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Finding the molarity of an acid in a titration
I was doing a question that asks the molarity of a 10 mL of tartaric acid that is titrated to phenolphthalein endpoint with 20 mL of 1.0 M NaOH.
I worked it out by equating the molarity of the NaOH times its volume to that one of the tartaric acid and got a value of 2.0 M but as tartaric acid is a diprotic acid does it affect this calculation or does the above method work with diprotic acids too?

Tartaric acid (2,3-dihydroxybutanedioic acid) is a diprotic acid; i.e. it is capable of donating two $\ce{H+}$ ions per molecule. Thus, $1\ \mathrm{mol}$ of tartaric acid can donate $2\ \mathrm{mol}$ of $\ce{H+}$ ions. Therefore, in order to neutralize $1\ \mathrm{mol}$ of tartaric acid, you need to add $2\ \mathrm{mol}$ of $\ce{NaOH}$.
The logarithmic acid dissociation constants of naturally occurring tartaric acid are $\mathrm pK_{\mathrm a1}=2.98$ and $\mathrm pK_{\mathrm a2}=4.34$; i.e. the second acid group of tartaric acid is roughly only as strong as acetic acid. Therefore, the $\mathrm{pH}$ at the second equivalence point is slightly alkaline (about $\mathrm{pH}\approx9).$ However, since you are using phenolphthalein, which changes from colourless to pink in the $\mathrm{pH}$ range of $8.2{-}10.0$, you are observing the correct equivalence point.

Note to OP - You have to be careful about pKa's... Phosphoric acid has 3 protons, but only two would be titrated with NaOH and phenolphthalein. $\text{pKa}_3$ for phosphoric acid is 12.319.


text:
Reactivity order in hydrogenation reaction of but-2-yne and butene isomers
Question 1:

The reactivity order towards hydrogenation of the following compounds is:

but-2-yne ($\ce{CH3-C\bond{3}C-CH3}$)
cis-but-2-ene ($\ce{CH3-CH=CH-CH3}$)
trans-but-2-ene ($\ce{CH3-CH=CH-CH3}$)
2,3-dimethyl-but-2-ene

Question 2:

Which of the following has the lowest heat of hydrogenation per mole:

cis-2-butene
trans-2-butene
1-butene
1,3-butadiene

In the first question, I understand that 2 > 3 (cis is more reactive than trans generally)  and 4th is the least reactive (steric hindrance). But I am not quite sure about the order of (1). I could think that alkynes are more reactive toward hydrogenation because they have more pi electron density because of which its adsorption on the surface of metal catalysts will be easier. Is that the reason?
I have also read that alkynes are more reactive because they have less number of substituents around them. Which logic is correct and why?
I am more confused in the second question since there are just so many factors to compare - more no of pi bonds (more electron density) (according to my above logic, it should increase reactivity), steric hindrance, increased reactivity of cis isomers. I can make out that a > b (cis is more reactive than trans)  and d > c (more electron density) in terms of reactivity toward hydrogenation but how do we compare the others?

This earlier answer may prove helpful.

Related: Which has higher rate of hydrogenation - methyl propene or trans-2-butene? and Are geminal disubstituted alkenes more stable than their cis/trans isomers?

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkynes/Reactivity_of_Alkynes/Catalytic_Hydrogenation

Monosubstituted alkene is more reactive than disubstituted because directly attached alkyl groups stabilize the system. Therefore 1 butene is more reactive.
Cis is more reactive than trans because Cis compounds are unstable. As they have similar groups on same side, it results in stearic crowding.
Thirdly, conjugated dienes are very stable beacuse they are resonance stabilized.
Thus order will be
1 butene will be most reactive, then Cis, then trans and then 1,3 butadiene..

Correct reason for 2nd question is heat of hydrogenation is inversely proportional to the stability of the compound. And (D) is most stable due to conjugation.

Good answer, I know a classic example of this is to compare the heat of hydrogenation of hydrogen with that of cyclohexene. The heat of hydrogenation of benzene is much less than three times that cyclohexene.

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titration of ascorbic acid
In this article. Authors use $\ce{KI}$ in potentiometric titration of ascorbic acid with permanganate and dont describe why! Anyvody know why should we use $\ce{KI}$ in potentiometric titration of ascorbic acid with permanganate?

Potassium iodide is used as an indicator. While there is ascorbic acid in solution, $\ce{KI}$ is protected from being oxidized by potassium permanganate. However, the moment ascorbic acid runs out, potassium iodide gets oxidized to iodine, $\ce{I2}$. This gives the solution a brownish color (blue if you add starch, which will also make it easier to see).

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text:
Something I don't understand about batteries
someone explained how a zinc copper battery works but there's already something I don't understand in the beginning.
Here is the reaction this person said happens in the battery
$$\ce{Zn + Cu^2+ -> Zn^2+ + Cu}$$
I understand how $\ce{Zn}$ transforms in $\ce{Zn^2+}$ (because it gives electrons to the wire, the $\ce{Zn^2+}$ then goes in the electrolyte) But I don't understand where the $\ce{Cu^2+}$ comes from. I understand that the $\ce{Cu^2+}$ gets electrons from the wire and turns into $Cu$ but the copper electrode that is in the battery is made out of $\ce{Cu}$ not $\ce{Cu^2+}$ so where does it come from?
It comes from the first answer here: https://physics.stackexchange.com/questions/247725/why-electrons-flow-through-a-wire-connected-to-a-battery

Copper ions are mostly floating around in the electrolyte. If you're just talking about a voltaic pile, there's enough oxidized copper around to supply the copper ions.

We manually put Cu ions on the electrolyte, it then reacts with the excess electrons on the electrode and makes Cu atoms.


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How can I interpret this solubility data?
I'm currently trying to make incense paper.  I've been instructed to mix ethanol 90% with potassium nitrate ($\ce{KNO3}$).  As ethanol 90% is hard to get where I'm located I'm currently using methylated spirits (95% alcohol) as an alternative.  The $\ce{KNO3}$ however is insoluble with the this.  If I dissolve the $\ce{KNO3}$ in water first then add to the methylated spirits the solubility improves but it is still very cloudy.
I went onto http://periodic-table-of-elements.org/SOLUBILITY to see if there were any suggestions.  The page had the below results for $\ce{KNO3}$ solubility with ethanol:

ethanol [$\ce{C2H5OH}$]: insoluble
ethanol [$\ce{C2H5OH}$] 50%: 3,7 (20°)
ethanol [$\ce{C2H5OH}$] 90%: 0,3 (20°)

I'm not sure what the 3,7 and 0,3 indicate though?  I'm not even sure what they refer to perhaps soluble parts per 1000 ?  If any one could help with what it means that would be great.  As you can guess I barely qualify as a hobbyist chemist.

SAFETY I'll point out that methanol and ethanol are both flammable. In a chemistry lab we'd do the mixing and drying in a fume hood to avoid a build up of vapors.

Yes, the website is a little bit ambiguous. However, there is a clue in the colour legend. The solubilities are indicated in units of grams per 100 grams of water. (This is also a very common unit for expressing solubilities.) Therefore, a solubility of "$3.7$" means that $3.7~\mathrm{g}$ of $\ce{KNO3}$ dissolves in $100~\mathrm{g}$ of whatever solvent it is listed under. This interpretation is readily confirmed by checking the values against this Wikipedia page.
As for "$50\%$", my educated guess is that it refers to a mixture of $50\%$ ethanol and $50\%$ water.
And lastly "$20^\circ$" refers to $20~^\circ\mathrm{C}$.

It may not be wise to use methylated spirits to make an incense. Also known as denatured alcohol, such solutions deliberately contain hazardous compounds to discourage consumption. Your spirits likely contain methyl alcohol, which is volatile (evaporates easily into gas form) and toxic. Excessive inhalation of the fumes from the solution can potentially poison you. I don't know your procedure to prepare the incense, but I would be wary of combusting or breathing something prepared using toxic liquids.
Here's some background from Wikipedia:
"Denatured alcohol, also called methylated spirits or denatured rectified spirit, is ethanol that has additives to make it poisonous, bad tasting, foul smelling or nauseating, to discourage recreational consumption. In some cases it is also dyed. It is usually blue in colour. Government provides it to health industries and schools so that there is no alcohol consumption in such areas, and only for experiment purposes. Pyridine and copper sulphate is added to it which is poisonous.
Denatured alcohol is used as a solvent and as fuel for alcohol burners and camping stoves. Because of the diversity of industrial uses for denatured alcohol, hundreds of additives and denaturing methods have been used. The main additive has traditionally been 10% methanol, giving rise to the term "methylated spirits". Other typical additives include isopropyl alcohol, acetone, methyl ethyl ketone, methyl isobutyl ketone, and denatonium."

Methanol burns just fine. Inhaling the combustion products is a lot different than drinking the methanol.

My main concern was the fumes, methanol is quite volatile yes?


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Are there any solids that undergo diffusion?

Are there any solids that undergo diffusion?

Introduction:
In school, my teacher is only talking about diffusion happening in liquids and I though that it may be able to happen to solids also.
Though I did think about how the concentration of the molecules in a solid was too concentrated for the molecules to pass through a cell membrane. What I am mainly wondering and would answer my question, is there a way that the molecules of a solid would break up into less concentration, in which they can undergo the process of diffusion.

I have found this class presentation: https://web.archive.org/web/20170517031823/http://home.anadolu.edu.tr:80/~esuvaci/egitim/Lecture2%20Diffusion.pdf

Diffusion also occurs in solids. For example, (intentional or undesirable) movement of impurity atoms or other crystal defects can be very important for the fabrication of semiconductors.
The classical experiment to demonstrate diffusion (and to measure the diffusion coefficient) in solids is the diffusion of gold in lead. A thin layer of gold that is labelled with $\ce{^198Au}$ is applied to the surface of a lead sample using electrolysis. The sample is heated, and the penetration of gold atoms into the lead sample is observed using the β radiation from $\ce{^198Au}$.
You could use a similar experiment to measure self-diffusion (e.g. of gold atoms in gold).


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What will the final product of this synthetic sequence be?

What is the major product formed in the following reaction?

NBS should do allylic bromination but where will it attack in the larger ring or smaller one ?

Potassium carbonate is not going to give you hydrolysis of a tertiary bromide. It will however close the halohydrin to form an epoxide (1). This intermediate undergoes Lewis acid catalyzed rearrangement to form the spiro compound that is choice (D).

Note that 1 or the carbocation could undergo certain elimination reactions to form (B) and (C). These are probably minor products.

Yes, this was what I initially thought too. The proposed SN1 reaction on the bromide does not work: the intermediate carbocation (if it is even formed to begin with) will be snapped up by the oxygen anyway.

looks very much reasonable.Thanks for pointing out @Zhe. I didn't consider epoxide formation because I thought steric hinderance might create some problem.

