text:
Why can't salt water and fresh water mix?
Why can't fresh water and salt water mix? What happens when they collide with each other? What are their differences?

Do you have any salt handy? Try and see for yourself. Chemistry is an experimental science, after all.

Strictly speaking, they do mix and if you take some salt water and fresh water and try it yourself as pointed out in the comments I think it would be pretty obvious that they do (eventually).
However, the initial "separation" that happens is because of the fact that salt water is denser than fresh water - as such it will sink through the water (while slowly diffusing). If left alone long enough, diffusion will cause the salt ions in the salt solution to mix with the fresh water, resulting in the homogeneous mixture.

I googled up for you a little bit. This large-scale phenomenon is quite interesting and has been discussed from different points of view.
Theological/physical:
https://www.quora.com/Quran-says-that-fresh-water-and-sea-water-never-mix-there-is-a-barrier-between-the-two-Whats-the-explanation-context-of-this-verse
Fluid dynamics:
https://physics.stackexchange.com/questions/51293/how-can-two-seas-not-mix
To quote from the physics answer:

There are two mechanisms for mixing at a liquid-liquid interface, firstly diffusion and secondly physical agitation.
Diffusion is negligably slow in liquids, it takes days for solutes to travel a few centimetres, so the mixing is dominated by physical agitation e.g. wave action, convention currents, wind mixing etc.


text:
π-Acceptor ligands

Which of the following are π-acceptor ligands?

$\ce{PR3}$
$\ce{Cl-}$
$\ce{NH3}$
$\ce{H-}$

I know that $\ce{PR3}$ is a π-acceptor, and that $\ce{Cl-}$ is a π-donor ligand from my lectures. I am uncertain however as to whether $\ce{NH3}$ and $\ce{H-}$ can be π-acceptor ligands. My first thoughts is that $\ce{H-}$ would not be a π-acceptor, while I have no idea how to tell if $\ce{NH3}$ is a π-donor or π-acceptor.

Like $\ce{PR3}$, $\ce{NH3}$ or $\ce{NR3}$ are π-acceptor ligands because they have an unoccupied σ* orbital, which can accept electrons from the metal's d-orbitals. For both phosphine and ammona, there is a backbonding $n_{M}$$\ce{->}$$σ^*_{N/P}$    interaction. These ligands can act as π-acceptors in much the same way as $\ce{CO}$, except that they have unoccupied σ* orbitals rather than π* orbitals. $\ce{NH3}$ is, however, a weaker π-acceptor than $\ce{PH3}$, but it is still a fairly high field-ligand. By electronegativity, you would expect the nitrogen ligand to hold the metal's d electrons more strongly, but in the case of ammonia vs. phosphine based ligands, the phosphorus ligand has bigger p-orbitals, which have better overlap with the metal's d-orbitals, so the $n_{M}$$\ce{->}$$σ^*_{P}$ interaction is stronger than $n_{M}$$\ce{->}$$σ^*_{N}$.
$\ce{H^-}$ is $1s^2$, so it has no low-energy p-orbitals to π-bond with. It is neither a π-acceptor or a π-donor. It is only a σ-donor.

By looking at ammonia's placing in the spectrochemical series, it is logical to deduce that the interaction with the N-H antibonding MO is rather weak since it is actually placed together with most of the sigma donors, instead of being placed with the other pi acceptors. I agree that this is likely due to the difference in size relative to the metal d orbitals.

$\ce{PR3}$ is π-acceptor (from metal d orbital to phosphorous d orbital electron cloud is transferred).
$\ce{Cl-}$ is π-donor (minus charge form σ bond with the metal orbitals, $\mathrm{e_g}$, $\mathrm{t_{1u}}$ and $\mathrm{a_{1g}}$ orbitals in $O_\mathrm{h}$ field. The lone pair of electron on $\ce{Cl}$ will try to form π-bonding with $\mathrm{t_{2g}}$ orbitals of metal, electron cloud is transferred from $\ce{Cl}$ to metal d orbitals).
$\ce{NH3}$ is σ-donor. The lone pair is used to form sigma bond with metal. No extra electron to give to metal and also no d orbital to accept electron from metal. $\ce{P}$ and $\ce{N}$ belong to same group but differ here.
$\ce{H-}$ is σ-donor; reason same as above.

It's not phosphorus d-orbitals, but rather P–R σ*! Please see https://en.wikipedia.org/wiki/Pi_backbonding#Metal-phosphine_complexes and references therein.

Alright; let me try.
In Cl-, the “HOMO” of the ion itself are the p orbitals; the two pi-bonding p orbitals act as a pi bonding (weak field) ligand.
In NH3, the HOMO is the 3a1 bonding MO (sigma donor) and the LUMO is the 4a1 antibonding MO. The LUMO+1 are the antibonding 2e orbitals (which involve the p orbitals). As the LUMO can’t act as a pi acceptor, it’s a sigma donor only.
In PR3, the HOMO is again a bonding MO similar to the 3a1 is NH3, BUT the antibonding “e” orbitals involving the p orbitals are lower in energy than the “a1” antibonding. As the LUMO is hence a pi acceptor, it’s strong field.
In H-, the case is similar to NH3, where the p orbitals are too high in energy.

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text:
Would consuming the Belousov-Zhabotinsky reaction be dangerous?
I'm wondering if the chemicals involved in the Belousov-Zhabotinsky reaction would be at all poisonous. If at all, do you know of another similar oscillating reaction that could safely be ingested?
[Edit] Seriously, I was wondering if it would be possible to invent a candy that changes flavors in waves. I know Turing came up with a mathematical demonstration of the possibility of this sort of reaction long before its discovery which makes me think that others exist.
Here is the apparent chemical reaction:
$$\ce{3CH2(CO2H)2 + 4BrO3- -> 4Br- + 9CO2 + 6H2O}$$

An oscillating reaction would be a terrible way to make a flavour changing candy and exceedingly complicated. You can't easily control the environment and the environment of the reaction matters to the speed of the cycles. Why not just make a candy with different layers with different flavours: solubility will do the job of changing flavours.

Well, to directly answer your question I think the fastest way to find out is to check the individual MSDS of each component. What you'll probably find is that it's a bad idea to eat these specific chemicals.
But perhaps more importantly, your real question is less about the safety of eating those specific components but the possibility of providing a solution/mixture that will oscillate between flavors. Now in purely theoretical terms (because I know of no system that can safely be ingested), I think it's possible. You would be looking out for a system that:

Starts with non-toxic compounds.
Has non-toxic intermediates.
At least three distinct steps that are all forward reactions where it goes $\ce{A->B->C->A}$ (you don't want equilibrium reactions because these tend to find a middle ground much faster)

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Production of sulfuric acid using sulfur dioxide
Is it possible to create sulfuric acid using sulfur dioxide directly dissolved in water? Usually, people oxidize the $\ce{SO2}$ to sulfur trioxide ($\ce{SO3}$) first. If this is possible, why is it not done industrially?

see this and this

If you try to dissolve sulfur dioxide into water, you'll get sulfurous acid instead of sulfuric acid: $$\ce{H2O + SO2 <=> H+ + HSO3-}$$
There's a very good reading material on the Contact Process to help explain why we have opted for the specific industrial conditions.

Sulfurous acid does not exist. It is an aqueous solution of sulfur dioxide.


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How many resonating structures of Carbon Monoxide are there?
Anywhere I look, I get 3 resonating structures for CO molecule, like in this answer.
However, according to the rules stated for drawing resonating structures in this site, I wonder why there can't be this resonating structure as well?

Although it will be very unstable (due to lesser number of covalent bonds, positive charge on more electronegative atom), yet a minute contribution of this structure must in present in the resonance hybrid of the CO molecule. Then why is this structure not shown anywhere?

For exactly the reason you have said. It contributes a very minor component to the overall structure. There are an awful lot of possible resonance structures but very few of them are significant contributors.

You have said it yourself. There are many possible resonance structures but most of them have a negligible contribution to the overall structure. Your proposed structure is one of them.
The major resonance contributors of carbon monoxide are: $$\ce{\!\overset{\ominus}{C}#\overset{\oplus}{O}\! ~<->~ \!C=O\! ~<->~\!\overset{\oplus}{C}-\overset{\ominus}{O}\!}$$

The carbon atoms in the last two contributing canonical forms are not in octet condition. Isn't that a problem?


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How to determine the percentage of iron(II) and iron(III) in a solution?

Outline a plan of an experiment to determine the percentage of iron
  present as iron(III) in a solution containing $\ce{Fe^3+(aq)}$ and $\ce{Fe^2+(aq)}$
  ions. You are provided with zinc, a standard solution of potassium
  dichromate(VI) and dilute sulfuric acid. How you would use the zinc and how you would calculate the
  final answer.

So I know that the following equations will be involved

$\ce{2Fe^3+ + Zn -> 2Fe^2+ + Zn^2+}$
$\ce{6 Fe^2+ + Cr2O7^2- + 14 H+ → 6 Fe^3+ + 2 Cr^3+ + 7 H2O}$

I am very unsure on the method to use, I know that I can titrate a sample of the solution against $\ce{K2Cr2O7}$ to eliminate $\ce{Fe2+}$ ions, but I am clueless on how to proceed after.

You are provided with $\ce{Zn}$, not $\ce{Zn^2+}$, so you have to find a reaction between $\ce{Zn}$ and either $\ce{Fe^3+}$ or $\ce{Fe^2+}$.

You are right, I'll change the equation.

So you have a mixture of $\ce{Fe^2+}$ and $\ce{Fe^3+}$ ions, and you know that Zn only reacts with one of them, so maybe you can use that to determine the amount of $\ce{Fe^3+}$ ions? And then what can you use the dichromate to do?

Well I can titrate the entire solution against the dichromate to find the amount of Fe2+ ions that reacted.

You’re already very close. You have realised that:

You can titrate the $\ce{Fe^{+II}}$ content with a potassium dichromate standard solution; and
You can turn $\ce{Fe^{+III}}$ into $\ce{Fe^{+II}}$ by adding metallic zinc.

Note that in the first bullet point of mine, your second reaction, you are creating $\ce{Fe^{+III}}$, hence iron(III) that was present before you started won’t be modified. Similarly for the other reaction and iron(II). That should lead us to following methodology, hidden in a spoiler tag:

 1. Titrate the iron(II) content in an aliquot of your solution. That gives you the content of iron(II) and only iron(II).

 2. Take another aliquot and reduce all iron(III) to iron(II). That will give you the total iron content.

 3. Substract the value from 1. from the value from 2. to get the content of iron(III).


text:
1,2,3-Trimethylcyclopropane stereoisomers
1,2,3-Trimethylcyclopropane has two stereoisomers. When one of the methyl groups is above the plane and other two are below the plane, then there are two chiral centres as well as a plane of symmetry. By definition this compound should have a meso isomer. However I read that this compound has zero meso isomers since it does not have any optically active stereoisomers.
So what is the exact definition of meso?

related http://chemistry.stackexchange.com/questions/35486/iupac-name-for-1-2-3-trichlorocyclopropane

The way that a meso compound is often described in introductory organic chemistry courses is a compound that has stereocenters but is not chiral (due to a symmetry element). This description works for most cases, such as those in the previous questions (see this or this). By this description, the stereoisomers of 1,2,3-trimethylcyclopropane are meso.
However, the IUPAC Gold Book definition of a meso compound is:

A term for the achiral member(s) of a set of diastereoisomers which
  also includes one or more chiral members.

