number of electrons released increases, the melting point increases again...until zinc releases 2 electrons for fully-filled stability.

For point 3-see this:-http://chemistry.stackexchange.com/questions/4766/melting-and-boiling-points-of-transition-elements


text:
Hybridization of Ester Oxygen
I'm on the email list of an Organic Chemistry help website, and they sent out this problem:
http://orgomadesimple.com/june-9-email-practice-problem/
This got me thinking:

What if the nitrogen here was replaced with another oxygen atom?
Would it still be $sp^2$ hybridized?
Can oxygen molecules hybridize?

Thanks in advance for your help!

Changing that nitrogen to an oxygen would make an ester, right?

It's similar, but not quite the same. I'd call it borderline.


text:
Can Oxybuprocaine hydrochloride react with lidocaine hydrochloride?
To anaesthetise the mucous membrane 2 separate drugs were given topically:
Oxybuprocaine hydrochloride and Lidocaine hydrochloride. The reason to use both of them seem to be beginning of action and its duration (the former starts faster but shorter, the latter starts later but longer acting).
If instead having them in separate bottles they will be mixed in one, would it cause any interactions between them?
oxybuprocaine:

There likely wouldn't be a chemical reaction between them. Why would you want them in one bottle?

@jerepierre It is more convenient and a part of actual (nondocumented) practice... I wanted to know does it affect the efficacy, thus I asked this question.

Looking at the functionality, oxybuprocaine looks more likely to react with itself rather than lidocaine. As it's a salt and also no doubt from numerous medical trials, this doesn't occur,  the anticipation would be that the two compounds would be fine.  The $pK_a$'s of the tertiary amines would also be expected to be similar, so one shouldn't neutralise the other.

Could you presume the reaction of oxybuprocaine with itself?

If it wasn't a salt, you could make the aniline cleave the ester given enough poke


text:
When is equilibrium analysis appropriate?
What is a good rule of thumb to use to determine if equilibrium is the appropriate assumption?  I assume it is mainly to do with the time scales of the problem, meaning if the reaction time scales are very short, we could probably assume that we get to equilibrium very quickly.  Is there an accepted metric for determining if the equilibrium assumption is valid?

Are you asking how long it takes to reach equilibrium?

bon - That is part of my question, but the other part is what other factors would prevent this assumption.  For example, if one of the system elementary reactions included an irreversible reaction, I would assume equilibrium analysis would not be ideal.

An irreversible reaction still goes to equilibrium. It's just the equilibrium is so far to the right that only a negligible amount of reactants will remain once it is reached.

Understanding this sort of time scale analysis is one of the key skills to learn in graduate chemical engineering transport phenomena classes: e.g., https://stellar.mit.edu/S/course/10/fa14/10.50/index.html.

Brian - I had tried looking into online course content.  Unfortunately the 10.50 website requires the mit kerbos certificate, and the OCW and DSpace versions don't have much of the course content posted.  Can you recommend any other online course resources?  I'm currently working through Dewitt (heat transfer & transport).

It depends on the context.
The accepted threshold is "when everything else is slow by comparison." For example, if you have a reaction that takes place in two parts, a Jones oxidation taking a secondary alcohol to a ketone and Diels-Alder (which is far slower), you can assume for modeling purposes that the oxidation will be in equilibrium whenever you're looking at how the Diels-Alder will interact with the oxidation.
By contrast, if you're reacting something like a $\ce{CN7}$ anion (which will react very very quickly, detonating with no provocation), the Jones oxidation can't be considered in equilibrium by comparison.

This sounds similar to the arguments for the partial equilibrium approximation.  Compare the relative reaction rates (timescales), the fast reactions are assumed in equilibrium and the slow reactions are assumed to be the rate limiting step.  Can a similar argument be applied to flowing systems, meaning if the timescales for the reactants to flow through the system are faster than the reaction timescales, I can assume a frozen flow, but if the flow is relatively slow I assume it is in equilibrium?  If so what's a good metric for "faster", 1 order of magnitude ?

That sounds about right. And the metric for "faster" is generally "as loose as I can get away with." Modeling multiple reactions not at equilibrium at once is quite a bit more complicated than just assuming one reaction is at equilibrium, so as long as the excuse exists, the assumption will generally be that one of the reactions is at equilibrium. There isn't a set quantitative cut off point though.

One clarification: If multiple phenomena have 'fast' time scales, then they will *inter-*equilibrate.  Also, one has to think a bit about the time scale on which one is interested in the dynamics of the system.  Any processes 'significantly faster' than this time scale can generally be assumed to always hold at mutual equilibrium.


text:
How to proceed when percentage "exchange" is given in a reaction?

If $3.7~\mathrm{mol}$ of $\ce{Cu}$ reacts with an excess of $\ce{Cl2}$ and the exchange is $45\,\%$, how many moles of $\ce{CuCl2}$ can at maximum be created?

So I start with write the reaction formula
$$\ce{Cu + Cl2 -> CuCl2}$$
and then balance it.
$$\ce{2Cu + 2Cl2 -> 2CuCl2}$$
I know that I have $3.7\ \mathrm{mol}$ of $\ce{Cu}$, which is $235.1202\ \mathrm{g}$ of $\ce{Cu}$.
I assume that $\ce{Cu}$ is limiting reagent and $\ce{Cl2}$ is an excess reagent.
This is where I get stuck, I don’t know how I’m supposed to proceed with the only information that the exchange is $45\,\%$.

Your "balancing" consisted of multiplying the whole equation by 2. Your original equation was balanced

We have this equation of reaction:
$$\ce{Cu + Cl2 -> CuCl2}$$
Which means that $1~\mathrm{mol}$ of $\ce{Cu}$ reacts with $1~\mathrm{mol}$ of $\ce{Cl2}$ to form $1~\mathrm{mol}$ of $\ce{CuCl2}$.
Therefore, we would expect $3.7~\mathrm{mol}$ of $\ce{Cu}$ to produce $3.7~\mathrm{mol}$ of $\ce{CuCl2}$ if there is excess $\ce{Cl2}$, which is stated in the question.

However, the exchange is only $45\%$, which means that only $45\%$ of $\ce{Cu}$ is converted, making we can only create $3.7\times45\%=1.665~\mathrm{mol}$ of $\ce{CuCl2}$.


text:
Why isn't Mg deposited on the cathode after electrolysis?
Well here's a question from my book:

A solution containing one mole per liter of each $\ce{Cu(NO3)2}$, $\ce{AgNO3}$, $\ce{Hg2(NO3)2}$ and $\ce{Mg(NO3)2}$ is being electrolysed by using inert electrodes. The values of standard electrode potentials in volts (reduction potentials) are:

$\ce{Ag+/Ag} = +0.80$,
$\ce{Hg2^2+/Hg} = +0.79$,
$\ce{Cu^2+/Cu} = +0.34$,
$\ce{Mg^2+/Mg} = -2.37$

The sequence of deposition of the metals on the cathode will be?

The answer to the above question is: $\ce{Ag}$, $\ce{Hg}$, $\ce{Cu}$.
Why isn't Mg deposited on the cathode? Is it because the reduction potential is negative?
If yes, what exactly is the reason?

