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Time taken in changing surface tension
I had read some where that in glasses the surface tension is high, but in a lota the surface tension is low.
My question how much time does the surface tension take to change? I mean that there was water in a glass but I put it into a lota in how much time will the surface tension of water change? Immediately?
This is a lota:

Everyone, I am so sorry for using the wrong word. It is a lota not a burbot. I have also attached a image of lota

@DakshShah No problem! Ask me about physicians and physicists on a bad day ;) Actually, the question whether the curvature of a lota bowl has an influence on the surface tension might be interesting, but unfortunately this is absolutely not my field of work.

@KlausWarzecha So you have no idea about this :( ?

@DakshShah No, not really. But I have an unscientific gut feeling that if there is an effect at all (which I don't know for this macroscopic level) it will be instantaneous. You might want to google for surface tension curvature or wait for some anwers from people more versed in this field.

I don't get it... Isn't surface tension a property of the water, not the container that it resides in? A meniscus will form in both this container and a glass cup, won't it? If the lota isn't doing anything but holding the water inside, I wouldn't actually expect any change to water's surface tension.

This is too much (macroscopic) physics for me, it might be better suited in physics.sx. If you do not get any answers here, it should probably migrated. @rch Wikipedia says:  If a liquid is in a container, then besides the liquid/air interface at its top surface, there is also an interface between the liquid and the walls of the container. I am no expert, that is why I am quoting the W.

@Michiel I felt that it concerned physics as well as chemistry so I posted it on both so that I get the physics reason and chemistry reason and then i can combine it both because there is no site like physics-chemistry

@Michiel So what do you suggest I do? It is already sort of rejected there. If you want to delete it here too. Go ahead!

@DakshShah - It is up to the moderators to decide. To be fair, I think the current question would have been ok on physics.se. The problem was, I think, that the short remark you added in your question over there (about health benefits of low surface tension drinks) ticked some people of. Imaginable, because think of all the toxic stuff that has a low liquid-air surface tension compared to water (e.g. methanol, chloroform)

@Michiel I think this is okay here for the time being.  I removed the part of the claim that we can't examine chemically.

Just as a quick answer, because apparently some people are in doubt so let me try to clean up the confusion.
The 'physics part'
Surface tension that people normally talk about is the liquid-air surface tension, in this case water-air, and I am pretty sure this is the surface tension you are referring to in your question. This water-air surface tension is a property of the water and air and has absolutely nothing to do with the physical shape or material of the container.
For completeness, I have to mention that the contact with the walls that have different chemical properties will change the shape of the water-air interface a bit, which you might misinterpret as a change in water-air surface tension but this is caused by the water-container and container-air surface tensions which indeed DO depend on the container.
The 'chemistry part'
What could * change the water-air surface tension is the exchange of some material out of the container into the water or the other way around. This would mean that you are not really looking at the water-air surface tension anymore, but rather at the (water$+$some solute)-air surface tension. Which might indeed be different, typically lower.
If this exchange is the mechanism then the time over which the surface tension will change is dependent on the kinetics of that process and without knowing what is being exchanged it is impossible to tell you what the time span will be, it could be seconds, it could be hours. Note that the explanation with stuff leaking out of or into the lota would mean that (assuming there will be equilibrium at some point) storing water in the lota and then drinking it from the glass would be the same as drinking it directly from the lota.

*I'm not saying it does I have no clue whether this happens with a lota

I don't think there is only water/air surface tension. OP sees a difference between lota surface and glass surface. Maybe, one should also consider solid/liquid surface tension (for a lota and then for glass) and compare it with air/liquid... I would be interested in knowing whether water spreads out over a lota surface better than over a glass surface.

@mannaia - I discussed the other surface tensions as well - 'For completeness, I have to mention that the contact with the walls that have different chemical properties will change the shape of the water-air interface a bit, which you might misinterpret as a change in water-air surface tension but this is caused by the water-container and container-air surface tensions which indeed DO depend on the container. '

@mannaia but judging by the question I really think the OP was thinking about the air-water surface tension. It would surprise me tremendously if water spreads better over the lota surface since (clean) household glass is already fully wetted by water


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Plotting a calibration curve for Gas-Liquid Chromatography
Question:

The given graph of abundance of different samples of butanol with respect to time is-

Textbook solution:

But why aren't the points like this:
$(20,5);(40,10);(60,15);(a,7)$
after which, to plot the curve we can use $y=mx+c$ and then find the value of $a$?

I get (60,16) for the third peak.

It's (40,11) in graph.

@hey, no the numbers are not on the correct lines.

@LDC3 Oh ok i see!

If I remember correctly, the amount of your compound isn't proportional to the peak height but to the peak area.
Back in the days, when plotters were just plotters (=no numerical integration of the peaks), people used to cut out the curves and weighted the paper pieces on a balance to determine the peak area.

I still do that!

If you have sharp, narrow peaks, the height is a good substitute for the area. But the area is more accurate.

You are actually correct, the formatting of the vertical axis in the solution graph has resulted in the vertical (Peak Height) values written below the actually gridline it is referring to - you can see the same for peak height values of 5, 10 and 15.
Using the linear equation formula $y=mx+c$ for the 3 known values yields a relationship of:
$$P = 0.25C$$
(Where $P$ is Peak height and $C$ is concentration)
Substituting the known peak height for the 4th value:
$$7=0.25C$$
$$C=28$$
Which is exactly where it is shown on the graph in the answer.

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What happens to water under high pressures without possibility of escape?
Knowing very little about the nature of water, wondering how it might behave at the centre of a planet or centre of an another massive gravitational body.
Could water take such pressures or might it break into separate hydrogen and oxygen to find something more accommodating for the pressure exerted, if such atmosphere would allow the volume even?
Finally, bonus points: could triple points in water play a role in keeping temperatures low under high pressures via fluid thermodynamics?

What happens to water under high pressures without the possibility of escape?
It depends on what you mean for high pressure and on the temperature, however, the water phase diagram can help you to understand what will happen. This is from Wikipedia User Cmglee:

You can see that at high-pressure water assumes a solid form (ice): you will have ice X at 100 GPa, and what is labelled as hexagonal ice XI in the diagram but is actually superionic water or ice XVIII at 1 TPa (temperature range 0 to 650 K).  These sorts of ices have different lattice and internal energies.
What would happen to water as pressures increase towards infinity without the possibility of the water escaping confinement?
Unfortunately, chemists don't deal with infinity, basically, however, there will be a point where the water molecule won't longer exist all the bonds will break.  After that probably you will have something called electron-degenerate matter have a look here, here is all about quantum mechanics, Pauli exclusion principle and black holes so on more Physic S.E. stuff.

