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How does KMnO₄ affect electrolysis?
I'm just blindly experimenting. I tried to dissolve some potassium permanganate in water before performing electrolysis.
I can't see much difference. How is $\ce{KMnO4}$ dissolved in water? How does it react during electrolysis?
I added salt and the electrolysis was so violent that the copper started to burn (probably reacting with something).

A lot of questions have been posed on electrolysis and electrochemistry on this forum recently, with many recurring themes. I will explain electrolysis in depth in order for a true understanding to come out of this; if you're in a hurry then skip to the last three paragraphs.
One of the most important things about learning electrochemistry is the abstract property of voltage. Voltage can be considered a driving force, which denotes which direction charges are going to flow in a system. A high positive voltage means that charges most definitely wish to flow in that direction, and a high negative voltage means that the charges wish to flow strongly in the opposite direction. If we really want charges to flow in the direction of negative voltage, we must supply energy to the system in order for that to happen—in other words, we're supplying energy to move a charged particle against an electric field (work must therefore be done to do this, as with mass in a gravitational field).
And so it is with $\ce{H2O}$! The reaction corresponding to our notion of electrolysis is:
$$\ce{2H2O -> O2 + 2H2}$$
At least two of the intermediate species in electrolysis are going to be $\ce{OH-}$ and $\ce{H+}$, and so the energy requirement for this process takes into account the energy required to move an electron from $\ce{OH-}$ to the anode, and also to move an electron from the cathode to $\ce{H+}$.The voltage requirement for this process is  $\pu{-1.23V}$. In reality the voltage requirement is closer to  $\pu{-2V}$ taking into account overpotential (the electrochemical equivalent to activation energy).
So this negative voltage tells us that this process will not happen spontaneously, and we will need to supply a voltage of  $\pu{2V}$ in order for this to happen. So far so good: we never observe electrolysis in a standard beaker of water just sitting in the lab!
Your question asked what happened to the $\ce{KMnO4}$ during electrolysis. $\ce{Mn}$ is quite happy in a $\ce{Mn^7+}$ and $\ce{Mn^4+}$ oxidation state. Therefore in solution, the following reaction can happen:
$$\ce{Mn^7+ + 3e- -> Mn^4+}$$
Note that the above is a simplification, $\ce{Mn^7+}$ and $\ce{Mn^4+}$ don't occur simply as isolated ions in solution. In our case, these oxidation states of manganese are tied up in molecules:
$$\ce{MnO4^- -> MnO2}$$
But the above isn't balanced, so we ask ourselves what other likely molecules are going to be produced in order to balance our oxygens. It's common for these redox systems to produce $\ce{H2O}$, so we add $\ce{H2O}$ on the right to balance the oxygens, and add $\ce{H+}$ on the left so we have enough hydrogens. Thus:
$$\ce{MnO4^- + 4 H+ -> MnO2 + 2 H2O}$$
The voltage requirement for this reaction is $\pu{+1.68 V}$, so this means that the electron transfer happens spontaneously. In other words, this reaction is more likely to occur than the electrolysis of water at a less than neutral pH. If you start your electrolysis of water using distilled water (neutral pH implied) then the reduction of $\ce{Mn^7+}$ doesn't happen initially. It's only when the electrolysis starts and $\ce{H+}$ is generated that the reduction of $\ce{Mn^7+}$ can happen; this is then favoured over electrolysis on thermodynamic grounds.
As a last point, it's unlikely your copper electrode was burning, but these can get very hot if the voltage is high enough and hence melt. Turn the voltage down as only $\pu{2V}$ is needed for water electrolysis.
Experimental hint: try using $\ce{NaOH}$ and $\ce{KMnO4}$ in combination and see if there is any change in your observations. Then try increasing the amount of $\ce{NaOH}$ relative to $\ce{KMnO4}$ to see if you observe any changes which would suggest my theory. Note that other manganese reactions (pdf via The Internet Archive) can still happen at lower voltages, which does complicate things.
To answer your question directly, nothing additional is observed using $\ce{KMnO4}$ as it reacts instead of behaving as a charge carrier for electrolysis. A quality of a good electrolyte is that it does not interfere with your reaction of interest.


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Can citric acid protonation produce benzene and water
Is this reaction possible: $\ce{C6H8O7 + 12 H3O -> C6H6 + 19 H2O}$?
Considering that the $\ce{H+}$ can be induced by ionization of the citric acid solution the benzene can be produced in this way?

Here's a drawing of citric acid and benzene.  You would have to spend a lot of energy in order to break a large number of bonds (carbon-carbon bonds, carbon-oxygen bonds and carbon-hydrogen bonds) to rearrange citric acid into benzene.  This would be unlikely to occur in a natural process and would require many steps in the laboratory.

No.
Since citric acid is a 5 carbon carbon chain, you would need to break a C-C bond and reattach the carbon on the end of the chain. This does not happen in acidic media.

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Preparation of NaOH Solution
How would one prepare a one molar sodium hydroxide solution?
I see a few issues with this:

$\ce{NaOH}$ absorbs carbon dioxide. So old NaOH samples might have a significant portion of $\ce{NaHCO3}$ or sodium bicarbonate. How would get get rid of the sodium bicarbonate?
I am told that $\ce{NaOH}$ is hygroscopic. So it absorbs water, apparently. What is the mechanism for this? I can see the hydroxide attacking carbon dioxide in the above bullet point, that's easy, but I can't really see $\ce{NaOH}$ attacking water. Also couldn't we circumvent the water absorption problem by cooking the $\ce{NaOH}$?

There is a difference in preparing a solution and determining its concentration. As you said, NaOH solutions absorb carbon dioxide over time. It is possible to remove the absorbed carbon dioxide by boiling, but it is just as easy to use the solution as is and re-determine its concentration when you use it.
The concentration of sodium hydroxide solutions can be determined by titrating a known amount of dried potassium hydrogen phthalate.
If you really need an exact concentration of the solution, make it slightly more concentrated, standardize it and then dilute it to get the concentration desired. It would only be accurate for a few hours.

I am told that NaOH is hygroscopic.

The powder and pellets of sodium hydroxide are hygroscopic, but the solutions of sodium hydroxide already have water, so it is unlikely more will be absorbed.

1)In short, you can't prepare exactly $1M$ solution of $\ce{NaOH}$. It is common to prepare roughly $1M$ solution and then titrate it to get an exact value of its molarity. For preparation of standards people usually avoid usage of $\ce{NaOH}$ and similar compounds that adsorb something from air. For exapmple, $\ce{NaHCO3}$ is a nice option.
2)$\ce{NaOH}$ granules adsorb water from air, forming a solution of the hydroxide in water. The driving force behind it is energy of solvation of $\ce{Na^+}$ and $\ce{OH^-}$ ions (as far as these words are applicable to concentrated solutions)

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When it comes to bond energy or molecular binding energy, do people mean enthalpy?
When calculating bond energy or molecular binding energy, which value should I use, enthalpy or total energy?
For example, A+B=C
Should it be C's enthalpy of formation minus (A's+B's)enthalpies of formation,
or C's total energy minus (A's+B's) total energies?

When in doubt, consult IUPAC Gold Book for definitions (italics below is mine).

bond-dissociation energy, $D$
The enthalpy (per mole) required to break a given bond of some specific molecular entity by homolysis, e.g. for $\ce{CH4 -> CH3^{.} + H^{.}}$, symbolized as  $D(\ce{CH3−H})$.

bond energy (mean bond energy)
The average value of the gas-phase bond dissociation energies (usually at a temperature of  298 K) for all bonds of the same type within the same chemical species. The mean bond energy for methane, for example, is one-fourth the enthalpy of reaction for: $$ \ce{CH4_{(g)} -> C_{(g)} + 4H_{(g)}} \, .$$ Tabulated bond energies are generally values of bond energies averaged over a number of selected typical chemical species containing that type of bond.

So, yes, conceptually you should use enthalpy. Note, however, that the enthalpy of a molecular specie in gas phase at, say 298K, is calculated as follows
$$H_{298} = E_{\mathrm{e}} + E_\mathrm{ZPE} + H_{\mathrm{trans}} + H_{\mathrm{rot}} + H_{\mathrm{vib}} + RT \, ,$$
where

$E_{\mathrm{e}}$ is the electronic energy;
$E_\mathrm{ZPE}$ is the zero point energy;
$H_{\mathrm{trans}}$, $H_{\mathrm{rot}}$ and $H_{\mathrm{vib}}$ are the temperature-dependent contributions to enthalpy from translation, rotational and vibrational motion, respectively.

And the problem (as Martin mentioned in comments) is that basically only the first term out of this five can be calculated accurately enough when the ultimate goal is to calculate energies (bond energies in our case) with the so-called "chemical accuracy" which is about $\pm 1$ kcal/mol.
For relatively large molecules harmonic frequencies and consequently $E_\mathrm{ZPE}$, as well as a rigid-rotor harmonic-oscillator approximation used to calculate thermal corrections, become inaccurate, and thus, the change of electronic energy should always be used as a reference.
$%edit$

In computational chemistry it is never wise to only use the enthalpy values. The goldbook is unfortunately not a very good address for these kinds of questions.

@Martin could you elaborate a little bit more? Do you mean that a rigid-rotor harmonic-oscillator approximation used to calculate thermal corrections is not suitable for relatively large systems?

Amongst others, yes. When systems go large, harmonic frequencies become more inaccurate. Tiny inaccuracies may lead to big errors. Numerical treatment can be a big problem. It may depend much more on implementation and approximation used in the different qc programms. The pure electronic energy should always be used as a reference.

@Martin, surely, that is a good point. I added relevant info to my answer.

These terms has more of historical value. In early theoretical chemistry they needed practical experimental values that can be reasonably inquired by the the methods, can cover wide range of (organic) chemistry and cannot be calculated by simple empirical equations. This kind of thermochemical data were just right for e.g. semi-empirical methods. That time I believe they were happy if they could compare electronic energies, and fit parameters for the electronic calculations, and didn't really care the other contributions.


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Why are metal oxides generally more stable than their corresponding elemental forms?
Why is the oxidation of certain metals such as iron favorable from a thermodynamic standpoint? What is it about these metal oxides that makes them inherently more thermodynamically stable than their starting, elemental forms? The only reason I can imagine is that iron, being a d-block element, has a low electronegativity and cannot stabilize valence electron density as well as oxygen, the second most electronegative element on the table.
Also, is oxidation always thermodynamically favorable?

The oxidation of gold is not favorable.  For many other metals the metallic bonding is weaker than the bonding with oxygen.
Let's look at the steps in oxidation:
Breaking metallic bonds (endothermic).
Ionizing the metal atoms (endothermic).
Electron affinity of oxygen (exo for the first, endo for the second).
Lattice energy of the oxide (exothermic).
The lattice energy is so high that the process is favorable.

Both the answers above are accurate and touch on my point here.
There are two factors that contribute to whether a reaction is spontaneous (if it will happen on its own) such as the one you described (iron with oxygen).