Your solution is aquaeous. There is another cation present. It very much has something to do with magnesium's potential being positive. Do you maybe see it? ;)

@Jan Water! Right. Didn't think about that. Thanks.

You are in aqueous solution. Everywhere where there is an aqueous solution, you have a certain concentration of a further cation: $\ce{H+}$. It has a potential of $\pu{\pm0 V}$ by definition, thereby being nobler than $\ce{Mg^2+}$.
Thus, hydrogen gas is ‘deposited’ on the cathode first and bubbles away while the solution gets removed.
It is not possible to deposit considerably less noble metals on the cathode in normal, aqueous solutions.

Good answer, I want to mention that spelling is "aqueous" not aquaeous. Happy to help.

@PrittBalagopal https://en.wiktionary.org/wiki/aqu%C3%A6ous#English

yup, it's an obsolete spelling.

Reduction takes place on cathode. Thus the element having most reduction potential will get reduced and get deposited. In your question the reduction potential of $\ce{Mg}$ is the least of all the ones that you have mentioned hence it will not get deposited.

It doesn't answer this.

Since the reduction potential is negative, the process becomes non spontaneous. Moreover, there are hydrogen ions already present having zero assumed reduction potential. Therefore, Mg won't get deposited


text:
For the half reaction HSO3- -> S2O6^2- why is it that 2e- are on the right hand side when balancing and not one electron?
$$\ce{HSO3- -> S2O6^2-}$$
The oxidation number of $\ce{S}$ on the left is +4 and +5 so when adding electrons to balance the oxidation number why is $\ce{2e-}$ added to the right instead of one?
Why is 2 placed in front of the $\ce{HSO3-}$?

This seems like a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed.

Also, HSO3 has a negative charge on it.

No it's not a homework question. I'm finished school and I am revising. I know that the balanced equation is 2HSO3-  --> S2O62-  + 2H+ + 2e- . I'm just confused about why there are 2 electrons instead of one when you balance it.

A "homework question" is any question whose value lies in helping you understand the method by which the question can be solved, rather than getting the answer itself. This includes not just questions from actual homework assignments, but also self-study problems, puzzles, etc.

@Jasper. What are the requirements for a balanced equation?

well usually when I'm balancing redox reactions, I find the oxidation numbers of the atoms and then balance the oxidation numbers using electrons. Then I balance the charges using H+ or OH- depending on whether its an acidic or basic reaction and balance o and h atoms using H2O if needed. I'm sure it's something simple but I don't understand why there are 2e- instead of just one on the right.

there's two s's on the right so you need two s's on the left so you balance it by adding 2 in front of HSO3-  so is that why you multiply the oxidation numbers by 2 so it becomes +8 on the left and +10 on the right?

@Jan oh woops i see that now just below where I'm typing. I don't really get the logic of multiplying the oxidation numbers by 2 as i thought the coefficient had no effect on the oxidation number?

@Jan thanks! I always over think these type of things :P

Why is $2$ placed in front of the $\ce{HSO3}$?

This is, of course, the elemental law of the conservation of mass. You have $x$ atoms that you put into your equation, you need to get $x$ atoms out again. While hydrogen and oxygen can be taken care of with water, you need to account for the sulphur immediately. Hence you need two sulphurs.

The oxidation number of S on the left is +IV and +V so when adding electrons to balance the oxidation number why is $\ce{2e-}$ added to the right instead of one?

The oxidation numbers are, of course, correct, but remember that in the first step you wrote a $2$ before the sulphur. Therefore you now have two sulphurs that you need to treat with the appropriate number of electrons — either by stealing or by gifting.
Since you have two sulphurs and they each lose one electron, you need to add $\ce{2 e-}$ to the right.

Note: The OP already understood what went wrong through hints in the comments. Unfortunately, OP decided not to write up a self-answer.


text:
Disregarding negative externalities and other barriers, what are some fuels that have a higher power density than gasoline?
Gasoline’s power throughput per kg is $12\,200\:\mathrm{Wh/kg}$ (Wikipedia.org). Are there any other fuels that have a higher power output that are not considered viable solutions as they are difficult to handle or have negative externalities?
Also what characteristics of their molecular structure give them the high energy throughput?

A table of energy released per mass of fuel (not counting other masses such as oxidizer) can be found in the energy density article on Wikipedia. Also of interest is that Wikipedia states (without citation) that the chemical reaction which produces most energy per total mass of reactants (fuel+oxidizer) is the combustion of beryllium metal, producing beryllium oxide.

This might be moved to an Answer.

I would suggest you take a look at what they use as rocket fuel.  There are a lot of fuels with higher power densities but they wouldn't work well on the highway.
The Navy has had great success with powering vessels on nuclear energy.  That is the ultimate top of the viable fuel density list.  Maybe one day fusion will become viable, then it will be at the top of the list.
As for the second part of your question, the characteristics of chemical fuels have to with free radicals.
Any viable fuel needs to be readily split into component parts (triggered by a spark or sudden compression or the like) that then combine in other configurations in an exothermic reaction so you get a net energy output.


text:
On adding sugar to water, does the entropy increase or decrease?
This question is a little confusing. According to my teacher, the entropy should decrease, but I think that sugar is crystalline first and later it dissolves, so the randomness increases and so the entropy should also increase. What is the correct answer?

Why do you only consider the sugar in your perspective?

@M.A.Ramezani I think that the entropy of both should increase, not only sugar.

While it may seem that randomness always increases when a crystal is dissolved into a liquid phase, it does not have to be that way.
Concerning sugar, the molecule has a large number of hydroxyl groups and is generally rather large when compared with the water molecule around it — much larger than your average sodium or chloride ions. Every hydroxyl will act as both a hydrogen bond donor and an acceptor, likely to two different water molecules, creating a cage of water that is rather ordered and large again when comparing to the sodium or chloride ion’s solvation cell. This will decrease the disorder (and hence entropy) of the free water so much, that it counteracts the increase in entropy of the crystal dissolving. (The effect is more pronounced with sucrose when compared to glucose due to the molecule’s size and the number of hydroxyls.)
On the other hand, the molecules in a sugar crystal are likely less ordered than the ions of a salt crystal. They will form extensive hydrogen-bonding networks, yes; but overall if you shift one row of molecules relative to its neighbour you won’t necessarily break the crystal as you won’t suddenly have ions of the same charge next to each other. And finally, the single salt crystal will break down into much smaller fragments than the sugar molecules.

So summed up, the effects are:

large solvation cells
crystal breaks down into larger (less disordered) fragments upon solvation
sugar crystals themselves are less ordered than salt crystals

Meaning that there is a lot less entropy to be gained.

So that means enough heat is released to compensate for the loss of entropy ?
Also I don't understand why sugar has such a high solubility even though each sugar molecule requires a large cage of water molecules ?

@olivieradam666 Well yes, if $\Delta G$ were not negative, sugar wouldn’t dissolve. Since $\Delta S$ seems to be negative, $\Delta H$ must be even more negative to outweigh $- T\Delta S$. The high solubility can maybe be explained by considering very high sugar concentrations more like sugar with dissolved water, but I'm not sure if any accepted theory has been established so far on when substances show high solubilities.