Could triple points in water play a role in keeping temperatures low under high pressures via fluid thermodynamics?
You can also note that there are other triple points like (100 k, 62GPa) this however doesn't directly affect the properties of water but are the properties of water that determine where are the triple points. In a close system, kinetic energy is constant so I think that fluid thermodynamics doesn't matter in this case is more related to equilibria.
More info about water phase diagrams here.

Thanks for this GM.. any idea what would happen to water as pressures increase towards infinity without possibility of the water escaping confinement?

@irth I've edited the question but I think you should try to ask it on physic S.E!

Thanks, @GM. Here's the question at PhysicsSE http://physics.stackexchange.com/questions/148914/water-behaviour-under-theoretical-near-infinite-pressure-conditions?noredirect=1#comment312460_148914 Seems like everyone is on the same page but it's still not clear at which point water's bonds will break and what the resulting reaction really looks like. Does h2o undergo fusion before the bonds break or does it undergo some exotic metamorphosis to pure neutrons, and what would the chain effect look like?

Your diagram is outdated.  Ice XI is now designated as a low-temperature, proton-ordered form of ordinary ice.  Instead the high pressure phase, with unusual properties is now Ice XVIII.

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How much Carbon Dioxide would be required to displace enough oxygen to prevent ignition?
In air, how much $\ce{CO2}$ would be required to prevent a match to light or a butane lighter to work?
I have a basement office that is unusable because of an apparent lack of oxygen, currently it is my guess that the oxygen is being displaced by $\ce{CO2}$.
A bit research suggests that it would take a reduction to around 14% for a safety match to fizzle out without any flame. But my naive estimate suggests 7% $\ce{CO2}$, and at that level isn't $\ce{CO2}$ deadly?
And this isn't even accounting for the displacement of nitrogen, surely that would mean the levels of $\ce{CO2}$ would be even higher.
Would a 6% increase in $\ce{CO2}$ displace oxygen by only small percentage say 2% as it's only a smaller ratio of the atmosphere? Or would the presence of $\ce{CO2}$ contribute to the lack of ignition compounding the lack of oxygen?
See my related question on the DIY stack exchange
UPDATE
An insepector from the local council has installed a data-logger, whilst he was in the basement he took a reading that I've attached below. When this reading was taken the air felt poor, but by no means the worst it's been, a candle wouldn't stay alight. An breathing during conversation seemed laboured. It looks like the assumption of displacement doesn't look correct, there's more to it.
Here's the readings:
Gas - Reading (Normal)
$\ce{CO2}$ 3.3% (0.04%)
$\ce{O2}$ 17.5% (20.8%)
Other 79.2% (unchanged?)

This is the wrong question, you should ask What's the min amount of oxygen necessary for ignition? or something similar in fact ignition could be prevent even with nitrogen and CO2 % is very low then nitrogen %.

My current theory is that it's CO2 that's displacing the oxygen, possibly from a nearby dairy farm slurry pit.

Okay, But you can have 50% of $CO_{2}$ that is 1500 times the normal value and still light a match if there is oxygen.So your question doesn't have sense..

So, as I said, the oxygen must be around 16% or lower!

So, the answer is a lot of CO2, how much would be required to displace enough oxygen to prevent ignition?

Okay that's the question!

@gm can I have an up-vote then, now we've got that cleared up

@GM Environmental Health arrived today with a meter for detecting Hydrocarbons! The air quality today was much better, this morning matches and lighters worked perfectly, very annoying the day they arrive, it's fine! I'm going to have to purchase a desktop air quality meter to try and rule out $\ce{CO2}$

So was it methane?

@GM no, they didn't detect anything, but the only had an instrument that detects Hydrocarbons! I've ordered a desktop indoor air quality monitor with data-logger that will monitor carbon dioxide up to 9999ppm. Well see what that shows. I'm wondering if $\ce{O2}$ is used up when the house is shut up (when it's cold) would the absence of Oxygen "sink" to the basement, or are we definately looking at an unknown gas displacing the oxygen, as apposed to depletion of oxygen?

No, I don't think it is possible gases concentration disproportion are always near to the source of disproportion, there are studies about it, maybe there is something that consume $O_{2}$ in your cellar but I'm afraid that source is in your cellar.

@DogEars Considering that it's a basement office, I would think that humidity may also play a factor as basements are usually more humid than above-ground open enclosures.

Most gases don't have an effect on the body directly. So if an inert gas displaces oxygen you don't notice until you lack enough oxygen to function (and you probably die). Carbon dioxide doesn't work like this. The body is very sensitive to carbon dioxide levels and will react if the level is elevated. CO2 is not inert. Other gases can lower the O2 concentration but CO2 will not do it in a way you don't notice.

What's the humidity? Vapor is a gas too... Have you ever try with an hygrometer?

@GM between 50-70% it does seem higher the higher the CO2

The parameter you are asking about is called the Minimum Oxygen Concentration (MOC), and is a measurement of the minimum amount of oxygen present in air to propagate a flame. The amount depends on the fuel being burned, and varies from about 4% for hydrogen to about 18% for some solid combustibles.
Let us consider your example of butane, and round some numbers to make our calculations simple. Butane has a MOC of 12%. Normal oxygen content in air is 20%, with nitrogen and sundries making up the remaining 80%. Assume uniform mixing and displacement of gases is equal. In order to reduce the oxygen content to 12%, you need to introduce enough CO2 to make up 40% of the air content:
$O2 + N2 = 100$  =>20 + 80 = 100%
$O2 + N2 + CO2 = 100$    => 12 + (12*80/20) + CO2 =100%
However, let us consider a couple of other very significant points.

An oxygen level of 12% is unsafe. Anything below 19.5% is unsafe. At levels of 12%, a person will have impaired judgement and coordination, and experience increased pulse rates. Prolonged exposure can have fatal consequences.
A CO2 level of 40% is fatal. Full stop. Levels above 7% can cause loss of consciousness and just 10% can be fatal.
CO2 is actually denser than air, so mixing of the main components is unlikely to be uniform. This means that higher CO2 concentrations will be found lower down in the room.

I hope you stay well clear of your basement until you have resolved this problem and have appropriate active ventilation installed, as well as possibly an oxygen monitor.
In the very least, get yourself a canary.