@Zhe I still am not clear why allylic bromination doesn't take place. My thought process was as follows: 1. Co-ordination of bromine of NBS and $\pi-$bond. 2. Formation of bromonium carbocation, with succinimide as a nucleofuge. 3. Attack by water on the bromonium resulting in the anti-stereochemistry. 4. Deprotonation by the extra pair on the succinimide. Is this correct?

@EashaanGodbole That doesn't give an allylic bromide. You're generating one of the two side products if I'm understanding your mechanism. In this reaction, the double bond is just more reactive, and bromonium formation will outcompete radical hydrogen abstraction.

Is NBS alone different from NBS in water? Here, I'd expect allylic bromination (proceeding through radical mechanism) to occur, instead of the observed HOBr addition. Please explain.

@schrodinger_16 probably depends heavily on conditions. But NBS is also quite a good as a $\ce{Br+}$ equivalent. For radical bromination, you probably need some kind of initiator, e.g. http://www.organic-chemistry.org/namedreactions/wohl-ziegler-reaction.shtm

Also, I would think that in the absence of anything else, the bromonium ion formation is probably reversible. In the presence of water though, it's readily trapped.


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What does mol L−1 mean?
I am researching how to Determination of Vitamin C Concentration by Titration for my science fair project and it says

Titrate the sample with $0.005~\mathrm{mol~L^{−1}}$ iodine solution. The endpoint of the titration is identified as the first
  permanent trace of a dark blue-black color due to
  the starch-iodine complex.

Does this mean to add iodine until it turns dark blue-black color?
What does $\mathrm{mol ~L^{−1}}$ mean?

Please have a look at molar concentration.

The units are mole per liter. $\mathrm{mol \times L^{-1}}$ is more conveniently written $\mathrm{mol/L}$. This set of units is referred to as molarity, and is a common measure of concentration for a solute dissolved in a solvent. The solute may be a solid (ionic or covalent), a gas or a liquid, but is commonly a solid. The solvent may be a gas or liquid, but is commonly a liquid. Together, the solute and the solvent form a solution.
Specifically, molarity is defined as
$$\text{Molarity} = \frac{\text{mol of solute}}{\text{L of solution}}$$
Note that the volume is of the total solution, not simply the solvent. In a dilute aqueous solution (a small amount of something dissolved in water), we may approximate the solution volume as simply the volume of solvent.


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What is the difference between the quantum numbers of the valence electrons of Carbon and the quantum numbers of Carbon?

Does the question requires something different than finding the correct sets of quantum numbers of C?
I'm using the diagram below, and I found that all possibilities fit. why are the unselected choices wrong?
It's certainly clear that the issue is in the orbitals ml, yet I can't figure out why.

Quantum numbers are a property of electrons, not atoms. Very loosely speaking each electron in each orbital can be identified with one set of quantum numbers (strictly speaking this violates the idea of quantum indistinguishability but let's not go there). So, "quantum numbers of C" doesn't make sense. There are six electrons in a C atom and each of them have their own set of quantum numbers.

I think the other two options are wrong because they violate the Hund's rule of maximum multiplicity. The orbitals with l=1 have opposite signs in each of the unselected options, but by the rule, you should have electrons having the same spin sign to have maximum multiplicity.

All the combinations listed are allowed ones for a $p^2$ configuration. (There are 15 combinations in total when Pauli exclusion is added). These combinations produce term symbols $^1\mathrm D$, $^3\mathrm P$ and $^1\mathrm D$, however, you are asked for the ground state which by Hundt's rules is the $^3\mathrm P$.
This has $M_l= m_{l1} + m_{l2} = 1,0,-1$ and spin $M_s= m_{s1}+m_{s2}= 1,0,-1$ which makes 9 combinations.
The $^1\mathrm D$ and $^1\mathrm S$ must have $m_{s1}+m_{s2}= 0$ but S has $m_{l1}+m_{l2}= 0$ and so one combination and D has $m_{l1} + m_{l2} = 2,1,0,-1,2$ making 5 combinations as $M_s=0$ only.
(The spin multiplicity superscript on the term symbol is $2M_s+1$).
The first entry in your table has $M_l=1, M_s=0 $ so belongs to the singlet state $^1\mathrm D$.
The second entry has $M_l=0, M_s = 1$ so  $\rightarrow ^3\mathrm P$.
The third entry has $M_l = 2,M_s=0$ so  $\rightarrow ^1\mathrm D$
The forth entry has $M_l = 0,M_s=-1$ so  $\rightarrow ^3\mathrm P$.


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What is the advantage of having aromaticity in biological compounds?
I am studying aromaticity, and apparently, aromaticity is common in biological molecules, just like 6-member rings (relieve ring strain and angle strain). I was wondering: is it because aromaticity makes the compound more stable to prevent unwanted reactions/modifications?
Thank you!
EDIT:
I finally found something interesting. Tyrosine, phenylalanine, and tryptophan interacts with cation-pi interactions in proteins to stabilize secondary structures. They are also over-represented in protein binding sites.

Orthogonal mechanism that comes to mind : evolution tried random molecules -> lower energy / more stable molecules were more likely to be tried -> evolution disproportionately tried aromatic molecules -> biological compounds have lots of aromaticity.

Personally I don't think this is representative of anything. Of the 20 amino acids only 4 have aromatic rings.

Ah, well, as for why they are over-represented in binding sites, that's a different question (and pretty  interesting, too).

Pi-stacking between the aromatic nucleobases in DNA helps stabilize the double helix structure.

Relared - http://chemistry.stackexchange.com/questions/60055/pi-electron-stacking-how-does-it-work


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Are 1st Electron Affinities positive or negative?
According to this ChemistryLibretexts website and this StackExchange answer,  1st electron affinites are negative and 2nd electron affinities are positive.

However, according to this Study.com website, 1st electron affinity (energy released) is positive.

Both contradicting websites address the point that when electrons are added to a neutral atom, energy is released. But the only problem is that one website says that when this exothermic reaction occurs, the 1st electron affinity is negative, while the other states that the 1st electron affinity is positive. Which one is right?

https://en.wikipedia.org/wiki/Electron_affinity tells it all

Well, I edited your question to look better. Don't use super sized font for writing paragraph. It is only used in headings. Looks like you were shouting...

I also don't get what you're saying about that question on SE. The question clearly states that the first EA is positive and the second one negative.

Well, the first electron that is taken up any atom other than a noble gas atoms or light atoms with half filled or full filled orbitals, is taken up with a release of energy, so ∆E should be negative. For the other atoms, they are already in the preferred electronic configuration, so they reluctantly take up electrons, tHus having a positive ∆E. In case of the second electron acceptance, there is repulsion of the electron by the negative charge gained by the atom, so it depends upon the situation. If the second electron acceptance leads to a stabler electronic configuration, like a noble gas, then the second ∆E may also be negative.

Well, the first electron that is taken up any atom other than a noble gas atom, is taken up with a release of energy || revisit electron affinity of II group elements.

@permeakra so mostly atoms with half filled and full filled orbitals are reluctant to accept electrons, aren't they?

yep. Large atomic size do help to lift the limitation, but still.

Yes, exactly the answer I was looking for! Thank you

This is sadly, one of those matters of "convention" that chemistry is so plagued by.
For most neutral atoms, the magnitude of $\Delta H$ for the following reaction is negative (i.e. energy is released upon the addition of an electron).
$$\ce{A (g) + e- (g) -> A- (g)} \tag{1}$$
The value of $\Delta U$ for the reaction is, in some places, called the "electron affinity". In other places, $\Delta H$ for the reaction is called the "electron-gain enthalpy".[1]
The convention that is more common (as far as I know) is to define the electron affinity as the negative of $\Delta U$ for the above reaction. This means that electron affinities, as defined in this way, are usually positive. This convention is adopted by most major introductory inorganic chemistry textbooks and is also what I learnt in undergrad. However, if you are supposed to use a different convention for your examinations etc., then do so.
The moral of the story is: in every book you read, always double check what is the sign convention. It can and will differ from book to book.

[1] To make things even more confusing, $\Delta H$ (the change in enthalpy) and $\Delta U$ (the change in internal energy) is used in different contexts. The difference between the two is given by
$$\Delta H(T) = \Delta U(T) + \frac{5}{2}RT$$
which may be understood using Kirchhoff's law since $\Delta H(0) = \Delta U$. (Alternatively there is a derivation in Housecroft & Sharpe's Inorganic Chemistry 4th ed., p 25.) Since $5RT/2$ is only roughly $6~\mathrm{kJ~mol^{-1}}$ at $298~\mathrm{K}$, this contribution is usually negligible compared to the usual magnitudes of electron affinities.

Is that difference between dH and dU is only accurate for ideal gases?

@electronpusher Yes it is for ideal gases. I should've mentioned.

To make things more confusing some authors use texts like "negative electron gain enthalpy" to refer to "electron affinity" and "negative electron affinity" to refer to "electron gain enthalpy". Glad I found this answer.

If we consider enthalpy change of e.g. atomized hydrogen ionisation and respective Delta (pV) term,  it may be funny to consider 1 mol of protons + 1 mol of electrons as an ideal gas.

Most places I can find say that electron affinity is negative when exothermic (energy leaves the system). This is also what I was taught in undergrad.
IUPAC says the opposite, that exothermic electron affinities are defined as positive (–ΔU or –ΔH). Wikipedia is the only site that agrees. IUPAC should be definitive for the convention. Perhaps what I was being taught was attempting to popularise what I would think is a more sensible convention, consistent with the standard convention for energies, especially in thermodynamics.
The contradictory conventions certainly does make it difficult for me to teach/tutor others.

When an electron is being added to a fully filled or half-filled shell, that electron would encounter repulsion due to the stability that already existed in the atom and therefore one would have to input more energy making the EA value positive.
However, when the ingoing electron doesn't encounter any repulsion, then energy would rather be given out making the EA value negative. This refers to any other case apart from the one in (1).

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pH Testing Without Cabbage/Disposal of Ferric Chloride
I have a year old bottle of ferric chloride in my basement for PCB etching that I realized that I was never going to use. What is the best way to make sure that it is neutralized without pH paper or cabbage for testing?

You can add powdered calcium carbonate (limestone) until the liquid component was clear upon settling or until the addition of calcium carbonate did not produce additional bubbling. Calcium carbonate is ideal because it is cheap, neutralizes acids to a non-corrosive pH with no need to measure, and is not itself caustic.


text:
Reaction of copper (II) ion reacting with sodium hydroxide
Based on an experiment, a solution that contained $\ce{Cu^2+}$ formed a precipitate with $\ce{NaOH}$. Is this how you write the equation?
$$\ce{2NaOH + Cu^{2+} -> Cu(OH)2 + 2Na+}$$
On the experiment: solution A is mixed with NaOH (solution turned white). Using a flame test solution A is color green. I figured solution A has $\ce{Cu^{2+}}$ ion.