@olivieradam666 See also this answer of mine, where Nicolau states in the comments that sugar's solubility increases linearly with temperature. Remember that $\Delta H$ is also a function of $T$, so likely the contribution of $\Delta H$ always outweighs $- T \Delta S$.

Hi.  This is wrong; sucrose dissolves endothermically.  See: http://webserver.dmt.upm.es/~isidoro/dat1/Heat%20of%20solution%20data.pdf   https://books.google.com.co/books?id=TYTnheXYWpUC&pg=PA208&lpg=PA208&dq=sucrose+enthalpy+of+solution&source=bl&ots=q-wp0rfB0V&sig=L80K8VrkezpNu2fHs5GOtZmqvvE&hl=es-419&sa=X&ved=0ahUKEwi8-f6e5KLVAhUIOj4KHb-pCQgQ6AEIczAJ#v=onepage&q=sucrose%20enthalpy%20of%20solution&f=false  http://pubs.acs.org/doi/abs/10.1021/ja01164a504
When we observe a linear increase in solubility with temperature, the simple explanation is that the process is entropically favorable!

The accepted answer is clearly lacking in emphasizing that no matter what consideration you might have the total entropy of the whole system will always increase (or very rarely remain the same). All questions of the type "does the entropy increase or decrease" in such or such situation should always be answered with yes, unless it is specified that we are only taking into consideration the entropy of a subsystem of any interaction. The answer omits to mention the heat released increasing the temperature (of either the sugar or a heat bath that it is in contact with). Thus entropy still rises.

The second law of thermodynamics states that a process in an isolated system will be spontaneous only if the entropy of the system increases as a result of the said process. Since sugar spontaneously dissolves in water, the total entropy of water and sugar has to increase when sugar dissolves.

You're the correct one here.
Entropy will go up when a solid becomes aqueous (or dissolved at all) or when liquids mix. This is because there are an increased number of microstates that are indistinguishable from each other when a solid is dissolved or when liquids mix.
(However, do note that if you dissolve a gas, entropy goes down)

I agree with @Jan that this is too broad a generalization, and it is simply not true in many cases. See Jan's answer for a more nuanced point of view. In short, OP is not correct here. The effect of the solute on the organization of the solvent was not taken into proper consideration.

The organization of the solvent is not sufficient to overcome the entropy of the dissolving sugar. This can be deduced from one simple fact: sugar dissolves faster when the solution is heated. Ergo, this implies first that the dissolving has an enthalpy >0. Second, the dissolving of sugar is a spontaneous process, especially as the temperature goes up. If entropy was less than zero, this would be impossible. (G = H-TS; stick deltas as appropriate). Ergo, entropy of the process MUST be greater than zero.

This is so over generalising that it is wrong. While still acceptable in a rough generalisation, it is by reading your comments mixing thermodynamics and kinetics that force me to think that you have confused ideas.


text:
Can the thermodynamic predictions of redox reactions based on E and dG contradict each other?
I am referring to the following three reactions. Two of these reactions are imaginary ($\ce{M}$ and $\ce{X}$ are imaginary).
$$\begin{alignat}{3}
\ce{X- + 2e- \;&<=> X^3-} \qquad &&E^\circ_{\ce{X^3-}/\ce{X-}}=0.9\ \mathrm{V}\qquad &&&\text{(a)}\\
\ce{2H+ + MO^2+ + e- \;&<=> M^3+ + H2O} \qquad &&E^\circ_{\ce{MO^2+}/\ce{M^3+}}=0.5\ \mathrm{V} \qquad &&&\text{(b)}\\
\ce{8H+ + MnO4- + 5e- \;&<=> Mn^2+ + 4H2O} \qquad &&E^\circ_{\ce{MnO4-}/\ce{Mn^2+}}=1.5\ \mathrm{V} \qquad &&&\text{(c)}
\end{alignat}$$
Now consider the reaction between a solution having equimolar amounts of $\ce{M^3+}$ and $\ce{X^3-}$ and a $\ce{KMnO4}$ solution. (Note that all the species are in their respective standard states, so there is no need for Nernst equation.)
To find out possible reactions, following steps are taken.
Equations $\text{(a)}$ and $\text{(c)}$;
$$
\begin{align}
&{-}\text{(a)} \times 5 + \text{(c)} \times 2 : \\
\ce{&16H+ +5X^3- + 2MnO4- -> 2Mn^2+ + 5X- + 8H2O} \qquad E^\circ_1\qquad\text{(1)} \\ \\
&E^\circ_1 = 1.5\ \mathrm{V} + (-0.9\ \mathrm{V}) = 0.6\ \mathrm{V} \\
&\Delta G^\circ_1 = -nE^\circ_1F = -10 \times 0.6 \times F = -6F
\end{align}
$$
Equations $\text{(b)}$ and $\text{(c)}$;
$$
\begin{align}
&{-}\text{(b)} \times 5 + \text{(c)}: \\
\ce{&5M^3+ + H2O + MnO4- -> Mn^2+ + 5MO^2+ + 2H+} \qquad E^\circ_2 \qquad(2)\\ \\
&E^\circ_2 = 1.5\ \mathrm{V} + (-0.5\ \mathrm{V}) = 1.0\ \mathrm{V} \\
&\Delta G^\circ_2 = -nE^\circ_2F = -5 \times 1.0 \times F = -5F
\end{align}$$
My questions:

Are there any errors in above calculations? If so, please suggest corrections.
According to $E^\circ$ values, reaction $(2)$ is more feasible than reaction $(1)$. Am I correct?
According to $\Delta G^\circ$ values, reaction $(1)$ is more feasible than reaction $(2)$. Am I correct?
Is it useless to use $E$ values to predict the feasibilities of redox reactions? Should we always resort to $\Delta G$?
Is it possible for predictions based on $E$ values and $\Delta G$ values to contradict?
If not how can these results be explained?

Standard potentials are written for reduction reactions and you wrote the first one for an oxidation reaction.

(a) is written as a reduction.

$\Delta G = -n {\mathcal F} \Delta {\mathcal E} $
$\mathcal F$ is Faraday's constant, which physicists currently believe has been constant throughout the universe for all time.  $n$ is the number of electrons transfered in an electrochemical reaction.  For a given reaction, $n$ is also constant in all places and times.
Thus, $\Delta G$ determines $\Delta {\mathcal E}$ and vice versa.

Are there any errors in above calculations?

Yes, there must be.

According to E∘ values, reaction (2) is more feasible than reaction (1). Am I correct ?
  According to ΔG∘ values, reaction (1) is more feasible than reaction (2). Am I correct ?

Both of these cannot be correct.

Is it useless to use E values to predict the feasibilities of redox reactions ?

No, it is just as good and exactly as informative as $\Delta G$.

Should we always resort to ΔG ?

No.  Feel free to resort to $\Delta G$ if you like units like kJ/mol instead of volts, but other than personal preference there is no reason to prefer one over another.

Is it possible for predictions based on E values and ΔG values to contradict ?