If the partial pressure of $O_2$ is the same, does the level of $CO_2$ affect the ability of things to burn?  If we replaced all of the $N_2$ with $CO_2$ but left the $O_2$ alone then it would be very bad for humans but wouldn't things burn just as well?  Of course, it would be very different if we replace the $O_2$ with $CO_2$ and left the $N_2$ alone .

As you know the atmospheric gasses concentration in dry air is very different, carbon dioxide normally is below 0.1%:

Assuming that Limiting oxygen concentration for ignite the wood of your match is 17 % and that the ratio between the atmospheric gas is preserved you have to find the factor ($x$) that multiplied by the oxygen content give you 17.
$$20.9646 \times~x = 17$$
Then you can multiply Argon%, Oxygen%, Carbon dioxide% etc. etc. for x and then find the % of the unknown gas subtracting to 100% the sum of all the gasses. In one formula:
$$100-((Ar\%\times x)+ (O_{2}\%\times x) + (CO_{2}\%\times x) \ldots)= Unknown~gas~\% $$
The result is this:

So the unknown gas could occupy 18.8% of the volume of your cellar. This means that if it is $CO_{2}$ you could have several problem:

Within 1 minute of  exposure to 20-30% $CO_{2}$, unconsciousness and
  convulsions occur in humans. blm.gov

If you calculate it with a LOC of 14% you have a $CO_{2}$ content of 33%, this content as stated long (credit to him) can kill you.

 However for what I've read in your question there are not enough evidence that you are dealing with carbon dioxide, what I think is sure is that your cellar is not a safe place to stay.

CO2 Maximum Levels:
The defined PEL limit per OSHA is 5000 ppm or 0.5%
The OSHA STEL/IDLH is 40,000 ppm or 4%
O2 Minimum levels:
The amount of Oxygen found in ambient air is approximately 20.9%, if O2 levels fall below 19.5% the air is considered to be oxygen deficient, and workers at this level would require oxygen enrichment of the work space.  A level of 18% or less is considered to be an IDLH event.
You need to have outside air introduced to be able to work in this environment.  Typically in a forced air HVAC you pull a small duct from outside the home to tie in with your air return.  I would add a floor level air return in the cellar to your furnace and add a fresh air outlet at the ceiling level to introduce new fresh air to the cellar.


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Why Isodecane (B.P.: 327 *C) has a higher boiling point than decane (B.P.: 174.1 *C)?
Branched chain molecules have lower boiling point than straight chain molecules because branched chain molecules have a smaller surface area where they can be attracted by other molecules. Then, why Isodecane (a branched chain molecule) has a higher boiling point than Decane (a straight chain molecule)?

If you mean 2-methylnonane with isodecane, your data is wrong.
2-methylnonane has a boiling point around 166 °C, which would support the rule of thumb that you mentioned.

:I am grateful to you. Actually this was the boiling point of isodecane given in my textbook. Now, I have come to know that the given information is wrong.

The only problem here is the units - the boiling point of isodecane is given in Fahrenheit and the boiling point of decane in Celsius.

Your answer has been flagged as low quality, which was disputed, but nevertheless you might want to expand your future answers a little more in order to avoid this. [answer] gives a short, helpful introduction.


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Does covalent character increase or decrease bond strength?
On one hand, covalent character increases the bond dissociation enthalpy (i.e., more energy is needed to be put in) and on the other it lowers the melting point of $\ce{Al2O3}$ to a value lower then would be expected with perfect ionic bonding.
So, does it decrease or increase bond strength?

You do realize that you are asking about 2 different energies; the molecular bonds (bond strength) and intermolecular forces (freezing point).

In complex terms we can't ever really describe something as been fully ionic or fully covalent but more of a mix of the two. For example, a HCl molecule will have a small amount of ionic bonding character even though we define it as covalent bonding. This is besides the point though... in this case, we generally assume ionic bonds are stronger than covalent.
If you increased the covalent bonding character in a species you are likely to decrease the ionic bonding character resulting in a lower melting point as forces of attraction are now weaker. However, we do see additional bonding occur in species such as vulcanised rubber where you get cross-linking between carbon atoms on adjacent chains. This then obviously increases the melting point as you have not reduced any ionic bonding character

Vulcanization of rubber actually increases the size of the molecules, so the melting point is expected to be higher.

There was a previous post concerning strength differences between ionic bonds here.
To summarize:

bond strength decrease in the following order covalent > ionic > metallic

Substances with covalent bonds tend to appear weaker than ionically bonded substances (i.e. exhibiting lower boiling points) because ionically bonded substances often form lattice structures from their constituent ions, creating a much more resilient substance, if you will.
You can see the difference in bond strengths, however, when covalent substances form lattice-like structures as well. These network covalent structures are far "stronger" than normal covalent structures because, like ionic substances, they have formed molecular networks that make them stronger. Some examples of these network covalent structures are carbon in the form of diamond or graphene and silicon dioxide (quartz). You will notice that quartz and diamond both exhibit a crystalline structure similar to an ionic substance like NaCl, which is due to their network covalent structures.


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Fluorescence Spectra, emission spectrum
What exactly does the excitation spectrum tell us about a sample in Florescence Spectroscopy, what information does it tell us? I know that we will get an output of concentration vs intensity but that's it.

I know that we will get an output of concentration vs intensity but that's it.

Actually, that's not what you are getting. Typically, you run an emission spectrum first to determine the maximum of the fluorescence emission (or any other point of interest in the emission spectrum). In the excitation spectrum, you monitor this particular emission line and its intensity while varying the excitation wavelength.

According to Wikipedia,

by analysing[sic] the different frequencies of light emitted in
  fluorescent spectroscopy, along with their relative intensities, the
  structure of the different vibrational levels can be determined.


text:
How to dissolve steel without affecting aluminium alloy
Disclaimer: I have very limited knowledge of chemistry
My problem is the following: I have motorcycle part made from some sort of Aluminium alloy - most probably 6005, 6061-T6 or 7005. And inside of this aluminium part there is steel part which is simply stuck there and it's really hard to get it out physically without damaging the aluminium part.
After a bit of googling, I've found out several ways of doing that:

By putting aluminium part with stuck steel part into water and start adding concentrated nitric acid until steel part starts to dissolve. But I've also read that dillute nitric acid (which will occur after mixing with water) will also dissolve aluminium or I'm wrong here?
By using some sort of Alum: either by using Potassium alum or Ammonium alluminium sulfate. Some guy on the forum suggested to dissolve above alum in hot tap water and then put aluminium part with steel part into resulting solution and let it stay there for a day. It should dissolve or at least damage steel part without affecting aluminium part which may ease physical removal of steel part.