As is pointed out in the question, none of the diastereoisomers of 1,2,3-trimethylcyclopropane are chiral, so by this definition, none of these compounds are meso.
The Gold Book also has a description of the meso prefix:

A prefix to indicate the presence of a symmetry element of the second
  kind (see chirality) in a chemical species which is usually one of a
  set of diastereoisomers that also includes a chiral member.

The usage of meso in this way, to make a particular call out for a member in a stereoisomeric set of compounds, fails for 1,2,3-trimethylcyclopropane. Because both of the stereoisomers are achiral, neither can unambiguously be called "meso-1,2,3-trimethylcyclopropane." Since that's the case, neither are meso.

Just as an aside, I like the more common description (has stereocenters, but achiral) better, because I think that's an interesting situation that deserves a name. Congratulations on finding a unique case that forces us to think about the precise definition of these terms.


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Why are non-spontaneous reactions driven in the presence of energy?
Here is my understanding. If you have an endergonic reaction (m) with $\Delta G_m=-m$ and you couple it to an exergonic reaction (n) with $\Delta G_n=n$, where $n>m$, then the endergonic reaction can proceed as the overall $\Delta G$ for the system is negative. I understand why having the endergonic reaction occurring is not a violation of the second law of thermodynamics and why it is made possible in that sense.
I was recently studying electrolysis and they talked about hooking up an electrochemical cell to a power supply and using electrical energy to drive a nonspontaneous redox reaction. If you include the power supply in the system, then the $\Delta G$ is negative, as the power supply gives a lot of energy.
This is my confusion: if the redox reaction is nonspontaneous (let’s call it R1), then it must be spontaneous if the opposite direction (let’s call it R2). However, in the presence of electrical energy… R1 is made spontaneous? Therefore it can occur? But what of R2? Shouldn’t that reaction be even more spontaneous, as it were? Shouldn’t the system try to minimize its energy by preferentially allowing R2 to proceed, even in the presence of an energy source? Thus, why would R1 occur? Why does shoving energy into a system cause a nonspotaneous reaction to occur when it would be more favorable to continue having it occur in the opposite direction?
I would also appreciate it if anyone could extend this to general energy coupling, say with hydrolyzing ATP and a nonspontaneous biological reaction. Why is the original spontaneous reaction not made more spontaneous?

You are incorrect in your assumption that the addition of a power source makes both the forward and the reverse redox reactions more spontaneous. An electrochemical cell operates on the same principle as a battery. On the anode of the cell, you will find a compound that is being oxidized, or losing electrons. These electrons then flow through some sort of a wire to the cathode, where the electrons are being used to reduce another compound. The electricity will flow in the "spontaneous" direction, in other words, from the compound that prefers to be oxidized to the compound that prefers to be reduced. The "voltage" across an electrochemical cell is determined by adding up how badly the reducing agent wants to get oxidized ($E°_{ox} = -E°_{red}$) and how badly the oxidizing agent want to be reduced ($E°_{red}$), and is a measure of how much the reaction wants to occur.
Now, lets try adding a battery in between the anode and the cathode. If the battery is added such that its anode points towards the cell's cathode, the battery will add to the voltage of the cell, causing the effective voltage for the reaction to increase. On the other hand, if the battery is placed such that its anode points towards the cell's anode, it will be creating a voltage in the other direction. When this happens, it becomes a question of whichever voltage is stronger. If the battery has a stronger voltage than the cell, electrons will flow in the opposite direction. This is what occurs in electrolysis, causing the originally non spontaneous process to become spontaneous. This results in charging the electrochemical cell. On the other hand, if the battery has a smaller voltage than the cell, the electrons will flow in the original direction - charging the battery as they pass through it.
While a power source can be used to either make a spontaneous process more spontaneous or to reverse the flow of electrons, it cannot do both at the same time.

Could you also explain how energy coupling—say, between the hydrolysis of ATP and a nonspontaneous reaction in the body—results in the driving of a nonspontaneous reaction? Why isn’t the spontaneous reaction made more spontaneous? I can see that a battery determines the direction of a reaction, but what if there’s no battery involved?

If there is no battery involved, the system acts like it normally does. The spontaneous reaction will occur. ATP is actually a really interesting compound. The terminal phosphate on the ATP will bond to an OH- group in some compound, turning the OH- into a phosphate group. The phosphate is a good "leaving group," allowing for some nucleophile to replace it on a carbon chain. Enzymes specificity causes ATP to bind selectively and drive endergonic processes.

So even in the presence of ATP, the spontaneous reaction occurs? Both spontaneous and nonspontaneous reactions occur?

ATP is used to make non spontaneous reactions spontaneous, but only spontaneous reactions occur.


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How are van der Waals parameters determined experimentally?
If the ideal gas law $PV=nRT$ is not enough, as a second approximation the van der Waals equation
$$\left(P+a\frac{n^2}{V}\right)\left(V-nb\right)=nRT$$
is applied. The constants $a$ and $b$ are van der Waals constants which take into account some of the attraction and repulsion  that the ideal gas law does not.
Solving for $P$,
$$P=\frac{nRT}{V-nb}-a\frac{n^2}{V^2}$$
and plugging it into the definition formula of compression factor $Z$ yields
$$Z=\frac{V_{m,real}}{(RT)/P}=\frac{V_{real}}{(nRT)/P}=\frac{V_{real}}{V-nb}-a\frac{nV_{real}}{RTV^2}.$$
This result, supposedly, helps in the measuring of constants $a$ and $b$. But how is it really done? What are the more rigorous definitions of the constants?

For a rigourous definition of $a$ and $b$ you can see this answer: http://chemistry.stackexchange.com/questions/40803/high-pressure-modification-to-the-van-der-waals-equation/42510#42510

You don't: While educationally useful (very much so), the vdW equation is not describing the actual behaviour of any real gas. To use it, you'd have to calibrate the parameters for the relevant pressure/temperature range. There are no universal values for a and b that make vdW much better than the ideal gas law.

At the critical point, the first and second derivatives of pressure with respect to volume are zero.
This provides two equations to solve for the two unknowns (a and b) in terms of the critical temperature and critical volume (or pressure).
See Critical Constants of the van der Waals Gas

That's correct, inside vdWaals theory, but hardly practically useful. Actual, sensible values for $a$ and $b$ are purely empiric, and vary quite a bit for different pressure/temperature ranges.


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Can nitrogens in urea be protonated?

Is it possible in very acidic solutions for the nitrogens ($\ce{N}$) to get another $\ce{H+}$, making it $\ce{R-NH2+-R}$? Since it has a free electron pair, it might be possible right? If so, at which pH?

Reagents/solutions that are acidic enough will protonate the urea group, just not where you think. The most basic site is the oxygen. Perhaps this is a surprise.
The lone pairs on the nitrogen atoms are delocalized (see resonance depiction below).

The oxygen atom has the most electron density. Below is an electron density map, where ${\mathrm{\color{red}{red}}}$ represents areas of high electron density and $\mathrm{\color{blue}{blue}}$ represents areas of low electron density. You can see the area around the oxygen atom is red and yellow, while the areas around the nitrogen atoms are green.

This image is from "The effect of intermolecular hydrogen bonding on the planarity of amides" by James A. Platts, et al. published in Phys. Chem. Chem. Phys., volume 2012, issue 14, page number 11944-11952, DOI:10.1039/C2CP41716B. The image is copyright the Royal Society of Chemistry and released under a Creative Commons Attribution 3.0 Unported license.
Protonating the oxygen atom leads to a cation where the positive charge is delocalized (better, more stable).

Protonating one of the nitrogen atoms leads to a cation where the positive charge is localized (worse, less stable).

@4C0G about pH=0. "the equilibrium constant of urea monoprotonation can be given
a value of 1.93 with an uncertainty of about 5%." http://pubs.acs.org/doi/abs/10.1021/j100135a013

Well, yes but no. The nitrogen’s lone pair is in resonance with the $\ce{C=O}$ π bond. They form a π system not unlike the allyl anion. This is usually depicted for the amide case, but there are no significant differences in the urea case.

Instead, the oxygen has an accesable lone pair perpendicular to the π system, so it is typically protonated first. This happens at low pH values; you typically need sulphuric acid or hydrochloric acid to get quantitive protonation.


text:
Why the products of combustion of Cu2S are Cu2O and SO2?
$$\ce{Cu2S + 2 O2 -> 2 CuO + SO2}$$
Why are not these oxides $\ce{Cu2O}$, $\ce{SO}$ or $\ce{SO3}$?
Is it connected to the stability of these molecules?

There should be CuO in the heading :)

Appreciate that.

The reaction gives different products when combusted at different temperature.

$$\ce{8Cu2S + 15O2 ->[\Delta] 6Cu2O + 4CuSO4 + 4SO2}$$

Copper(I) sulfide react with oxygen to produce copper(I) oxide, copper sulfate and sulfur dioxide. This reaction takes place at a temperature of 500-600°C.Here

$$\ce{2Cu2S + 3O2 ->[\Delta] 2Cu2O + 2SO2}$$

Copper(I) sulfide react with oxygen to produce copper(I) oxide and sulfur dioxide. This reaction takes place at a temperature of 1200-1300°C.Here

Sulfur monoxide is very unstable and it is unlikely to exist in Earth's atmosphere but it exist in interstellar spaces. Sulfur trioxide will not form from sulfur dioxide as it requires an oxidising agent.

Right reaction is $$\ce{2Cu2S + 3O2 -> 2Cu2O + 2SO2}$$
$\ce{Cu2O}$ is because copper in this compound has oxidation state +1 and $\ce{2Cu + O2 -> 2CuO}$.
$\ce{SO2}$ is because $\ce{SO}$ is unstable and $\ce{S + O2 -> SO2}$.
Upd. This answer isn't right. See comments of @Jan under it.

Your explaining arguments don’t really make sense …

Why they don't make sense?

Because all they do is explain the simple redox process coming from the elements. I can do that, too: $\ce{C + O2 -> CO2}$, yet there are many reactions that cleanly produce $\ce{CO}$, never $\ce{CO2}$. $$~$$ It’s like saying ‘the result of mixing yellow and blue is green, because plants are green.’ It has nothing to do with each other.


text:
Can I use the mono ammonium phosphate found in fire extinguishers to make crystals?
I recently opened up a old decompressed fire extinguisher out of curiosity. After some research I discovered that the dry powder inside, mono ammonium phosphate, can be used to make beautiful crystals. However, I found this powder to be extremely hydrophobic and could never get its dissolve in water at all. I mean this stuff is amazingly hydrophobic. Its fun to see how it reacts in water. But why cant it dissolve and what kind of preparation of mono ammonium phosphate will dissolve and become crystals?