No.

If not how can these results be explained ?

UPDATE: Arithmetic error.
The choice of $n$ is arbitrary.  Once you pick $n$ for a given half reaction, it determines the choice of $n$ for the other half reaction, but that's it.  If you then consider a separate cell with separate reactions, you could write that new cell's equation with an arbitrary $n$.  The $\textbf{mol}$ in $\Delta G$'s units of $\frac{\textrm{ kJ}}{\textbf {mol}}$ is $\textbf{mol}$ of reaction, and thus the value of $\Delta G$ depends on how you write the reaction.  For example you could write equation (b) + (c) as
$$\ce{10M^3+ + 2 H2O + 2 MnO4- -> 2 Mn^2+ + 10 MO^2+ +4 H+}$$
Now, $n$ is twice as large, so $\Delta G$ is twice as large.  But $\Delta \mathcal E$ is unchanged.  This is because $\Delta \mathcal E$ has a basis of "$\textbf{mol}$" of electrons, not of moles of a reaction of arbitrary scale.  In reality the "$\textbf{mol}$" of electrons is converted to a different units (Coulombs) to make things more confusing.  (Remember, a volt is a joule per coulomb.)
So contrary to what I said above, the sign of $\Delta G$ but not the magnitude of $\Delta G$ is fixed, meaning that whether a reaction is absolutely spontaneous can determined by either $\Delta G$ or $\Delta \mathcal E$, but to compare the relative spontaneity of two reactions, either the $n$ values of the two reactions must be the same and then $\Delta G$ values can be used, or else without worrying about $n$ you could just look at $\Delta \mathcal E$.

Thanks for the answer. Can you please point where the arithmetic error is ? Could it be in the provided $E^{\circ}$ values of imaginary reactions ?

Doesn't $\Delta G$ has the unit kJ or J because, $\Delta G = -n (mol) \times E (JC^{-1}) \times F (C mol^{-1})$ ?

(Not too good with MathJax, so sorry about the lack of elegance. If someone could suggest a good tutorial, I'd appreciate it.)
Your methods are all correct, but it seems you are not comparing apples to apples. The $\Delta G$ of a particular reaction is going to scale with the amount of reactants you have. For example, the $\Delta G$ of $\ce{10N2 + 10H2 -> 10NH3}$ is going to be 10 times that of $\ce{N2 + H2 -> NH3}$. The same reasoning can be applied to your equations. So, in your first calculation of $-(a)\times5+(c)\times2$ the $\Delta G$ of $a$ and $c$ are 5 and 2 times greater than what they should, respectively. This is not a fair comparison.
What you need to do is to normalize your equations such that the same numbers of electrons are being transferred. You can do this with any integer, but it's probably easiest to just normalize to 1. In your first calculation, normalizing to an electron transfer of 1 leads to:
$$\ce{8/5H+ + 1/5MnO4- + 1/2X^3- -> 1/2X- + 1/5Mn^2+ + 4/5H2O}$$
$$E^\circ= 1.5\ \mathrm{V} + \left(-0.9\ \mathrm{V}\right) = 0.6\ \mathrm{V}$$
$$\Delta G^{\circ} =−nE^\circ F=-1\times0.6\times F=-0.6F$$
Applying the same normalization to the second equation yields:
$$E^\circ= 1.5\ \mathrm{V} + \left(-0.5\ \mathrm{V}\right) = 1.0\ \mathrm{V}\\
ΔG^\circ=−nE^\circ F=-1\times1.0\times F=-1.0F$$
The standard potential and the $\Delta G$ of the second overall reaction of half reactions b and c are in fact more negative than the first overall reaction. In case you're wondering why $E$ does not scale with the size of the reaction it's because $E$ is proportional to $G/n$. As you scale the size of the reaction, both $G$ and $n$ will grow or shrink in the exact same proportion and thus $E$ will always stay the same.

Welcome to Chemistry.SE. Our own in house tutorials are here and here. The second link has further reading suggested.

Also,  this page and this one are good complements to what @Ben suggested.

Yea, those links help a lot. Can't wait to make use of them next time. Thanks editors for cleaning up my garble.


text:
Has a non-carbon-based form of life been discovered since 2010?
In 2010 NASA announced the discovery of microorganisms that, after being treated in laboratory, could keep growing only using arsenic, instead of the (though very small amount of) phosphate they also had in nature.
However, this was confuted by further studies in the following two years.
Has there been success in finding "alien" life on Earth, somewhere else, since then?
Edit: the NASA article from back then says what gilleain sais below. I naively hoped a couple of articles from different (not scientific, but not bad) journals were not modifying the facts. Thus the question actually is if non-carbon-based life has been found.

Wait...You tag this [tag:organic-chemistry] and exclude Carbon explicitly from your question?

@M.A.Ramezani Oops. I thought I could call it still organic, since we're dealing with life. Like unconventional organic chemistry, rather than inorganic. Though I've removed the tag.

Not only was it refuted, but the original claim did not go so far as to say that it was 'non-carbon-based' life, only that the arsenic was used in place of phosphate in the DNA backbone.

@gilleain Correct, the NASA article from back then says what you say. I naively hoped a couple of articles from different (not scientific, but not bad) journals.

@M.A.Ramezani Would it have to be called inorganic, then?

Yes...But that tag in unimportant, since the main tag here is [tag:chemical-biology].

Just because an organism might need a nutrient, whether arsenic or cobalt (which homo sapiens needs for making RBC's) does not mean that the organism is not carbon-based. What had been announced was that an organism used arsenic to replace phosphorus, not carbon, which has since been amended.
There is also the question whether the GFAJ-1 microbe is a facultative or obligate user of arsenic, or simply excludes arsenic. Though the organism can survive high concentrations of arsenic, it appears to do so by discriminating between As and P and using the phosphorus in its metabolism.
There are organisms such as Sulfurospirillum arsenophilum and S. barnesii that are at least facultative As users, deriving energy from oxidation or reduction of arsenic compounds, but the arsenic is not replacing carbon, much less phosphorus, in their biochemistry.

Yes, your answer and gilleain's comment made me understand what some not scientific  articles didn't. I meant to rephrase the question: has non-carbon-based life ever been found on Earth? Thank you.

No... however,, when we finally create self-constructing, self-repairing machines (with additional qualities we use to define "life"), will we have created non-carbon-based life? See https://en.wikipedia.org/wiki/Artificial_life and https://en.wikipedia.org/wiki/Golem for example.

Indeed I had thought of that. But that's not even chemistry, is it? :D

Not necessarily, as I can see. Gratitude.

Possibly silicon "chemistry"... both for IC's and aluminosilicate clay ;)


text:
Oil refinery - How many plates are needed for a 10 degrees difference in temperature?

I am having some trouble going about this question. My first thought is to set up an ASTM separation equation to figure out the difference, but I do not know how to incorporate the reflux ratio which is 0.5.
ASTM: TBP 50% --> 550 (heavy oil)-300(light oil)= 250 degrees
reflux ratio is 0.5. Do I multiply the ratio by number of plates and then take $\frac{\pu{250 ^\circ C}}{\text{reflux}\times \text{number of plates}}$ to get the answer of 10 degrees?