The thermodynamic properties of the reaction (like the net heat exchange, exothermic vs. endothermic, etc.)
The entropy of the reaction. This second one is often missed or not thought of.
You see, the processes involved in summing the relative bond strengths, ionization energies/affinities, and lattice energies, to name a few, are examples of thermodynamic calculations that you can use to predict if a reaction will occur spontaneously. In this case, as was already stated above, the reaction between iron and oxygen is very energetically beneficent for both substances because the oxygen, in essence, ionizes the iron and forms an incredibly strong lattice. So strong it's more stable energetically than Fe itself. Another result of this stability is that the oxidized Fe is less likely than pure iron to react with atmospheric water so it's less prone to rusting, another benefit.
But that's not even all! That's just number one!
Remember how I mentioned entropy (that's the dispersal of energy in a system. Iron is a Dblock atom as you said. When it gives up electrons in pairs to oxygen it starts to become more stable because its Z.eff (effective nuclear charge, look it up) increases on its remaining electrons and as the oxygen gains octets, it becomes just a little more stable because its entropy (energy dispersal which is a good thing) is increasing!

The result is an increasing entropy and a decreasing internal energy for the system. These are two independent factors that often come in pairs in chemistry. But not always. The evaporation of sweat for example is endothermic and wouldn't be spontaneous is not for it's subsequent huge increase in entropy.
These two principles are the circumstances that spawn spontaneous reactions and explain why iron is more stable as an oxide.

When you think about it, metals tend to want to lose electrons to finish with full outer shells.  When they bond with oxygen, they do lose those electrons, ending in a more stable electron configuration.  These ions don't go off and violently react with other things because they're ionically attracted to the oxygens they've just given their electrons to.  So all their electrons are happily occupied, meaning they have no need to react and share them off.  Also, because of oxygen's high electronegativity, it is unlikely to give those electrons up to another element, causing a reaction.

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TLC: Why is a more polar eluent more effective at displacing the points?
If the points are non-polar, I don't see how making the eluent more polar will make them travel faster. Anybody care to explain?

Most likely you are using "normal phase" TLC plates. The stationary phase is silica gel, which has free hydroxyl groups making it a polar surface. In TLC, there is a competition for the plate between the analytes and the eluent. As you make the eluent more polar, the eluent will interact more with the plate leaving the analytes to be carried with the mobile phase.


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How does the vacuum in atoms persist?
From very elementary chemistry I understand that atoms are mostly empty space, with a few tiny little objects whizzing around.
There must then be a vacuum in between these particles. I always thought that nature abhorred a vacuum, and rushed to fill it.
Why doesn't everything clump together in one little ball and get rid of the vacuum? Or are my premises wrong?

There is no vacuum. The space inside atom is by no means empty; it is occupied by delocalized electrons.

nature abhorred a vacuum, and rushed to fill it Thats not nature, what you're saying is air filling up the vacuum due to pressure differences.

I think what you are asking is why electron do not fall in nucleus.
For this you should have general idea of quantum mechanics. Because in the microscopic world nature follows quantum mechanics equations and not classical mechanics equations. Quantum mechanics equations include electromagnetic fields, and their solutions are stable and allow for the existence of atoms, which is what we experimentally observed to start with.
This is general idea of answer given to question physics stack exchange. You can see the detailed answer over in Physics S.E..
This Link will  explain diffrent theories which failed to explain why electron do not fall in nucleus.

Well, the glib answer is that nature doesn't abhor a vacuum or anything else. Abhorrence is an emotional state and the universe, as far as we can tell, is not run by emotions. And the space inside the atom has electrons moving through it so one could argue that it's not a vacuum, at which point, we have no abhorrence, no vacuum and no question.
But that doesn't help. Freddy's answer explains why the electrons don't fall into the atomic nucleus, which is basically because of physics that we've only come to understand in the last hundred years.  Gases can't get inside an atom because molecules are surrounded by electrons, which are repelled by the atom's electrons; a gas can only fill a vacuum if it can move there.

Without quantum mechanics, it becomes a bit sloppy, but the general principle is easy to grasp:
The nucleus is positive, and surrounded by a number of negative electrons matching the charge of the nucleus.
Once the electrons get pulled too close to the core, they start repelling each other strongly, so there is some equillibrium distance.
Why doesn't the core/nucleus pull in the electrons one by one? Well, they fly "in vacuum", so they get ever faster when they're pulled close to the nucleus. The nuclues can't catch them either, because he has not way to store or discard the huge kinetic energy the electron has. So the whole system is stable, in rather the same way the planets and comets surround the sun.
With a bit of QM, there is the additional point that pairs of electrons fly in a formation ("orbital") that keeps them at an extra distance, and all pairs of electrons have a different formation. That makes sure the whole system stays symmetric and thus stable. Electrons are individualistic, they hate to be indistinguishable from the others. Which of course they are. ;-)

There's also the fact that electrons aren't little round balls.

@bon  I thought they were yellow. What's you point?

pairs of electrons fly in a formation is perhaps slightly misleading. They don't fly at all they just exist in space with a probability of being at any one point.

The non-quantum mechanics explanation does not work. Planets have no problem crashing into the sun (what we see is the result of most matter falling into the sun, and some matter, planets, having just the right orbit to stay put). Also, the electron repulsion idea fails for one electron systems (e.g. hydrogen atom, which is common in the universe, https://iopscience.iop.org/article/10.1088/0004-637X/698/2/1467/pdf).


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Why does alcoholic KOH prefer elimination whereas aqueous KOH prefers substitution?
What is the basic difference between aqueous and alcoholic $\ce{KOH}$? Why does alcoholic $\ce{KOH}$ prefer elimination whereas aqueous $\ce{KOH}$ prefers substitution?

Related : https://chemistry.stackexchange.com/questions/4054/why-does-water-favour-nucleophilic-substitution-over-elimination

$$\ce{R-OH + OH- <=> RO- + H2O }$$
In alcoholic solution, the $\ce{KOH}$ is basic enough ($\mathrm{p}K_{\mathrm{a}} =15.74$) to deprotonate a small amount of the alcohol molecules ($\mathrm{p}K_{\mathrm{a}}= 16–17$), thus forming alkoxide salts ($\ce{ROK}$). The alkoxide anions $\ce{RO-}$ are not only more basic than pure $\ce{OH-}$ but they are also bulkier (how much bulkier depends on the alkyl group). The higher bulkiness makes $\ce{RO-}$ a worse nucleophile than $\ce{OH-}$ and the higher basicity makes it better at E2 eliminations.

I don't get how the alcoholate salts are formed. Where is the 'R' group coming from?

@InternetGuy $\ce{ROH}$ is the notation for a generic alcohol, where $\ce{R}$ is usually an alkyl chain. Thus, when $\ce{KOH}$ deprotonates an alcohol, you get an alcoholate salt $\ce{ROK}$, i.e. $\ce{KOH + ROH <<=> ROK + H2O}$.

Hmm, isn't $\ce{RO^-}$ often considered a stronger nucleophile that $\ce{OH^-}$?  (or did you factor in solvent interactions as well?) :-)

Does that mean that in ethanolic KOH, it is the ethoxide that is the primary basic species involved in proton abstraction from the substrate. Does that also mean that the concentration of the alkoxide ion would be factored into the rate law?

In carbylamine reaction, why $\ce{OH-}$ is used in mechanism instead of $\ce{EtO^-}$?

$\ce{OH-}$ acts as a nucleophile. Reactions carried out in alcohol tend to be elimination reactions, and reactions carried out in water (aqueous) tend to be substitution reactions.
If water were used as a solvent in an elimination reaction involving $\ce{KOH}$, the equilibrium would be shifted towards the reactants (water reacting with product), so substitution is favored.

The reason is quite straightforward-

$\ce{OH^-}$ is a weak base and a stronger nucleophile specially under polar protic conditions. Hence substitution occurs.
$\ce{RO^-}$ is a strong base owing to the inductive effect (+ I effect) of the R group. Hence under alcoholic conditions, $\ce{RO^-}$ extracts the $\beta$ hydrogen of the halides and gives an alkene.

Aqueous KOH is more solvated by water and not able to abstract H+from  substrate but act as nucleophile but alc KOH is less solvated and alkoxide ion can abstract acidic hydrogen

See, we know that nucleophiles which are in polar protic solvents are solvated and therefore nucleophilicity is directly proportional to the size of nucleophile 'cause they will be less solvated. And in polar aprotic solvent the anions are not solvated so they will show their usual nucleophilicity behaviour and is inversely proportional to the stability of anion.
Hence, in alcoholic medium the $\ce{OH-}$ is less solvated and acts as a strong base and eliminates acidic $\ce{H+}$ whereas in aqueos medium $\ce{OH-}$ shows nucleophilicity and prefers substitution.


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Does CO2 dissolution in different salt concentrations affects the amount of carbonic acid formed?
We have a school experiment, where I must report on the following question:

Does salinity affect the solubility of $\ce{CO2}$ in saltwater?

So we selected four different concentrations of salt (in the range of 1-4 %) in 100 mL of water and added 5 g of $\ce{CO2}$ and measured the time it took to dissolve completely. We then used titration to figure out the amount of $\ce{CO2}$ (reacting with water to give weak carbonic acid) that is dissolved.
My questions are:

Will the amount of $\ce{CO2}$ that reacts with water depend upon the salt concentration?
Will the amount of titrant used to figure out the amount of $\ce{CO2}$ dissolved also change depending on the salt concentrations?

If you expect an answer, you should give your guesses (ooops, hypothesis) to the questions.

It seems like you did the experiment. Why even ask the question? Just analyse the data you got.

100 mL of water and added 5 g of $\ce{CO2}$ and measured the time it took to dissolve completely

This statement doesn't make sense.  Firstly, if you know it dissolved completely, why try to determine how much dissolved?  Secondly, the only way that much CO2 would dissolve is if you did the experiment at high pressure.
If you want literature data to compare your experiment to, see Fig. 6c or equation 6 of Behaviour of the CO2-H2O system


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Identifying probable compound using its colour & reactants
This is quote from OxyChem document on Hypochlorites:

Steel tanks in caustic soda service become passivated with a
  gray-black film forming on the metal. Thereafter, caustic soda at less
  than 130° F will dissolve very little iron from the tank. The
  protective film should be preserved whenever practical. Therefore,
  when sediment is rinsed from a tank, only the bottom should be washed.
  Caustic Soda solution should then be put into the tank as soon as the
  washing is finished. Otherwise a soft rust will form and contaminate
  any future storage.

I cannot understand what can be this gray-black film chemical formula. Plus is he referring soft rust to common Iron (II) Oxide when it comes to O2 rich atmosphere. May be thats why he instructs to refill cleansed container as soon as possible?


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How to separate HCl and NaOH from an aqueous solution of NaCl?
At home I make experiments on salt (as it is easily available at home).
One day I came to know that on mixing salt and water I will get Hydrochloric Acid and Sodium Hydroxide. So I became really excited, but my excitement broke into small pieces because I don't know how to separate $\ce{HCl}$ and $\ce{NaOH}$ from the solution of $\ce{NaCl}$ and water.

I don't think you can. And if you can then certainly not at home.

The solution can be electrolyzed to produce chlorine and sodium hydroxide solution.  The chlorine can be added to water to produce HCl and HOCl.