I think you aren't using the word "hydrophobic" properly

Sprinkle it with alcohol.

Fire extinguisher powder can be made hydrophobic by additives like stearates or silicones. However this is rather done with sodium/potassium hydrogencarbonate, potassium sulphate, or sodium/potassium chloride powders (BC class). For monoammonium and ammonium phosphate powders (ABC class) I wouldn't expect them to be hydrophobic. Are you sure about the chemical composition?

If you google for MAP, you'll find out that it is quite soluble in water, ranging from 370 to 401 g/litre of H2O at temperatures around 25 C. Thus, what you report contradicts literature data on MAP's solubility. So, maybe the powder you found is not MAP, it's something else (CaCO3 comes to my mind, because it is insoluble in water and yields CO2 when heated, but I may be wrong).

sanity check, does not react to vinegar -> not carbonate. Separately, litmus paper shows acidic, as expected for ammonium phosphate and sulfate

see https://en.wikipedia.org/wiki/ABC_dry_chemical "specially fluidized and siliconized monoammonium phosphate powder" - it's the anticaking agents that make it hydrophobic. Ammonium salts being hygroscopic on their own, i figure forming a lump inside extinguisher is no good


text:
Does this reduction mechanism of an diazonium via stannic chloride sense?
Is this mechanism possible.
I didnt find any sources of a mechanism of this specific reaction.

That’s not even tin!

Agreeing with Jan's comment, there are a few problems, and they all have to do with your reducing agent. That's not tin!
Tin is $\ce{Sn}$ and not $\ce{Zn}$ (zinc). Your mechanism is using $\ce{ZnCl2}$, zinc chloride. Let's say you made a typo and replaced $\ce{Sn}$ with $\ce{Zn}$. Your electron configurations match tin and not zinc. Then your mechanism is good. $\ce{SnCl2}$ is a reducing agent and could do this reaction. $\ce{SnCl2}$ is not stannic chloride, but stannous chloride. Stannic chloride is $\ce{SnCl4}$. Regardless $\ce{SnCl2}$ would work. $\ce{ZnCl2}$ would not work. The zinc in $\ce{ZnCl2}$ is already in zinc's highest common oxidation state. $\ce{ZnCl4}$ does not exist.

Oh I see i meant Sn (in german) its called Zinn (so i got the letter wrong)

I expected something similar was the origin of the confusion. If you swap $\ce{Zn}$ for $\ce{Sn}$, your proposed mechanism looks fine.

I also worked with something different as second mechanism: http://www.fotos-hochladen.net/view/reductionofdiumjq0i3yz1.png


text:
The ionic radius of aluminium(III) vs. the covalent radius of neon?
I was wondering if it is a valid comparison to look at the ionic radius (noble-gas-like configuration) of a metal and the covalent radius of a noble gas. Specifically, $\ce{Al^{3+}}$ and $\ce{Ne}$.
This website gives me the impression that for comparable radii definitions, neon is slightly smaller than $\ce{Mg^{2+}}$ but larger than $\ce{Al^{3+}}$.
Someone I know is wondering why we should/shouldn't just use the atomic radius of neon, rather than the covalent radius, to compare to the ionic radius of metals like aluminum and magnesium (yes, even though they are on the next period).

Covalent radius of neon? I wonder how it was obtained, not experimentally I'm afraid.


text:
Why is H2O2 named hydrogen peroxide?
Why is $\ce{H2O2}$ named hydrogen peroxide and not dihydrate dioxide?

Hopefully you meant dihydrogen...

Oxide is a term reserved for the oxide anion $\ce{O^2-}$. This anion is typically characterised as rather stable since the very electronegative element oxygen is in its maximum negative oxidation state. Dioxide means, that there are two oxides present in whichever structure we are looking at.[1] A typical example for a dioxide would be manganese dioxide, also called manganese(IV) oxide $\ce{MnO2}$. As shown in the image below, it crystallises in the rutile structure, like titanium dioxide $\ce{TiO2}$.

Figure 1: Crystal structure of rutile, $\ce{TiO2}$ which is equivalent to β-$\ce{MnO2}$. Red spheres represent oxides, grey ones the metal in question. Image taken from Wikipedia, where a full list of authors is available.
Hydrogen peroxide, on the other hand, as is evident from the structure in figure 2 below, contains not two $\ce{O^2-}$ units but a single $\ce{O2^2-}$ unit, composed of two oxygens bound to each other. This means that oxygen’s oxidation state is $\mathrm{-I}$ rather than $\mathrm{-II}$ and that hydrogen peroxide is readily reduced. (It can also easily be oxidised back into $\ce{O2}$. The functional group $\ce{O-O}$ is labelled peroxide and it would not make sense calling it dioxide since it does not feature oxides.

Figure 2: Structure of hydrogen peroxide. Red spheres are oxygen, white ones hydrogen. Image taken from Wikipedia, where a full list of authors is available.

Aside from that, IUPAC also allows for nomenclature to entirely ignore the molecular structure and name entirely by molecular composition. In that compositional nomenclature, dihydrogen dioxide is an acceptable name like dihydrogen monooxide would be for water. Note, however, that this only tells us the number and type of atoms involved while not giving us any clue to how they are connected. It is thus typically better to refer to a peroxide as a peroxide to emphasise the actual functional group present.

According to IUPAC, dioxidane, hydrogen peroxide and dihydrogen dioxide are all acceptable names for $\ce{HOOH}$. See 2005 Red Book IR-6.2.2.1 (p 86), and 2016 Brief Guide to the Nomenclature of Inorganic Chemistry (table 3).

The structure of $\ce{H2O2}$ is $\ce{H-O-O-H}$, and an $\ce{- O-O -}$ functional group is called a peroxide, by definition. The peroxide functional group is attached to a hydrogen atom, so it's called hydrogen peroxide.
Calling it a dihydrogen dioxide (hydrate is water, not hydrogen) would not only be long but also not show the type of reactivity the substance has.
So it's just more convenient to call it hydrogen peroxide, rather than dihydrogen dioxide.

Not just $O-O$ for peroxide?

This is incorrect. $\ce{OOH}$ is the hydroperoxide group. Peroxides are $\ce{O-O}$ compounds. Compare dibenzoylperoxide $\ce{Ph-CO-O-O-CO-Ph}$.

My bad. I didn't know that

Please update your answer accordingly @name.disp.


text:
Polysytrene factories and hypothetical health concerns
Background/Research so far
I'm actually not much of a chemist at all; but I'm doing research for a book I'm writing.
The events of the story begin from a factory that produces/supplies jewel cases for cheap CDs. This factory then pollutes the city's water by reckless dumping. Basically I'm a psychopath and I want to give my city's citizens harmful affects that are revealed after 15 years. Maybe their livers melt or something.
I learned that these jewel cases are made from a plastic called polystyrene.
From this website, I found this:

Polystyrene (PS, #6): styrene, ethylbenzene, benzene, ethylene, carbon
  tetrachloride, polyvinyl alcohol, antimony oxide, and tert-butyl
  hydroperoxide, bensoquinone

I thought maybe the pollution could be caused by leftover ingredients for the plastic used in the process.
I then examined carbon tetracholoride ($\ce{CCl4}$). This seems really harmful and may cause long term damage if accidentally ingested, especially to the liver. So this sounded exactly what I am looking for. However, I'm not sure how realistic/true this is because further research showed that $\ce{CCl4}$ is, in fact, so harmful that it's rarely used anymore. So it seems like it may be unrealistic to use in the story, because maybe the above article is inaccurate and $\ce{CCl4}$ is not that prevalent.
Benzene also is an obvious one, but it seems to mostly cause blood cancer (leukemia) through ingestion. For reasons that occur later in the story, isn't really isn't going to work. I was kind of looking for a specific organ failure/problems/cancer and not those that effect blood and/or skin.
Ethylene also looks promising because it's so close to ethylene gylcol, and ethylene glycol would really mess people up (it is my understanding is that it's basically antifreeze). But further research indicates that ethylene isn't very harmful at all.
I also thought of cleaning agents for polystyrene instead of actual ingredients. And it sounds like $\ce{CCl4}$ comes into play again because it's used as a "dry solvent," according to Wikipedia in "Historical Uses".
Question
I'm starting to fall short in research and I want my story to be as facutal as possible. So I'm wondering which chemicals would be the most realistic in terms of use in the practical manufacture of polystyrene that could be dumped a really cause problems health-wise for a city.

It is good to see the some authors want to build realism into their fiction: this clearly isn't common. And you have done extensive research already.
The problem is that industrial chemistry is very heavily regulated. So, while it is common for dangerous substances to be used, it is rare to find places where they can plausibly be misused. And the standards for harmfulness are very low. Even some things at the harmful end of the official scale are not that harmful.
The biggest problem you will have when it comes to realism isn't really to do with polystyrene manufacture at all. A factory that makes products with polystyrene won't usually make its own polystyrene but will buy in the plastic raw material as polystyrene beads which can be mixed with other ingredients (like colours) and melted in moulding machines which produce the product. So none of the chemicals used to manufacture the plastic will be used anywhere while making products from the plastic.
If you were to assume that the raw polystyrene was made locally (which isn't that realistic) then you might find plenty of chemicals that could be misused as you have already realised. CCl4 is one of the best candidates but i rarely used any more. Benzene is used in manufacture so is also plausible; styrene (the monomer that is used to make the plastic) is also nasty and possibly the best candidate you haven't considered.
However, if you want to stick with broad realism and a story that uses plausible things associated with a jewel case factory, then you need to consider a different approach. But it might be easier because you can make some things up that even experienced experts can't question. You could, for example, imagine that your factory adds some special ingredient to its products to create unusual colours or effects (maybe the cases glow in the dark, are unusually strong or contain inks that allow patterns to be printed deep into the plastic by lasers). The ingredients to create this effect could be anything you want (and you probably don't have to be too explicit about exactly what it is: give it an obscure acronym). This ingredient could plausibly have any bad properties you want without being unrealistic. Possibly the product will have been invented recently to give the products a special edge and nobody else has yet observed the side effects.