Could you please edit in what you have done.

This is a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed.


text:
What is the purpose of UV-VIS Spectroscopy?
I understand the purpose behind infrared spectroscopy and mass spectroscopy, but UV-VIS doesn't make much sense to me.
Through research, I see that it has to do with electron orbitals, and it has a relation to Beer's Law.
But what exactly can I learn about a molecule from its absorbance/transmittance of UV/Visible light?

Emission and absorption in the visible and UV can uniquely identify the material in question.

@JonCuster How so?
I mean if I have an unkown compound and run it with a UV VIS spectrophotometer, wouldn't I have to go through thousands of other compounds to see if the spectra match to identify the sample?

And you don't have to do that for IR and mass spec?

It's another way to identify chemicals

@JonCuster Sorry, I haven't taken organic chemistry. I was just curious about the subject so forgive my ignorance. But doesn't the IR yield info on functional groups at least.

Often, IR energies match to vibrational states in a molecule.  Vis/UV will probe electronic states.  You might very well need both to identify something.  Consider how useful color vision is to rapidly distinguishing, say, chlorophyll from xylophyll.  All the different phosphor materials have different absorption and emission lines, lifetimes, etc...  Often, just as for IR or mass spec, you will need to use some other info to help you out.

The broad absorption bands of the electronic transitions observed in UV/Vis spectroscopy/spectrophotometry, as well as the myriad types of electronic transitions that might lead to absorption at a given wavelength, indeed make the technique much less powerful for identification of unknown compounds as compared to, e.g., IR, NMR, microwave, and mass$^*$ spectroscopies.
Jon Custer is correct in his comment to the question that UV/Vis spectra can provide key information in identifying unknown species.  However, I know of only two approaches to obtaining predictions of a given UV/Vis spectrum:

Experimental spectra of compounds suspected to be structurally related to the unknown species
Quantum chemical computation of the electronic transition spectra of candidate species

The former approach requires the relevant experimental data to exist; the latter is not trivial to perform accurately. Both suffer from significant uncertainties: How similar is a reference experimental spectrum to that of the unknown species? Can I trust that my quantum computation will give realistic band locations and strengths?

In my experience, the strength of UV/Vis lies in its facility for inexpensive, convenient, routine, non-destructive quantitation of known solutes, with known absorption spectra and/or extinction coefficient(s) at given wavelength(s).  The following list of strengths is almost certainly incomplete:

The instrumentation and consumables are relatively inexpensive.
No auxiliary cooling, gas supply, etc. are required.
Preparation and analysis of fixed-volume samples is trivial: pour liquid into cuvette; place cuvette into instrument; measure.
Continuous analysis of dynamic systems (e.g., chromatographic eluates and ongoing reactions) is possible using a flow cell.

A simple example is described here, unfortunately paywalled.
A wide variety of configurations is presented at a commercial vendor here.
(No endorsement intended. There are numerous other suppliers out there; I chose this one because it had a good variety shown in schematic.)

Interference from most solvents is minimal, particularly in the visible range.
Beer's Law in its linear form has a relatively high dynamic range for most UV/Vis-active solutes$-$that is, the concentration where absorbance deviates from linearity is much greater than the minimum concentration detectable:

$$
\frac{C_{\max}}{C_{\min}} \gg 1
$$

Another implication of Beer's Law is that interference(s) from a small number of absorbing co-solutes are relatively straightforward to handle, either by measuring at isobestic points or by measuring multiple wavelengths and solving the linear absorbance equations for the unknown concentrations. For example, if two species $\ce{X}$ and $\ce{Y}$ are present and absorbances are measured at wavelengths $\lambda_1$ and $\lambda_2$, the concentrations can be found by solving the following two expressions:

$$
A_{\lambda_1} = \epsilon_{\ce{X},\lambda_1}\cdot C_\ce{X} + \epsilon_{\ce{Y},\lambda_1}\cdot C_\ce{Y}
$$
$$
A_{\lambda_2} = \epsilon_{\ce{X},\lambda_2}\cdot C_\ce{X} + \epsilon_{\ce{Y},\lambda_2}\cdot C_\ce{Y}
$$

The steady-state nature of the measurements allows for 'deep' examination of a particular sample.

The full UV/Vis spectrum for a given volume of sample can be measured, giving a rich "fingerprint" of the particular solution and the solutes in it
The absorbance at a particular wavelength can be tracked over time, allowing for detailed kinetic analyses

At least in the visible range, the quality of the substance being measured is directly perceptible to the human senses. This is actually quite viscerally satisfying, to see that a sample anticipated to contain significantly more of a solute does indeed produce a noticeably stronger coloration of the solution.

While IR and microwave spectroscopies share some of the above advantages, they are far less suited for analysis of solutes, as the solvent absorbance is often much more likely to confound measurements.  Further, the instrumentation and sample preparation are generally much more involved.
Mass spec is terrifically expensive and complicated, is an intrinsically destructive method, and is less well suited for direct quantitation since there is no way to know a priori what fraction of the injected sample is actually making it to the detector and thus careful calibration is required.
NMR, though non-destructive, also requires quite expensive equipment that is complicated to operate properly, and interpretation of the spectra requires quite a bit of practice.
I know too little about EPR/ESR, EXAFS, XANES, and others to comment on them here. I believe the above strengths hold in comparison to those methods as well, however.

$^*$ Technically, MS is mass spectrometry, not spectroscopy, since the measurement is of discrete ion impacts on a detector, as opposed to absorbance of portions of the electromagnetic spectrum by a detector.


text:
Ring expansion in cyclic compounds

My attempt
In the first case:
$\ce{H+}$ adds to the $\ce{OH}$ group, giving us a carbocation. The carbocation thus formed is exceptionally stable due to back bonding. I wonder why would it go under ring expansion even though the strain is not a factor here as the ring strain in a cyclobutane ring is ~$26.3\ \mathrm{kcal/mol}$, and that in a cyclopropane ring is ~$27.5\ \mathrm{kcal/mol}$.
In the second case:
Again the $\ce{H+}$ adds to the $\ce{OH}$ group, giving us a tertiary carbocation with seven hyper-conjugating structures. Why would it go under ring expansion to give secondary carbocation with just two hyper-conjugating structures? I believe is based on ring strain in this case, as the ring strain in a five-membered ring is ~$6.2\ \mathrm{kcal/mol}$, while the ring strain in a six-membered ring is ~$0.1\ \mathrm{kcal/mol}$.

Source: Advanced Problems In Organic Chemistry, MS Chouhan, 11th edition; Chapter - Hydrocarbons (Alkenes); Question 180 in latest edition

The mechanism for the first part of the question is here: Benzene from cyclopropane-1,2,3-triol in acidic medium

You overlooked the strain from the sp2 carbon in a small ring. The ring strain you have listed is based on sp3 carbon. sp2 supposed to have a planar triangular geometry (120°, compared to 109° for sp3) and will be more strained in a 3 or 4 membered ring. 6 membered ring will be best followed by 5 membered ring.
Also, you have discussed a lot about the stability of carboncations. Acid catalyzed elimination/rearrangement reaction is a thermodynamic condition. The stability of product is more important than the stability of intermediate.