Also, the chlorine can be burnt with the hydrogen appearing at the cathode.

One day I came to know that on mixing salt and water I will get Hydrochloric Acid and Sodium Hydroxide [...]

No, you won't!
In theory, electrolysis of a sodium chloride solution and photolysis of the combined gases formed (hydrogen, chlorine) will provide you with hydrogen chloride, as I have outlined here.  In reality, you might want to avoid this - unless you're planning to jump out of the genetic pool ;)

The reaction you are speaking of if NaCl + H2O -> HCl + NaOH, witch is the oppisate of a neutralization reaction of sodium hydroxide and hydrochloric acid and generly goes in the oppisate direction you are speaking of. in theory you could find a way to produce the reverse reaction but it would be hard and not cost efferent, the best way to produce hydrochloric acid and sodium hydroxide is by electrolysis of molten salt which will give chlorine and sodium.


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Extracting silver from Tollens test
When we add tollens reagent (diamminesilver(I) cation, $\ce{[Ag(NH3)2]+}$) to glucose it forms a silvery mirror on the walls of the container. Since the sides are coated with actual Silver,

is there any possible reagent/method to take the silver out from the sides of the test tube.?
will it be in its pure form or will there be organic/inorganic  impurities?

Silver's pretty soft—it's possible to just scrape it off with a metal spatula or something. Also, not all of the silver deposits on the glass. You might be able to adjust the conditions of the reaction to make more silver precipitate into the liquid where it's easier to get at. You can, of course, redissolve the silver in nitric acid or the like.
Since the byproducts of the reaction are soluble, it should be pretty pure, depending on how pure the starting materials are. If other easily reducible metal cations are present, you might get some impurities from that.

As a side-note, this isn't a particularly efficient way of getting elemental silver. The typical films produced in a test tube probably contain a couple hundred milligrams at most. Even a whole undergrad lab's worth of tubes is probably not worth recovering.

1) Silver is pretty easily dissolved in concentrated nitric acid (attention, nitrogen oxides fumes). The resulting solution may be gently evaporated to get rid of excess of nitric acid and silver nitrate may be then reduced with zinc and then washed from excess zinc with diluted sulfuric acid. Alternatively, silver nitrate may be used as is for similar tests in future.
2)As Michael noted, it is pretty pure.


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Is it possible to make meta-phenol?
Normally phenols have the $\ce{OH}$ group at the 1 or 4 position. Can they be made at 2 or 3 positions, where nitro groups and $\ce{-CN}$ can attach? If not, why not?

The enumeration of the phenyl carbons is arbitrarily chosen to start at the functional group. If you are not talking about isotopically labeling a phenol molecule, then the enumeration will not change.

I guess by "meta-phenol" you mean a phenol that has another group besides the $\ce{OH}$ group attached in the meta position of the aromatic ring, right?

Are you interested in Gallic Acid? http://en.wikipedia.org/wiki/Gallic_acid

Related: Why does m-cresol exist, given that -OH group is activating and should be o-, p-directing?

The hydroxyl groups in phenols are ortho-/para-directing. So, if you want to introduce a group in the meta position with respect to a hydroxyl group you will have to "overwrite" the ortho-/para-preference. This can be done by starting out with meta-directing group which can later be transformed into $\ce{OH}$ in the position where the hydroxyl group shall be. Then you only need to introduce the group you want to be in the meta-position by a simple electrophilic aromatic substitution reaction and subsequently convert the meta-directing group into $\ce{OH}$.
An example for this procedure would be to start out with a carbonyl group at the position of the $\ce{OH}$ group, then nitrate the aromatic ring in the meta position and employ the Baeyer-Villiger oxidation to introduce a $\ce{OH}$ group:

Or you start out with a nitro group at the position of the $\ce{OH}$ group, then introduce chlorine in the meta position of the aromatic ring, reduce the nitro group to an amine, which is then coverted into an diazonium salt which in turn is then transformed into a $\ce{OH}$ group using a Schiemann-type reaction:


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Home-made way to detect lye in solution
I believe that electrolysis of salt water creates lye ($\ce{NaOH}$ - sodium hydroxide). In salt water, there's $\ce{Na+}$, $\ce{Cl-}$, $\ce{H+}$ and $\ce{OH-}$ as of ionts.
So at cathode, you get $\ce{Na2}$ and that should react with water into $\ce{NaOH}$.
I want to test, if this is true - how do I do it with household items? The solution didn't react with aluminium.

I found more general question asking about measuring pH using household items.

The aluminium probably did not react because of its highly shielding oxide layer.

I scratched it. I think the solution is not concentrated at all. I've left a drop on my skin and nothing happened.

A will filter out the copper chloride and procceed with electrolysis.

Where did you make copper chloride now? This is starting to confuse me...

I used (intentionally) a copper anode. Copper reacts with chloride. First it forms yellow stuff that dissolves in water. Later on, I can see orange chunks that do not dissolve. I didn't talk about this in the original question because it doesn't seem to be important.

The easiest way to measure the pH of a solution is to use an acid-base indicator. Limited to household supplies only, I would suggest a solution made from red cabbage, which should turn blue-greenish in the alkaline region.
The ions you actually get from your reaction is a higher concentration of $\ce{Na+}$ and $\ce{OH-}$, due to the fact that you are removing $\ce{H2}$ and $\ce{Cl2}$ (from $\ce{H+}$ and $\ce{Cl-}$ ions respectively) via the performed electrolysis. Under no circumstances would you generate a $\ce{Na2}$ species, you would rather write it as $\ce{2Na}$. Also, in aqueous solutions the $\ce{NaOH}$ most certainly dissociates. So speaking of the formation of solid $\ce{NaOH}$ from $\ce{Na}$ in water is quite a stretch.
Bottom line: Since the salt is already dissolved when you start the electrolysis, all you get is a change in concentration of the ions involved. The thereby induced change in pH can be measured observed using an acid-base indicator such as the juice of the red cabbage.

So what happens to $\ce{Na+}$ ionts when I dry the water? I don't think I'm gonna get pure sodium...

If you evaporate all the water, you will be left with a mixture of NaCl and NaOH. They need water to exist in their ionic form.

How does the sodium come in contact with chlor? I never meant to think that solid NaOH will appear in water by the way. But since that's what I get after drying the water, it's just about the same.

Electrostatic interactions. And it comes in contact with chloride, not chlorine. (Chloride is the anion of chlorine.)

It is easy and cheap to buy pH testing strips from an aquarium store or from Amazon.  I bought SEOH brand strips from Amazon.com and have been happy with them.  I'm sure many others are just as good.


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Addition of combined gas law constant
There are two containers connected to each other but separated by a cork. If container(1) has a gas $G_1$ at a volume $V_1$,  pressure $P_1$ and temperature $T_1$ and similarly in container(2) there is a gas $G_2$ at volume $V_2$, pressure $P_2$ and temperature $T_2$. If the cork between the containers is removed and the temperature changed to $T_3$ what is the pressure of the mixture inside the system of containers.
My textbook gives it as:
$\frac{P_1V_1}{T_1}+\frac{P_2V_2}{T_2}=\frac{P_3V_3}{T_3}$
where $V_3$ = $V_1$+$V_2$
$P_3=\frac{T_3}{V_3}\left(\frac{P_1V_1}{T_1}+\frac{P_2V_2}{T_2}\right)$
I don't understand how that comes out. Can we just add the constants(of combined gas law) like that?

There are no constants in the equations you posted.

Caution: This answer is exclusively from my perspective.
Moments after uploading this question I realized the answer myself. For those who strike upon this question. I 'think' the answer goes like this:
The combined gas law states that :
$\frac{PV}{T}=nR$ ----------  [or $PV=nRT$ ideal gas equation] (where 'n' is the number of moles of the gas in consideration)
Now therefore:
$\frac{P_1V_1}{T_1}+\frac{P_2V_2}{T_2}$ = $n_1R+n_2R$ = $R\left(n_1+n_2\right)$
Now
$n_1+n_2=n_3$ (NOTE: $n_3$ is not the number of moles of a single gas but the gases $G_1$ and $G_2$ combined)
This can be said as the volume gets added up. Therefore further referring to Avogadro's Law (of equal volume) we the latter equation yields the equation in question.
PLEASE DO CORRECT THIS ANSWER IF IT IS FAULTY.

It looks good to me.


text:
Dihedral angle of gaseous and crystalline HOOH
Why does hydrogen peroxide exhibit a dihedral angle of $111.5^\circ$ in the gaseous state? And a dihedral angle of $90.2^\circ$ in the crystalline phase?

I know that in general, there is likely to be more hydrogen bonding in the crystalline phase than in the vapor phase.
But what's up with 1) the bond angles themselves in the first place and 2) the bond angle expansion in the gas phase? How can one rationalize them?
I know that there exist lone pairs on the oxygens and lone-pair/lone-pair repulsion should probably be minimized, but what it is about these conformers that minimizes energy? I'm guessing that we'd also want to minimize eclipsed and gauche interactions ... Newman projection time?
It seems that there is only one Newman projection possible for $\ce{HOOH}$; i.e. we'll have the inevitable lone-pair/lone-pair interaction, and the lone-pair/hydrogen interaction. I guess all we can do is minimize these interactions; i.e. the energy-minimized conformer is probably not the one with them all eclipsed. However, why does it seem that the energy-minimized conformer for vapor $\ce{HOOH}$ is one that puts the lone-pairs so close to each other in terms of dihedral angles?

another answer http://chemistry.stackexchange.com/questions/58353/

You've already answered your own question - hydrogen bonding.
First, don't think of it as to why is the gas state different than solid-state.  Rather, think about why is the solid-state different than the gas state. In the gas state, there is very little intermolecular interaction (very, very few species do not follow the ideal gas law to many significant digits). The gas state geometry is close to what is expected for $\mathrm{sp^3}$ tetrahedral geometry and may be considered normal. The link below is a paper discussing the gas phase microwave spectroscopy of $\ce{H2O2}$ in far greater detail than I can hope to here. The description of the torsional rotation energies was not trivial.
Internal‐Rotation in Hydrogen Peroxide: The Far‐Infrared Spectrum and the Determination of the Hindering Potential by
Robert H. Hunt, Robert A. Leacock, C. Wilbur Peters, Karl T. Hecht, The Journal of Chemical Physics 1965, 42, 1931-1946, https://doi.org/10.1063/1.1696228,
mirror
When condensed to the solid-state, molecules get close enough to allow intermolecular forces (in this case hydrogen bonding to a much greater extent than van der Waals). Additionally, the energies of the crystal lattice must be considered.  In this case, th

Why might the angles be what they are in the first place?

Is this answer incomplete?

In an attempt to improve the given answer and also address other issues in the discussions of other duplicates of this question, I have optimised (PBE-D3/def2-SVP using NWChem 6.6) the $\ce{H2O2}$ structure in three different point groups: $\ce{C2}$, $\ce{C2_h}$ and $\ce{C2_v}$.
From frequency calculations, only the $\ce{C2}$ geometry is a minimum, the other two geometries having an imaginary vibrational frequency each.