Could you please advise which of above solutions is better? And in case of using alum - which of them should I buy - potassium or ammonium one?
Thanks in advance for any help!

The part of the steel insert that is stuck to the aluminum is not accessible to water.  Let's look at linear coefficients of thermal expansion,
http://www.engineeringtoolbox.com/linear-expansion-coefficients-d_95.html
Aluminum is 22.2 ppm/K. Steels are smaller, the largest being 304 austentic at 17.3 ppm.  Ambient temp is around 24 C, a 400 F kitchen oven is around 204 C.  That should not be hot enough to lose temper in the alloys.  If the aluminum hole is an inch in diameter, worst case increasing the temp 180 K times 4.9 ppm/K net gives you 882 ppm or 22 microns of slack.  The steel insert pushes out.

Thanks for your response! Looks like it's easier than I thought :) However, for more complex cases like broken steel screws inside aluminium parts which cannot be extracted - do methods in my question at least make sense?

According to http://chemwiki.ucdavis.edu/Under_Construction/Chemguide_%28Jim_Clark%29/Physical_Chemistry/IV._Equilibria/B._Redox_Equilibria/The_Electrochemical_Series
you are in bad position for any type of etching process. Aluminum will be oxidized preferentially over iron.

@UncleAl tried to heat aluminium part yesterday and managed to push steel part out - thanks again for your help! Accepting your answer

@ssavec thought I didn't understand the article, thanks for mentioning that any type of etching won't work in my case :)

@Yura: As Uncle Al wrote in the other answer, there can exist specific conditions, under which you can etch iron and keep aluminum "intact". It will result in something like elox surface. Which you probably do not want inside your motor.

"However, for more complex cases like broken steel screws inside aluminium parts which cannot be extracted" Iron touching aluminum is a shorted electrolytic cell. The aluminum will preferentially erode in a wet corrosive environment. Hot concentrated nitric acid will passivate an aluminum surface by oxidation. Even if it preferentially goes after the steel, that is extremely hazardous - including inhalation.
Screw removal/extractor sets are tapered left-threaded hardened drill bits.  As they drill into the broken stub they unscrew it.  It still helps to heat the piece to equilibrium to disengage aluminum from steel.  You don't want galling as you force out the steel.
http://www.harborfreight.com/12-piece-screw-extractor-set-40349.html
http://www.sears.com/craftsman-3-pc-screw-out-174-damaged-screw/p-00952154000P
http://www.liquidwrench.com/
Add a drop of silicone oil, Kroil, WD-40, Liquid Wrench, or Mobil 1 0w20 before starting.  Silicone ends up everywhere.  This is a disaster in a high vacuum lab.  A machine shop is different.

yes, already got screw extractor sets and used them with partial success. Just was curious whether it's possible to apply chemical compounds in this case which will react with steel but not aluminium... Regarding nitric acid - yes, I've read about it and I'm not so eager about using it because of hazards. The other solution with alum seemed more appropriate for me, but it seems it won't work as well... Thanks a lot for your time and help!

This is an old post but these methods sound like overkill. If I couldn't get it out I would drill it out and re-tap the hole or insert a helicoil. Similar to what is done when a spark plug strips an aluminum head.
Ask any old shop guy. They have been dealing with problems like this long before anyone thought of dissolving the steel.

Yes, now I'm just re-tapping the hole and use bigger bolts if it's impossible to drill out the bolt. I was just curious if it's generally possible and feasible, since this looked quite effective to me :)

If steel is really stuck in a blind hole in an aluminum part, immerse in a heated alum solution for several days. There wil be very little solid ferrous material left but a crust.

The alum in water method works wel;. It is safer, cheaper, and can be accelerated with a small amount of heat, such as from a low wattage incandescent lamp.


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Why does tetrachloromethane have a higher boiling point than trichloromethane?
London dispersion forces (LDF) are present in all molecules, whether polar or non-polar. Molecules also exhibiting dipole-dipole interactions (in addition to the LDF) must have stronger forces of attraction than those molecules which exhibit only LDF.
Then, why does tetrachloromethane (carbon tetrachloride), which is a non-polar molecule exhibiting only London dispersion forces, have a higher boiling point ($\pu{77 ^\circ C}$) than trichloromethane (chloroform) ($\pu{61 ^\circ C}$) which is a polar molecule, exhibiting dipole-dipole interactions?

You also need to account for the difference in dispersion forces between the two molecules. Chlorine is much larger than hydrogen. Therefore tetrachloromethane has a larger molecular surface area which increases the intermolecular interaction strength. In this particular case, it outweighs the weak dipole interactions present in trichloromethane.

Chloroform numerically has a relatively strong dipole moment, but the negative charge is distributed over three large chlorine atoms.
The positively charged hydrogen atom finds nothing it could really be attracted to.

That's a little incomplete. Chloroform has a stronger dipole and that does matter. But it also has weaker london forces than carbon tetrachloride and those outweigh the strength of the dipole forces (it isn't that the dipole forces don't exist).

@matt_black Ja, see other answer. I should probably just have commented there.

tetrachloromethane has a higher mollecular mass than that of trichloromethane i.e  tetrachloromethane has a Rmm=154 and trichloromethane has a Rmm=118.5
thus the heavier the mollecule the greater the forces of attraction between the mollecules hence a high boiling point and vice versa

What about Polarity of molecules? Why are you not considering them as well ??


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What type of intermolecular forces will dominate Diethyl ether?
Is Diethyl ether (also known as ethyl ether) a polar molecule? What type of intermolecular forces dominate it? Dipole-Dipole Interactions, London Dispersion Forces or Hydrogen Bonding? Please Explain.

Take a look at these pictures. http://en.wikipedia.org/wiki/Diethyl_ether  Based on only the figures, can you make some deductions as to what it might be?  Please offer some reasoning and we'll be more inclined to help you.

Most organic molecules are polar by definition - they have a dipole moment (ethers included). In the context of solvents, polar refers to solvents with higher polarity, and non-polar to solvents with lower polarity. It's relative. Diethyl ether is at the lower end of the polarity scale and so is generally considered a 'non-polar' solvent .
This is what I read recently about Diethyl ether's polarity. Is it correct?