Then kindly explain why the following contraction takes place?here

I will expect the carboncation after ring contraction reaction to be trapped by some nucleophile instead of elimination product. With no $sp^{2}$ carbon center involved in the product, the ring strain data will make sense in this case. Thus, resonance structure stabilization  will overcome the ring strain. This reaction should be a reaction somewhat under kinetic condition with careful control, and it may not happen under general thermodynamic condition(just heat up without any nucleophile), where I will expect ring opening to happen.

i agree  with you for the first case. But can you please elaborate on the second compound.As you said the stability of the product is of major concern and not the intermediate. So in case of thermodynamic product, whenever we have the chance to expand the ring from 5 membered to 6, the process will be favoured ?


text:
Derivation and application of the Henderson-Hasselbalch equation

Write down an expression for the dissociation constant, $K_{\mathrm{a}}$, of a very weak monobasic acid $\ce{HA}$ in an aqueous solution in terms of the concentrations of $\ce{H+ (aq), A- (aq) and HA (aq)}$ in the solution.
Hence, show that

$$
\mathrm{p}K_{\mathrm{a}} = \mathrm{p}\ce{H} - \mathrm{log}_{10} \frac{[\ce{A- (aq)}]}{[\ce{HA (aq)}]} \text{ where } \mathrm{p}K_{\mathrm{a}} = - \mathrm{log}_{10} K_{\mathrm{a}}
$$

At a particular temperature, $\pu{2.00e-3 mol}$ of the acid $\ce{HA}$ was dissolved in water and the solution diluted until the volume was $\pu{75.00 cm^3}$. When $\pu{25.00 cm^3}$ of a $\pu{0.004 mol dm^{-3}}\,\ce{NaOH}$ solution was added to that acid solution, the $\mathrm{p}\ce{H}$ of the resulting solution was found to be 6.0. Calculate the dissociation constant, $K_{\mathrm{a}}$, of the acid $\ce{HA}$ at that temperature.

Below you can see what I have tried with my current answers.

For 1, I got $$K_{\mathrm{a}} = \frac{[\ce{H+ (aq)}][\ce{A- (aq)}]}{[\ce{HA (aq)}]}.$$
For 2, I got the same answer.

But I am stuck on 3. These are the things I already know:

Moles of NaOH added: $(25/10000) * 0.004 = \pu{0.001 mol}$
Moles of acid HA present initially: $\pu{0.002 mol}$

How can I proceed further?

The reaction taking place is the quantitative reaction:
$$\ce{AH + OH- -> A- + H2O}$$
The initial concentrations are:
 $$\ce{[OH- ]_0}=\frac{0.001}{0.1}=0.01  \mathrm{M}$$  $$\ce{[AH ]_0}=\frac{0.002}{0.1}=0.02 \mathrm{M}$$
The equilibrium concentrations:
 $$\ce{[AH ]}=\frac{0.002-0.001}{0.1}=0.01 \mathrm{M}$$
 $$\ce{[OH- ]}=10^{-8}  \mathrm{M}$$
 $$\ce{[A- ]}=0.02-0.01=0.01 \mathrm{M}$$
Using the equation of question (ii):
$\ce{pH = p}K_a=6$

Hey, how did you take  [OH−]=10^−8M ?

If the pH=6, then [H3O+]=10-6M, and [OH-]=10-8M

But do we need that thing. I feels it  is unnecessary

Of course, it is not necessary here. But please see the answer as I've added another method to find Ka (using the concentration of OH-). So, you have two ways to find Ka


text:
Why is H2S2O7 stronger than H2SO4?
I've read that for an oxyacid, the oxidation number of the central atom shows the power of that acid, but here, the oxidation numbers of sulphur in sulphuric acid and pyrosulphuric acid are both the same, i.e. +6. That is confusing! How should I proceed now?

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Thank you. That's not my homework. I've read dat for an oxi-acid the oxidising number of the centre atom shows the power of dat acid. But here in this case,oxidising numbers of sulpher in sulphuric acid and pirosulphuric acid both are same,+6. That makes me confused! :/

From the homework policy link above includes not just questions from actual homework assignments, but also self-study problems, puzzles, etc.

H2S2O7 is stonger than H2SO4 due to conjugate base of olium is more stable than conjugate base of sulfuric acid.Negative charge is delocalize by two S=O bond for olium but in sulfuric acid it one .

@Totan Nandi It is more stable but mesomeric stabilization is  practically the same. It's inductive effect that matters.

@NewsteinKing thank you for writing your attempts and thoughts.

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Miessler_Fischer_Tarr)/06%3A_Acid-Base_and_Donor-Acceptor_Chemistry/6.03%3A_Brnsted-Lowry_Concept/6.3.07%3A_The_Acidity_of_an_Oxoacid_is_Determined_by_the_Electronegativity_and_Oxidation_State_of_the_Oxoacid%27s_Central_Atom

To answer this question at a more relatable level, typically the strength of oxoacids can be determined from the number of "lone" oxygen atoms in the compound--that is, the number of oxygen atoms that do not have a corresponding hydrogen. What usually happens is the central atom in an oxoacid will draw electrons from the bonded oxygen atoms, thereby making them more electronegative. This increases the polarity of the O-H bond, increasing acidity. In your example, H$_2$SO$_4$ has 2 lone oxygen atoms; while in H$_2$S$_2$O$_7$, 5 lone oxygens exist. We therefore suspect that the latter is more acidic. Although there are exceptions to this rule, for a general chemistry course it typically is valid.

Rule with oxo groups works for every central atom separately, and accounts rather for mesomeric than inductive effects.

$$\ce{H2SO4 ⇌ H2O + SO3}$$
$$\ce{SO3 + H2SO4 ⇌ H2S2O7}$$

The mutual electron-withdrawing effects of each sulfuric acid unit on its neighbour causes a marked increase in acidity. Disulfuric acid is strong enough to protonate "normal" sulfuric acid in the (anhydrous) sulfuric acid solvent system.
From the Wikipedia entry on Disulfuric acid


text:
How to detect lead paint on a bicycle?
Recently I started renovating a bicycle. The frame has been painted probably over 30 years ago, and despite holding well, big patches of rust are showing. So it's time to give the bike a new paint job. And before that, I need to sand the frame and remove the current paint.
Sadly it came to my attention that some bikes around that age have lead paint. Toxic, obviously.
I'm by no means a pro chemist. Far from that actually: if you show me a bunch of formulas, I might understand it, but I'm not skilled for manipulations neither properly equipped.
Provided that I can scratch a bit of that paint with sandpaper, what's the easiest way to test it for presence of lead? Apparently some lead test kits exist, but they're incredibly hard to find here.

Nice question! But don't you think it's a little late to worry? Unfortunately, if it is lead paint, then you've been exposed to it for 30 years.

Good point! I've owned that bike since last year only, so I really know nothing of its history. Hopefully its previous owners were not intoxicated, but as I'm about to sand it, I want to make sure I'm not inhaling lead dust. And also that I'm not releasing that into the environment. On top of all that, I'm curious :)

technically you can, but it requires special equipment.