Figure 1: $\ce{C2}$ geometry (angle in degrees, distances in angstrom)

Figure 2: $\ce{C2_h}$ geometry (angle in degrees, distances in angstrom)

Figure 3: $\ce{C2_v}$ geometry (distances in angstrom)
The results are summarised below.
            | Relative electronic | Dipole moment |  Dihedral
Point group |  energy (kcal/mol)  |    (Debye)    | H-O-O-H (°)
---------------------------------------------------------------
 C2         |       0.000         |     1.697     |  119.274
 C2h        |       0.022         |     0.000     |  180.000
 C2v        |       0.305         |     3.205     |    0.000

So the gauge structure is indeed the most stable found, as expected, with a dihedral angle of 119.3° (7.8° above the one cited by the original question).
NBO second-order stabilisation energies are as follows (there are two in each case, only one is shown):
            |          |            |    E(2)     | E(j)-E(i) | F(i,j)
Point group |  NBO(i)  |   NBO(j)   | (kcal/mol)  |  (a.u.)   | (a.u.)
---------------------------------------------------------------------
 C2
            |  LP(O1)  |  BD*(O2-H) |    0.98     |   0.60    | 0.022
            | BD(O1-H) | BD*(O1-O2) |    0.63     |   0.68    | 0.018
            | BD(O1-H) |  BD*(O2-H) |   <0.50     |   ----    | -----
 C2h
            | BD(O1-H) |  BD*(O2-H) |    1.20     |   0.99    | 0.031
            |  LP(O1)  |  BD*(O2-H) |   <0.50     |   ----    | -----
            | BD(O1-H) | BD*(O1-O2) |   <0.50     |   ----    | -----
 C2v
            | BD(O1-H) | BD*(O1-O2) |    1.17     |   0.67    | 0.025
            |  LP(O1)  |  BD*(O2-H) |    0.85     |   0.95    | 0.025
            | BD(O1-H) |  BD*(O2-H) |   <0.50     |   ----    | -----

In this table, LP(O1) stands for a lone pair ($\ce{n_{O}}$), BD*(O2-H) is a anti-bonding orbital ($\sigma^*_{\ce{O-H}}$), etc..
From delocalisation energy contributions alone, we would expect the $\ce{C2_v}$ geometry to be the most stable.
Natural steric analysis shows the following for the repulsion between lone pairs in the oxygens:
            |          |          |  dE(i,j)
Point group | NLMO(i)  | NLMO(j)  | (kcal/mol)
----------------------------------------------
 C2         |  LP(O1)  |  LP(O2)  |    6.84
 C2h        |  LP(O1)  |  LP(O2)  |   16.68
 C2v        |  LP(O1)  |  LP(O2)  |   16.57

This seems to be in better agreement with the energy differences found.

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


text:
Determining if reaction is endothermic or exothermic using equilbrium constant

For the following reaction
$$\ce{W(s) + 4 Br(g) <=> WBr4(g)}$$
at $\pu{900 K},$ $K_p = 100$ and at $\pu{2800 K}$, $K_\mathrm{p} = 5.$
Determine if the reaction is endothermic or exothermic.

At higher temperatures, the equilibrium has shifted towards the reactants since $K_p$ is significantly smaller. However, how do we determine if a reaction is endothermic or exothermic from this information?
If the reaction was endothermic, an increase in temperature would push the equilibrium to products to counteract the increase in heat. However, the equilibrium has been pushed to the reactants thus signifying the reaction is exothermic. But this assertion is on the assumption that the forward reaction is exothermic.

Think about it this way.  In an exothermic reaction that is at equilibrium the forward reaction has a lower activation energy.  So, if you increase T the forward reaction rate increases but the reverse rate increases even more.  That's because the rates of reactions with higher activation energy are more sensitive to temperature.  With the reverse rate higher more reactants will be made until the concentration reactants reaches a point where the rate of the forward reaction equals the reverse rate.
In an endothermic reaction increasing T increases the forward rate more than the reverse, producing more product until the increased product concentration causes the reverse rate to equal the forward rate.
The equilibrium constant can be thought of as a ratio of products to reactants needed to equalize the rates of two reactions with different activation energies.

Just to be clear, you're referring to the link between chemical equilibrium and chemical kinetics where $K_{c}=\dfrac{k_{f}}{k_{r}}$. Since $K_{r}$ increases more than $K_{f}$ at higher temperature, $K_{c}$ decreases. So since the $K_{c}$ decreased at higher temperature, the reaction is exothermic. Correct?

That is correct.

I think you're looking for the van't Hoff equation. Since you have the equilibrium constants for two temperatures you can integrate the equation and get $\Delta H^{0}$ from it
\begin{align}
  \frac{\mathrm{d} \ln K}{\mathrm{d} T} &= \frac{\Delta H^{0}}{R T^{2}} \\
  \int\limits_{\ln K(T_{0})}^{\ln K(T_{1})}\mathrm{d} \ln K &= \frac{\Delta H^{0}}{R} \int\limits_{T_{0}}^{T_{1}} \frac{1}{T^{2}} \mathrm{d} T \\
  \ln \frac{K(T_{1})}{K(T_{0})} &= \frac{\Delta H^{0}}{R} \left( \frac{1}{T_{0}} - \frac{1}{T_{1}}\right) \\
  \Delta H^{0} &= \frac{R \ln \frac{K(T_{1})}{K(T_{0})}}{\left( \frac{1}{T_{0}} - \frac{1}{T_{1}}\right)}
\end{align}
which will enable you to decide whether the reaction is exothermic or endothermic.


text:
Protonation of nitric acid
I found that reaction: HNO3 + 4.1633363423443E-17 H3O = NO2{+} + OH{-}
Please help me understand " 4.1633363423443E-17" term.

The title actually should be "What does XXXX E YY formula mean?" which would quickly leave this question off-topic.

Could you provide the reference where you saw this?

That come from http://www.webqc.org/balance.php when I tried to ballance the hypotetical reaction HNO3+H3O=NO2{+}+OH{-}

It seems to me like you were asked to balance a chemical equation and you used some computer program to do it for you instead.
4.1633363423443E-17 is the result of the limitations of floating point arithmetic (ie that computers only have finite precision). The real answer should be 0 (which you could have figure out by balancing it yourself).


text:
Dissolving Time Function for Concrete
So I was just exploring some ideas on my own and am having trouble analytically solving the following:
Let $D(n,m,T)$ be the time it takes to dissolve an n x n x n cube of concrete in m molar HcL at a temperature of T.  Find $D(n,m,T)$. Over the temperature range of $32 ^o < T < 273^o$ in fahrenheit.
Mathematically how does mechanism of dissolving work? Intuitively I understand the exchange of $\ce{H+}$ and $\ce{OH-}$ but still don't understand how to use this to generate said function.


text:
Is there way to detect surface area of nanoparticle by electrochemical technique?
I want to know other than SEM is there any other technique or instrument to detect the surface area of particle esp eletrochemical way.

if you have a reference point, you can either titrate some surface groups or measure change in concentration of some dissolved compound due to adsorption on particles' surface.

Dynamic light scattering is commonly used to measure nanoparticle size: http://en.wikipedia.org/wiki/Dynamic_light_scattering
Whether you can measure this sort of thing electrochemically depends a lot on the type of nanoparticle. Conductive nanoparticles, you might be able to estimate the surface area by measuring adsorptive behaviours of various things, like CO for Pt and Pd: http://www.sciencedirect.com/science/article/pii/S1388248113000933 Non-conductive particles might be measured by how they impede diffusion to a surface. Electrochemical techniques could suffer from problems related to the shapes of the nanoparticles, as well.
The most straightforward ways of examining the surface features of nanoparticles would be electron microscopy or AFM.
Without further details of your application, it's difficult to recommend a specific technique. If the nanoparticles are being used as an electrode, the apparent electrochemical surface area may be all you care about, even though this is not necessarily the same as the geometric surface area.

But you cannot measure surface area with DLS can you? If I have two equally sized nanoparticles, with a different surface area (i.e. roughness/porosity) I get the same signal from DLS right? Or is there a way to do this?

@Michiel No, you're right. DLS is looking at mainly size, though you can get some information about shape as well. If he's interested in the surface area of individual nanoparticles, electron microscopy or AFM are going to be the go-to techniques. It may be possible electrochemically, but the apparent electrochemical surface area can be very different from the actual surface area under certain conditions.


text:
About the way chemicals and molecules are represented
The standard way of representing molecules with letters like $\ce{C}$ for Carbon, $\ce{O}$ for Oxygen etc.; and a lot of lines and double lines like a graph from Graph theory where each letter is a 'vertex', the lines like edges. I forget what this chemical notation system is called. Yet it is 2-dimensional. The formula for '$\ce{H2O}$' ,water doesn't indicate the 'angle' of 'separation' between the two Hydrogen atoms that are bonded with the Oxygen atom. Similarly a formula for some molecule like a protein might not indicate all the 'involved angles' in 3-dimensional space that 'allow' it to be folded in a precise way. So does the 2-dimensional notation system for molecules miss details of how a molecule is twisted in 3-D space?  Also can two distinct molecules 'look' the same in 2-D notation?

I see some inconsistencies there. $\ce{H2O}$ does not even represent topology, do you mean $\ce{H-O-H}$ ?

Below is 2 types of sugars; glucose and mannose.

If you know how to interpret the lines, then you can determine which one is which.

Very useful info. , but can two distinct molecules have exactly the same 2-D representations, with the same array of lines?

@user128932, any schematic representation is not going to be fully precise. Molecules can exist in various conformations with disparities that no generic type of projection or formula can encompass down to an arbitrary level of detail. In skeletal formulae, for example, bond angles of tetrahedral atoms are rendered as $120^{\circ}$, making the mental reservation that this is not usually the true angle. However, no two distinct molecules have the same skeletal formulae, as long as the stereochemistry is correctly depicted with wedges/dashes.

@GregE. Not entirely true. First, chemical formulas do not have generic notion to include electronic configuration, in rare cases it may lead to ambiguity. Also, topological features (like knots) may be impossible to figure from 2-d structure if drown without explicitly preserving them. Last, skeletal formulas fail in case the molecule has significant amount of non-covalent interactions, like ionic and dipole interaction. Luckily, all above is in full effect mostly in cases of 1) very big molecules 2) active particles 3)when it doesn't matter.

So 2-D representations of molecular formula do not necessarily capture topological features in 3-d space important in protein folding which is very important in cellular 'mechanics'. Why do most chemistry and biochemistry books ( that I have seen) tend to focus on the 2-D notation system for molecules when explaining molecular 'activity'?

Are there two molecule that are identical except one is the mirror image of the other yet according to their 2-D representations this 'chiral' quality ( if that's the right word) is hard to see?

If they are mirror images of each other, it is usually easy to see the reflection. If one is rotated, the reflection might to harder to see.