Okay, so diethyl ether has a small dipole.  So, now what about LDF and H-bonding?  You're getting closer.

As Diethyl ether has a very low polarity, if any, so, H-bonding is not possible, I suppose. LDF are present in all either polar or non polar molecules, so they are present here too. As it is slightly polar so it may also exhibit weak dipole-dipole interactions.

Carbon-donated hydrogen bonding is a very real thing but these are very, very weak non-covalent interactions.  So yes, I think your reasoning that H-bonding isn't possible is fine.  LDF is present in every chemical system but these are weak as well (there are exceptions where LDF may be significant in the binding energies of certain dispersion-dominated molecules).  Quick question: Is diethyl ether soluble in water?  If so, what force that you listed would most likely contribute to this?

Ethers in general can actually act as hydrogen bonding acceptors; just not as a donors.

@CTKlein I'm curious if diethyl ether can accept one- or two- hydrogen bonds.  I'm just going by the picture on Wiki.

@LordStryker good question... I presume that it is in principle possible given the two lone electron pairs but that it won't happen frequently enough to matter. Additionally it would depend on the molecule that the ether is interacting with. E.g., two water molecules might be able to both donate to the oxygen whereas larger molecules may be sterically hindered but I don't really know.

As you have already figured out, diethyl ether is a small dipole due to the electronegativity difference between carbon and oxygen, which can also be seen in this electron density plot. As both dipole moment vectors point roughly in the same direction, they do not cancel each other out like in the case of carbon dioxide, so the molecule has a net dipole moment and is therefore (weakly) polar.

London dispersion forces, which result from short-lived dipoles induced by fluctuations in the electron shell of molecules, are also present. They are responsible for the weak interactions between the alkyl chains, like in other alkanes and non-polar molecules. The permanent dipoles can also interact with each other, in the form of intermolecular attractive interactions between centers with $\delta+$ and $\delta-$ partial charge (and likewise repulsion between centers of equal partial charge).
Hydrogen bonding between C-H groups and oxygens, although possible in principle (see comments to your question) and, for example, found in protein structures, can be neglected in this case due to the low number of donors and the weakness of the interaction (< 17 kJ/mol). Nevertheless, the ether oxygen can act as a hydrogen bond acceptor in combination with better hydrogen bond donors like OH groups, which explains why diethyl ether is sparingly soluble in water despite its low polarity (69 g/l at 20 °C (source)). On the other hand, the absence of hydrogen bonding between ether molecules becomes apparent in the much lower boiling point of diethyl ether (34.6 °C, source) compared to water.

Clarifying a portion of the answer in the "Best Answer" you cannot have hydrogen bonding with a C-H bond. You only have hydrogen bonding with F-H, O-H, and N-H bonds. The hydrogen bonding in proteins usually results from amide hydrogen bonding shown in the image below


text:
Electrochemical half cell without redox couple
In a book about electrochemistry the situation of a cell is described of which one half cell has an "undefined thermodynamic potential" as no redox couple exists:
$$
Pt/Cu^{2+}\ (0.01\ M),\ Cd^{2+}\ (0.01\ M),\ H_2SO_4\ (1\ M)
$$
Estimating the potential where reduction of Cu sets in by thermodynamic data when a potential is applied,
$$
Cu^{2+} + 2e \rightleftharpoons Cu,\ +\ 0.340\ V\ vs. NHE\\
Cd^{2+} + 2e \rightleftharpoons Cu,\ -\ 0.4025\ V\ vs. NHE\\
O_{2+} + 4H^+ + 4e \rightleftharpoons H_2O,\ +\ 1.229\ V\ vs.\ NHE,
$$
they argue that there is a rest potential range $0.340\ V ... 1.229\ V\ vs. NHE$ where no current is to be expected. Beyond these values current flow is expected. I don't understand this, because I understand $0.340\ V\ vs. NHE$ to be the standard potential of copper reduction under standard conditions, but the concentration of 0.01 M is not "standard" (= 1 M), right?
On the other hand, I'm afraid I have a general lack of understanding a system where one species of a redox pair is missing, because argueing in terms of Nernst's equation for such systems always leads to infinitely large (or small) potentials. In the case of copper above Nernst is not applicable in my opinion (correct ?), but what does the general relation
$$
 \lim_{[R] \rightarrow 0} (E_0 +  \frac{RT}{nF} \ln\frac{[O]}{[R]})=\infty
$$
for the reaction $O+ne\rightleftharpoons R$ mean? And how can one figure out the rest potential range for a half cell on thermodynamic basis (neglecting kinetics)? What does the situation for copper looks like where [R] does not appear in Nernst's equation? When the solution in the book is right, why?

Nernst's equation is not applicable for systems where are more ions in the solution. If the concentrations of ions are different, than there will be potential shift. If no-one else provides the answers, I will try to do my best next week.

Thanks for your efforts in advance! But I must add that I would even not understand the situation if there were only ONE electroactive species (for example only $H^+$ coming from H2SO4, for which the missing partner is $H_2$; no $Cu^{2+}$, no $Cd^{2+}$) in solution.

Nernst's equation
The Nernst Equation can be derived from the Gibbs energy $\Delta G = nFE $. The free energy for a cell reaction:
$$\Delta G = \Delta G^\circ - RT \ln Q$$
can be written as:
$$E = E^\circ - \frac {RT}{nF} \ln Q$$
for reaction
$$\ce{O + $n$e -> R}$$
$$E = E^\circ -  \frac{RT}{nF} \ln\frac{a_\mathrm R}{a_\mathrm O}$$
where $a_i$ is activity of species $i$.
Activity
In a solution, the activity of $i$ is defined: $a_i=\gamma_i \frac {C_i}{C^0}$ where $C_i$ is the concentration of solute, $C^\circ$ is standard concentration ($1\ \mathrm M$), and $\gamma_i$ is unitless activity coefficient.
The equation can be simplified in:
$$E = E^{\circ'} -  \frac{RT}{nF} \ln\frac{[\ce{R}]}{[\ce{O}]}$$
But for the reaction
$$\ce{Cu^2+ + 2e- -> Cu}$$
the Nernst's equation will be:
$$
E = E^{\circ'} -  \frac{RT}{nF} \ln\frac{[\ce{R}]=1}{[\ce{O}]}
$$
because activity of solid (deposited copper) is 1.
If you try to calculate the potential of $0.01\ \mathrm M$ solution you will obtain the difference between standard concentration and the potential at $0.01\ \mathrm M$.
$$\begin{align}
E &= E^{\circ'} -  \frac{0.02559\ \mathrm V}{2} \ln\frac{1}{0.63}\\[6pt]
&= 0.34\ \mathrm V -  0.0059\ \mathrm V
\end{align}$$
The computed potential ($0.0059\ \mathrm V$) is lower than the standard potential and they are probably neglecting it.