Wow thanks for the awesome answer! I'm a full time engineer so I don't mind diving in deep with research. I didn't consider that the factory that supplies the plastic would be different from the factory that produces the jewel cases, but that makes sense. I really like your point with the factory adding an additional ingredient to the production of the jewel cases in such a way to produce a new feature. This really adds a ton of flexibility in what I can integrate with the story. Thanks!


text:
Derivation of the Lever Rule
To derive the Lever rule, the textbook I am reading sets:
$$nz_A = n_\alpha x_A + n_\beta y_A\\ nz_A = n_\alpha z_A + n_\beta z_A$$
$x_A =$ liquid phase mole fraction of component $A$
$y_A =$ gas phase mole fraction of component $A$
$n_A=$ amount of phase $A \ \text{in mol}\ (A=\alpha \ \text{or} \ \beta)$
$z_A=$ overall composition
"Overall composition" sounds to me like $z_A=x_A+y_A$, but the equation suggests $x_A = z_A = y_A$ which doesn't make sense so I think I'm missing an assumption that is made or I'm misunderstanding the equation.


text:
Synthesis of Almotritpane - mechanism for educational purpose
I purposed a full mechanism of the almotriptane synthesis.
I wanted to help my fellow students and myself.
Could you give me any tips? Are there any mistakes? I posted the SnCl2 reduction already, but I proposed a second mechanism and cant decide which one is right.

Mechanism of the Fischer-indole synthesis                                               http://www.fotos-hochladen.net/view/fischerindoleigtlcb3yfk.png

proposed mechanism of the methylation http://www.fotos-hochladen.net/view/methylationprin3ypgqkt2j.png

Is it correct to say that your question is about DMF followed by Methylation with MeI vs Formaldehyde followed by borohydride reduction?

No DMF is only the solvent for the reduction with Raney-Nickel. It should explain the whole synthesis and the reactions with mechanism.

trying to figure out what the question is because you mention second mechanism, not sure if you meant the section labeled "2) proposed reduction mechanism" or the last section above "1) proposed net reaction".

Also, for the last step with Borohydride you would only get one methyl group and not two. You would have to do it twice, don't think you can do it in one pot because the formaldehyde in excess would react with the Borohydride first.

Like this http://www.fotos-hochladen.net/view/methylationprij7upc2rn0y.png ?

second mechanism is the 3rd one my fault ... the 2 proposed mechanisms are 2) (first) 3) (second)

Not too familiar with the reduction for that, sorry. But I did find a better way to make your dimethyl instead of doing the borohydride reduction twice: https://en.m.wikipedia.org/wiki/Eschweiler%E2%80%93Clarke_reaction

Wow, thats a really good reaction. Thank you.

Like this: http://www.pic-upload.de/view-30917534/methylation_primary_amine_methanal_methanoic_acid.png.html


text:
Can hot food ever emit x-rays or gamma rays?
I was just wondering, if heating food up is the result of increasing the energy of bends and stretches in the bonds of the molecules, is it ever possible for tiny amounts of x-rays and gamma rays be emitted?
When we give a molecule enough energy, its electrons can jump to higher orbitals and then return to ground state, releasing EMR with a frequency proportional to the energy gap. So if we heat food, is there a chance that some electron received enough energy to jump up to a really high energy level?

Sure, but you'd be hard-pressed to call it "food" at that point, as it'd be dissociated into atomic substituents at best.

Agreed! However, this is worth the "Mythbuster" treatment. If we cannot generate x-rays at reasonable food temperatures, what would it take and what would we have left? See my answer below.

Your description in terms of bound states of an isolated molecule inside the food doesn't really work. Food is solid or liquid (condensed matter), so the molecules don't exist in isolation. If you did have, say, an isolated water molecule (in steam), it would only have bound states with energies up to some limit, which would be on the order of 1 eV. Any states with higher energies would be states in which the molecule had been ionized or dissociated. As explained in Ben Norris's answer, condensed matter emits a continuous spectrum of radiation, so it's a completely different deal.

Can "hot" food ever emit x-rays or gamma rays? // Seems like a Freudian slip. I'm sure vegetables from Chernobyl and Fukushima would be "hot."

While not quite related to your question, if you really want to create X-rays in your kitchen, you'd have more success using scotch tape. Though your kitchen would have to be particularly well equipped to include the necessary vacuum chamber.

What about the radioisotopes in the food?

I joined this stack just to upvote this question and to leave a link for those who like videogames. How about a nice cup of Nuka-Cola Quantum for that extra Cherenkov radiation kick?

@BenCrowell, why does condensed matter emit a continuous spectrum of radiation? Is it what happens when molecules are given energy that would free the electrons from their bound state?
I was thinking more of food at its normal temperatures, but everyone's given so many interesting alternative ways for food (and scotch tape) to be radioactive.

In theory, yes, you can heat objects to a high enough temperature to emit x-rays or gamma rays. You cannot do this to food, and you certainly cannot do this in your kitchen (or probably any kitchen).
Let's take the lowest energy x-ray out there and see what it would take. X-rays range in frequency from $30 \times 10^{ 16}$ to $30\times 10^{10}$ hertz. The energy of one photon of 30 petahertz radiation is:
$$E=h\nu = \left(6.626\times 10^{-34}\mathrm{\ J\cdot s}\right)\left(30\times 10^{16}\ \mathrm{s^{-1}}\right) = 1.988 \times 10^{-16}\ \mathrm{J}$$
This is not a lot of energy! However, a single photon is boring. Let's consider a mole of photons. This will also ease comparison with other phenomena, whose energies are listed per mole of events.
$$1.988 \times 10^{-16}\ \mathrm{J} \times 6.022\times 10^{23}\ \mathrm{mol^{-1}}=1.197\times 10^8\ \mathrm{J\cdot mol^{-1}}$$
In theory, if you could pump that much energy into something, you should get some high energy photons out. In practice, it does not work that way. Other stuff happens first. To simplify our example, let's just consider 1 mole of water (18.0 grams) and heat it up. The fate of basically any other matter will be the same, but the energy required will vary a bit.
First, adding energy heats the water. If we start at room temperature $\left(20\ ^\circ\mathrm{C}\right)$, it takes $80\ ^\circ\mathrm{C}\times 18\ \mathrm{g}\times 4.184\ \mathrm{J\cdot g^{-1}\cdot ^\circ C^{-1}}=6025\ \mathrm{J}$ to heat that water to boiling. It takes 40.66 kJ to convert the water into gas. Neither of these puts a big dent in our energy. It takes further energy to heat the water vapor again, but let's see how far we need to take it.
Once we get enough energy into our sample of water, the molecules start to fall apart.
$$\ce{ H2O(g) -> 2H(g) + O(g)} \ \Delta H^\circ =+920\ \mathrm{kJ\cdot mol^{-1}}\ \Delta S^\circ =0.202\ \mathrm{kJ\cdot mol^{-1}\cdot K^{-1}}$$
By fixing $\Delta G=0$ at equilibrium, we can solve for a temperature at which this reaction becomes spontaneous:
$$T=\dfrac{\Delta H}{\Delta S}=\dfrac{+920\ \mathrm{kJ\cdot mol^{-1}}}{0.202\ \mathrm{kJ\cdot mol^{-1}}\cdot K^{-1}}=4596\mathrm{K}$$
We need to heat our water vapor up an additional 4218 K, which takes $18\ \mathrm{g}\times 1.996\ \mathrm{J\cdot g^{-1}\cdot K^{-1}}\times 4218\ \mathrm{K}=151.5\times 10^3 \mathrm{J}$.
So, we now have pumped nearly 200,000 J into our water sample, atomized it, and heated it to approximately 5000 K. We are now close to the temperature of the outer layers of the sun! Surely we have enough energy at this temperature to produce x-rays. Nope. At 5000 K, we produce minimal x-rays. Most of the radiation is in the visible, UV, and IR (think about what we get from the sun). Below is a plot of black-body radiation as a function of temperature (image by Wikipedia user Darth Kule and released into the public domain):

Okay, so we are far beyond the reality of what can happen in a conventional oven (or almost any reasonable heat source used for food). At this temperature, we can use the Planck Law to calculate the power output ($I$) of x-rays at the temperature. We can also do this at some normal temperatures and for gamma rays. This model is a little goofy, since food is not a black body, but we will at least calculate the max x-ray and gamma ray output.
Rather than grinding through all the maths, I'll just put in a table of some temperatures and watts. 1 watt is not a lot of power. Most lightbulbs produce light in the kilowatts.
$$\begin{array}{|c|c|c|c|}\hline
\mathrm{T\ (K)} & \mathrm{P_{x-ray}\ (W)} & \mathrm{P_{gamma}\ (W)} & \mathrm{notes} \\ \hline
378 & \approx 0 & \approx 0 & \text{boiling point of water} \\ \hline
550 & \approx 0 & \approx 0 & \text{approximate common highest temperature on residential ovens}\\ \hline
700-800 & \approx 0 & \approx 0 & \text{temperature range for wood-fired ovens, tandoors, etc.}\\ \hline
5770 & 4.26\times 10^{-129} & \approx 0 & \text{temperature of the photosphere of the sun}\\ \hline
1.57\times 10^7 & 10.4 & 7.87\times 10^{-54} & \text{estimated temperature of the center of the sun} \\ \hline
\end{array}$$
So, if you could heat your food to the temperature of the sun, it would produce minuscule x-ray radiation. It would also no long resemble food.

"This model is a little goofy, since food is not a black body" It'll be pretty damned black at the kinda temperatures you're talking about! *baddum-tsh*

@DavidRicherby not anymore. Hydrogen as well as oxygen are transparent, and any hydrogen-oxygen plasma will be too dilute to significantly interact with photons as well.

I think that, very strictly speaking, any black body above absolute zero has a chance of emitting an x-ray/gamma photon, because the Planck distribution is never zero for any finite wavelength. Of course, due to exponential suppression, the probability of emitting such a photon is ridiculously small. I believe I have read that a black body at $\mathrm{37\ ^oC}$ can be expected to emit a visible photon every 1000 years.

"Most lightbulbs produce light in the kilowatts." - What kind of lights do you have in your house? o_O

@marcelm Probably the average film studio equipment :)

A more useful comparison might be: How many seconds do we have to be exposed to a phosphere-tempered steak to have the equivalent of an X-ray snapshot at the doctor's?

Does the table at the end list power emitted per unit mass? Unit area? I first understood that whole center of the sun produces 10 W of x-rays, which doesn't seem right. :P

Maybe the units in the table ought to be in spectral radiance (W/(srm^2Hz))...?

"if takes" should probably be "it takes" and "t heat" should probably be "to heat", but I don't have the reputation here to suggest such a tiny edit.

@NicolauSakerNeto "any black body above absolute zero has a chance of emitting an x-ray/gamma photon, because the Planck distribution is never zero for any finite wavelength." That's right. But we have to note that the Plank distribution is a classical distribution and not a quantum statisical model. So I guess it doesn't apply all the time. At low temperatures the total energy emitted is lower and the energy emitted in X-ray range will be very low probably lesser than the energy of a single photon.  So I guess there won't be enough noticeable X-ray photons emitted to have significance.

I wonder what happens if we heat the H and O atoms past 4596K. Would they break up into neutrons and electrons and that's what shifts the blackbody curve to the left?
@shre_sudh_97  Does a single photon with half its usual energy still count as EMR?

Would a photon with half its energy still be called an x-ray?