@user128932 Just imagine one of the formulae in the answer drawn without wedges, only with plain lines. This would be perfectly valid and common formula representing both structures (and many more).


text:
How do I round a calculation involving both addition/subtraction and multiplication/division to the correct number of significant figures?
I'm just learning about significant figures (sig figs) in my chemistry class, and I'm confused about the rules. An example to my problem: if I had the following expression:
$$\left ( \frac{2.378 - 1.2}{1.03} \right )$$
I've thought of 2 ways to approach this. Which one is correct (or is neither correct)?
Approach 1:
Round each intermediate operation as you go. This is what my textbook does for its examples, but it itself suggests "Approach 2" below. I'm just listing this approach here as an option since addition/subtraction and multiplication/division have their own rules for rounding (hence, maybe do intermediate-operation rounding).
1) Round the subtraction part.
$$\left ( \frac{1.2}{1.03} \right )$$
2) Do the division and round.
$$1.2\,\,\, (2\ \mathrm{sig\ figs})$$
Approach 2:
In actual practice, the book says:

In a series of calculations, carry the extra digits through to the final result, THEN round.

I think it means this:
1) Do the subtraction, but keep the *exact* number while noting that the rounded number was supposed to have 2 sig figs.
$$\left ( \frac{1.178}{1.03} \right )$$
2) Do the division with the exact number and the denominator, and since this step produces the final result, NOW you round, rounding to 2 sig figs per the note in Step 1.
$$1.1\,\,\,(\text{2 sig figs})$$
The problem here is that these two approaches produce 2 different results. So, to repeat, which one is correct (or is it neither)? The textbook doesn't provide an example like this. I found two different websites, but they too conflict with each other.
http://www.occc.edu/kmbailey/chem1115tutorials/Sig_Fig_and_Math_Answers.htm
(From the solution to problem #3, I should use "Approach 1".)
http://fabice.com/misc/significant_figures.html
(It quotes, "In a composed operation, intermediate results are not rounded." So that means I should use "Approach 2?".)
Compared to the other questions on this forum, this question seems silly, but I find the concept of sig figs important if I want to understand chemistry at all.

According to me first one is better.

According to me the second one is better. But that depends very much on the problem at hand. In praxi for every result you obtain you should also do an error analysis, which in the end gives you a value range, how trustworthy your number actually is. In this case, both of you values would be in this range. Rounding has been and will always remain a problem in the sciences.

So based on what you said, Martin, there's more to consider than just the type of operation involved, and that uncertainty should be somehow considered with every operation. Thanks, I'll look more into that.

This question appears to be off-topic because it is about math, not about chemistry.

I'm voting to close this question as off-topic because this question is more about maths.

This is a common chemistry problem. It does not seem off topic.

There are two videos that come to mind, which seem relevant to this question.

Approach 2 is better than approach 1.
This approach corresponds with methodology employed in both web links you provided.
The basic 'rule' is to do your calculation keeping all digits as you go right until to the very end (without rounding parts as you suggest in approach 1). This ensures your calculation retains maximum accuracy. You leave the 'rounding' til the very end, that is when you report your answer, only 'so many' digits are 'significant'.
To use your example, when calculating the subtraction part: 2.378-1.2 = 1.178, in approach 1, by 'rounding' this down to 2 significant digits (1.2), you are reducing the accuracy of your calculation before you finish.
The correct (most accurate) way to calculate the result is to keep all significant digits in your calculation until the very end, that is, 1.178 / 1.03 = 1.14368932 and then report only the 'significant' digits. In this case, since the number 1.2 only has 2 significant digits, your final answer can only have 2 significant digits. That is, the final answer is only as precise as the least precise measurement (in your case, 1.2), so in reporting the final answer for your calculation, you can only rely on 2 significant digits, so you report 1.1 as your calculated answer.
If you re-examine the answer provided in your link (http://www.occc.edu/kmbailey/chem1115tutorials/Sig_Fig_and_Math_Answers.htm) in which you state problem 3 follows approach 1, you will find that the solution provided in that link actually follows approach 2, similar to what I described in the above paragraph. That is, all decimal places are retained in the calculation until the very, when the final answer is reported to the number of significant figures corresponding to the least significant measurement.

Frankly, I don't get why textbooks bother including the first method. All it does is inspire questions like these. At least put the preferred method first


text:
Can organic acids be concentrated and used like concentrated sulfuric acid and similar mineral acids?
Is it possible to make organic acids concentrated like inorganic acids like concentrated sulfuric acid and similar mineral acids?

mostly because they are usually solid. Some that are not, are used, like acetic acid.

@permeakra depending on your climate zone, also acetic acid is solid. ;-)

".. used like inorganic acids ..."  For what? There are not so many cases where I would say "throw in any inorganic acid, no matter which".

They can. Vinegar is typically 4-18% acetic acid by mass, but of course concentrated acetic acid also exists with a concentration >90% by mass. It is corrosive and flammable, but it surely exists.

Some can and some can't maybe? Not all inorganic acids can be concentrated to >90%.

@fp, well some inorganic acids, say, carbonic acid, also can not be concentrated. But in general one could concentrate both inorganic and organic acids as well as any other substances.

Glacial acetic acid is 100% acetic acid. There are many organic acid available at 100%: phthalic acid, ascorbic acid, citric acid, etc...

There are a number of very important inorganic acids which are either gaseous (all hydrogen halides) or unstable (carbonic acid, nitrous acid). No organic acid of the same description springs to mind.


text:
How to get sodium from sodium aluminate?
The sodium aluminate is formed when aluminium or aluminium hydroxide reacts with water solution of lye. After drying the water you'll end up with white crystals - sodium aluminate.
I was wondering if it's possible to get pure Na out of this compound - by heating or a redox reaction.

Generally, it's quite difficult to get elemental alkali metals from their compounds. If your aim is to get sodium, you should consider reacting sodium hydroxide with magnesium (violent reaction) to form magnesium hydroxide and sodium.

@AnonymousShadow I heard of that reaction. I also saw it done with aluminum as well. The problem is that it requires you to grind sodium hydroxide. When you do so, it's gyroscopic properties take on so fast that it's immediately wet and the reaction is not quite possible.


text:
Atomic Mass of ¹²C before 1961
Question:

Prior to 1961, the atomic mass unit was defined as 1/16 the
  mass of the atomic mass of oxygen; that is, the atomic mass
  of oxygen was defined as exactly $\pu{16 amu}$. What was the
  mass of a $\ce{^{12}C}$ atom prior to 1961 if the atomic mass of oxygen on today’s scale is $\pu{15.9994 amu}$?

I don't fully understand the logic behind the question. Here's my thinking. Atomic mass of all substances were related to $\pu{16.0000 amu}$ prior to 1961. Therefore, we can make the following expression where $X$ is mass of $\ce{^{12}C}$
$$\dfrac{X}{\pu{16.0000amu}}$$
Today, all atomic mass is related to $\pu{12.0000 amu}$ of $\ce{^{12}C}$. We can model the relationship as $$\dfrac{\pu{12.0000amu}}{\pu{15.9994amu}}$$
What the solution manual does is make the two equations equal to each other and solves for $X$.
$$\dfrac{X}{\pu{16.0000amu}}=\dfrac{\pu{12.0000amu}}{\pu{15.9994amu}}$$
How are we justified in doing this?

Here is another equivalent way of thinking about the problem. The mass of oxygen on the old scale is 16 amu. The mass of oxygen on the new scale is 15.994 amu. When you divide them:
$$\frac{16\ \textrm{old}}{15.9994\ \textrm{new}} = 1.0000375 \frac{\textrm{old}}{\textrm{new}}$$
Basically, taking a ratio of any value in the old scale to the equivalent value in the new scale (or vice versa) gives you a conversion factor. So for every new amu, there is 1.0000375 old amus. The question is asking you to convert 12 new amus to old amus. Just by looking at the units, it should be obvious what to do:
$$\frac{16\ \textrm{old}}{15.9994\ \textrm{new}} * 12\ \textrm{new}=12.00045\ \textrm{old}$$
If you explicitly label old and new, it's easy to see that the new units cancel out.
Incidentally, if you keep track of your significant figures, you can see why this redefinition of the amu didn't cause too many problems for chemists—even to five sig figs, the change is negligible.

In the old system the atomic mass unit was abbreviated as $\mathrm{amu}$. This system is now obsolete. To avoid conflict, the new system, the unified atomic mass unit uses the symbol $\mathrm{u}$.

Yeah I know, but I've seen amu used as the unified atomic mass unit in general chemistry textbooks a lot so I used new and old for this example to make it clearer. Personally, I prefer Da to u, if for no other reason than "daltons" is easier to say than "unified atomic mass units".

Think about what happened to the scale. For $\ce{^{16}O}$, it decreased from $16.0000\,\mathrm{amu}$ to $15.9994\,\mathrm{amu}$. So the scale was stretched out (since 0 needs to stay constant). That would mean that $\ce{^{12}C}$ would need to decrease from some number greater than $12.0000\,\mathrm{amu}$. With $0$ being constant, there is no addition factor (like when converting from Fahrenheit to Centigrade); it is just a ratio from one scale to the other.
$$\frac {\text{unknown}}{16.0000\,\mathrm{amu}}=\frac {12.0000\,\mathrm{amu}}{15.9994\,\mathrm{amu}}$$


text:
Most Comprehensive Chemistry Book(s)?
Although I am now taking a pretty fundamental chemistry course, I am really looking to gain an extremely comprehensive knowledge of the subject, specifically, in inorganic and organic chemistry. I am looking for two texts, keeping in mind that the most important thing is comprehensiveness and the most information about the most topics, not ease of readability. I have the time and dedication to do it, but I really don't want to start with a book that doesn't cover everything and I have to fill in the holes as I go or individually seek out topics it misses.

No matter what the cost like $1000 of dollars?

For the inorganic chemistry I recommend Holleman-Wiberg's text, also exists in english. It is reasonably readable and very thick and contains a lot.
Holleman-Wiberg's Inorganic Chemistry. If you prefer, the original is in German and updated very often.
For the organic chemistry, besides being thick also very nicely readable,  Organic Chemistry by Clayden, Greeves and Warren
Besides that, you would probably need some nice general chemistry book, but I have no good personal experience in this area.

For Organic Chemistry, definitely "Organic Chemistry by Clayden, Greeves and Warren".

Sorry to say it (I love the Clayden, I really do) but in terms of comprehensiveness the Clayden would not be my first choice for Organic Chemistry. Here, the Smith and March is the gold standard.

@Philipp: of course I fully agree, how could I forget about "Jerry". Needless to say, I think it is not the book you can read cover to cover without reasonable background. That's why Clayden immediately popped in my mind, as the question was stated.

Here is the list some books which i fill it is good and detailed:

IB Chemistry Course Book 2014 edition: Oxford IB Diploma Programme
General Chemistry (Dover Books on Chemistry)
Comprehensive Chemistry by Dr. N.K. Verma, Prof. S.K. Khanna and Dr. B. Kapila


text:
Why won't increasing the temperature increase the solubility of a gas in a liquid solvent?
If I raise the temperature of a solution of a gas in a liquid solvent, why won't the solubility increase?
Increasing the pressure will increase the solubility, but increasing the pressure doesn't mean increasing the temperature?  What is wrong with this argument?

A gas has bigger entropy when it's free, than in a solution. By increasing the temperature the molecules have bigger energy and they tend to be in a state with more freedom. Therefore, the solubility would decrease.