Looks good to me.

Yes, the reaction is correct however it may be preferable to either include all or no spectator ions.
Use either a net ionic equations (omit the Na+), molecular equation (include the copper compound full) or complete ionic equation.
For example, if the compound is copper sulphate the equations would be (without states)
Net ionic
$$\ce{Cu^2+}+\ce{2OH^-->Cu(OH)_2}$$
Molecular equation
$$\ce{CuSO_4}+\ce{2NaOH->Cu(OH)_2}+\ce{Na_2SO_4}$$
Complete Ionic Equation
$$\ce{Cu^2+ +SO_4^{2-}}+\ce{2Na^+ +2OH^-->Cu(OH)_2}+\ce{2Na^+ +SO_4^2-}$$
Edit
In the case where you don't know what copper compound it is, use a net ionic equation. As a side note, copper hydroxide is not white. You might want to check your calculations.
If you're wondering what it is, check what other compounds will emit green.

Usually a ppt is indicated in a chemical reaction as: $$\ce{Cu(OH)2 v}$$ or $$\ce{Cu(OH)2_{(s)}}$$ Also dissolved ionic species can be indicated so: $$\ce{Na2SO4_{(aq)}}$$.


text:
Why is water considered a pure substance instead of a mixture?
Why is water considered a pure substance instead of a mixture?
Wouldn't it, when the hydrogen atoms combine with the oxygen atom, be considered a mixture, because the atoms were combined to form a new molecule?

A mixture is when two or more substances combine physically together.
However, in water, two hydrogen atoms combine with one oxygen atom chemically, forming a new substance that has properties different from hydrogen alone or oxygen alone.
For example, if you combine iron powder and sulfur powder physically (just mixing them together without applying heat), you can find that the mixture retains the properties of the original components, i.e. you can still use a magnet to attract the iron inside the mixture.
However, if you heat it up, the iron and the sulfur would combine chemically, and a new compound would be formed, which we call "iron sulfur" ($\ce{FeS}$). This is a new compound and loses the properties of the original compounds. For example, it is not attracted by magnets.

Therefore, water is not  a mixture; it is a compound and it is pure.

A mixture contains different compounds/elements. Pure water only contains 1 type of molecule/compound - H2O, making it a pure substance


text:
1,4 addition to α,β-unsaturated carbonyls?
This is a numbering problem.
This is a 1,2-addition to an α,β-unsaturated carbonyl:

This is a 1,4-addition to an α,β-unsaturated carbonyl:

I don't see why the second reaction would be called 1,4. I'd call it rather 3,4, so obviously I am not getting the numbering rule. Help?

start counting from the oxygen.

It is called 1,4 - addition because when you go through the mechanism of this reaction, then actually you will find that the carbonyl bond will break and oxygen will get negative charge.
Then nucleophilic will attack on beta carbon, i.e. carbon at 4th position and proton will migrate to negative oxygen which is at first position. Double bond will form between the carbonyl carbon and  carbon at 3rd position.
Thus, 1,4-addition takes place. But in the last step, keto-enol tautomerism takes place. And proton transfer takes place from oxygen to the carbon alpha to carbonyl. Therefore, final product is again saturated carbonyl compound.

Say you were to add a methyl cuprate to acrolein (this example is made-up but the general idea stands). The reaction is given in the scheme below:
$$\ce{H2C=CH-CH=O + MeCuMgBr -> Me-CH2-CH=CH-OCuMgBr}\tag{1}$$
The initial intermediate is not the ketone but the 4-substituted but-1-en-1-olate. The metal has been added to position 1, the organic residue to position 4. Hence 1,4-addition.
Only after aquaeous workup will the but-1-en-1-ol be formed which can tautomerise to give the aldehyde.
You may think of this as a 1,4-addition followed by 1,3-hydrogen rearrangement. Or as a 1,4-addition followed by C-protonation of the resulting enolate.

1,2-addition happens directly on the carbonyl, so the negative appears on the oxygen and the nucleophile adds to the carbon. If you're number atoms, these are 1 and 2. For 1,4-addition, the oxygen could be labelled 1 and the carbon where the addition happens would be labelled 4.
After the addition, you might have the negative species attack an electrophile, but the attack will not necessarily occur at the 1 position on the oxygen. 1,2- and 1,4- refer to the initial attack.


text:
Explain the function of this Galvanic Cell in detail

So I failed my test because of this question. It was titled Galvanic Cell although it looks more like an electrolytic cell with the external voltage and everything.
What was asked: explain the function of this cell in detail.
some keywords were also presented (not mentioning them all) but there were Gibbs, Faraday, K, and -nFE
I got too confused with the left side while Ni2+ + 2e- -> Ni (s) on the right side looks like a normal reduction in the cathode. What is going on in here?
Standard Reduction Potentials at 25 degrees Celcius:
**Cr2O7- (aq) + 14 H+ (aq) + 6 e- -> 2 Cr3+ (aq) + 7 H20 (l)  +1.33 V
                              Ni2+ (aq) + 2e- --> Ni (s)  -0.25 V**

Pt is an inert electrode that will not be involved, what is the potential of the various chromium species?

How about providing some reduction potentials so that we don't have to look them up?

EDITED: original post

Can you double-check that reduction of dichromate to manganese (II)?

Double-checked. IDK why I had manganese there. Fixed now.

in the case of electrochemical cells/Galvanic cells: Oxidation at the anode (the A and O of electrochemistry, german people will know what I mean), reduction at the cathode (in electrolysis its the other way round)
$EMF = E_{red} - E_{ox} = E_{cat} - E_{an} > 0$ for the reaction to run without external voltage (<0 in the case of electrolysis) "the higher redox potential always oxidizes the lower one"

In your case this means running the cell without applying any voltage gives an $EMF = 1.33 - -0.25 V = 1.58 V$
reduction at cathode:
Cr$_2$O$_7^{2-}$$_{(aq)}$ + 14 H$^+$$_{(aq)}$ + 6 e$^-$ $\rightarrow$ 2Cr$^{3+}$$_{(aq)}$ + 7 H$_2$0$_{(l)}$
(in the electrolyte -6e charges)
oxidation at anode:
3 Ni$_s$ $\rightarrow$  3 Ni$^{2+}_{(aq)}$+ 6e$^-$
(in the electrolyte +6e charges)
(already multiplied by 3 for the total reaction)
total:
Cr$_2$O$_7^{2-}$$_{(aq)}$ + 14 H$^+$$_{(aq)}$ + 3 Ni$_s$  $\rightarrow$ 2Cr$^{3+}$$_{(aq)}$ + 7 H$_2$0$_{(l)}$ + 3 Ni$^{2+}_{(aq)}$
Now there is however a clear symbol in the electric cycle standing for a voltage source "V". This is indeed confusing and I would ask what was the meaning of that if the figure was actually titled "Galvanic Cell" which is the other way round. So if it would't be for the title and we would consider running the cell with external voltage, this would then turn around all reactions which I wrote before.
Other things that might be relevant:

K$_2$SO$_4$ is responsible for the charge transfer between the cells. Since one cell produces positive charges in solution and the other one negative ones the electrolyte has to establish that the overall charge in both cells stays neutral. The salt bridge connects the electric cycle, without it, the cell won't run for long
the Gibb's energy is related to the EMF. $\Delta G = -n\cdot F\cdot EMF$, so again a positive $EMF$ gives a negative $\Delta G$ which means the reaction runs "alone" without external voltage, the Gibb's energy provides the driving force. n is the number of electrons involved in the redox reaction (6) (F is the Faraday constant, a natural constant)


text:
How to name compounds with more than one spiro atom and cyclic compounds with more than two fused rings?
The nomenclature of the bicyclic and spiro (mono) compounds are easy to understand. But when the number of spiro atoms or the number of rings exceeds 2, the nomenclature turns very complex. I cannot understand the method of numbering the carbons and the method of writing those bracketed numbers with dots and superscripts.
I have read the IUPAC Recommendations 2004, but the text given there is really hard to understand (for me only, I hope). I want to know the method of writing those numbers with the superscripts inside the bracket for spiro and cyclo compounds. Diagrams with arrows or a detailed text with one particular example, will be helpful.

Related: How to name tricyclo carbon compounds?

I have also read the post tagged by @Loong but it only quotes from the IUPAC book, and it is not clear.

Bicyclic compounds

Locate the bridgehead carbons, the carbons part of two rings.
Now, identify the three connections between the 2 bridgeheads, called bridges: arrange these in decreasing order of number of carbon atoms (excluding the bridgehead). Say these three numbers are m, n, o [m > n > o].
Write the name of your compound as: bicyclo[m.n.o]"m+n+o+2"alkane.
In case there is a substituent or a functional group is present, you need to number your carbons: start from of the bridge-head, through the longest bridge to the shorter and then the shortest (if any).
Other rules of IUPAC apply as such.

Spirocyclic compounds

Locate the bridgehead carbon, the only carbon part of the two rings.
Now, spot the two ring systems having having 2 or more carbons: arrange these in increasing order of number of carbon atoms (excluding the bridgehead, again).
Say these three numbers are p,q [p < q].
Write the name of your compound as: spiro[p.q]"p+q+1"alkane.
In case there is a substituent or a functional group is present, you need to number your carbons: start from of the atom next to bridgehead in the smaller ring, through the bridgehead carbon to the bigger ring.
Other rules of IUPAC apply as such.

It appears that you have not understood the question properly. I can understand the naming of bicyclo and mono spiro (i.e. with one spiro atom) compounds; I have asked for the method of naming tricyclo, dispiro, trispiro and other such compounds with more than 2 bridges or spiro atoms. Please modify your answer accordingly.

That really does not answer the question.

Spirocyclic compounds:
Smaller rings are numbered first, followed by the shorter bridge between the two bridgehead carbons in the middle ring (bridge heads named in this counting itself); the other (longer) bridge of the bridge heads is named last
Named as:
Dispiro[w.x.y.z]alkane
w = number of carbons in the smaller terminal ring
x = number of carbons in the smaller bridging ring
y = number of carbons in the longer terminal ring
z = number of carbons in the longer bridging ring

Tricyclic compounds:
Consider naming as a normal bicyclic compound
The secondary bridge is named with attachment of locants as superscript, as if it were a normal primary bridge of bicyclic compound
Named as:
Tricyclo[w.x.y.x^m,n]alkane
w = number of carbons in the longest primary ring between main bridgeheads
x = number of carbons in the longer primary ring between main bridgeheads
y = number of carbons in the primary bridge between main bridgeheads
z = number of carbons in the secondary bridge, with the locants of secondary bridge (secondary bridgehead) as superscripts


text:
In the reaction between salicylic acid and acetic anhydride to produce aspirin, why is it necessary to quench the reaction?
I understand that water is added to hydrolyse acetic anhydride into acetic acid, but what is the point of this?