It has nothing to do with what you were going for, but there is a small, but non-trivial amount of x- and gamma-ray output for most food and so the answer is trivially "yes".
In particular any food containing potassium will have the usual admixture of K-40 (with it's 1.3 and 1.5 MeV gamma lines).
When I worked in a lab with a low-background, high-sensitivity germanium detector I used to try to bring a banana in my lunch on days we were expecting a visitor so I could demonstrate the detector without having to do any paperwork concerning possible exposure of the visitor to radiation: just put the banana in a clean ziplock bag, put that in the detector, start the DAQ, and watch a nice peak grow on the screen.  Broccoli works just as well, of course, but it's broccoli.
But of course none of this has anything to do with heating the food.

This is a for-sure reason to use to say that our food emits high energy EMR.

Not even close! Your food would likely be vaporized, and leave an awful mark on your kitchen table, before it could contain enough energy to be producing xrays.
When I was a youth, I used to do a lot of 'microwave experiments', as in domestic microwave ovens. I would be known for buying a used microwave from a thrift shop for around $10 and use it to produce 'microwave plasma' with (which was essentially ionizing carbon, analogous to the way neon gas would ionize if you stuck a neon gas filled tube in the microwave and turned it on).
During the course of this, I witnessed a lot of different materials arcing, melting, vaporizing, or ionizing in domestic microwave ovens. At some point, probably after cooking up filamented (incandescent) light bulbs, I started thinking and also wondered about what kind of energies produce xrays. Could I be irradiating myself with with harmful rays or high-energy particles being emitted off of all these arcs? Some of the lightbulbs glowed a crazy green color right before the end of its life, and I remembered reading reading that old-timey xray bulbs produced a distinct green light when producing xrays.
So I looked into it, and it turns out... There is no way that my 1000 watt domestic microwave oven was producing xrays, even if you consider that most lightbulbs are evacuated (under a mild vacuum).
Apparently, if I wanted to start creating x-rays in my own back yard (i didn't), I was going to need at least 2 things:

More power. A lot more, namely voltage. I dont know what kind of voltage was being induced during arcing, but xray tubes range from 30 to 150 kilovolts (kV)!
A much stronger vacuum. Xray tubes have whats called a 'high vacuum', which is about 10^-4 Pa (or 0.00009869233 atm) . In comparison, standard incandescent lightbulbs sit at around 70 kPa (0.7 atm), or 'low vacuum'.

Now people will love to start in with 'Well technically, it could', point out that we can calculate this, and proceed to work out a value that requires hundreds of decimal places before you see a significant digit, but just because the equations don't hit a limit and break down at some point, does not mean its ability to describe something physically real doesn't.
Other quick interesting facts about microwaves:

As stated above, black carbon will ionize in the microwave, such as that from burnt wood or black toner from copy machines.
While it is true that when the magnetron tube is on, the radio waves are at 2.45 Ghz, the magnetron itself turns on and off at 60 cycles per second. This explains the noise that the plasma makes. When the magnetron tube i on, the plasma expands, when the magnatron tube is switched off, shrinks. The plasma expanding and contracting at 60 Hz is why it makes a loud, 60 Hz noise.
Glass is not very good at absorbing microwaves... until its molten. Once glass becomes molten, it will absorb microwaves appreciably, causing it to not only stay molten, but grow, eventually turning all the glass molten. Also this is how some glass manufactures keep their glass molten more efficiently, once already melted.
Microwave assisted chemical synthesis is the new 'green' way to synth, requiring, less to no solvents, sometimes allowing for two homogenized, dry solids to react, higher yields, lower activation energies or no catalysts and in only a fraction of the time to boot.

Re, "More power... xray tubes range from 30 to 150 kilovolts"  Voltage is not power, and you can generate harmful levels of X radiation with equipment that uses no more power than your microwave oven.

Re, "Xray tubes have... In comparison, standard incandescent lightbulbs..."  The operating principle of an x-ray tube is completely different from the operating principle of an incandescent light bulb.  A light bulb works by heating a tiny bit of wire until it is so hot that it's black-body radiation is visible.  An X-ray tube works by forming a beam of high-energy electrons, and colliding them with a metal target to produce Bremsstrahlung radiation.  The high vacuum is needed for the electron gun in the X-ray tube.  A light bulb doesn't need vacuum, it just needs an inert gass fill.

@Adam White, would you mind explaining more why the equation breaks down in describing real phenomena? "Just because the equations don't hit a limit and break down at some point, does not mean its ability to describe something physically real doesn't".

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


text:
What is the numbering of substituted benzene derivatives according to IUPAC nomenclature?

According to the lowest locant rule shouldn't the numbering of the compound 1 start from the carbon at the 4th position, thus giving the IUPAC name as 1,2-dimethyl-4-hydroxybenzene?
For compound 2 they could have prioritized the functional group, but they have prioritized the lowest locant rule instead.

The functional group has the highest priority.

Agreed. But i can also find some other examples where the functional group is'nt given the priority.

Halides and nitro groups aren't considered to be functional groups in nomenclature, they are substituents, while a hydroxy group is a functional group e.g. you would never  say phenyl chloride or hydroxybenzene, while chlorobenzene and benzenol (phenol) are preferred names. Understand the difference?

There are several criteria for the numbering of locants. The most important simplified criteria for the numbering for substituted benzene compounds are:

lower locants for the principal characteristic groups, which are expressed as a suffix or implied by the name of the functional parent compound (If there is a choice, the principal characteristic groups are assigned in accordance with the order of seniority of classes of compounds.)
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording of the nomenclature rules (taken from Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)) reads as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

The first compound given in the question has the functional parent compound phenol. The name ‘phenol’ implies the presence of the characteristic group $\ce{-OH}$. Since there are no other characteristic groups and low locants are assigned first to the principal characteristic group, the locant ‘1’ is assigned to the  substituent group $\ce{-OH}$. Therefore, the systematic name for this compound is 3,4-dimethylphenol.

The second compound given in the question does not have any principal characteristic group – neither expressed as a suffix nor implied by the name of a functional parent compound. Therefore, low locants are assigned to the prefixes ‘chloro’, ‘methyl’, and ‘nitro’, all considered together as a set in a series of increasing numerical order. Thus, the systematic name for this compound is 2-chloro-1-methyl-4-nitrobenzene and neither ‘1-chloro-2-methyl-5-nitrobenzene’ nor ‘3-chloro-4-methyl-1-nitrobenzene’ since the locant set ‘1,2,4’ is lower than ‘1,2,5’ or ‘1,3,4’.

Lowest set locant rule is considered after prioritising the functional group (i.e. giving it the first position).
so answer of the first question is 3,4-dimethylphenol.
In the second question the answer given in the book is wrong . It should have been 3-chloro-4-methylnitrobenzene


text:
why is laser printer toner magnetic?
I had an old toner cartridge lying around, so I took it apart and emptied the toner into a jar. I had heard that one could make a ferromagnetic fluid using toner powder and oil, so I mixed a little powder with some kerosene that I had on hand. Turns out, kerosine does something wierd to the polymer part of the toner, and instead of becoming a solution, the toner powder coagulated into a rubbery mass that had a consistency similar to chewing gum.
When I put a powerful magnet from a hard drive next to it, the ball was attracted to it, which surprised me a little, as I had seen on wikipedia that toner had no magnetic components (any more) So, the questio is, Why is printer toner powder magnetic? Is it because carbon is paramagnetic?
Also, what happened to the polymer when kerosine was added?
EDIT
A clarification-
I think the polymer is a resin (based on what the kerosene did to it)

The toner might contain magnetite or iron powder. Both are magnetic.

Some toner is magnetic but not all toner. Some was designed to be machine readable but that is far less common now. You cannot assume from one example that it applies to all toner as many people trying to make ferromagnetic fluids have found out.

One of the major ingredients in modern printer toner cartridges (for laser printers) is magnetite, Fe3O4. It is used a charge control ingredient, known as a triboelectrification agent. 'Tribo' comes from the Greek meaning 'to rub' (carfeul if you google that one), and essentially refers to the creation of static charge through friction. This is required to have the printer particles stick to the paper. You can read about triboelectrification here. Magnetite, of course, is magnetic, and this is why your toner can produce a ferromagnetic fluid.
Other ingredients in toner cartridges include waxes, polymers and pigments. Many formulations are water or alcohol based and the toner is actually prepared as an emulsion which is filtered and dried. In order to get this material dissolved into kerosene, you probably need to add an emulsifying agent. Oleic acid is commonly used to achieve this, as it is a long chain acid that has both polar (hydrophilic) and non-polar (hydrophobic) properties.

So what is the kerosene doing to the polymer? When I added kerosene to the powder it pretty much instantly clumped up into a gummy mass. Is the kerosene polymerizing the resin?

Unlikely to be polymerising. More likely it is simply coagulating, but there will be some components that have partial solubility in kerosene; waxes and and some resins. Overall, it is probably a water based or hydrophilic formulation.

by water based, do you mean like acrylic paint (which becomes water insoluble when dried)?


text:
Equation to calculate the triple point
Is there any equation that can calculate the triple point of a substance? Because a method for calculating the critical temperature and the Boyle temperature of a substance that I posted in another question I recently found that it also seems to be able to predict the triple point for some substances.
I searched the Internet and couldn't find any other such equation, in fact I came across a site that said there was none. Is this true?  My equation predicts it from the Lennard-Jones parameters. Source:

Here is an excerpt from a document that I'm putting together. It also predicts the reduced triple point temperature to be 0.5

Well, there is no general relation between these temperatures.

Oh, you copied those L-J constants... Well as you can see this constant isn't equal to triple point temp.

@Mithoron What constant?

I agree.  We can't predict the triple point from just liquid and vapor characteristics alone.  The properties of the solid phase must enter in.  In the case of water, you have many phases each of which would potentially make its own triple point with liquid and vapor, unless you know which solid phase to pick.

Is only true for non polar molecules in the hard sphere liquid model limit.
The chemical potential must be equal for all phases:
http://www.sklogwiki.org/SklogWiki/index.php/Hard_sphere_model
compare with chemical potential for a van der Waals gas.
Or look here:
http://www.sklogwiki.org/SklogWiki/index.php/Lennard-Jones_model
and compare the critical point with the triple point

In principle, you can use the integrated Clapeyron and Clausius–Clapeyron equations to calculate $P$ vs. $T$ co-existence lines and find the crossing point which is at the triple point. You need the $\Delta H$ for fusion and evaporation and the densities of the solid and liquid phases.
The equations are $$ p_\text{liq-vap}=\exp\left(\frac{\Delta H_\text{vap}}R\left(1/T_1-1/T\right)\right)$$ and for the solid vapour line use $\Delta H_\text{fus} +\Delta H_\text{vap}$
For the solid-liquid line use $$ p_\text{sol-liq} = P_1 + \frac{\Delta H_{\text{fus}}}{\delta V}\left(\ln\left(T\right)-\ln\left(T_1\right)\right)$$
where $\delta V = M\left(1/d_\mathrm l-1/d_\mathrm s\right)$ where $M$ is the molecular weight and $d$ the densities of liquid and solid. $T_1$ and $P_1$ are temperature and pressure at the triple point. Thus you have to iterate $T_1$ and $P_1$ by fitting (say by least squares) to the calculated curves to whatever data you have to find the triple point. Tricky!
Naively, I am surprised that using Lennard-Jones parameters would work as they describe interaction potential between two atoms and not in a solid or liquid.