text:
How long burned tobacco/cigarette smoke remain toxic?
It's hard to evaluate Environmental Tobacco Smoke lifecycle, it can certainly react, be absorbed by organisms, or be diluted in rain, and be emitted again from the ground.
But in the end, I'm interested to understand (as a non expert) if the toxicity (cytotoxicity, genetoxicity) of the compounds decrease over time (especially how are they eliminated) in a global outdoor environment where we suppose emissions would stop
Several resources list the components Chemistry and Toxicology of Cigarette Smoke, Comparison between Marijuana and Cigarette Smoke, CAEPA (page6):
Gas and particle phase contain roughly the same constituents as explained in the first paper, some compounds are 1,3-butadiene, acrolein, ammonia, aromatic amines, benzene, CO, isoprene, nicotine, nitrosamines, PAHs (polycyclic aromatic hydrocarbons), pyridine, toluene, hydrogen cyanide...
What could be their half-life in outdoor environment?

Tobacco smoke is never toxic the chemicals are too low  concentrations- only irritative. Otherwise it would be kill too many smokers immediately.

long-term toxicity? without much medical/biologic knowledge, when I see the differene between the skin, the voice of a smoker and a non-smoker, I imagine there are consequences on any other cells, neurons, and biological processes

Tobacco is a plant. The residuals are very much the same as those after a wildfire. There is no relevant long term toxicity except from direct inhalation.

@Karl I meant cigarette smoke, which don't contain just pure tobacco. Maybe should edit title

Toxicity of tobacco combustion products in different phases
For answering your question I think is better split it in two. First thing you need to understand that toxicity depends on the method of administation of the toxic agent. Tobacco smoke is a physical state of the tobacco combustion products. It is an aerosol originating from combustion where its particles are liquid, gas and solid (tar). Quoting Chemistry Central Journal 2012, 6:98

Cigarette smoke is a complex and dynamic aerosol consisting of at least 5,600 chemicals and toxicants found across two phases, the particulate (tar) and vapour phase.

(I would like to add the gas phase...) so when the product of the combustion of the tobacco are in this physical state you have to look at a different method of administration in this case vapour inhalation, Dust and Mist Inhalation and gas inhalation these are the most potentially dangerous method of administration. It seems that after about 40 minutes the liquid and solid part of the aerosol deposit (of course this depends on the condition, T, wind etc. etc.) so is probable that after this period of time the toxicity of tobacco combustion products decrease considerably. However this is not true for all the combustion products. For example: carbon monoxide is recognize tobacco combustion product and it is a gas so it wont deposit after this time and it will be "removed by reactions with OH radicals (85%), by soils (10%), and by diffusion into the stratosphere" very slowly.
Benzene is a good example of how the toxicity depends on the method of administration. LC50 inhalation value of benzene is 10000 ppm TFLo Dermal is 0.92 mL/kg so it is a big difference if you wait that the areosol deposit or if you inhale it.
If you are interested in rate of decomposition of benzene it seems that biodegradation of benzene can reach 0.95% at day according to Chiang.
Regarding 1-3 butadiene according to William A. McClenny and Donald Whitaker:

i.e., a 10-6 lifetime risk level for cancer due to inhalation exposure
of 0.03 $g/m^{3}$ (12.4 pptv at 20 °C and 1 atm pressure).1 This compound
is very reactive in the ambient atmosphere with a short
atmosphericlifetime, estimated to be 2–3 hr.

The half-life of acrolein is 14.4 hr.
You can also find a very complete EPA report EPA/600/P-98/001F
October 2002.
However determine the half-life of all the compounds is a very complex task because is related to how the ecosystem respond to these compounds so to the actual capability to decompose them through different mechanism is strictly related with thousands of factors.

Thanks, a good answer in my opinion, so this is more a matter of time than distance, (even if distance in general help with dispersion).
I'm pretty worried by smokers, and feel the harm is proportionnal to the quantity inhaled and have no lower limit (where our body would handle this extra-toxicity), so in my opinion we are all in danger. I wonder how the combustion of simple natural tobacco/marijuana is so concerning (additives in industrial cigarettes don't show a significant difference)

@dvok you are welcome.  I agree with you, probably compered with other major dangerous behavior a passive smoker doesn't have to worry much but I think that indeed smoke can cause damage not only to the smoker...

@dvok P.S. however distance is very important because the smoke diffuses so I would say that is a matter of both.


text:
How does the equilibrium shift when the products are solid and gas when increasing pressure?
$$\ce{CaCO3(s) <=> CaO(s) + CO2(g)}$$
How will the amount of each element change if the pressure is increased?
My guess is that the equilibrium will shift to the left, because the right side has 1 gas molecule. I know my solution contradicts with Le Chatelier's principles, but there is no gas on the left. So how will the the amount of the elements on the right decrease and the solid on the left increase since pressure only has an effect on gases.

Consider the equilibrium expression for this reaction.
$$\ce{CaCO3(s) <=> CaO(s) + CO2(g)}\quad K_c^{'} = \frac{[\ce{CaO}] [\ce{CO2}]}{ [\ce{CaCO3}]}$$
The concentrations of solids and liquids are constant.  They are the molar densities. Since $\ce{[CaO]}$ and $\ce{[CaCO3]}$ don't change, they are moved to the left hand side and "folded into" the equilibrium constant.
\begin{align}
K_c^{'} &\cdot \frac{\ce{[CaCO3]}}{\ce{[CaO]}} = \ce{[CO2]} \\
K_c &= \ce{[CO2]}
\end{align}
Therefore, as long as solid $\ce{CaO}$ and solid $\ce{CaCO3}$ are present along with $\ce{CO2}$ gas there will be an equilibrium.  Only changes to the concentration of $\ce{CO2}$ will cause a shift in the equilibrium.
You asked how will the amounts change if the pressure is increased.  The pressure of $\ce{CO2}$ is increased by either adding more $\ce{CO2}$ or by reducing the volume of the container.  Adding more $\ce{CO2}$ will increase the concentration of $\ce{CO2}$ momentarily, which will shift the equilibrium to the left, using up some $\ce{CaO}$ and making $\ce{CaCO3}$. The pressure of $\ce{CO2}$ can also be increased by reducing the volume of the container.  Again, the concentration of $\ce{CO2}$ is increased, which increases the reaction with $\ce{CaO}$ to make additional $\ce{CaCO3}$.  The $\ce{CO2}$ pressure gets closer to what it was originally, as predicted by Le Chatelier's principle. Since the molar densities of CaO and $\ce{CaCO3}$ are constant, they don't appear in the equilibrium expression. That is why only changes to the pressure (concentration) of $\ce{CO2}$ affect the position of the equilibrium.
If the pressure in the container is increased by adding an inert or non-reacting gas, nothing happens to the amounts of $\ce{CO2}$, $\ce{CaO}$ or $\ce{CaCO3}$.  The added gas won't affect the partial pressure of $\ce{CO2}$.