Anyway, you can safely use abrasive blasting as long as you use proper protection (a well fitted respirator and protective goggles.)

@permeakra Any guidance as to what respirator I should use? I have FFP2 NR D here, would that be sufficient?

@aspyct there are various types of respirators. You need something with separate valve to breath out, like this http://www.amazon.com/3M-Facepiece-Respirator-Respiratory-Protection/dp/B008MCUVN4/ref=pd_sim_328_61?ie=UTF8&refRID=0XZAV1K7E2TG1XVMJZ5Q ; and

@aspyct and filters specifically tuned to stop particles and not organic vapors, like http://www.amazon.com/3M-Advanced-Particulate-Respiratory-Protection/dp/B009POHG2M/ref=pd_rhf_ee_s_cp_35?ie=UTF8&refRID=0ARJDAW2VGQ1GQM6ZKF1

@aspyct since the paint is solid, you need a very efficient way to filter out solids. However, very efficient: common 95% efficiency grade is likely unsuitable.

@aspyct also, note: protective goggles, with no ventillation, like that of a diver; or at least with filtered ventilation.

Thanks for these useful precision, @permeakra. I'll get the necessary protections :)

@permeakra sounds like a good answer!

Two chemicals are used in lead test kits: sodium sulfide or a rhodizonate salt. Sodium sulfide should be easy to obtain (or synthesize), and is useful for light-colored paint, as it produces a gray or black color change. Potassium rhodizonate can be synthesized from inositol, but you might be better off getting sodium rhodizonate, e.g. from Sigma-Aldrich for ~US$35 for 5 grams.

Btw, what if there is lead in the paint, but not in pure form (I mean it would be a molecule with lead and something else)? Would potassium rhodizonate work as well to detect it? Or would lead even be toxic in that configuration? Also, how am I supposed to get rid of the vinegar-potassium rhodizonate mix after the test? Just discard it in the sink?

If it is made of Lead (II, IV) oxide, "Red Lead", it will dissolve in dilute solutions of $\ce{H2O2}$. Lead carbonate will dissolve in a hot solution of hydrogen peroxide and acetic acid. Lead chromate will form chrome red in a solution of $\ce{NaOH}$.


text:
Condition For NaHCO3 to react to form CO2 with organic acids
My professor told us the following:

$\ce{NaHCO3}$ will react to evolve
  $\ce{CO2}$ whenever the acidic nature of the reactant exceeds the
  acidic nature of benzoic acid.

I want to know why this is. And is the given statement valid? He didn't give us the reason.  And in that case should $\ce{NaHCO3}$ react with 1,2-dihydroxy
cyclobutanedione to evolve carbon dioxide?

$\ce{NaHCO3}$ is the basic sodium salt of carbonic acid, a diprotic acid which exists only in small amounts in aqueous solution, because the equilibrium
$$\ce{CO2 + H2O <=> H2CO3}$$
lies mainly on the educt side (this is also the reason why $\ce{CO2}$ evolution is observed). When this fact is taken into account, a $\ce{pK_{a}}$ value of 6.3 at 25 °C is obtained for the first protolysis step (source). Benzoic acid has a lower $\ce{pK_{a}}$ value of 4.19 and is therefore able to liberate carbonic acid from its salts, like any other acid that is significantly stronger than $\ce{H2CO3}$.
$$\ce{C6H5CO2H + NaHCO3 -> C6H5CO2Na + CO2 + H2O}$$
Just in case that you mean 3,4-Dihydroxycyclobut-3-ene-1,2-dione, which is also known as squaric acid, then a reaction with $\ce{CO2}$ evolution is likely. Squaric acid is a fairly strong diprotic acid, with $\ce{pK_{a}}$ values of 1.5 and 3.4, respectively. The acidity can be attributed to the stabilization of the dianion by resonance:

If carbon dioxide evolution with 1,2-dihydroxy cyclobutanedione is possible depends on the $\ce{pK_{a}}$ value of this compound. However, I would expect it to be less acidic than squaric acid.

One could call the squaric acid anion aromatic...

@Martin-マーチン , How is squaric acid supposed to be aromatic without satisfying the Huckel (4n+2) pi electron rule?

@Aaron Hückel's rule is strictly applicable to monocyclic hydrocarbons only. Extending it to systems with heteroatoms is done frequently and  yields remarkable results by coincidence. (There really is not much rigour to it.) However, since the squaric acid anion has a very high symmetry (D4h) it works with the same premise. Depending how you would like to count the electrons in the pi-system, you can come up with 2, 6, or 10; all of which fit Hückel's scheme.

I'm sorry but thophene, furan, pyrrole etc all follow Huckel's rule, although they have got heteroatoms(not by coincidence). The lone pair on the heteroatom is considered to be two of those considered pi electrons, since they are delocalised.

@AaronJohnSabu no, these compounds do not follow Hückel's rule (by definition). They are indeed considered to have 4n+2 pi electrons, but there is more to these rules than this number. Just keep in mind that there is more to aromaticity than Hückel's rules.


text:
At the eutectic point, what is the effect on the melting point of adding impurities?

At the eutectic point, what is the effect on the melting point of adding impurities?

Increase the melting point
Decrease the melting point
No effect on melting point

I know adding impurities decreases the melting point of a compound in comparison to the pure compound. But, the eutectic point is the lowest temperature where the new compound (the compound + impurity) melts. I'm indecisive between 1 and 3.

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Thank you for adding your thinking - the question is much better (and quite interesting).

Phase diagrams for ternary mixtures are pretty complex (the impurity being the third member). See http://demonstrations.wolfram.com/BasicTernaryPhaseDiagram/, https://www.uwgb.edu/dutchs/Petrology/Teutect.htm and http://www.tulane.edu/~sanelson/eens212/ternaryphdiag.htm, for example.  That said, an impurity that is insoluble in the mix might have no appreciable effect on the MP.