Acetic anhydride is contaminating your final product. Quench simply means to destroy excess reagent before continuing. It might not matter too much in this particularly reaction since you're about to purify via recrystallization anyway, but imagine if the next step involved adding a very expensive nucleophile. That nucleophile would add to acetic anhydride and be wasted.


text:
Do non-English speaking countries use the same element symbols?
The question does sound pretty absurd, but hear me out first. The Periodic Table of the Elements, as I know it, is supposed to be a common standard adopted by the global scientific community.
However, I see a hitch when it comes to a total world-wide acceptance of such a standard.
Every Periodic Table of the Elements that I've seen in print uses the Latin Alphabet, and quite a few chemical symbols are assigned based on the Latin words for the elements ( Fe- Ferrum, Sn- Stannum, Au- Aurum, Cu- Cuprum, etc etc).
Now this appears to be a rather clever compromise reached by the English-speaking countries (the U.S.A, the U.K, Australia, etc) and countries that don't really use English, but use the Latin Alphabet (Germany, France, Spain, etc, etc). Obviously, using a "dead-language" to assign chemical symbols helps eliminate resentment that could rise by favoring one language (say, German) over another (say, English). Well, this appears to be the same logic behind the use of Latin in biological nomenclature too.
Now here's the catch;
How do countries that don't use the Latin Alphabet deal with this?
Alright, it seems credible to believe that Middle-Eastern countries (Arabic) and African countries would use the standard, Latin Alphabet Periodic Table that we use, without complaint. I guess the same goes for former European colonies too.
But what about:
1) Countries like Russia (Cyrillic) and Japan (Kanji), which have made significant contributions to the modern Periodic Table? Since they did contribute quite a bit to the Table, they probably "feel" they have the right to use their own customized Periodic Table using Cyrillic or Kanji characters.
2) Countries that vehemently oppose cooperation with Western countries, and would obviously dismiss the Latin Alphabet Periodic Table as an 'instrument of the West'. For example North Korea.

Element names might be in a different language, but the one/two letter symbols are universal.

Our element symbols are the same as yours. Element names are in Russian, but that's another story.

Chinese periodic tables make nice use of their one-character names.

Ivan (and Zhe) are right. The Chinese characters are simply the Chinese names of the elements, they are hardly meant as a replacement for the symbols. Any Chinese version of the Periodic Table will most likely have the standard symbols printed along with the Chinese characters. (And of course, you can see that perfectly clearly from the one DHMO linked.)

Even western European countries don't all use the same names for elements. But they all use the same symbols. And the structure doesn't change according to what language you speak because it depends on the electronic properties of elements which don't change with the language you use to describe them.

In Danish, sodium (Na) is natrium, potassium (K) is kalium, and tungsten (W) is wolfram. Actually, "tung sten" means "heavy stone/rock" in Danish and Swedish (W was discovered in Sweden).

I remember seeing a printout from a Chinese journal in a lab I was working in. I asked the guy if he understood Chinese. ‘No, but I can understand the reaction equation.’

Come to think of it, the majority of our element names (although written in Russian) are the cognates of yours. Some are not; these are mostly the elements which were known since antiquity. They often also happen to have different names in English and Latin: ferrum/iron/железо, stannum/tin/олово, stibium/antimony/сурьма and the like. There are also examples of elements whose English name is a cognate of Latin, but Russian is not (cuprum/copper/медь). There are also opposite examples (natrium/sodium/натрий).

In Israel and Hebrew we use the latin letters for the symbols. However, for written text we will spell out the name of the element in hebrew rather than use the symbol. For example: אני אוהב עופרת לארוחת בוקר instead of אני אוהב Pb לארוחת בוקר. ("I like lead for breakfast" instead of "I like Pb for breakfast")

I am surprised that the comment section here is almost informative as the answers themselves.

The International Union of Pure and Applied Chemistry (IUPAC) is the body that standardises these symbols, along with names and symbols for new elements.
IUPAC is affiliated with 60 different national chemistry organisations, including the National Committee of Russian Chemists, the Chinese Chemical Society, and the Science Council of Japan.
Historical precedent combined with the fact that (as far as I'm aware) the vast majority of scientific literature is published in English are obviously large factors here, but there's no massive incentive for any given scientific user to rock the boat on this, and there's a disincentive to training scientists in a system which would leave them unable to easily read scientific literature (even if you're an isolationist liebox like North Korea).
On a side note, Japan in particular recently gained the distinction of having an element synthesised there named after it (Nihonium), and while I'm not an active follower of nucleosynthetic proceedings, it wouldn't surprise me particularly if we saw a Chinese-named element some time in the next decade.

Both North Korea and South Korea use the periodic table. The names, however, are different. Here is a periodic table in North Korea

The title translates to "Mendeleev's atomic periodic table." The names of the elements are different for some.
In South Korea
Hydrogen is known as and pronounced as soo-saw.
Phosphorus is known as and pronounced as in.
Potassium is known as and pronounced as kallium.
etc.
But some are the same.
In South Korea
Magnesium is known as and pronounced as magnesium.
Argon is known as and pronounced as argon.
Fluorine is known and pronounced as fluorine.
I haven't taken classes in North Korea or visited any research facilities. But it is likely that they have further modified the names  to be "less western".
All of these are names, and do not replace the symbols used in the periodic table.

They even have their own OS: https://en.wikipedia.org/wiki/Red_Star_OS
Most of their education is brainwashing though. A university degree there is not very transferable.

While interesting, it should be pointed out that the question asked specifically about symbols, not names.


text:
Which haloalkane would yield 2-methyl-2-butene only, upon dehydrohalogenation?
I guess this alkylhalide, in picture below, would be the answer but this alkyl halide may also give 2-methyl-1-butene because there are 2 beta-carbons and therefore 2 beta-positioned hydrogens I.E.if Hydrogen on the leftside of alpha-carbon  leaves with chloride then formation of 2-methyl-1-butene would take place in stead of 2-methyl-2-butene but I need a alkylhalide which would always give 2-methyl-2-butene upon dehydrohalogenation.

You should switch the positions of the chlorine and hydrogen to be eliminated.
For your given molecule a secondary carbo-anion intermediate (at the 3/green carbon) will have to compete with the formation of a primary carbo-anion (at the 1/red carbon) intermediate to provide the desired product.

If you use 2-chloro-3-methyl-butane your desired product will form a much more favorable ternary carbo-anion intermediate (at the 3/green carbon) which will compete again with a primary carbo-anion intermediate(at the 1/red carbon).

Oh god, can't understand it. Complicated.. but I'll try or you extend it if you think it can help

@SufyanNaeem Do the images and colors help?

+1. Yes its lovely! But why tertiary carbon cation is more favourable?

@SufyanNaeem Because the more electron rich a carbon, the more stable a cation will be on it. since the methyl groups are more electron donating than the hydrogens a cation will be more stable on a more substituted group.

@Jan Simple you mean an anion but mix the two up and type cation. I have corrected the error.

@A.K. Note that tertiary anions are the least stable anions, not the most stable ones. Compare the reactivity of tBuLi with n-BuLi or sec-BuLi. But that’s probably not of too much relevance since your elimination would proceed via $\mathrm{E2}$ or $\mathrm{E1}$.

2-Chloro-2-methylbutane is correct molecule only
You will get upon dehydrohalogenation the major product as 2-methylbut-2-ene always; however, it won't be exclusive product. The other one i.e. 2-methylbut-1-ene will also be there albeit in smaller amounts
The % of the major product however could be improved using high polarity solvents, which favour SN1. Free carbocation will preferably give more substituted alkene.

But the book want me to give a conpound which gives 100% 2-methylbut-2-ene. So what should I do?

get another book. it is called a second opinion!


text:
Formation of tetrahydrofuran in the reaction of n-butanol with lead tetraacetate
When n-butanol reacts with lead tetraacetate it forms tetrahydrofuran.
And lead tetraacetate is a good oxidising agent so why doesn't it form n-butanal?
So when does cyclisation take place and why?

The reaction of lead tetraacetate, $\ce{Pb(OAc)4}$ with alkanols, $\ce{R-OH}$, is believed to proceed via an alkoxy-lead acetate, $\ce{R-O-Pb(OAc)3}$.
In this intermediate, the bond between the $\ce{Pb}$ atom and the oxygen of the alkanol cleaves homolytically, this yields an alkoxy radical, $\ce{R-O^.}$ and $\ce{Pb^.(OAc)3}$.
If the alkoxy radical has a hydrogen atom in the $\delta$ position, intramolecular hydrogen atom abstraction may occur.
By this $\ce{CH3CH2CH2CH2O^.}$ is converted to a carbon-centered radical, $\ce{^.CH2CH2CH2CH2OH}$.
The later is oxidized by $\ce{Pb^.(OAc)3}$, which leads to $\ce{Pb(OAc)2}$
and the cation $\ce{^+CH2CH2CH2CH2OH}$.
This cation undergoes cyclization to furnish tetrahydrofuran.


text:
Proof of Concentration dependence of molar conductivity
If we go by the simple electron model, Conductivity/Concentration=carrier charge*mobility and mobility is only dependent on temperature for electrons. My question is regarding strong electrolyte, I can somewhat understand the mean activity coefficient calculation using ionic strength, but still, how does the √c dependence of molar conductivity comes about in Kohlrausch Law?

http://pubs.acs.org/doi/pdf/10.1021/j150551a038 any way I can view the rest of this? other than the first page?

Yes, you can pay ACS money or find a library that has an electronic subscription.

can you give me some idea as to from where the √c dependece comes for strong and degree of dissociation relation for weak comes from?


text:
What chemical is responsible for the distinictive odor of Scotch tape?
Scotch Tape has a distinctive odor which is noticeable if you hold a piece close enough to smell. What chemical is responsible for this odor? The wikipedia page doesn't seem to mention composition.
Is the odor due to the the kind of plastic used in the strip of tape itself? Or is it due to the adhesive?

You might be smelling ozone and nitrogen oxides created during ESD by pulling the tape off the roll. See http://www.adhesivesmag.com/articles/86652-protecting-adhesive-tape-systems-from-static-electricity

This is a long answer, but according to this:

While some pressure sensitive tapes are still prepared with natural
  rubber, the majority are now made using mostly synthetic materials.
  The backing for cellophane tape usually consists of cellulose acetate,
  a synthetic derivative of cellulose, which comes from wood pulp or
  cotton seeds. The cellulose is chemically treated with acetic acid and
  anhydride, and the side that won't receive an adhesive coating is
  treated with a release agent that enables the tape to be wound and
  unwound without sticking together. Although this compound varies among
  manufacturers, some commonly used substances include stearato chromic
  chloride and polystearic carbonate. Prior to the application of the
  adhesive, the adherend side of the backing may be primed with a
  solvent or aqueous dispersions such as nitrile rubber or chlorinated
  rubber. To produce the final adhesive substance, some manufacturers
  use as many as 29 raw materials that go through various stages of
  production. However, the generic adherend is made up of acrylic
  resins, petroleum byproducts that are broken down into alcohols and
  acids before being fused into a polymer compound. This compound is
  then mixed with mineral spirits or a hydrocarbon solvent, creating an
  aqueous emulsion (a solution in which the microscopic resin particles
  are held suspended) that is applied to the backing.