I don't think that your guess contradicts Le Chatelier's Principle, which states that introducing a change will prompt the reaction to oppose that change.
Increasing the pressure will lead to a decrease in $\ce{CO2}$ as it converts back into the starting reactant, which will lower the pressure as the gas is converted into $\ce{CaCO3}$.


text:
MaVa = MbVb - Limitations and Uses
We know that for strong acids with a $1:1$ mole ratio between all particles involved are suitable for use with the aforementioned formula for titration problems. I want to know the limits on this formula, that is:
Consider if we had $$\ce{H2SO4(aq) + 2KOH(aq) -> 2H2O(l) + K2SO4(aq)}$$
Since the moles of $\ce{H+}$ = moles of $\ce{~^{-}OH}$ in order to neutralize, how do we apply $M_aV_a=M_bV_b$ to this situation, if at all possible?
Secondly, does this formula work if we had weak acids? I would think not, because with weak acids we don't have the benefit of complete dissociation and thus not a $1:1 = \text{acid}:\ce{H+}$ ion ratio nor a $1:1 = \text{base}:\ce{~^{-}OH}$ ratio.

Regarding the ionization extent of sulfuric acid solutions:
$\text{1)}$ The first ionization is complete at most molarities (until you start reaching extremely high molarities - i.e. $\text{18 M}$). So sulfuric acid is a strong acid.
$\text{2)}$ The second ionization (i.e. the ionization of the hydrogen sulfate ion, $\ce{HOSO3^-}$, can be complete at low starting concentrations of hydrogen sulfate ion. This is because $K_a(\ce{HOSO3^-}) = 1.2 \times 10^{-2}$ is rather sizable!
So we can generalize these two cases of sulfuric acid solutions:
For very dilute sulfuric acid solutions:
$[\ce{H3O^+}]=2M_i - [\ce{HOSO3^-}] = [\ce{HOSO3^-}]+ 2\times[\ce{SO4^{2-}}]$, in which we generally take $[\ce{HOSO_3^-}]$ as negligible (i.e. = $\epsilon \to 0$).
For non-dilute solutions of sulfuric acid, we only have these relationships:
$$[\ce{H3O^+}]=M_i +[\ce{SO4^{2-}}] = [\ce{HOSO3^-}]+2[\ce{SO4^{2-}}]$$
As we can see, the first relationship (one for very dilute solutions) follows from the second because in a very dilute solution, $[\ce{HOSO3^-}]\approx 0$ - i.e. it has fully ionized into $[\ce{SO4^{2-}}]$.

Ionisation is adding or taking an electron. What you are describing is dissociation as in heterolytic bond cleavage resulting in ions. For your second equation it is better to write $$\ce{[H3+O]} = 2 M_{i} - \ce{[HOSO_{3}^{-}]}.$$

Yes, ionization can also be used in that context - i.e. ionization of vaporized sodium in the Born-Haber cycle. But textbooks also use ionization to refer to the loss of protons by acids.

I agree on your edit for the second equation; that would be a more accurate way of getting at the hydronium ion concentration.

Unfortunately you are correct, as ionisation provides the base definition for dissociation, too. Well, well, well, we are all here to learn.

Actually, you can. $\ce{H2SO4}$ liberates $\ce{2 H+}$ ions which is neutralized by the $\ce{OH^-}$ ions. The $K_\mathrm{a}$ for water is $10^{-14}$, this means that the concentration of those ions are very low compared to the $\ce{K^+}$ and the $\ce{SO4^{2-}}$ ions. You just need to use the stoichiometry of the equation properly to apply it to $M_aV_a=M_bV_b$.
You can use weak acids and bases if you allow them time to react. Most of them will react in less than a second, but I seen a few that take several seconds. In fact, the best way to determine the concentration of sodium hydroxide solution (since it can absorb carbon dioxide from the air) is to weigh dried potassium hydrogen phthalate (a weak acid) to determine the concentration.

If you apply this to a weak acid (equal amount of substance, mol), you will not get a neutral solution. Since at $\ce{pH} = 7~(25^\circ{}C)$ not all of the acid/base is dissociated.

Manual titrations usually use pH indicators. Of the dozen listed on wikipedia, one has a pH change near 7. Most titrations using NaOH use phenolphthalain, which changes near 8. At this pH, the amount of undisassociated acid (from a weak acid) is insignificant. http://en.wikipedia.org/wiki/Ph_indicator

They use phenophtalein, because it changes exactly (at least for $\ce{H3C\bond{-}COOH}$) at that point where $\ce{n_0(HA)=n(OH^{-})}$. The solution is not neutral. If it is neutral at pH 7, then for most weak acids like $\ce{HSO4-}$ there is a significant amount undissociated. This is why you can use the as buffers.

@Martin $H_2SO_4$ is a strong acid, both protons. Look it up.

$\text{p}K_a(H2SO4)=-3$, $\text{p}K_a(H3O+)=-1.7$, $\text{p}K_a(HSO4-)=1.9$ Hence the second deprotonation is incomplete, especially with excess hydronium.

@Martin It is unlikely that you would stop a titration with a pH of 3 (where most of $H_2SO_4$ is disassociated). How much $HSO_4^-$ would still be in solution at a pH of 7? $1.9 = \frac {{[H^+]}\times {[SO_4^{2-}]}}{[HSO_4^-]}$ or $1.9 \times {[HSO_4^-]} = 10^{-7} \times {[SO_4^{2-}]}$ So, even starting with a 1 M solution of $H_2SO_4$, you will have ${[HSO_4^-]} = 10^{-7}$ M. A very low concentration.