But does my method work?I mean it appears to to me at least but I dunno what a trained chemist would say. If it does it appears to be simpler than the method you posted, just a simple proportionality.

From the list you show, the comparison between calculation and experiment is sometimes good, but sometimes it is poor.  You do need to understand the reason for this before you can proceed further.

I have no idea. I would like a qualified individual to help me develop the theory further but so far nobody has been interested.

Note that the integrated equations you gave are oversimplifications - for example, you have assumed that the solid and liquid have negligible volume relative to the vapor. They're pretty good approximations, but will not give exact values for triple point even if solved as described.

@Andrew You are free to answer with the exact equations if you wish to :)


text:
Compound with a bright green emission when electrically excited
I rolled back the edit because I really am looking for the following:
I am looking for an element or compound that would have a bright green emission when electrically excited and is a gas at between 1 torr and 4 torr. If such a compound could be nontoxic, that would be beneficial.
The pressure range is important.
I'm familiar with the emissions spectra of all the noble gases, but none of them work for me. I want to find such a compound to use in a Geissler tube, which will be excited with about 50kVac at 300 to 500kHz. I am capable of pulling a vacuum down to 0.2 torr, and I normally work between 1 torr and 4 torr.
I know of a Xenon/nitrogen mix that produces the desired color, but it's too fussy for me to achieve, and Xenon is too expensive for this purpose.
Relatedly, I was wondering why the Xenon/nitrogen mixture emits a green light. Neither Xenon or Nitrogen has much green at the appropriate pressure, but somehow the combination, within a narrow range of Nitrogen partial pressure, produces a bright green.  What causes this color when the two compounds are mixed together, and how can I predict the color of a given mixture of gases at a given pressure?
I know of the bohr model and balmer series etc, but I don't understand how to apply that to compounds or mixtures of compounds. Is there some other way to predict the spectra produced by mixtures of gases?
It's not easy gettin' green!  (Apologies to Kermit)

Mother nature produces green northern lights using oxygen atoms and nitrogen molecules. I wonder if this has been ever done in a laboratory.

I'm aware.   One question would be whether the actual composition of the atmosphere is different at that altitude.  The other is, when pumping oxygen through a vac pump using oil, "Do ya feel lucky punk?"

It may be difficult to control a xenon/nitrogen mixture, but the reason why it is used may be because, all things considered, it is the best way to achieve the result you want. If there were cheaper or easier ways to do it, those who want a green glow would already use it. I'd be interested to see whether there are some cheap but poorly performing alternatives, though.

That would assume that the solution space has been thoroughly explored, and that this information had been preserved.    I don't believe either of those is really true.

I did find this which explains the xenon/nitrogen/oxygen mix:  https://www.quora.com/Is-xenon-the-perfect-element

While it is not a gas, copper and barium compounds come to mind. If you could introduce a volatile copper compound, you could get something remotely like a mercury vapor lamp, except with copper.

Many commercial discharge lamps do something like this but they work at high pressures using metal halide compounds of, for example, rare-earth halides with narrow emission lines.

Also, if it is possible to use copper electrodes, then it might be a very low-tech solution for introducing copper ions into the discharge. No idea about the pressures and gas mixtures required though.

My pieces don't have electrodes as such.  The excitation is AC coupled through the glass, so the temperature inside is not much elevated over ambient.

Is there somewhere an index of the emission lines/bands for compounds?  Seems like this should exist.  If I can find out what compounds will give me the right emission bands, then I can sort for which will be in the gas state at the appropriate temperature and pressure.

I am not aware of such an index, but I never looked for one so I would suggest googling. But I just had another idea: boron compounds also have green emission lines. You could try trimethylborate, although it could break down and form deposits. Diborane could be more stable, since it cannot form soot or boron oxides, but it is toxic and extremely flammable.

Thanks, I will give the trimethylborate a try.  I attempted to buy 100ml but got turned down because I'm not a lab.  I read the MSDS and it really doesn't seem that bad.  I can make it easily enough.

I tried TrimethylBorate, but no luck at any pressure down to a few millitorr.


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Axial Chirality in Carbodiimides
Do carbodiimides exhibit axial chirality or do the nitrogens invert rapidly enough, as in trisubstituted amines with different alkyl groups?

Yes, carbodiimides exhibit axial chirality similar to allenes. While there are significant differences in the allene - carbodiimide systems (see below) there are also some similarities. For example, just like in amines and imines the barrier to racemization in generally carbodiimidesis low, typically in the 6-10 kcal/m range.
Anet did a study on 1,3-diisopropylcarbodimide ("Energy barrier of racemization in diisopropylcarbodiimide", Frank A. L. Anet, J. C. Jochims, C. H. Bradley
J. Am. Chem. Soc., 1970, 92 (8), pp 2557–2558, DOI: 10.1021/ja00711a064).

In the chiral compound each isopropyl group would contain two diastereotopic methyl groups split by the isopropyl proton so we would expect to see 2 doublets. On the other hand, if racemization is rapid all methyl groups become equivalent and we would then expect to see only one doublet. Experimentally it turns out that around -150 C the doublet broadens into 2 doublets giving rise to a barrier height for racemization around 6.7 kcal/m.
Important Notes:
1) Just like with amines and imines, placing very electronegative groups on the carbodiimide nitrogens (fluorine, for example) raises the barrier to racemization.  Incorporating the carbodiimide structure into a small ring (see here for an example) can also raise the barrier to enatiomerization.  Both techniques have been used to permit isolation of chiral carbodiimides.
2) Carbodiimides are not linear, but rather the $\ce{N-C-N}$ angle is around 170°.  This is attributed to the terminal nitrogens being more electronegative than the central carbon and consequently affecting the hybridization of the central carbon (e.g. changing it from $\ce{sp^1}$ to $\ce{sp^{x}}$ where 1            < x < 2)
3) Racemization in carbodiimides can be achieved by two different mechanisms

by traditional nitrogen inversion involving a planar nitrogen atom, and/or
by traditional cis-trans isomerization about the $\ce{C=N}$ double bond

both mechanisms are thought to play a role in carbodiimide racemization.

Could you elaborate note 3? How does racemization occur by planarization of nitrogen, as it is already planar?

I just mean the traditional nitrogen inversion transition state where the lone pair exists in a p-orbital and the other 3 ligands attached to nitrogen all lie in the same plane as the nitrogen. In the case of an imine or carbodiimide, 2 of the ligands are the pi bond. Think of the pi bond using the bent-bond (2-membered ring) formalism. Now the planar transition state occurs when the 2 bent-bonds and the R group attached to nitrogen and the nitrogen all lie in the same plane.

But aren't they already in a plane? The nitrogen is sp2.

No, the pi-bonds to nitrogen are in a different plane (orthogonal plane) than the plane containing the $\ce{N-R}$ bond. Again view the pi-bond as a two-membered ring, that might make the 2 planes easier to visualize.

@Marko Geez, you can call it linearisation here...

@Mithoron, if it is just like linearisation, how is that mechanism different from traditional cis-trans isomerisation about the C=N double bond?

@Ron Could you please draw both isomerisation process?

See this old Ron's answer   http://chemistry.stackexchange.com/a/26319/9961

I've edited the picture.  The "circle" between the C and the N on the right represents the pi bond or 2-membered ring bent-bond description. Does that help clarify the 2 processes?

Now I see what you meant. It's much easier to understand with diagram. Thank you very much.

Glad it helped.


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Condensation of Ethylene Glycol
I'm looking to reflux a 250 mL of ethylene glycol solution. Would an Allihn condenser be suitable; if so what length?

My guess is that entirety of the glycol would condense in the first sphere, so you may safely take the shortest one. Using an air-cooled condenser (a glass tube of a proper length) is also an option and is preferable to just simplify the things, though I'm less sure that everything will condence in the first one or two spheres of the condenser in the case.

Ethylene glycol was a boiling point of 193 °C using any water cooled condenser will crack. Use a air cooled condenser. It is just a glass rod. The length of tube depends on how much you want to condense and how quickly you want it to come over. Remember visible condensation should always take place in the condenser. If it goes further down the tube then some hot vapours will be lost to atmosphere if your system is open. If under vacuume the ethylene glycol will boil at a lower temperature but you should use a cold trap at minus 40 °C or below to catch vapours which come over. Remember liquid ethylene glycol is toxic and the vapours are more dangerous check the MSDS. I suggest that you should work with a closed system in a fume cupboard for this job or you might never see your grandchildren. Good luck.

I refluxed propylene glycol with water condenser. It worked just fine.


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Deducing orbital degeneracy in geometries apart from octahedral or tetrahedral
How do you determine the crystal field splitting pattern when given the structure of a complex?
It's quite simple for structures like octahedral or tetrahedral, but I find that the reasoning behind the patterns for the trigonal bipyramidal structure, for example, is somewhat arbitrary.
Continuing the example of a trigonal bipyramidal structure, the $d_{xy}$ and the $d_{x^2-y^2}$ orbitals are supposed to be degenerate, but the $d_{x^2-y^2}$ fully overlaps with one of the ligands on the complex, while the $d_{xy}$ partly overlaps with two, and I don't think we have enough information to assume that the two "parts" add up to one.
Do I just have to accept that this is how it works (as in, is this just a "limitation" of the crystal field theory)? Or is there some aspect that I am missing/misunderstanding?

The easiest way to do it is to look at what other people before you have done.
If you look at the $D_\mathrm{3h}$ character table, you see that $(x^2 - y^2, xy)$ transform together as $E'$.
I don't think there is a way of looking at diagrams of the orbitals and concluding that they are necessarily degenerate. As you said, there isn't "enough information to assume that the two parts add up to one".
It isn't really a limitation of crystal field theory, either. Crystal field theory isn't about looking at how much orbitals overlap with each other and saying "oh, I think those look about the same, they're probably degenerate". If that was the case then in an octahedral complex you would not be able to tell why $\mathrm{d}_{z^2}$ and $\mathrm{d}_{x^2-y^2}$ are degenerate, since the former only really overlaps with two ligands and the latter overlaps with four.
The degeneracy is predicted by symmetry, the application of which is integral to, and implicit in, crystal field theory. There is no CFT without symmetry considerations.

I think you might just be confusing the orientations of the five d orbitals. I don't think you're misunderstanding any aspect of crystal field theory. You're just missing that dxy overlaps with a ligand. I drew a picture.