In other word, there are many candidates.  I consider the top possibilities of those listed above to be: compounds composing the cellulose, the nitrile-based rubber compounds, and the semi-volatile organics composing both the adhereant and the base material.  With so many compounds going into a single product, it is likely the mixture that gives the distinctive smell rather than any individual compound.


text:
Why doesn't CO2 repel itself?
In a molecule like $\ce{CO2}$, the oxygens are more electronegative than the carbon, and so the electron cloud is denser around them. Then, isn't the oxygen going to be a negative dipole? The picture I have in my mind is a positive magnet sandwiched between two negative magnets. So my question is – why don't $\ce{CO2}$ molecules repel each other?

Related: Why is O=C=O nonpolar?

There is no such thing as negative dipole. As for the molecules repelling each other: why, they surely do, much like any other stable molecules.

Carbon dioxide is a linear molecule. Therefore, you have two directly opposite dipoles (each point from the C to one O). These cancel each other out and CO2 is actually nonpolar, as detailed in the answer to this question. Then, two CO2 molecules can interact and attract each other through london dispersion forces.
That all said, even in the hypothetical case where those dipoles didn't cancel each other out, and you somehow did have electron density at two different ends of the molecule, I still wouldn't expect CO2 molecules to all repel each other. In that thought experiment, your carbon would still be strongly positively-charged, so you would still have multiple CO2 molecules clustering via the electron density on the oxygens interacting with the partial positive charge on the carbons of other molecules.

Carbon dioxide has no permanent electric dipole. Atoms do not behave like the small bar magnets you suggest. However, the answer your question 'why don't $\ce{CO2}$ molecules repel each other' is both 'yes' they do and 'no' they don't as I try to explain.
Between one atom and another or between two molecules there is always a dispersion force even when there is not a permanent dipole. Although quantum mechanical, the origin of this force can be understood intuitively as follows. Even though the average dipole moment is exactly zero, at any instant in time, the positions of the electron may not be exactly symmetrical and so an instantaneous dipole is produced. This dipole generates an electric field that propagates at the speed of light and polarises nearby atoms or molecules, and this induces a second instantaneous dipole. The resulting induced-dipole induced-dipole interaction  produces an attractive force that has a finite time-average i.e. it is not zero.
Experiments on many gases,  for example $\ce{CO2}$, show that the ideal gas law is not obeyed exactly. To describe the experimental data   van-der-Waals extended the ideal gas law to become $(p-a)(V-b)=RT$ where b accounts for the finite volume of a real gas and a accounts for the attractive potential, one cause of which is described.
At a given temperature there is a distribution of molecule speeds, (or energies) as described by the Maxwell-Boltzmann distribution. A few molecules have a low speed, and a few very high speeds, and obviously, most fall between these limits. The distribution looks sort of like a lopsided bell shaped curve. When two molecules collide and they are travelling relatively slowly wrt one another then their mutual energy may not be sufficient to overcome the attractive dispersion energy and so may pair up for a short while until, perhaps, the pair collide with another molecule and separate. This reduces the overall pressure because it reduces the effective number of molecules in the gas that can collide with the walls.
When the relative collision energy is far larger than the attractive dispersion energy, the potential is no longer limiting and the molecules can exchange energy and momentum as collide one another and separate.


text:
Are Lewis Structures the same throughout solids, liquids, and gases?
For example, is the lewis structure for water constant even when it is in a solid, liquid, or gas phase?

Lewis structures are a representation of the atomic connectivities and electronic lone pairs present in an individual molecule of a substance. The molecule of the substance does not change in these states, so the Lewis structure would remain the same.
What does change in these states is the way individual molecules interact with each other. If you were to draw a diagram of how multiple Lewis dot structures of the molecule interact with each other, that would be different for these different states.


text:
How can sulfate be nonpolar covalent if it is a polyatomic ion?
$\ce{SO4^2-}$ is a polyatomic ion as well as a nonpolar covalent compound. How can it be ionic and covalent at the same time?
Same applies to sulfite $\ce{SO3^2-}$. It's a polyatomic ion as well as a polar covalent bond. How?

A great deal of compounds are ionic and covalent at the same time. There is nothing strange or unusual about it.

These two species have bonds within them, and (non)polar covalency refers to those internal bonds. The species as a whole can be an ion, and yet not have net polarity on its bonds.
Sulfate is tetrahedral, so each individual dipole vector would be opposed and cancelled, making the overall molecule non-polar (in the same way that this question and answer about $\ce{CO2}$ details).
Sulfate has negative charges which could make a Lewis dot structure appear to have a net dipole, there is resonance that distributes that charge throughout the molecule, negating the net effect.

Sulfite, on the other hand, is trigonal pyramidal. In this geometry, not every bond is opposed and thus there is a polarity within these covalently-bonded atoms that constitute the ion.

Polyatomic ions, usually negatively charged, are groupings of atoms bound together by covalent bonds. These groupings can carry extra electrons, giving them a negative charge, or have fewer electrons than the combined valence count of the atoms present, giving them a positive charge. Thus, despite its charge, a polyatomic ion is still composed of covalent bonds. As in uncharged molecules, those covalent bonds can be either polar or nonpolar, and the ion, as a whole, can be either polar or nonpolar, depending on the spatial arrangements of the bonds. Therefore, it is possible for a charged ion, such as sulfate, to be both ionic, and nonpolar covalent - meaning the electron arrangements among the covalent bonds are equally and symmetrically distributed.


text:
What is a typical monolayer formation time for liquid phase adsorbate on a solid surface?
Let's say a solid is submerged in a solution of species with functional groups and a self-assembled monolayer is forming. What are the typical time scales for the formation of a continuous monolayer? (seconds, minutes, days..?)
 (source)
For an adsorbate in gaseous phase and high vacuum, a rough approximation on Wikipedia is ~1 sec.

Depends on concentration and other parameters. For thiols on gold, for example, a typical concentration is 1-5 mM, and formation time is 24-48 hours; "Longer assembly times tend to result in better monolayer packing." (ref)

To emphasize the variability: I do research where I use ethanethiol + mercaptoethanol to make a gold surface hydrogen-bonding or non-hydrogen-bonding. I dip the gold-coated material in pure thiol for 30sec. We haven't characterized the packing arrangement, but our materials are noticeably non-polar (ethanethiol) or polar (mercaptoethanol).


text:
Why is there oxidation in the anode?
In a battery, I don't understand what makes the element in the anode separate from the anode to become a positive charge in the electrolyte and leave some electrons inside the anode.
Also, why does the molecule separated from the anode leave electrons on the anode?

What do you mean by ‘separate from the anode’?

So you’re basically asking why they react? The diffusion is because one can dissolve the other can’t. There would be no diffusion if there was no reaction.

For example in a zinc-copper battery, zinc^2+ ions separate from the zinc electrode and go in the solution. This leaves a negative charge in the zinc electrode which gives it a potential

Again, the ions don’t ‘separate’. The metal reacts giving away electrons to become an ion which is then free to dissolve and diffuse.

We define anode to be the electrode where oxidation takes place.
Oxidation signifies loss of electrons and Reduction signifies gain of electrons.
For example, in this electrochemical cell,

Copper has more reduction potential than Zinc. So it will be easy for copper to reduce (or gain electrons) and if copper is reducing (or gaining electrons) then there must a source of these electrons. These electrons are supplied by Zinc (which gets oxidized).
At anode, $Zinc(s)$ in losing electrons and converting into $Zn^{+2}(aq)$ ions and at cathode, $Cu^{+2}(aq)$ ions are gaining these electrons lost by $Zinc(s)$ and converting into $Cu(s)$.
These reactions occur because they are spontaneous $(\Delta G =-ve)$. Spontaeous reactions occur in nature by themselves without any external driving force. A system aims for least energy $(\Delta H = -ve)$ and maximum entropy $(\Delta S = +ve)$ and its the combination of both these factors that result in spontaneous processes. $(\Delta G = \Delta H - T\Delta S)$.

The driving force is the transfer of electrons from one substance to another. This is favored by thermodynamics. In a galvanic cell or battery, the completion of the circuit provides a path for the substance at the anode (oxidation) to give electrons through the circuit to the substance at the cathode (reduction).


text:
Why is Prussian blue ‘blue’ in colour?
Potassium ferrocyanide $(\ce{K4[Fe(CN)6]})$ is used for the qualitative analysis of ferric $(\ce{Fe^3+})$ salts. It precipitates as Prussian blue $(\ce{Fe4[Fe(CN)6]3})$.
How to explain the blue colour of Prussian blue? May the colour be due to the charge transfer between a ligand and the central atom?

Prussian blue contains both $\ce{Fe(II)}$ and $\ce{Fe(III)}$. Mixed-valence compounds, such as the one in question, absorb certain wavelengths of visible light resulting due to the phenomenon of inter valence charge transfer (IVCT).
General Scheme for IVCT is:
$$\ce{L_nM^+-bridge-M'L_n + h\nu ->L_nM-bridge-M'+L_n}$$
In this case, orange-red light around $680\ \mathrm{nm}$ in wavelength is absorbed, and the reflected light appears blue as a result.
Additionally, changes colour from blue to colourless upon reduction. This change is caused due reduction of the $\ce{Fe(III)}$ to $\ce{Fe(II)}$, and thereby eliminating the IVCT responsible for Prussian blue's namesake blue colour. This is also why Prussian Blue exhibitis electrochromic behaviour.
For a more detailed description read this and this. Alternatively, you can look it up in a textbook.

From the wikipedia article of potassium ferrocyanide:

The intense blue color of Prussian blue is associated with the energy
  of the transfer of electrons from Fe(II) to Fe(III). Many such
  mixed-valence compounds absorb certain wavelengths of visible light
  resulting from intervalence charge transfer. In this case, orange-red
  light around 680 nanometers in wavelength is absorbed, and the
  reflected light appears blue as a result(...)Prussian Blue is
  electrochromic—changing from blue to colorless upon reduction. This
  change is caused by reduction of the Fe(III) to Fe(II) eliminating the
  intervalence charge transfer that causes Prussian blue's color.


text:
"Gasoline" in Other Languages
I speak Russian and the term used for gasoline is "бензин", pronounced like benzene. I've also noticed that in Italian, it can be called "benzina". What's the significance behind this, if any? I think gasoline contains benzene, but that seems like a very minor part of the overall composition.