When a gas dissolves in a liquid, bonds are formed between the gas particles and the liquid particles. The formation of these bonds liberates energy. Increasing the temperature of the solution will shift  the equilbrium backwards. Hence less gas can dissolve and solubility
decreases.

In reference to Henry's Law,

"At a constant temperature, the amount of a given gas that dissolves
  in a given type and volume of liquid is directly proportional to the
  partial pressure of that gas in equilibrium with that liquid."

Thus, increasing partial pressure of gas above the liquid increases solubility of gas, which does NOT mean that temperature of the solution is increasing. There is a distinction to be made between the dynamics of the solution and the gas above the solution.


text:
What would the Electronic Configuration of Ru²⁺ be?
I'm doing homework for Inorganic Chemistry, and I am a little confused.
The Question is:

What is the electronic configuration of $\ce{Ru^{2+}}$?

If I were to remove from the highest energy level it would be $\ce{[Kr]~5s^1~4d^5}$
leaving $6$ unpaired electrons.
If I were to take into account electron pairing it would be $\ce{[Kr]~4d^6}$ with $4$ unpaired electrons.
What would be the correct approach?

I don't know if it's kosher to provide the answer like this but it turns out (after i got my homework back) that the correct answer was;

$Ru^{2+}: [Kr]$ $4d^6$.

The reasoning provided that was behind it is "When the d or f orbitals start to fill up, the energy of these orbitals is lowered such that they have lower energy than the corresponding s orbitals."

You've already figured out that the electron configuration of $\ce{Ru}$ is $\ce{[Kr]\,4d^7\,5s^1}$.
All you need to know know is the order of energies for the $\mathrm{4d}$ and the $\mathrm{5s}$ orbitals. You might want to have a look at the Aufbau principle (and possible pitfalls/exceptions) and another thread on this site.

As mentioned above i attempted an answer based on the order of energies (I considered that 4d is higher energy than 5s), here is a link for reference. I've considered the Madelung Rule here but it doesn't seem to apply.

There is another simple version of Aufbau principle. n+l rule, where n is principle quantum number and l is azimuthal quantum number. For 4d n+l= 4+2=6 and for 5s n+l=5+0=5, so electron will enter 4d orbital first and then it will fill 5s if 4d is filled or half filled. Note: This principle is for filling the orbital only. But for removing electron it must be removed from the outer shell first, in this case 5s. So, electron will be removed from 5s and then 4d and so on.
NB: Order of energy should only be applied for filling the orbital and for removing electron it must be removed from the outer shell and so on.


text:
Why is one lobe of an sp3 hybridized orbital smaller than its other half?
A hybrid sp3 orbital is drawn with one lobe smaller than its other half, the latter which is of equal size when drawing the p orbital. Why is it so?

When combined at a given atomic center, any two atomic orbitals add in a vectorial
manner. For example, consider the orbital $\phi$ defined by $\ce{p_{x}}$ and $\ce{p_{y}}$ atomic orbitals as
\begin{align}
\phi = c_1 \ce{p_{x}} + c_2 \ce{p_{y}}
\end{align}
The orbital addition can be pictured like this

for the two cases $c_1 = c_2 > 0$ and $c_l = -c_2 > 0$, respectively. The ratio $c_1 / c_2$ controls how much the orbital $\phi$ is tilted away from the $x$ (or $y$) axis.
The mixing of atomic orbitals with different angular momentum quantum number is
also controlled by a vectorial addition and leads to various types of hybrid orbitals
some examples of which are shown here:

The example a) would be an $\ce{sp}$ hybrid orbital. It gets its sense of direction from the $\ce{p}$ orbital used to construct it because the $\ce{s}$ orbital is isotropic. Its shape comes about because the parts of the orbitals that have an opposite sign (this is hinted at by the shading in the pictures) cancel each other out to some extent. So, if the left lobe of the $\ce{p}$ orbital is negativ and the right one is positive and the $\ce{s}$ orbital is also positive then the left half of the $\ce{s}$ orbital overlaps with the left lobe of the $\ce{p}$ orbital in a destructive manner and they cancel each other out, such that only a small portion of the initial $\ce{p}$ orbital lobe remains. But the right half of the  $\ce{s}$ orbital overlaps with the right lobe of the $\ce{p}$ orbital in a constructive manner and so the right lobe of the initial $\ce{p}$ orbital becomes larger.

It might be worth pointing out that the positive and negative signs refer to the wavefunction and have nothing to do with charge. As the probablility of finding an electron is given by (wavefunction)squared, both ends of the p orbital have a significant probability of being where the electron is, and are therefore both have a delta negative charge.

@steveverrill Good point. I never said in my answer that the "negative" or "positive" parts would have anything to do with charge, but it's definitely worth to point out explicitely that it doesn't have anything to do with that.

So it's basically a glorified interference pattern.

The reason is when one s orbital hybridizes with three p orbitals the electron density is concentrated between the line joining the nucleus of the s orbital and that of p orbitals at one lobe of the $sp^3$ hybrid orbital hence the electron density concentrated appears to be bigger than the one with lower electron density.

This answers half the question.

'p' orbital has an 100% p-character , but 'sp3' has 75% p-character and 25% s-character in addition . Therefore, it is the presence of the s-character which gives one of the lobes a kind of spherical shape , thus, making it look comparitivly smaller as to the other lobe.
(P.S.- s-round shape ;  p-dumbbell shape)


text:
Why does water break salt really
A kid asked me this question just the other day, and I said that the positive part of $\ce{H2O}$ attracts the negative $\ce{Cl}$ ion and the negative part of the $\ce{H2O}$ attracts the positive Na ion.
Then he asked me several questions which stumped me:

Why doesn't $\ce{NaCl}$ dissolve water instead (break water into $\ce{H}$ and $\ce{O}$)?
If $\ce{H2O}$ is electrically neural, why would it attract any ions?
Why doesn't $\ce{NaCl}$  stick together, how can one know if one molecule will attract the ions hard enough that the ions breaks away?
Why is Na positive and $\ce{Cl}$ negative?

Hopefully someone can help me answer the above questions as I have not done chemistry in years!

I think this answer should address all your questions
http://chemistry.stackexchange.com/a/11147/7402

See: Why is salt so hard to remove from water?

1.) The covalent OH bond strength in water is slightly stronger than the NaCl ionic bond strength.
2.) Water might be neutral but each atom exhibits a partial charge.  Oxygen sucks up most of the electron density away from the hydrogen atoms leaving O with a partial negative charge and hydrogen with a partial positive charge.  This gives rise to the famous hydrogen bonding behavior of water.
3.) You would have to analyze the energies of each of the interacting components to know.
4.) Cl is more electronegative than Na due to the larger number of protons in the nucleus.  This means that Cl will suck electron density away from Na as opposed to the other way around.

The main reason about why water breaks salt is related to the relative permittivity $\epsilon_r$ of  water compared to its value in air. In fact,$\epsilon_r= 80$ in water, while its value in air is $\epsilon_r= 1$.
Now, if we look at the formula of the Coulomb force, which is responsible of the electrostatic attraction between ${\rm Na}^+$ and ${\rm Cl}^-$ :
$$f = \frac{q.q'}{4\pi \epsilon_0\epsilon_r r^2}$$
(Where q and q' are the charges of  ${\rm Na}^+$ and ${\rm Cl}^-$, and r is the distance between the two punctual charges).
We notice that, this force is 80 times bigger in water than it is in air.
So, the attraction between the two charges is  80 times weaker in water than in air.C


text:
Amine group - electron donating or withdrawing group?
Is the amine group a donating or withdrawing group? I think it depends on resonance. If the amine group can participate in resonance, the amine group is inductively donating. Resonance depictions below.

You are right, when the amine group takes part in resonance through its free electron pair it donates elctrons - but not inductively but through conjugation. When it can't conjugatively use its electron pair the rather high electronegativity of $\ce{N}$ comes into play and the amine group exerts an inductive electron-withdrawing effect on neighboring carbon atoms.

When the atome of nitrogen participates through its free electron pair in resonance, the amine group plays the role of a donating group by conjugation. When it cannot participate in conjugation through resonance, the amine groupe plays the role of inductive electron-withdrawer group, due to the higher electronegativity of nitrogen atom compared to carbon atom.
When the amine group plays the role of electron donating group through conjugation, its inductive withdrawing effect (still existing) is weaker and masked by the first effect.


text:
Dehydration of alcohol with Lewis acid aluminum trichloride
I am told that this method yields ONE equivalent of water, but I seem to get two equivalents of water. Also does this mechanism actually work? I've never seen anything like it, but it seems to make sense. Would another Lewis acid such as boron trifluoride also work?

One of your two $\ce{H2O}$ molecules comes from the reactants, so it "doesn't count".

What do you mean it doesn't count?

@Dissenter It means that the $\ce{H2O}$ isn't really generated from the reaction, but exists already. You added a $\ce{H2O}$ in the second step.

In the reaction $\ce{{}^{$i$}Pr-OH + H2O ->[\ce{AlCl3}] C3H6 + 2H2O}$ the $\ce{H2O}$ on the reactant side and one of the $\ce{H2O}$s on the product side cancel, so you get $\ce{{}^{$i$}Pr-OH ->[\ce{AlCl3}] C3H6 + H2O}$.
I think the reaction would also work with other Lewis acids. But it will rather proceed through the $\mathrm{E}1$ mechanism in any case. And it is likely that you'll get $\mathrm{S}_\mathrm{N}1$ side reactions.

What do you mean by e1 and sn1 side reactions? That a hydrogen will be abstracted? But by what? Also what sn1 products? Alcohol is a poor leaving group.

@Dissenter The $\ce{OH}$ group is a bad leaving group, yes. But the Lewis acid turns it into a better leaving group, much as protonation would. As for $\mathrm{E}1$ and $\mathrm{S}_\mathrm{N}1$: You should know about them from your Organic chemistry courses. They work by first letting the leaving group leave, thus forming a carbocation, which is then either attacked by a weak nucleophile or eliminates a hydrogen.

So you're saying the OH group is likely to leave by itself and then water attacks as a nucleophile and the electrophile/Lewis acid does nothing?

@Dissenter No, the OH group is coordinated by the Lewis acid (in a similar manner as a proton would bind to it - similar but definitely not the same) and will thus be a good leaving group. Then the water can act either as a nucleophile or as a base to effect either a $\mathrm{S}_\mathrm{N}1$ reaction or an $\mathrm{E}1$ elimination.

Okay I see what you're saying, will edit answer to reflect what you said.

@Dissenter Basically, you're right now, but you have your charges wrong in some of you structures, e.g. it must be $\ce{F3B^{\ominus}-OH}$. Furthermore, with secondary alkyl halides you will likely get not only $\mathrm{E}1$ products or $\mathrm{S}_\mathrm{N}1$ products but a mixture as both mechanisms will compete with one another.

Hey, good catch!

But it's a coordinate covalent bond ...?

@Dissenter It's the same situation as in $\ce{NaBH4}$. There you have $\ce{H3B^{\ominus}-H}$ and rather than $\ce{H3B^{\ominus}-H^{\oplus}}$.