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How to measure saturated salinity and mineral levels for flaked sea salt?
I am a sea salt enthusiast looking to make the perfect flakey sea salt. I start with sea water and boil it down to a thick brine. Then with less heat I reduce the brine more slowly and inverted salt flake pyramids are made. I am looking for tools/instruments that will help me measure/understand the content of the brine (mineral content and salinity). Sometimes I get perfect flakey salt pyramids like the photo attached and sometimes I get I get more chunky results. The temperature of the brine is always the same but I get different results at times. This leads me to think that its the content of minerals in the saturated brine solution.
Are there chemistry tools/instruments I can use to understand the actual chemical/mineral makeup of the brine that is turning to salt.

What's your budget like and how quantitative do you want the results to be? There's a range of analytical methods that might work for you (acid testing for chlorides and sulfides of alkalis and alkaline earths, flame testing, ICP-MS, and HPLC, for example). See this Wikipedia link for some examples that might work for you on the cheap end of the spectrum.

Thanks for your input Todd. I had a look at the link you attached and couldn't find much info that was helpful. Is there any way to measuring NaCl, calcium, and magnesium, separately in a completely saturated brine solution? Thanks for your help!

The technique that comes to mind is HPLC, and you'd do serial dilutions of the brine until it doesn't saturate your cation and anion columns, and then back-out the concentrations of the relevant species in the original solutions. Are you looking for a more DIY approach, or is sending samples to a lab for cations/anions within your budget?

I'm looking for a more DIY approach as I'd like to be able to analyze each batch.

My unresearched thought is that you might want to go to a fish store (especially a fish store that specializes in salt-water fish) and ask them what sorts of equipment and test kits they have available (or know of that they can order) to test for main salts you're probably dealing with (NaCl, NaI, KI, MgCl2, CaCl2 would be my guesses).
You might also want to pay attention to the pans you're using. Surface roughness, bubbles, or manual jostling can nucleate crystals suddenly and less controllably.

Why not use salt (NaCl) bought from a store, or even better at a defined purity grade from a chemical vendor?
When you use sea salt, you get lots of other ions but more importantly you get colloidal particles, organic substances, and other compounds that will affect the crystallization. Chunky results sounds to me like a colloid particle that started crystallization in all directions at once.
If you want perfect crystals, do it perfectly, i.e. with a pure solution, in a clean and new glassware or polished stainless container. If you want perfect crystals you have to reduce the new-age-ness of it ;)
As for understanding, nobody fully understands crystallization, but a lot of things are known. There are books on this topic, check amazon and search for "crystallization". As for instruments, x-ray fluoroscopy and diffraction wouldn't be amiss but you will get very far with an optical microscope, I know I can watch crystals through a microscope for hours and still find something interesting. Barring XRF, XRD and microscope, and within a household budget there are salinity measurement probes and conductance measurement probes which will be able to tell you something of the brine such as it is.

"Whole" sea salt will be a lot more than just $\ce{NaCl}$.

What to measure??
Commercially available sea salts on the market today vary widely in their chemical composition. Although the principal component is sodium chloride, the remaining portion can range from less than 0.2 to 10% of other salts. These are mostly calcium, potassium, and magnesium salts of chloride and sulfate with substantially lesser amounts of many trace elements found in natural seawater. Though the composition of commercially available salt may vary, the ionic composition of natural saltwater is relatively constant. https://en.wikipedia.org/wiki/Sea_salt
The conclusion from Wikipedia seems to be that the composition of natural seawater is relatively constant (but complex!), while the dry commercial sea salt is quite variable, probably due to differences in production. Commercial table salt is essentially pure for your purposes.
You are "looking to make the perfect flakey sea salt". NaCl tends to crystallize out as cubic crystals, so you are trying to control the crystallization by means of other active agents, which might be at a very tiny concentration. Understanding the chemistry in such depth that you could publish in a respected journal would be possible, but probably not feasible, and maybe not even all that interesting.
Here's what I propose: You have some situations where you take seawater and get some desirable flakey crystals. You will also have some liquid brine left over. Is the active agent all deposited with the solid salt, or is active agent a major component, and retained in the concentrated liquid brine? Test: 1) Recrystallize the solid salt. Does it retain its flakey crystal structure, or have you fractionated the active agent into the retained liquid?  2) Take the liquid brine left over from the first crystallization, add the exact amount of ("pure") table salt that was removed, dissolve with heat, and cool to crystallize again. Do the new crystals have the flakey structure, indicating that the active agent was retained, or do you get simple cubes? The two experiments should give some idea how the active agent behaves. #2 can also be modified by adding more or less pure salt to determine how effective the active agent is in producing flakey crystals.
Hopefully, the active agent will go to one place and not the other, and you will have established some means of control that satisfies your need, For example, if all seawater has the active agent, just at slightly different concentrations, you could take your seawater, precipitate out the first (non-flakey) crop of salt, then add the liquid retained to a fresh batch of seawater and boil it down. In this way, you will have increased the level of impurities. If, on the other hand, the active agent (or lack of an anti-active agent!) has gone into the solid salt, add that to your next batch of seawater to produce flakey crystals.
Another possibility is that the flakeiness is not dependent on the analysis, but on some procedural difference (stirring, rate of temperature decrease, size of batch, seeding, some biologics that come from the seawater that do or do not decompose under your boiling down - seawater is foamy from animals living and peeing in it). In that case, even the best analytical procedures might miss the clue.

There's no good way to quantitatively or semi-quantitatively analyze for Na, Ca or Mg at home. You could investigate Ion Selective Electrodes and see if they'll give you the desired accuracy (ThermoFisher has Cl, Na, & Ca I.S.E.s which generally can be simply plugged in to a pH meter and then calibrated). Two other things worth mentioning: you need to filter ocean water - bacteria, algae, and all sorts of other organisms will be present in varying amounts, not to mention silt, and other colloidal inorganics. There also could be differences in dissolved gasses present. So, you need to standardize your handling of the water - from where you get it, to where and how you store it, to how you filter is, and how you process it. Don't assume that mixing it - or shaking it won't affect it. Try to figure out what you're doing to get the crystals you want. It's unlikely, imho, that the Na, Ca, Mg content varies much (although you should treat my opinion as a layman's - I've no direct experience with sea water variability.) I AM assuming you know enough to understand what near-by rivers and recent rains and obviously wind and tidal currents might do to change the composition of the surface water (not to mention air temperature and sunlight). Anyway, one of the cheapest easiest ways to determine dissolved solids is using a specific gravity bulb. (as long as temperature is held constant) You simply float the bulb into the solution and read from the scale what the density is. Finally, what container are you heating this in? Be aware that aluminum should not be used! And I'd prefer glass to stainless steel, but as long as there's no visual signs of corrosion, ss is probably ok. Also, while I'm sure you want a "natural" product, seed crystals of store-bought sodium chloride might make the crystallization more reproducible. Chemists are taught to scratch the glass side of a beaker to start crystallization. Scratches are also a site which may "catalyze" crystal formation, and since any semi-commercial (?) process can't use a new polished pan for each batch, maybe scouring the pan consistently before each batch is the way to go. HTH, these are mostly random thoughts of an industrial chemist kibitzing in an area he has little experience in.

While there is a lot of information here, much of it appears irrelevant to OP's question, and the monolithic single paragraph format makes it very hard to read and understand.


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TM hydride chemical shift
Why is the chemical shift for hydrides in TM complexes so strongly shielded, to the point where it's negative?
Initially, I thought it was because hydrides are $\ce{H-}$, but isn't the charge meaningless once it coordinates to a metal centre?

You essentially answer this question yourself. The chemical shifts of transition metal hydrides are often largely negative because they are highly shielded. That's what chemical shifts are a measure of - the amount of shielding from the external magnetic field. A very small hydride nucleus buried in a great big fluffy cloud of d-shell electrons of a transition metal is incredibly highly shielded from the base magnetic field.
The negative number of the chemical shift is arbitrary; what is relevant is that chemical shifts to the right are more shielded, and those to the left are less shielded.
Read the information about chemical shifts on this great online resource.


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Why is caesium considered the most reactive element and not fluorine?
Some people say caesium is most reactive element. I thought it to be fluorine as it is the element that reacts with almost all elements (except couple of inert gases). But caesium won't react many of the elements in the periodic table, will it?
I just want to know which element is considered most reactive and the basis for such consideration?

Reactive is a very broad word, there is no good basis for comparison.

There is no such thing as "most reactive" element (or, for that matter, "most stable").

There is no one answer.  Cesium and fluorine react in qualitatively different ways and thus cannot really be compared against each other.

Disclaimer; I don't know these "some people", I can only guess towards their reasons.
My guess: Because fluorine is so reactive that it does not exist... Joking aside, difluorine viciously reacts with whatever, even noble gases. however due to it being a gas and forming stable bonds, usually passivates stuff, it doesn't... explode. (well, mix it with a reactive gas, say hydrogen, or an element that does not passivate enough and you will experience its reactivity when you get covered with the remainder of it immediately following the explosion. Take care.) You just can't in a reasonable manner show its reactivity as well as you can with Cs. Cs comes as a soft metal (under oil, mind you), it does not "passivate", it will react with most everything, dissolve in everything and melt and cover everything else, this is usually followed promptly by intense fires and/or explosions. It seems more reactive, because it kinda is. It reacts wildly with more stuff than fluourine does.
Of course, I still hold fluorine to be the most reactive, just carefully prepare ClF3 and see if there are any limits to what it can incinerate violently. There are none.


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Why is partial pressure used as a metric for oxygen in blood?
I know haemoglobin basically acts as an oxygen buffer. But when we talk about non-bound oxygen in blood (or water), we're referring to the amount of O2 in solution (aqueous state). From a biochemistry point of view, the concentration seems like a good metric to me.
So why is it that medical papers report blood oxygenation in terms of partial pressure pO2 instead of concentration?
What does a partial pressure inside a liquid even mean? I tried interpreting the arterial pO2 value of 11% atm (i.e. about 11kPa). When using the formula P = CRT at 310K (body temp), I found an oxygen concentration of about 4.3 mM. This value is 20× higher than the maximum O2 concentration for water according to Henry's law. For those more familiar with DO values in mg/L, it would be 137.6 mg/L.
According to Henry's Law there is a limit to the amount of O2 that can dissolve into water at a given pressure and temperature. I know the lungs experience a slightly lower partial pressure than the atmosphere. So how can the oxygen concentration be so high? Does their metric include the oxygen bound by haemoglobin (which would be misleading at the least), or is my calculation off somewhere?
Or does a pO2 value correspond to the steady-state equivalent concentration at that pressure, in which case the arterial concentration in my example would be 0.14 mM (based on this table in wikipedia).
This similar question didn't answer my question:
What is the state of aggregation (gas, liquid) of oxygen in blood?

The equation P=CRT is the relationship for the oxygen concentration in the gas phase.  To get the mole fraction of oxygen the the liquid phase (blood), you first use the Henry's law constant.  Here is a link that can help:  https://www.google.com/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=henry%27s%20law%20constand%20for%20oxygen%20in%20blood
If, after applying Henry's law, you end up with a concentration in units of x ml per unit volume, this result implies that x is the volume that the mass of dissolved gas (solute) would have it were an actual gas at the same pressure as that in the gas phase.