I'm voting to close this question as off-topic because it is about etymology rather than chemistry.

I vote not to close it. We chemists might be more suited for answering it than etymologists. On a side note, to make things worse, Russian for benzene is "бензол", pronounced like "benzol", as if it were some sort of alcohol, which is pretty confusing.

I can't speak about the etymology of the word, since I don't speak any of the languages in question (plus, it it is purely concerned with that, I believe this question is off topic here).
However, if you are interested in the contents of gasoline, I can tell you a little something about them.
Gasoline (or petrol if you're not from North America) is typically a mixture of hydrocarbons with 4 to 12 carbon atoms per molecule. It is a comprises of alkanes, cycloalkanes, alkenes, or in industry parlance parrafin, naphthenes and olefins.
Moreover, as you suspected, it does contain aromatics in general and benzene in particular, and most countries set limits on their content in the fuel. For example, in the EU the benzene content should be under $1\%$ (v/v) for automative grade gasoline$^{[1]}$ Benzene, Toluene, and Xylene (BTX) are important petrochemicals, and valuable chemical feedstocks, and most are often recovered during catalytic reforming.

EU: Fuels: Diesel and Gasoline


text:
Configuration of chiral center in cocaine

I couldn't get the absolute configuration for the chiral center I marked. I got R, but according to my answer key it is apparently S. Would really appreciate an explanation.

The simplest explanation is that answer keys occasionally have mistakes. Note that the chirality where the benzoate group is attached has the $S$ configuration, as does the bridgehead chirality center to the front. It is possible that if this was an assignment or exam question, your instructor previously gave cocaine with a different chirality center marked on an older version and forgot to update the key for the different chirality center.

You are actually right, that stereocenter does have R stereochemistry. The four substituents to the chiral center in question are:

1 - $\ce{CO(O)}$ = $\ce{C(O)3}$
2 - $\ce{C(O)(C)(H)}$
3 - $\ce{C(N)(C)(H)}$
4 - $\ce{H}$

These are ranked and numbered according to the Cahn-Ingold-Prelog priority rules. Adding the numbering to the structure:

Orienting the lowest priority group away from us and following the direction of the numbering, we confirm that it does indeed have R stereochemistry:

ChemSpider returns the following name when searching for cocaine:

(1R,2R,3S,5S)-2-Methoxycarbonyltropan-3-yl benzoate

And the following image:

The structure is identical to the one given in your assignment and the carbon in question is position 2. The name gives carbon 2 as R-configured.
This answer was meant to supplement ringo’s answer which shows how to derive the correct CIP priorities which clearly give R as well.


text:
Hydrolysis of Phosphorus Trichloride
During hydrolysis of a halide of carbon via $\mathrm{S_N2}$ mechanism, the incoming neutral water molecule attacks the partially positive carbon centre from the side opposite to the halide side and forms a transition state where the hybridisation of centre is $\mathrm{sp^2}$. The nucleophile and the leaving group (halide) thus are attached to opposite sides of the same p orbital (now elongated). Ultimately the nucleophile kicks off the halide from the back.
But when we are carrying hydrolysis on $\ce{PCl3}$ (phosphorus trichloride) why does the water attack an empty d-orbital of the central phosphorus atom and not do the same mechanism as it does in the previous case mentioned. Why can't the attack take place like above leading to formation of an elongated p orbital?

I don’t know where you heard about the water molecule interacting with d orbitals before displacement takes place, but the idea is wrong. Phosphorus’ d orbitals are energetically too far removed to take part in bonding. Remember that according to the aufbau principle, 3d orbitals are higher in energy than 4s! They exist and you can excite electrons into them but that’s it (and you need high energies for that excitation).
Rather, the oxygen of water interacts with the $\sigma^*_{\ce{P-Cl}}$ orbital. Since phosphorus uses an orbital with a very high p-content, it has a notable lobe on the far side allowing for interaction. An intermediate tetracoordinated phosphorus species is formed which displaces the better leaving group (chloride). Water can lose a proton at any point in this mechanism.
The mechanism is basically identical to that of a carbon-centred $\mathrm{S_N2}$ reaction except that the intermediate breaks down much faster in carbon-based reactions.

Thank you ! The d orbital concept was given to me by my school teacher, thus I couldn't corelate it to SN2 . But now it's clear.


text:
Calculation of dispersion energy in Gaussian 09
How can I calculate the dispersion energy using Gaussian 09? Where is it printed in the output file when using wb97xd/cc-pvdz sp? If it isn't printed by default, what flags can be used to turn printing on?

Depending on the functional you use, it either contains some dispersion correction or not. For those which do not you can manually add it with EmpiricalDispersion=... if it is parameterized. If not, you can specify the parameters via the command line. See the manual for detail. The omegaB97XD DFA already contains the dispersion; it is parametrized that way. You will not be able to find a value for the dispersion correction.

The value of the empirical dispersion correction is printed when the print level is turned up using #p in the route line. For any method that contains an empirical dispersion correction, search for R6Disp or Dispersion energy, where other parts of the line will vary depending on the type of correction (-D2, -D3, or -D3(BJ)):
R6Disp:  Grimme-D2 Dispersion energy=       -0.0174233223 Hartrees.

This line is not printed with # or #n at the start of the route section.
For wavefunction methods that include dispersion via electron correlation, the dispersion contribution is non-separable. For a method like B2PLYP-D3, the printed "Dispersion energy" is the empirical correction, not the MP2 contribution!

This is very good to know and I have to admit I did not know about it. It is one of the neverending shortcomings of the Gaussian manual.


text:
Can HI/P reduction be used to reduce nitroalkanes and nitroarenes to alkyl amines and aryl amines?

Can $\ce{HI/P}$ reduction be used to reduce nitroalkanes and
  nitroarenes to alkyl amines and aryl amines ?

I learnt from my textbook that heating the nitroalkanes and nitroarenes with $\ce{H2}/\ce{Pt}$ is sufficient for the above conversion. But I am not sure whether $\ce{HI}$ and red $\ce{P}$ reduction will give the same products. Also, I could not find related information about this on the Internet.

Related: http://chemistry.stackexchange.com/questions/21638/reduction-of-%CE%B1-%CE%B2-unsaturated-nitro-compounds

I've never seen the $\ce{HI/P}$ method being used for the reduction of aliphatic or aromatic nitro compounds.

I learnt from my textbook that heating the nitroalkanes and nitroarenes with $\ce{H2/Pt}$ is sufficient for the above conversion.

I've reduced a lot of nitroarenes to their corresponding amines by catalytic hydrogenation under standard conditions in ethanol (or methanol) over $\ce{Pd/C}$. In the cases I examined, heating wasn't necessary.


text:
Can terminal enols be reduced by using LiAlH4?

According to the solution given in my textbook, no reaction takes place in this case.
But I am not quite sure about it. I feel that the enol can easily tautomerize to aldehyde and we know that aldehydes are reduced by $\ce{LiAlH4}$ to give alcohols.
So, will the above reaction take place? What will be the final product after reduction occurs?

What textbook? Is it worth posting a picture, so we can verify if the textbook is indeed incorrect (or perhaps there is some caveat)?

Not only can that enol easily tautomerize to the aldehyde, the aldehyde is the major component at equilibrium at room temperature (by a lot!). The equilibrium constant for the similar system of ethanal/vinyl alcohol (one carbon shorter than yours) is $K \approx 3\times 10^{-7}$ (source: Wikipedia page on keto-enol tautomerism):
$$\ce{ CH3CHO <=> CH2=CHOH}\\
K = \dfrac{[\ce{CH2=CHOH}]}{[\ce{CH3CHO}]}\approx 3\times 10^{-7}$$
Thus, for every molecule of enol in this system, there are $\sim 3 \times 10^6$ molecules of aldehyde. $\ce{LiAlH4}$ will reduce the aldehyde and then LeChâtelier's principle will drag the enol along for the ride.


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how to modify a transport device with an internal combustion engine with different atmospheres
How to modify a transport device with an internal combustion engine with:

the methane-nitrogen atmosphere
carbon dioxide atmosphere
the atmosphere of argon and chlorine.

Which substances will react? How will they feed? How effective will it be?

Welcome to Chemistry.SE! Feel free to take a [tour] of the site, and see [ask] for details about asking a great question. Please note that we have a policy which states that you should show your thoughts and/or efforts into solving homework problems. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed. Please [edit] in your reasoning or thoughts on this.

Related: Is oxygen necessary for burning? and Can oxygen burn in hydrogen?.

Each of the cases will strongly depend on the non-atmospheric natural
resources available on each planet.
For example in an atmosphere of $\ce{O2}$ + inert gases, it would be desirable to have deposits of hydrocarbons or $\ce{H2}$ gas to be mined and pumped into your fuel tank to burn with the atmospheric $\ce{O2}$.  In each case proposed below it is assumed that the non-atmospheric reagents would be available as deposits on the planet or that they could be synthesized in a way that they are acting as batteries, storing the energy that went into their synthesis.  That said, here are some possible simplistic solutions for the different scenarios.
Scenario 1: $\ce{CH4}$ + $\ce{N2}$ atmosphere:
Here you could simply reverse the typical internal combustion engine process used in Earth-vehicles. Rather than filling your tank with hydrocarbons to be mixed and burned with atmospheric $\ce{O2}$, you would fill your tank with (probably liquid to keep the volume small) $\ce{O2}$ to be mixed and burned with atmospheric $\ce{CH4}$.  Just as with conventional internal combustion engines on Earth, the reaction products would be $\ce{H2O}$, $\ce{CO2}$, and heat.
Scenario 2: $\ce{CO2}$ atmosphere:
As $\ce{CO2}$ is already thoroughly oxidized, you would need to fill your tank with a good reducing agent.  If there were deposits of $\ce{H2}$ gas trapped within the planet, then you would mine the $\ce{H2}$ and fill your tank with pressurized $\ce{H2}$ gas (the resources required to form and store liquid $\ce{H2}$ just wouldn't be worth the trouble, though there are other ways to concentrate $\ce{H2}$ then liberate it when needed).  The $\ce{H2}$ would then be mixed and burned with $\ce{CO2}$, ideally giving off $\ce{CH4}$, $\ce{H2O}$ and heat as products.
Scenario 3: $\ce{Ar}$ + $\ce{Cl2}$ atmosphere:
First to address is the composition of the vehicle, or at least the parts that would come into contact with the $\ce{Cl2}$ and other reactants and products, I propose a $\ce{SiO2}$ based coating for all components that would be in such contact.  For a gas-phase ignition scenario, I propose a diesel type of concept (ignition heat from compression of the mixed gases) to avoid the need for electrical ignition, metal wires, etc.
One possibility is to use the $\ce{Cl2}$ analogously to how $\ce{O2}$ is used in internal combustion engines, namely as the oxidant.  A hydrocarbon, $\ce{CH4}$ to keep it simple, would act as the reductant.  This reaction would ideally produce $\ce{CCl4}$, $\ce{HCl}$ and heat.  The atmosphere is already toxic and corrosive so the $\ce{HCl}$ exhaust is probably OK.
In the same manner as the $\ce{CO2}$ atmosphere, $\ce{H2}$ is another possibility.  In this case, the ideal reaction product is just $\ce{HCl}$.
One final possibility is worth considering here.  In all other scenarios I've only proposed fluids as fuel for the combustion.  But $\ce{Cl2}$ reacts very exothermically with $\ce{Fe}$ metal.  And the ideal product, $\ce{FeCl3}$, is also a solid.  So introducing a fine iron dust in some carrier gas could actually have the advantage of ease of collecting the exhaust do be disposed of at a dedicated treatment center.  That way we wouldn't pollute the air for the $\ce{Cl2}$-breathing natives.
Note that most of the proposed fuels here are composed of some of the most abundant elements in the universe.