I know, if we do consider it as a coordinate covalent bond strictly even then the formal charges should be +/- 2

$\ce{BF3*OEt2}$ in $\ce{CH2Cl2}$ can be used too, at least for the dehydration of tertiary alcohols [DOI].
As for the moles of water yielded, please see Philipp's answer and use your common sense on whether the formation of a second molecule of water out of nothing is likely.

What do you mean out of nothing? Do you think there is no solvent?


text:
Resonance stabilization and size of ligand atoms
I am told that for these two molecules, one of them is not as resonance stabilized as the other. Apparently it's the chlorine one, and it's because of the mismatch in the size of chlorine and carbon. This makes sense prima facie, but can someone confirm this is true? I can see how carbon and nitrogen have similarly sized p orbitals and thus have good overlap, and how chlorine and carbon might have poor p orbital overlap, like a BB pellet with a basketball.

Yes, your analysis is accurate. The nitrogen is donating electrons through a 2p orbital which is similar in size to the carbon and oxygen 2p orbitals. The chlorine is donating electrons through a 3p orbital, which is significantly larger than the second row elements' 2p orbitals.

Yes, this is true. The $\pi$ overlap between a valence $\ce{p}$ orbital of a second row element and one of a higher row element is much smaller than the $\pi$ overlap between two second row elements. One reason is that the bond length is generally higher when higher row elements are involved (due to their size). Another reason is, that the $3 \ce{p}$ or $4 \ce{p}$ orbitals are much more diffuse than the $2 \ce{p}$ orbitals. If you want to read deeper into the topic have a look at:
[1] W. Kutzelnigg, Angew. Chem. Int. Ed. Engl., 23, 272 (1984).
[2] N. C. Norman, Polyhedron, 12, 2431 (1993).


text:
Chemical nomenclature of ions
I'm a very beginner in chemistry. The book I'm studying says that an ion is an atom, or a collection of atoms, that has an electric charge, then it writes about the difference between molecules and ions. What I'm wondering is, if I consider a unit of, say, sodium chloride (by unit I mean one atom of sodium plus one atom of chlorine), should I say "a molecule of s. chloride" or "an ion of s. chloride"?

It is said: one formula unit of sodium chloride.


text:
According to what criteria was the periodic table of elements ordered?
I was wondering why the elements in the periodic table were disposed the way they are. I understand, of course, that they are put in increasing atomic number fashion, but I'd like to know more about the topic. Could you link some reference that examine in detail the structure of the periodic table?

It was realized that (what is now) every 8th element exhibited similar chemical behavior and the periodic table was arranged as such.  It was later realized from quantum mechanics why these behaviors were so similar (the discovery and understanding of orbitals, electron configuration, etc.).

Elements were ordered by the increase of their atomic weight, not numbers.

sounds like a duplicate of this: http://chemistry.stackexchange.com/questions/15883/how-was-mendeleev-able-to-develop-his-table/15884#15884 Maybe you mean something else, in which case please clarify the question.

I'm not asking how Mendeleev found out atomic weight, but what are the other chemical properties to consider when you build the periodic table. If increasing atomic weight were the only criterion, it would just be one long string of elements, or a rectangle.

@Nicol All the Noble gases have virtually zero reactivity with anything we know of.  We know why this is now (because their valence space is filled) but back then nobody knew WHY.  They just knew that there were some elements that had the same chemical behavior (in this case, atoms that don't react).  Rinse and repeat for other groups in the periodic table.

I should note that my example using the noble gases works but may perhaps be misleading.  When the 'law of octaves' and the concept of 'periodicity' came about, the noble gases were not known.  These elements were discovered after Mendeleev had arranged the periodic table with the elements he knew at the time.

My comment was an addition of LordStryker's, of course that atomic weight wasn't the only criteria.

I do not address the question of history, of Mendeleyev's investigations, that is answered in How was Mendeleev able to develop his table?
Modern understanding of basic principles follows. Let us have a piece of periodic table
  ... A0 B0 ...
  ... A1 B1 ...
  .............

Element "B0" must have the next atomic number after "A0", i.e. one more electron. It is different chemically, although may have (albeit not necessarily) similar physical properties.
Element "A1" must be chemically similar to "A0", although physical properties may differ significantly (as a general rule, ".1" elements are denser than ".0"). It happens that for each element there exist its "analog" in the next period, that has similar chemical properties (based on similar outer-shell electron configuration). This pattern, once noticed, permitted for good predictions (search for Mendeleyev's "eka" elements in Internet for more information).
Periods have different lengths. Analog's atomic number may be 8, 18, or 32 ahead off the prototype's atomic number (the number of electrons expended to "raise" all the configuration one shell up). This difference is due to quantummechanical effects.

The vertical columns are families, which are elements that share similar characteristics. For example on the far right side of the periodic table are the alkali metals, known for being extremely reactive. and on the opposite side are the noble gases, which are known for reacting to very few substances. The horizontal periods are organized by the atomic number; the number of protons in an particle.
Dimitri Mendeleev purposely left spaces in the periodic table because he knew there were elements that had not yet been discovered, that would need a place on the periodic table.


text:
Silver identification of the group I of cations
I was wondering if this actually happens in a chemical reaction
$$\ce{AgCl_{(s)} + 2NH_4OH_{(aq)} \rightleftharpoons  Ag(NH_3)_{2(aq)}^+ + Cl^{-}_{(aq)} + 2H_2O_{(l)}}$$
My teacher wrote something like this but on Internet I saw that it yields to $\ce{Ag(NH_3)_2Cl}$ and water, so can someone explain it to me.

Both formulas for diamminesilver(I) chloride essentially mean the same. Writing it as
$$\ce{Ag(NH_3)_{2(aq)}^+ + Cl^{-}_{(aq)}}$$ emphasizes that the compound is dissolved in water as discrete ions (the cationic silver complex and chloride anions). One can also write $\ce{Ag(NH_3)_2Cl}$ instead because it is known that the compound exists in the aforementioned way in aqueous solution. In the latter case, it would still be more precise to use brackets to indicate the complex, i.e. $[\ce{Ag(NH3)2}]\ce{Cl}$, so that it becomes clear that the two $\ce{NH3}$ are the ligands of silver.

Now everything makes sense, I'm sorry but how does the color change after the reaction?

@user162485 The white precipitate of AgCl will dissolve to form a colorless solution of diamminsilver(I) chloride.


text:
What are the correct resonance structures of bromoethene?
Can anyone explain why my textbook only gives the following resonance structures?

Write the important resonance structures for each of the following:

I also put two resonance case where the double bond electrons are moved on the individual carbon atoms.

@Freddy Thank you for taking care of this site. Please do not use markup in the title field, see here for details. In short: It looks ugly in the search engines and results may not be found.

@Martin Sorry i didn't take that in consideration.

Your resonance structure is perfectly alright. But the resonance structure with the negative and positive charges won't be very important and here is why: For one thing, it contains seperated charges and usually those resonance structures are most important that contain as few charges as possible. This is simply a matter of electrostatics - seperating charges and holding them apart from one another costs energy. A second reason has to do with the "electronic origin" of this resonance charge: It describes a conjugation between a free electron pair of $\ce{Br}$ and the $\pi$ system of the double bond - more accurately put the filled free-electron-pair-orbital can interact with the $\pi^{*}$ orbital of the $\ce{C=C}$ bond to form two new orbitals: an in-phase combination that is lower in energy and will contain the two electrons from the free electron pair and an out-of-phase combination that is higher in energy and will remain unoccupied. This stabilizes the system. The stronger the overlap between the free-electron-pair-orbital and the $\pi^{*}$ orbital of the $\ce{C=C}$ bond and the closer they are in energy the better the stabilization of the system through this conjugation is and the more important is the second resonance structure. But in the case at hand the $\pi$-overlap between the bromine orbital and the $\ce{C=C}$ bond is very weak (the reason for this you can find here) and so the conjugation is very weak, too. Thus, this resonance structure is not very important.

"I also put two resonance case where the double bond electrons are moved on the individual carbon atoms." What about those? Sorry if I was not clear, but that was my main question :O!

@yolo123 Do you mean each carbon gets one electron from the double bond or one carbon gets two (thus becoming negatively charged) or the other gets none (thus becoming positively charged)?

Just to add a little factoid, separated charges on a MOLECULE can be favorable under certain conditions.  Amino acids such as glycine are known to be zwitterions (neutral molecules that contain separated charges) in solution across a wide pH range (~3-11).


text:
How can we measure the atomic radius of an atom?
The definition of atomic radius is half the average internuclear distance of an atom.
But many times I've been told that it is not possible to handle a single atom. Thus how can we measure the internuclear distance of an atom?

As you told it is not possible to measure the inter nuclear distance of single atom. But here internuclear distance does not mean diameter of a single atom but it means distance between nucleus of two atom of same element.

This internuclear distance can be determined by two methods:

X-rays method
Spectroscopy method

Note: Atomic radius is not a set value for a given atom. Eg. $\ce{Fe^{+2}}$ ion have different atomic radius than  $\ce{Fe+}$. Moreover,  $\ce{Fe^{+2}}$<  $\ce{Fe+}$

I think it is important to note also that atomic radius is not a set value for a given atom.  For example, Fe2+ ions have a different atomic radius than Fe3+ or Fe in metal.  When we talk about atomic radii in minerals (which are crystalline) then there can be many different radii possible.  You can find tables of these radii in mineralogy texts where they've been determined as the spherical size of the atom that fits into its site in the crystal.

I think OP what to know the practice method of determining atomic radius, but we can add this as note. :-)

Also, I'd rather stress that atoms do not "look" clear-cut as in the picture, but rather blurry, so even the specific atomic radius of an atom in a particular state is somewhat less definite than it might seem.

@ZSG - It is more basic than that. The $\ce{Fe^{3+}}$ ion really has a different sizes in $\ce{FeCl_3}$, $\ce{FeBr_3}$, and $\ce{Fe_2S_3}$ for example.


text:
Is the Ocean acidification inconsistent with the Bjerrum plot?
$\ce{CO2}$ dissolves in water and goes through chemical reactions and exists at different concentrations of,

dissolved $\ce{CO2}$,
carbonic acid,
bicarbonate, and
carbonate.

The Bjerrum plot says that a high concentration of carbonate results in an increased pH (i.e., basic). But many diagrams in the Internet about ocean acidification show that as carbonate is produced, it becomes more acidic (although they also shows as carbonate is produced $\ce{H+}$ is also produced).

Source
I don't know how to reconcile this kind of figure and Bjerrum plot as I feel they are not consistent.

 source
I think I see the confusion. You are right that directly adding carbonate, perhaps in the form of $\ce{Na2CO3}$, would increase the pH. However, ocean acidification, as shown in your figure, is from atmospheric $\ce{CO2}$ dissolving in the ocean. The relevant equilibrium: $$\ce{CO2_{(aq)} + H2O <=> H2CO3_{(aq)} <=> HCO3^{-}_{(aq)} +H+_{(aq)}<=>CO_{3(aq)}^{2-} +H+_{(aq)}}$$
The Bjerrum plot is simply a graphical representation of this equilibrium. As you add $\ce{CO2}$, you push the equilibrium towards the right, forming $\ce{H+}$, lowering the pH. Conversely, when you add $\ce{CO3^{2-}}$, the equilibrium move towards the left, consuming some $\ce{H+}$ and raising the pH. Although adding $\ce{CO2}$ leads to an increase in the absolute concentration of carbonate, the ratio of concentrations goes the other way.
You can use this two ways:

If you have a change in pH, you can move along the x-axis to predict how the concentrations of the carbonate species change.
If you change the ratios of the species, e.g. by adding $\ce{CO2}$, you can predict how the pH will change.