Lets state my assumption that we can represent the free energies of the mixtures in fairly regular thermodynamics of phases ways (as I will get to here quickly). I will also assume fairly limited solid solubilities (think something like the Au-Si system).
OK, just take a system of A and B. The Gibbs free energy of a binary mixture (either solid or liquid) is:
$G(T) = x_{A}G_{A}^{sol/liq} + x_{B}G_{B}^{sol/liq} + x_{A}x_{B}H_{AB}^{sol/liq} + RT(x_{A}\ln x_A + x_{B} \ln x_{B})$, with $x_{A} + x_{B} = 1$, $G_{A,B}^{sol/liq}$ being the Gibbs free energy of pure A (or B) in the solid (or liquid) at that temperature, and $H_{AB}^{sol/liq}$ being the enthalpy of mixing of A and B.
For a eutectic, the free energy of the liquid drops below the tie-line connecting the free energies of the solid phases, and it clearly does this in such a way that the eutectic point has a composition that is intermediate to the compositions defining the solid tie-line. In a perfectly symmetric and regular system, this puts the eutectic at $x_{A} = x_{B} = 0.5$, but at least keep in mind that liquid composition has reasonably large values of $x_{A}$ or $x_{B}$, certainly larger than the minority composition in either solid end-member.
Now, we add a third element. The Gibbs free energy formula becomes longer, since it now takes in to account the binary and ternary interactions.  So, one gets:
$G(T) = x_{A}G_{A}^{sol/liq} + x_{B}G_{B}^{sol/liq} + x_{C}G_{C}^{sol/liq} + x_{A}x_{B}H_{AB}^{sol/liq} + x_{A}x_{C}H_{AC}^{sol/liq} +  x_{B}x_{C}H_{BC}^{sol/liq} + x_{A}x_{B}x_{C}H_{ABC}^{sol/liq} +  RT(x_{A}\ln x_A + x_{B} \ln x_{B} + x_{C} \ln x_{C})$, with $x_{A}+ x_{B} + x_{C} = 1$. This is pretty ugly, but we just added a little bit of C, so $x_{C}$ is small.
Now, back to the assumptions on the system being considered. For the liquid, $x_{A}$ and $x_{B}$ are reasonably large ($\sim 0.5$). In solid A, $x_{A}$ is near 1, $x_{B}$ is small(ish), and $x_{C}$ is really small.  In solid B, $x_{B}$ is near 1, $x_A$ is small(ish), and $x_{C}$ is small. Really small.
What has really changed? Well, for $x_{C}$ being really small, you can throw out all terms linear in C. The only thing left (!) is the entropy of mixing terms, where adding a little C results in a change of  $RT x_{C} \ln x_{C}$. But, this gets applied to both solids (A and B), and also to the liquid. All the free energy curves shift down by the same amount on an absolute scale. By geometry this means that the tie line connecting the two solid phases and the liquid at the eutectic point does not shift to first order in $x_{C}$.
The eutectic point will not shift.


text:
Why allene cannot be described with an allyl system?
Why is it impossible in allene to have two pi-bonds in the same orientation?
=> Why isn't allene planar? (4 $p$ electrons are in two 2x2 $p$-orbitals)
From my chemical intuition I would guess that there are 4 electrons in a $p$-system with three orbitals, but frankly I don't know how to allocate the lone pair electrons of $\ce{CS2}$ into the $sp$ and $p$ orbitals.
=> Why does $\ce{CS2}$ have another geometry than $\ce{H2CCCH2}$?
I've tried to rationalize this by allocating electrons to different hybrid orbitals (especially in $\ce{CS2}$ I was not sure how many electrons do we have in the 2x3 $p$-orbitals):

What is the hybidization of the central carbon in each compound?

Because C-C-C and S-C-S are both planar, I think both central carbon should be sp-hybridized, but I'm not sure.

Yes, that's right. Based on your diagram for sp-hybridization, those atoms have two sp-hybrid orbitals and two p-orbitals available for bonding. What is the orientation of those orbitals?

As depicted in the scheme the p orbitals must be perpendicular. How does this help?

'Why is it impossible in allene to have two pi-bonds in the same orientation?'
'the p orbitals must be perpendicular' Does this not give you the answer?

If you could agree, that $\ce{CO2}$ can be looked at instead of $\ce{CS2}$, then this bundle of question and answers will certainly help you: http://chemistry.stackexchange.com/q/10976/4945 (You still get my +1)

Why isn't allene planar?

Let's start by looking at the structure of allene and seeing what happens if we make it planar.

(image source)
As the above image shows, the 3 carbons in allene are linear (form a straight line) and the terminal hydrogens lie in orthogonal planes (e.g. the planes containing the hydrogens are rotated 90° one from the other as shown in the bottom half of the drawing).  Allene has two orthogonal pi bonds (e.g. the planes of the two pi bonds are also rotated 90° one from the other).  Each one of those pi bonds contains 2 electrons.  The terminal carbons are $\ce{sp^2}$ hybridized, while the central carbon is $\ce{sp}$ hybridized.
Now let's twist one of the terminal methylenes by 90°.  It doesn't matter which one, so let's just say we twist the one on the right, and twist it so that its red p-orbital lobe lines up with the red lobes on the other double bond.  Now our molecule still has the 3 carbons arranged in a line, but all of the atoms - including the hydrogens - now lie in the same plane; we've made our molecule planar.
To accomplish this we broke one pi bond.  In our planar molecule we have 3 pi electrons (2 from the original pi bond on the left and 1 from the terminal methylene p-orbital that we rotated) in the 3 aligned p orbitals that exist in the plane of the screen.  We are also left with 1 electron in the p-orbital on the central carbon that is perpendicular to the screen (the p-orbital that was part of the double bond on the right side of the molecule).  So we took a molecule with two stable double bonds, rotated a p-orbital 90° and wind up with a planar molecule that has 3 aligned p-orbitals with 3 pi electrons and one electron by itself in a single p-orbital.  This is a biradical, an allyl-like (it's just allyl-like because the 3 carbons in our molecule are still linear, whereas the 3 carbons in the allyl system are bent) radical and a radical in a p-orbital.  Converting 2 double bonds to a biradical is energetically uphill, it does not happen at room temperature. Allene prefers to keep its 2 stable double bond structure rather than be a planar biradical molecule because the former geometry is much more stable.

Why does $\ce{CS2}$ have another geometry than $\ce{H2CCCH2}$?

It doesn't. Carbon disulfide and carbon dioxide have the same structure and hybridization as allene.  It is just a little harder to notice because we don't have hydrogens on the ends of those molecules to "tip us off" that the two double bonds are orthogonal.
Here is a picture of the orbitals in carbon dioxide (it's exactly the same in carbon disulfide)

(image source)
Just like in allene, we have 3 atoms arranged in a linear fashion and the central carbon is $\ce{sp}$ hybridized; there are two pi bonds that are orthogonal to one another.

Now I understand allene. Is in acetone the hybridization of the oxygen atom also $\ce{sp}$ (orthogonal lonepairs) and not $\ce{sp^2}$ (trigonal)?

It is hard to know for sure just what the hybridization is for an atom like the oxygen in acetone.  It has two few substituents to let us figure out the hybridization.  However, most textbooks treat the acetone oxygen as being $\ce{sp^2}$ hybridized.  An $\ce{sp^2}$ sigma bond, a p-orbital for the pi bond and 2 more $\ce{sp^2}$ orbitals for the two lone pairs.

If we assume the oxygen in $\ce{CO2}$ to be $sp^2$ hybridized I would expect the totally four lonepairs in plane and a $\Pi$-system  that's like allyl anion with four pi-electrons. The Frost-Musulin diagram (making here for simplicity the assumption of the same coulomb and resonance integral for oxygen as for carbon) tells us that the total energy of two ethylenes ($4\alpha+4\beta$) is lower in energy than a allyl anion ($4\alpha+2\sqrt(2)\beta$). This might be the reason why $\ce{CO2}$ has a $sp$ hybridization and not a $sp^2$ hybridization, correct?

Yes, very nice analysis!  When I answer a question, many times I don't know the level of understanding the questioner has, so I'll strive for simplicity.  What you wrote is a nice mathematical description of what I meant by "Converting 2 double bonds to a biradical is energetically uphill".