text:
What is physically responsible for different compounds with the same types of atoms?
I only have a GCSE in Chemistry but I have discovered a lot of resources myself. One thing I cannot find an answer to is this: What is physically responsible for the different compounds you can get with the same types of atoms.
For example, if you were to mix a bunch of carbon and hydrogen what exact process determines if you get methane, butane, ethyne etc.? Is it the different concentrations of carbon and hydrogen? Then what is the physical explanation? If there is 4 times more hydrogen than carbon might a carbon atom more likely find itself surrounded by four hydrogens by chance and then bond with them to form methane? Similarly if there is equal carbon and hydrogen is it just morel likely that ethyne will form? If this is the case then in general if you mix carbon and hydrogen in different concentrations would you always get different concentrations of methane, butane, ethyne as products just by chance? For example if there is 4 times more hydrogen than carbon then is there still a small chance that at least some ethyne will form?

Statistics plays a role, but it is very heavily weighted by the "energy landscape" of a mixture of materials under a given set of conditions, which is an extremely complex, very high-dimensional potential energy surface.

It's hard to get my head around the concept of energy landscapes and potential energy surfaces. Is there a way you can sum up what you said in simpler terms? Are there any trends which can roughly be followed even if they aren't accurate?

In general its a huge question but specifically (a) you will need to determine what products are energetically feasible for a given ratio of reactants (b)   what energy are you are going to  give the reactants to make them react and (c) probably most important, of the products that can be formed, chemical  kinetics (reaction rates) will determine these and they may not be the therm0dynamically most favourable

Okay, I'm sure there's a much more direct, down-to-earth answer that you'd find more helpful, but I'm going for something different. To stop myself from writing a book chapter, there will be some hand-waving and cut out details. With more time and thought it could be unpacked further, but hopefully I'll at least be able to give you a taste.

Right, let's forget chemistry and try to deduce the product for the following simple reaction:
$$\ce{H2 + O2 -> ?}$$
To start off, we establish some basic rules:

Molecules are formed from the combination of integer amounts of atoms
Atoms cannot be formed or destroyed (i.e., the number of atoms is a conserved quantity)
There is a quantity (let's call it potential energy) inherent to each kind of free atom and molecule
The Universe tends towards the lowest possible potential energy

With nothing else to guide us, we are forced to consider every possible product (isomers neglected for simplicity):
$$\ce{H2 + O2 -> H + O + H_2 + HO + O_2 + H_3 + H_2O + HO_2 + O3 +  H4 +...}$$
Now, if the point is to minimize the total potential energy of the system, we can develop a heuristic (which will turn out to be wrong, but entertain the idea for now). First, find the specific atom or molecule among the listed products with the lowest possible energy,$^1$ then make as much of it as allowed by stoichiometry. When no further can be made, look for the next lowest energy molecule, then make as much of that as possible, and iterate the process until no lower energy molecules can be made.
That's nice and all, but without knowing which molecules have lower or higher energy, you don't get anywhere. Here, we either resort to some hefty quantum mechanical calculations, or empirical data. By doing some thermochemical experiments (measuring enthalpies of formation), it turns out that $\ce{H_2O}$ is the molecule with the lowest energy of the list by quite a margin, so it's going to be the main product. Indeed that is what is observed empirically when hydrogen is burnt in oxygen.
So we know that water is the lowest energy product for the reaction, but what happens if we mix $\ce{H2}$ and $\ce{O2}$ with a ratio different from that in water? Say you have two moles of $\ce{H2}$ and two moles of $\ce{O2}$, does that mean you now get $\ce{H2O2}$ or potentially some other products? In fact, this is not the case. The heuristic above still applies to minimize the potential energy of the system, so $\ce{H_2O}$ is preferentially produced. This shouldn't come as too much of a surprise, because the low energy of $\ce{H_2O}$ is an inherent feature of the molecule, and is entirely independent of how much $\ce{H2}$ and $\ce{O2}$ you decide to mix in an experiment.
Then what happens to the excess $\ce{O2}$? Again following the heuristic, it will seek out the lowest energy molecule available. Since there's no more hydrogen left to react, it has a more limited number of possibilities. It could conceivably still react and turn into other molecules (for example, $\ce{O3}$ is perfectly acceptable), but it turns out that of the remaining candidates, $\ce{O2}$ is the molecule with the lowest energy, so the reaction just stops there, and we get a mixture of $\ce{H_2O}$ and left over $\ce{O2}$.
Okay, we've made some progress, but we need to talk about some important complications. In the real world, at any temperature above absolute zero, you can actually retain a small amount of molecules with a higher potential energy. All of those lower energy molecules can randomly gain a bit of thermal energy, and every so often that causes them to react and turn into a higher energy molecule. If the lowest energy molecules are much lower in energy than anything else (as is the case for $\ce{H_2O}$ among the combustion products of $\ce{H2}$ and $\ce{O2}$), then this effect is very small and can be neglected. However, in many cases there isn't such a clear-cut winner, and you end up getting a mixture of different products with similar energies. This is especially true as you get into more complicated chemistry.
Unfortunately, even after all this we're still missing a very big piece of the puzzle. Everything I've mentioned so far is a part of thermodynamics, and that alone cannot always predict the product of a reaction. Allow me to prove it. According to everything I've argued, $\ce{H2}$ and $\ce{O2}$ will always react to form primarily $\ce{H2O}$ since it has the lowest energy, and $\ce{H2O2}$ can never be a significant product. But hydrogen peroxide is made in very large amounts industrially, so how do they do it? A quick search on the internet provides the answer, the anthraquinone process, whose net reaction is simply:
$$\ce{H2 + O2 -> H2O2}$$
What gives? What we've been missing is the concept of kinetics. For many transformations in nature, chemical or otherwise, even if the final state has a lower energy than the initial state, they are often separated by an energy barrier. Even if there can be a net release of energy, the transformation is hindered because to start off it needs a net input of energy.
This is the case even for the reaction $\ce{H2 + O2 -> H2O}$. If you just mix the gasses together at room temperature, they will not react even after a century. Either some initial energy needs to be supplied (a flame, a spark, etc.) or the energy barrier needs to be decreased by using a catalyst (Pt), after which the reaction will proceed in a fraction of a second.
When you consider all possible reactions, and all the energy barriers separating the different molecules, you get the very high-dimensional potential energy surfaces I mentioned in my comment. All our work in Chemistry boils down to finding reliable paths in (or making our own by actively modifying) the crazily complex energy landscapes and exploiting them for our purposes.
$^1$ Technically, the lowest amount of potential energy per amount of atoms the molecule is made of. You can think of it as the lowest potential energy per gram of molecules.

Here's something which might also help. Is there a case where all this potential energy and kinetic stuff doesn't skew the results, and you can tease out the purely statistical effect? Sure, let me present a typical case.
Consider the following transesterification reaction (poorly drawn because I don't currently have access to ChemDraw):

Both sides of the equation have an alcohol and an ester. The alcohols are very similar to each other, and the same can be said for the esters. Because of this, there's barely any difference between the potential energy of the products and the reagents.
Now consider taking a mole of the starting  ester and alcohol. In the beginning, there is only ethanol and no methanol, so statistically you'd expect the reaction to happen in the direction shown. As the reaction progresses, the amount of ethanol available is reduced, disfavouring further generation of products, and the presence of methanol now actually allows the reaction to also go the other way around. By pure statistics, you'd expect the reaction to stop when there's about the same amount of methanol and ethanol, half a mole each, and this is what is observed. In short, simple transesterification reactions tend to have equilibrium constants very close to 1 (e.g. this article).

Thanks for the answer! Very comprehensive. So you mentioned that you'd have to do a load of quantum calculations to find out the molecule with lowest energy. Are there any basic rules of thumb which aren't entirely accurate but hold some truth to determine molecules with lower energy? For example, is bond energy anything to do with it? I'd imagine that double bonds are higher energy and atoms might prefer to make single bonds perhaps?

Without quantum calculations, the rules of thumb are pretty much what make general chemistry. The bond energies are very important, yes, but not all-encompassing (e.g., some reactions will happen spontaneously even though they absorb energy and make higher energy products). There is no clear rule whether multiple bonds or single bonds are better, at least for the second row elements, though heavier elements tend to favour single bonds. For example, $\ce{N2}$ is an exceptionally low energy molecule, and it has a triple bond. Meanwhile, $\ce{P2}$ also has a triple bond, but is very reactive.

Ultimately, you can't run. If you want to learn chemistry, you'll have to familiarise yourself with how the elements behave by reading some of their basic properties including typical reactions, and looking at tabulated empirical values of multiple properties (ionization energies, electron affinities, electronegativities, proton affinities, equilibrium constants, etc) to help compare different compounds.

Yeah I wish I had time to learn it all before my stage 0. I'm doing a PhD and my background is more informatics / complex systems. I have to write a detailed research proposal for my stage 0 which is due very soon. They want a super high level of specificity, to the point where they expect you to have the answers to a lot of your research before you've actually done it! I'm doing a computational chemistry course next semester and I'll be teaching myself a lot of chemistry. But the stage 0 is before that! Hopefully I can convince them. Thanks anyway!

The products you can have from a reaction depend on many things, not only their concentrations. Also, you can hardly mix only free atoms of different elements. You are thinking on a situation where you mix carbon free atoms with hydrogen free atoms, and that is not the usual situation.
The answer is about conditions, for instance, one important factor could be pressure: in situations with high pressures, it is more likely to form compounds that have less volume per mass unit.
In the end, the compounds formed depend on a combination of two factors: thermodynamic and kinetics. Atoms will form the compounds in which they have the minimum energy, as much as the energy barriers can be overcome.