I notice now that the equilibrium looks funny inlined like that. Note that the last part is just the products from bicarbonate losing its proton and doesn't need H+ to get there.


text:
Determining the redox half-reactions of single displacement reactions
For the following double displacement half reaction, the half reactions can be determined easily using the oxidation numbers.
$$\ce{MnO4- {(aq)} + IO3- {(aq)} -> MnO2 {(s)} + IO4- {(aq)}}$$
We know this is the reduction half reaction because the oxidation number of $\ce{Mn}$ changes from +7 to +2:
$$\ce{MnO4- {(aq)} -> MnO2 {(s)}}$$
and we know this is the oxidation half reaction because the oxidation number of $\ce{I}$ changes from +5 to +7:
$$\ce{IO3- {(aq)} -> IO4- {(aq)}}$$
However, for the following reactions, this strategy cannot be used. So how do we determine the half reaction equations for the following reactions:

$\ce{Mn(OH)2 {(s)} + H2O2 {(aq)} ->T[Base] Mn(OH)3 {(s)}}$
$\phantom{iiiiiiiiiiiiiiiii}\ce{MnO4^2- {(aq)} ->[Acid] MnO2 {(s)} + MnO4- {(aq)}}$
$\phantom{iiiiiiiii}\ce{IO3- {(aq)} + I- {(aq)} ->T[Acid] I3 {(aq)}}$
$\phantom{iiiiiiiiii}\ce{P {(s)} + PO3^3- {(aq)} ->T[Base] HPO3^2- {(aq)}}$

Let's take the first equation in the question by example:
$$\ce{Mn(OH)2 (s) + H2O2 (aq) -> Mn(OH)3- (s)}$$
In this equation ion $\ce{Mn(II)}$ is oxidised into $\ce{Mn(III)}$, and oxygen in $\ce{H2O2}$ (oxidation number -1) is reduced into ion hydroxide (oxidation number -2):
$$
\begin{align}
\ce{Mn^{2+} &-> Mn^{3+} +e^- \tag 1}\\
\ce{H_2O2 (aq) + 2e^- &-> 2OH^- \tag 2}
\end{align}
$$
If we multiply the first equation by 2 and add the equations $(1)$ and $(2)$:
$$\phantom{iiii}\ce{2Mn^{2+} + H_2O2 (aq) -> 2 Mn^{3+} + 2OH^- \tag 3}$$
By adding four ions hydroxide to the two sides of equation $(3)$:
$$\ce{2Mn^{2+} + H_2O2 (aq) +4OH^- -> Mn^{3+} + 6OH^- }\phantom{iiiiiiiiii}$$
By rearranging the equation $(3)$, we find:
$$\ce{2Mn(OH)2 (s) + H2O2 (aq) -> 2 Mn(OH)3 (s) \tag 4}\phantom{iiiiiii}$$
We do the same for the other equations.

Well, I guess those are just examples where you have to know more about the chemistry:

For the first one, I see the question about $\ce{H_2O_2}$ acting weird in redox-reactions here and on other forums quite often. It can do both, reduce and oxidize. And there is a disproportionation of $\ce{H_2O_2}$ reacting with itself which is catalyzed by things like $\ce{MnO_2}$ or $\ce{KI}$. So in these cases, it's all about the redox-partner or catalyst and the pH-environment.
For the next one, this is a classic disproportionation reaction. In your case, this depends on the situation. Of course, if permanganate was given and the note that acid is added your teacher may also ask you for the formation of $\ce{Mn_2O_7}$. If the topic of the lecture or exam is redox-reactions however and no other partner is given, and the acid is written on the arrow instead of a redox-partner, then you could think about a disproportionation. To predict these at home, so if you have access to literature or the internet, the so-called 'Frost-Ebsworth'-diagram is of help. You can check them on the net and learn how to draw them. I use them all of the time when I think of the possible site reactions my redox-reactions could do. It helps you to decide whether a species in solution is unstable and will do disproportionation or not. If you draw one for the correct pH-environment here you can predict it and it will also come up with the answer given in your example.
The same should be true for the last equation, also I didn't try that on my own yet. I know that $\ce{P_4}$ will react with water and does disproportionate under basic conditions to phosphane and some oxidized species.
For the third one, you'd have the oxidized iodine and the reduced iodine synproportionate to form elemental iodine, which, in the presence of iodide and in water form the brown solution of triiodide. At least this would seem logical if there was a negative charge in your equation. Unless there is another molecular $\ce{I_3}$ species I haven't heard about yet (if someone has info on that I'd be interested).


text:
Carbon decrease from 1-octyne to 1-hexene
I am to turn 1-octyne to 1-hexene.
Method 1

1) Reduction using sodium metal in pure ammonia to an alkene.
2) Halohydrin formation; the -OH installs at the more substituted carbon due to its greater ability to stabilize positive charge (this always confused me; the more substituted carbon is both more positive in character and has more electron-donating methyl groups?)
3) Conversion of OH to a good leaving group (-OTs) using a sulfonate ester and a weak base, pyridine.
4) E2 elimination using NaOH and heat. Because of resonance and sheer bulk and negative induction the OTs(-) should make for a better leaving group than Br-. So the internal alkene is formed.
5) Two-part one step ozonolysis; creation of aldehyde.
6) Reduction of carbonyl with hydrogen and transition metal to terminal alcohol.
7) Conversion of terminal alcohol to a better leaving group using tosyl chloride and pyridine again.
8) Use of LDA to deprotonate and create the terminal alkene.
Method 2
Also we can take another route with the brominated alkene; we can use OsO4 and basic hydrogen peroxide to create a 1,2 (vic) diol, which can be cleaved by periodic acid, and this creates a primary alcohol, which we convert to a good leaving group using tosyl chloride and pyridine, and we can eliminate. Same number of steps, but just something different that I thought of. Well, perhaps slightly simpler since there isn't a need for a two-step process (ozonolysis).

"Oxidative cleavage of alkynes proceeds via cycloaddition to metal oxides. Most famously, potassium permanganate converts alkynes to a pair of carboxylic acids." |  http://en.wikipedia.org/wiki/Hofmann_rearrangement | https://en.wikipedia.org/wiki/Hofmann_elimination

So I have to make an amine first @permeakra? Seems like a good route but we I'm only employing methods I know already. I will however keep what you mentioned in mind for future stuff.

Is oxidative cleavage of alkynes and carboxilic acid reductions already known?

Yes to oxidative cleavage; no to COOH reductions.

Than reread $\HIO4 cleavage, it results in aldhydes.

Oops, good catch! Messed up.

No it did not. Reread your writing.

Is that the only thing I screwed up on?

Klaus Warzecha's answer suggests a more economical method than any I could think of or find in the literature. If you're limited to the standard set of OChem I/II reactions, however, one simple reaction sequence I can think of is (1) oxymercuration-demercuration, (2) Baeyer-Villiger oxidation (giving hexyl acetate), (3) ester hydrolysis (or hydride reduction with, e.g., excess LAH), (4) conversion of alcohol to good leaving group (by tosylation, $\ce{PBr3}$, $\ce{SOCl2}$, $\ce{HX}$, etc.), (5) elimination of the leaving group.

Once you have the alkene (intermediate 2), just react with HBr to give the secondary bromide. That avoids the need to convert -OH to -OTs and it avoids any selectivity issue in the elimination.

Another alternative: isomerize the terminal alkene to 2-octene with sulfuric acid. Ozonolyze and proceed as you have outlined.

This might be a shorter route that employs common synthetic techniques you're probably already familiar with.

The route begins by forming the acetylide anion, followed by alkylation with pentyl bromide.  Oxidation of the resultant acetylene will produce the two possible carboxylic acids.  These two acids should be readily separable by distillation (C6 b.p. 206C; C7 b.p. 223C).  Further reaction (reduction, tosylation, elimination) of the hexanoic acid fraction would produce 1-hexene.
Non-Subverting Route

I think you've studied radical allylic oxidation (maybe it was radical allylic bromination that you posted about?  Well, same idea.).  In any case, it is a well-known reaction (here are some examples).  You might get a mixture of the propargylic alcohol and ketone - no matter, carry them both along to the next oxidation step. In this second step we oxidize the triple bond and any remaining propargylic alcohol to the alpha-ketoacid, which will decarboxylate (see the first few sentences in this reference) under the reaction conditions.  This yields hexanoic acid which can then be converted to the desired olefin.

I think this is a good approach, but it subverts the spirit of the question, as you throw away most of the carbons from the starting material.

Regarding your comment, it's perfectly fine, it doesn't upset me, but I'd like to share my thoughts. There's a fine line between subverting the spirit of  a question and thinking outside the box.  I used the chemical properties of A to get to B.  If a student ever correctly answered one of my exam questions in a way I hadn't anticipated, the student would receive full credit.  I never thought of it as subverting my intent.  Either I hadn't written the question in a definitive enough manner, or (more likely) the student was smarter than me.  In any case, I appended a "non-subverting" route :)

I totally see where you are coming from, and if I was in the position as the grader, I would have to view it the way that you do. I only pointed that out because (1) it's a practical limitation and (2) I could see other instructors either wording the question in such a way to avoid this method (hopefully) or not receiving the answer in the same way you do. Thanks for adding the alternative route. Another thoughtful answer, as I've come to expect from you.

In principle, this can be realized in two steps:

Oxidative cleavage using catalytic amounts of $\ce{RuO2}$ and excess of oxone converts 1-octyne to heptanoic acid. (DOI)
Heating a mixture of acetic anhydride and a carboxylic acid in the presence of $\ce{Pd}$ catalyst turns the acid to a terminal alkene lacking one carbon atom (DOI). Admittedly, the high temperatures needed might be a problem here and may force you to perform the reaction in a sealed tube in an autoclave.

+1, interesting techniques and nice answer. I was thinking along the same lines: oxidative cleavage of the alkyne to give heptanoic acid (with, e.g., acidified potassium permanganate) followed by decarboxylation. I had some version of the Hunsdiecker reaction in mind for the latter step, followed by elimination of the halogen.

Your second step will not work as written. Reaction of an alkene with aqueous HBr will produce the bromide (not the bromohydrin) with Markovnikov regioselectivity and the possibility for carbocation rearrangement. To get the bromohydrin, se $\ce{Br2/H2O}$
Your fourth step will be challenging also. There are two leaving groups, and LDA is a very strong base. There are regioselectivity and chemoselectivity problems, and unless the reaction is controlled well, you will also get 1-octyne.
Better to install just one functional group in step 2 (oxymercuriation, while old-fashioned, gives you just 2-octanol with little to no rearrangement), and then your elimination in step 4 will behave better.

Why if I created a fluorohydrin?