That's what I did in the last calculation (0.14mM) using the value for water instead of blood. But that assumes steady state and still leaves me wondering why they choose to use partial pressure as a metric, as electrodes measure the concentration (amount) of O2.

If that's what you did, then that would be correct.  As far as motivation for using different measures of concentration, not being a medical person or biologist, I can't answer that.  Sorry.

Concentration is moles/volume, after applying Henry's Law I got 0.14mM. In a water solution that would be about 2.5ppm.

The Henry's law coefficient is typically expressed in terms of all kinds of oddball units.  So, whatever concentration it leaves you with, you need to work it into concentration units that work for you.

My own attempt at an answer:
Based on the answer from @Chester Miller it seems my last suggestion was accurate: the pressure values actually do correspond to an equivalent concentration at that pressure. So in fact a solute in a solution doesn't really have a partial pressure, just an equivalent partial pressure (EPP). This EPP value corresponds to the partial pressure of the gas outside the liquid--or the theoretical partial pressure should the liquid not be in contact with a gas.
In other words, the partial pressure in liquid is a bit of a misnomer. When we say e.g. the partial pressure of CO2 in the blood is 5%, what we mean is: that is the theoretical CO2 pressure required to achieve that concentration of CO2 if the liquid medium was in a steady-state exposure to a gas containing CO2.
This implies you can just convert between a concentration and its equivalent partial pressure by using Henry's constants (if known).
Now for the reason: in models where there is exchange between a gas and a liquid, the partial pressure determines the chemical potential driving the diffusion from one phase to the other--not the concentration. So one can more easily compare EPP values of a solute to determine the diffusion exchanges. This is typically relevant with lung-blood gas exchanges, which explains most of the physiology science papers use of partial pressure.
Unconfirmed part:
Lastly, although the PP drives the gas-liquid diffusion, once the gas is dissolved in the liquid it diffuses in a concentration-dependent manner. This implies at the tissue level in humans diffusion should be concentration dominated, the solubility differences between e.g. lipids and water shouldn't play a major role.
In contrast, if a gas interfaces with both lipid and water compartments, then the partial pressure dominates and the solubility becomes important, as it could allow high and low concentrations to exist near each other.


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Iron (II, III) oxide and concentrated nitric acid
According to the Russian Wikipedia, the reaction goes as follows:
$$\ce{Fe3O4 + 10HNO3 -> 3Fe(NO3)3 + NO2 + 5H2O}$$
Since $\ce{Fe3O4}$ is often written out as $\ce{Fe2O3*FeO}$, why don't we have instead
$$\ce{Fe3O4 + 8HNO3 -> Fe(NO3)2 + 2Fe(NO3)3 + 4H2O}$$
Why does the reaction evolve gaseous nitrogen dioxide?

Because nitric acid is a powerful oxidizer, and when it sees something to oxidize, it does not think twice.

@IvanNeretin - thank you! Can one write an answer based on this section of the electrode potentials table, I wonder. NO3(-) more actively wants to become NO2 gas than Fe(3+) wants to become Fe(2+).

@IvanNeretin - I want to try and write an answer to make myself memorize the reaction.

You don't memorize reactions. Also, I'm not sure whether you should memorize the electrode potentials table (I, for one, do not). I just have a vague feeling about some oxidants being "strong" and the others "weak", and that's enough for most situations.

@IvanNeretin - so this reaction cannot be explained using the potentials table?

Why, of course it can. The table tells you just the same as I did, only with more confidence.

"Guess" is not quite the right word either. Do you really mean $\ce{Mn3N2}$, the nitride? OK, this seems to be an acid-base reaction, $\ce{HCl}$ is an acid, and $\ce{N^{3-}}$ is a mighty strong base, so...

Sorry, Mg3N2 + HCl. I looked it up, and it produces HN4Cl - I would never have guessed (or reasoned) that out. (0:

Again, the reasoning is pretty straightforward, and there are just no gaps to be filled with guesswork. It is acid+base. What is the acid here? What is the base? What happens between the two?

@IvanNeretin - but why not just NH3? Why all the way to HN4Cl? What if we have not enough Cl? Will there be a combination of three products: NH4Cl + NH3 + MgCl2?

What is NH3? What do you know about its properties? Will it react with HCl? Well, of course, if we don't have enough of the latter, there will be all sorts of "underreacted" ("недореагировавших") products.


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Material to change its color by temperature condition meets over time
There are thermochromic materials which their color changes reversible by temperature. Simply they show their current temperature by color. We can say their color is a function of current temperature:
$$c_t = f(temp_t)$$
Where $c_t$ is color in time $t$ and $temp_t$ is temperature in time $t$
I wonder if there is similar material which their color changes irreversible by history of temperature they meet over time. In comparison to thermochromic material, we can say their color changes by integrations of temperature:
$$c_t = f(\int_0^t temp_x dx)$$
Where $c_t$ is color in time $t$ and $temp_t$ is temperature in time $t$.
This seems to be more general and simpler than thermochromic materials as they are like regular chemical reactions such as oxidation, ... happens every day.
Are there any industry suited solutions out there based on this behaviour which shown in the second equation?

Compounds that irreversibly record the thermal history of the surface they are attached do do exist and others are being developed.
One class, used in harsh industrial environments is described on the Sensor Coating Systems site.
There are chemical systems used to detect high temperatures in refrigerated food, but most don't seem to be single chemicals. There are certainly many devices in development, but how many involve single chemicals is unclear.


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How to find the concentration of an of ammonia solution by titration with nitric acid?
I am stuck with this question I am doing:
To determine the ammonia ($\ce{NH3}$) concentration in a cleaning product, a chemist diluted 20 mL of the product to 500 mL. Of the 500 mL solution, 25 mL was titrated with 0.151 M $\ce{HNO3}$. The endpoint occurred at 26.2 mL. Calculate the concentration of the cleaning fluid.
[Ans: 3.956 M]
What I have done so far:

I worked out the moles of $\ce{HNO3}$ that the $\ce{NH3}$ must have reacted with, using the concentration and volume they gave me. I got 0.0039562 moles.
I then wrote the balanced equation and worked out that the stoichiometric ratio of $\ce{HNO3:NH3}$ must be 1:1. Therefore, I worked out that the amount of moles of $\ce{NH3}$ reacting with the $\ce{HNO3}$, must also be 0.0039562 moles.
From here, I initially worked out the concentration of $\ce{NH3}$ reacting with $\ce{HCl}$ and got 0.158 M, but then I realised this wasn't what the question was asking. I tried using 0.02 mL as the volume instead (as this was the very initial volume of $\ce{NH3}$) but still got the wrong answer.

Can someone please tell me what I have done wrong/ guide me in the right direction.
Edit: I have just tried it again, and this time I used the concentration I previously found (0.158 M) and a volume of 0.5 L to work out the number of moles in the 500 mL dilution of the product. I used this value which was 0.079 moles and a volume of 0.02 mL which was the original volume of cleaning liquid, to work out the concentration. I seem to be getting the right answer, but is my working correct?

Yes, although your presentation is a bit confusing.
To be clear,

You determined the quantity of $\ce{HNO3}$ consumed during the titration
$$0.151\ \mathrm{M} \times 26.2\ \mathrm{mL} = 3.956 \  \mathrm{mmol}$$
Then given this titration equation:
$$\ce{HNO3 + NH3 -> NO3- + NH4+}$$
You deduced that the $25~\mathrm{ml}$ sample contained $3.956 \  \mathrm{mmol}$ of $\ce{NH3}$, hence a concentration of
$$\frac{3.956\ \mathrm{mmol}}{25\ \mathrm{mL}}=0.1582\ \mathrm{M}$$
From which you calculated the amount of $\ce{NH3}$ in the $500\ \mathrm{mL}$ sample:
$$0.1582\ \mathrm{M} \times 500\ \mathrm{mL} = 79.12 \  \mathrm{mmol}$$
And finally since the original $20\ \mathrm{mL}$ sample also contains $79.12\ \mathrm{mmol}$ of $\ce{NH3}$, that gives you the original concentration:
$$\frac{79.12\ \mathrm{mmol}}{20\ \mathrm{mL}}=3.956\ \mathrm{M}$$

What I think you should have done in place of 4 and 5 is:
4'. Since the $25\ \mathrm{mL}$ sample is taken from a solution diluted from $20\ \mathrm{mL}$ to $500\ \mathrm{mL}$, the original concentration is:
$$0.1582 \ \mathrm{M} \times \frac{500\ \mathrm{mL}}{20\ \mathrm{mL}}=3.956\ \mathrm{M}$$
Which spares you one calculation's work and potential errors.


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How does Dalton's Law relate to scuba diving?
My assignment is to create a report that emphasises on the important of chemistry more specifically the gas laws in scuba diving. I know that Dalton's Law is when the total pressure of a gas mixture equals the sum of the partial pressures that make up the mixture but i don't know how it relates to scuba diving. Can someone give me some hints?

In scuba diving it is essential to consider the partial pressures of different gases for the following reasons:

The partial pressure of oxygen should be greather than 0.16 bar to avoid hypoxia (at the surface this partial pressure is obviously 0.21 bar).
Oxygen is toxic at high partial pressure. For the dive one want to stay as close as possible to 1.3-1.4 bar (maximum), while decompression is made with maximum 1.6 bar.
Nitrogen is narcotic at high partial pressure and we usually try not to go higher to partial pressure similar to what we have at 30-35m.

As an example consider a deep technical dive to 80m (9 bar). If you have air in your tank (21% oxygen, 79% nutrogen), the partial pressure of oxygen will be
$$pp_{\ce{O2}} = p_\text{tot} \times x_{\ce{O2}} =  9\times0.21=1.89 > 1.6,$$
thus you are in serious danger because of oxygen toxicity (visual changes, nausea, anxiety, irritability, confusion, dizziness, tonic-clonic seizure, ...) which can lead to death. If you want to have an optimal oxygen partial pressure of 1.4 you will need to reduce your oxygen content to 15% (less than air). But since the partial pressure of this new mixture at the surface is 0.15 < 0.16 you should not start your dive with this tank.
In conclusion, what it is really important for your body is the partial pressure of the gas you are breathing, not the absolute pressure. Dalton's law is necessary to compute the different partial pressures.
This argument is quite advanced for scuba diving an you will have to search for technical diving in order to find the necessary informations. You may also want to look at other interesting physical phenomena that are important for scuba diving (Boyle's law, Charle's law, ...).

I would say that it's not just Dalton's law in effect here. Henry's Law is the main factor in solubility of gases in a solvent (in this case blood) .

@shre_sudh_97 There is plenty of physics and physiology behind diving! (One of my favourite phoenomena is the isobaric counterdiffusion.) That's why I suggested the OP to look at technical diving articles and different phenomena in order to get the broader picture. However Dalton's law is necessary to compute the partial pressure at different depth in order to choose the optimal gas mixture for that depth.