I disagree that atoms will always form the compounds in which they have the minimum energy: that is very much not the case in many examples. For example if you have species, A, B, and C, and compound AB is lowest energy but can (even rarely) dissociate back into A + B, but--if AC is formed even for a second, AC is removed from the system (such as as a solid or as a gas), then over time AC will eventually be the primary product, even though it is not lower energy than AB. Perhaps this is what you meant by "as much as the energy barriers can be overcome", but this is not clear.

Could you explain a few terms for me? I'm unsure what volume per mass unit means exactly, is that molar volume? By that did you mean smaller compounds are created at higher pressure? So in terms of thermodynamics and kinetics do you mean that for example at high temperatures molecules/atoms move more quickly so then bonds are broken from higher energy collisions and others can form etc? I have heard about the minimum energy thing. Does this mean that single bonds are prefered over double ones because double ones have more potential energy?

Yeah, by energy barriers I mean that a reaction can be kinetically controlled. If molecules/atoms have no energy enough, then will rather form a  less stable compound, as you explained.

Also, when I was talking about the pressure in gas reactions, I meant that the compounds formed at high pressures are those occupying less volume. See if you had the reaction: 2 SO2 + O2 <--> 2 SO3, you would get:

you would get at high pressures, more SO3, because two molecules of SO3 occupy less volume than two of SO2 and one of


text:
Why Gibbs free energy is equating to 0?
I have studied thermodynamics, in which the infamous Gibbs equation came.
$$ \mathrm{d}G = \mathrm{d}H - T\,\mathrm{d}S $$
Then my book said the criteria of spontaneity based on gibbs energy is at constant pressure and temperature.
Since $T\,\mathrm{d}S = q = \mathrm{d}H$ (since at constant pressure $\mathrm{d}H = q$), shouldn't $\mathrm{d}G$ all always be equal to $0$?

This is quite a different kind of dS.

Can you please elaborate?

dS in TdS=q is the enthropy change of a system upon introducing some heat. dS in Gibbs is the enthropy change during a chemical reaction.

That is not the Gibbs Free Energy at constant pressure and temperature.
Alphabetic soup definitions:

$G$ - Gibbs free energy
$H$ - enthalpy
$T$ - temperature
$S$ - entropy
$P$ - pressure
$V$ - volume
$w$ - work
$q$ - heat
$U$ - internal energy

Definition of $G$ is $$G = H - TS$$
Definition of $H$ is $$H = U + PV$$
The total differential of $G$ is
$$dG = dU + PdV + VdP - TdS - SdT$$
The definition for the change in $U$ is $$dU = dq + dw$$
There are two kinds of work: pressure-volume work (pv) and non-pressure-volume work (non-pv)
$$dw = dw_{\mathrm{pv}} + dw_\mathrm{non-pv}$$
Thus, substituting we get
$$dG = dq - PdV + dw_\mathrm{non-pv} + PdV + VdP - TdS - SdT$$
If the process is (approximately) reversible:
$$T =\dfrac{dq}{dS}$$
Substituting $q = TdS$ and simplifying, we get:
$$dG = VdP - SdT + dw_\mathrm{non-pv}$$
At constant pressure, $dP=0$, and at constant temperature $dT = 0$.
Therefore:
$$dG = dw_\mathrm{non-pv}$$
Hence, Gibbs free energy is sort of a measure of useful work we can get out of a system.
$$w_\mathrm{non-pv\ by\ system} = - w_\mathrm{non-pv\ on\ system}$$

Um what do u mean by pv and non pv?

Pv work means pressure work, like F.ds is work, P.dv is work. Any other sort like electrical or gravitational work comes under non pv

Hello @Mrigank, so I can define the Gibbs free energy also when the temperature and the pressure are not constants?| Sorry for the banal question.

Although the answer @ELiT gave is the perfect answer, I would like to answer my own question. What a silly mistake I made!
$\mathrm{d}S = \frac{q}{T}$ only for reversible processes. In reversible processes, what I said is true, as $\mathrm{d}S_\text{total} = 0$, thus $\mathrm{d}G = 0$. But $\mathrm{d}S \neq \frac{q}{T}$ for non-reversible processes. Hence I cannot substitute that value here.

Please use MathJax to typeset mathematical expressions!

@getafix Sorry. I am on my mobile, and it is really hard for me to do so. Ill edit it once my computer is fixed


text:
Product of a organic reaction with KOH
In the below reaction (reaction of 1-bromo-2-phenoxyethane with KOH at hugh temperature):
$$\ce{Ph-O-CH2-CH2-Br ->[\ce{KOH}][200\ ^\circ C]P}$$
According to me KOH will help in displacing bromide ion and making a carbocation. This carbocation would be attacked by the double bonds of the benzene ring. So the product should be benzotetrahydrofuran:

But the product is given as phenoxyethene (phenyl vinyl ether):

I could not understand how is this possible.

Related: http://chemistry.stackexchange.com/questions/16978/why-is-a-strong-base-needed-for-an-e2-reaction-and-not-a-an-e1-reaction

First, let's take a look at your proposed carbocation intermediate:
$$\ce{Ph-O-CH2-CH2+}$$
This is a primary carbocation, which makes it very unstable. It's unlikely to form, so the reaction can't involve going through this stage. Primary halides are much more likely to undergo $\ce{S_N2}$ or $\ce{E2}$ reactions. Whether it's the former or the latter will depend on the nucleophilic and basic charachter of the other reagent. Bases (specially if sterically hindered) tend to favor $\ce{E2}$, while nucleophiles of low basicity tend to favor $\ce{S_N2}$. However, many reagents may and will give a mix of both products.
With all this information, since $\ce{KOH}$ is a very strong base and the reaction happens at high temperature, the result will be the $\ce{E2}$ product, in your second picture.


text:
Measuring pH of pasty solutions/mixtures
I'm trying to measure the pH of a very viscous paste mixture and have been told to dilute the sample for more accurate results, because the viscosity of the sample may prevent the sensor to detect the real pH of the sample.
Does this sound right to you? Wouldn't diluting the sample alter the pH?
Thank you

Yes, this does sound right. And yes, of course it would change the pH. The advice was probably given in the hope that you'd be able to account for that change. This, BTW, is not always easy and not even always possible.

Related: How can I measure the pH of high-viscosity fluids like dough? and How does dilution affect result of titration of jam?

Do you know the acids/bases in your system? Are they weak acids/bases?

Diluting the sample does alter the pH, but it does so in a predictable way.
When we are measuring pH, we are measuring the concentration of hydrogen ions (or hydornium ions, or whatever ) via the negative logarithm:
$$\mathrm{pH} = -\log_{10}\left([\ce{H+}]\right)$$
Diluting a solution has a predictable effect on concentration, since the total amount (in moles) of $\ce{H+}$ will not change by the addition of more solvent. This relationship is summarized by the formula (where $c$ represents concentration, $v$ represents volume, and $1$ and $2$ represent the solution before and after dilution, respectively:
$$c_1v_1 =c_2v_2$$
Let's take an example. Suppose you have 1.000 mL of a viscous acidic aqueous solution. The viscosity of the solution will impede easy measurement of pH because of the kinetics of mass transport. If you dilute this solution with water to a total volume of 10.00 mL, you now have a less viscous solution with the same number of moles of $\ce{H+}$, because sources of $\ce{H+}$ were neither added nor removed.
If this new diluted solution has $\mathrm{pH = 3.03}$, then $$[\ce{H+}] = 10^{\mathrm{-pH}}=10^{-3.03} = 9.33\times 10^{-4}\ \mathrm{M}$$.
The concentration of the original solution can now be determined:
$$c_1 = \dfrac{c_2 v_2}{v_1} = \dfrac{\mathrm{9.33\times 19^{-4}\ M)(10.00\ mL)}}{\mathrm{1.000\ mL}}=9.33\times 10^{-3}\ \mathrm{M}$$
The pH of the original solution is:
$$\mathrm{pH = -\log_{10}(9.33\times 10^{-3}) = 2.03}$$
Note, that since I chose to dilute by a factor of 10, the pH of the dilute solution is 1 pH unit higher than the original solution.
This approach works for solutions of strong acids and strong bases and will work over much of the pH range, but may get a little awkward around neutral. The approach breaks down for weak acids and bases since changing the concentration changes the position of equilibrium. For weak acids, diluting likely results in additional ionization events. However, even without knowing the acid/base and its equilibrium constant, you could perform multiple dilutions and do some curve-fitting and get back to the original pH.


text:
What is Fajans rule?
when reading about polar covalent bond I came across the rule called 'Fajan's Rule' but could not find the explanation nor the derivation

https://en.wikipedia.org/wiki/Fajans'_rules

Fajans' rules (note the difference) were formulated in 1923 by Kazimierz Fajans.
They are a method for predicting ionic vs. covalent that predates electronegativity (by three decades) and make use of ionic and atomic radius data that was becoming available through x-ray crystallography.
To use Fajans' Rules, assume your binary compound is ionic and identify the potential cation and anion.
By Fajans' Rules, compounds are more likely to be ionic if: there is a small positive charge on the cation, the cation is large, and the anion is small. For example, $\ce{NaCl}$ is correctly predicted to be ionic since $\ce{Na+}$ is a larger ion with a low charge and $\ce{Cl-}$ is a smaller anion.
Compounds are more likely to be covalent if: there would be a large positive charge on the cation, the cation would be small, and the anion would be large. For example, $\ce{AlI3}$ is correctly predicted to be covalent since it would have a small cation with a high charge and a large anion.
Note that Fajans' Rules have been largely displaced by Pauling's approach using electronegtivites. However, the remnants of Fajans' Rules are found in Hard-Soft Acid-Base Theory, which predicts bonding properties based on polarizability (which is based on size and charge). Binary compounds having a soft acid and/or a soft base are often covalent.
$$\begin{array}{|c|c|c|c|}\hline
\mathrm{compound} & \mathrm{Fajans} & \mathrm{Pauling} & \mathrm{HSAB}\\ \hline
\ce{NaCl} & \mathrm{low + charge \\ larger\ cation \\ smaller\ anion \\ionic} & \mathrm{3.16-0.93 = 2.19\\ ionic} & \mathrm{hard\ acid \\ borderline\ base \\ ionic}  \\ \hline
\ce{AlI3} & \mathrm{high + charge\\ smaller\ cation \\ larger anion \\covalent} & \mathrm{2.66 - 1.61 = 1.05 \\covalent} & \mathrm{hard\ acid\\ soft\ base \\covalent} \\ \hline
\end{array}$$

Fajans' rule states that a compound with low positive charge, large cation and small anion has ionic bond where as a compound with high positive charge, small cation and large anion are covalently bonded.
For high charge, small cation will have more polarizing power.
Where as larger is the size of anion, more will be the polarization of anion. Because if this electron cloud of anion is more diffused. This makes the anion easily polarizable.