Agreed on the low concentration (still much bigger than autoprotolysis), but that was not the question. At $2M_AV_A=M_BV_B$, starting at $c_0(\ce{H2SO4})=1\frac{\text{mol}}{\text{L}}$, with $\text{p}K_B(\ce{SO4^{2-}})=12.08$ and $c(\ce{{}^{-}OH})=\sqrt{K_B\cdot c_0}$ and $K_W\approx14$ at $T=298.15~\text{K}$, you will find $\text{p}H=7.7$ and that is not neutral.

@Martin You're right, it has nothing to do with the OP's question. But if you look above, I did answer the OP's question in that you can use the equation even with weak acids or bases using the correct stoichiometry. You assumed that the titration ends with a pH of 7.0. It doesn't. Phenolphthalein changes color between pH 8 and 9. Yet it is most commonly used with sodium hydroxide titrations. The endpoint is chosen where the pH changes rapidly with a little addition of titrant so that minimal error is introduced.

You cannot use this equation for weak acids or bases. It is a cringe if you want to get close to neutrality $\text{p}K\in(6;8)$. Since the moles of H+ = moles of −OH in order to neutralize, how do we apply MaVa=MbVb to this situation, if at all possible? The answer is obvious: You don't.

Also the Ka of water is not 10^-14.

@Dissenter From Wikipedia: From these data it can be deduced that Ka = Kw = $10^{−14}$ at 24.87 °C. Since $[H_2O]$ is a constant. Where do you see that Kw is not $10^{−14}$? http://en.wikipedia.org/wiki/Acid_dissociation_constant

Kw is 10^-14 but Ka is not. Kw/[H2O] = Ka. Ka is a quotient; simply because the concentration of water is roughly constant doesn't mean we don't divide by the concentration of water.


text:
What's the mechanism for the oxidation of primary alcohol to carboxylic acid with HNO3?
Can someone help me find the mechanism for this reaction. I've been looking for it, but I
haven't found it. Thank you.

Welcome to Chemistry.SE. I do not think this is a homework question, but a few people have voted to close your question because it violates the homework policy. Would you be willing to improve your question with some additional details, like what your thoughts are or where you have tried looking?

http://connection.ebscohost.com/c/articles/25511593/oxidation-nitric-acid-aliphatic-alcohols-diols-carboxylic-acids
http://pages.towson.edu/ladon/orgrxs/reagent/oxidizer.htm
It is a dangerous thing to do for the easy detonation of nitrate esters.  Activated and stabilized alcohols, e.g., benzoin to benzil, are reasonable.

The reaction starts with the formation of an organic nitric acid ester (as these compounds are dangerously unstable, as Uncle Al already pointed out, I would not recommend to use nitric acid for this oxidation). The negatively charged nitrate oxygen abstracts an alpha proton from the alkyl group, and $\ce{HNO2}$ is eliminated, yielding the aldehyde. Nucleophilic attack of nitrate on the carbonyl group and subsequent elimination of another equivalent $\ce{HNO2}$ finally yields the carboxylate.


text:
Fluorescence spectrocopy, sensitivity
lets say we increased the excitation wavelength higher than it should be, what effect would this have on the sensitivity on our analysis, or changing the emission wavelength to a higher value than it would be.
My guess is that the higher wavelength leads to a lower frequency and so lowers the sensitivity but I have no idea what changing the emission wavelength would do?

How would you change the emission wavelength?

sorry I mean if we were to raise the wavelength of the florescence spec used to excite the molecules.

@KlausWarzecha - You could also choose to monitor the emission in a single wavelength experiment at wavelengths $\ne \ \lambda_{max}^{em}$. You may want to do this, for example, to avoid matrix effects.

@BenNorris Absolutely true! Excitation spectra measured at different wavelength plus repeated experiments after dilution aren't the worst idea. The world is full of dimers ;)


text:
Reduction of an Oxime with LiAlH4
I want to know the mechanism of this reaction. My intuition says $\ce{H-}$ will attack on $\ce{C}$ attached to $\ce{N}$ and hydrolysis afterwards would produce $\ce{-NH-OH}$. But the answer says that the amine will be formed.


text:
Wax and oil polish on wood
Wood countertops are frequently sealed, when a non-hardening, food-safe finish is desired, with some magic cocktail of waxes and oils. Typical components might include mineral oil, beeswax, and carnauba wax, mixed in some unspecified ratio and then sold at premium prices.
Not being content paying for a bottle of magic goo, I'm seeking a scientific understanding of how these products actually work.
These products claim to (over time) build a coating of wax on the surface which is more durable than simply mineral oil would be. If the product is a mixture of oil and wax which is quite thin, how can it build a layer of wax on the surface of the wood with repeated applications? Does the wood selectively absorb the oils? When subsequent applications are made, are the thinner oils able to diffuse through whatever wax has built on the surface?
What are the relevant concepts in play here, and in what directions might I research to gain a deeper understanding?

Mineral oil serially penetrates the wood to render its surface hydrophobic.  Nothing eats mineral oil and it does not easily oxidize (go rancid).  It is also the solvent vehicle to transport beeswax (plasticizer, compatibilizer, internal soft sealant) and the hard wax surface sealant. Nothing polymerizes (re linseed or tung oil) or chemically bonds to the wood.  All three are FDA GRAS.
Like dissolves like.  While the surface will be sealed against polar materials and provide no food for bacteria and fungi of itself (and absence of water is a good biostat), it can imbibe fats and oils.  Butcher block surfaces and such are best made of dense oily woods that are naturally sealed against water.  Top them off.  Lignin is a fair antibacterial and antifungal agent.  Polyethylene cutting boards have higher sustained levels of viable bugs than do wood cutting boards.
http://faculty.vetmed.ucdavis.edu/faculty/docliver/Research/cuttingboard.htm


text:
What is the pKa of the hydronium, or oxonium, ion (H3O+)?
Although the wikipedia page on Hydronium indicates a $\mathrm{p}K_\text{a}$ of −1.74, I noticed in the discussion of this page that the subject seems debated (cf. http://en.wikipedia.org/wiki/Talk:Hydronium#pKa.3F) since alternative (apparently more rigorous) reasoning leads to $\mathrm{p}K_\text{a} = 0$ for $\ce{H3O+}$.
Also, many demonstrations lead to $\mathrm{p}K_\text{a}$ and $\mathrm{p}K_\text{b}$ of 15.74 for $\ce{H2O}$ and thus, using $K_\text{w}=[\ce{H+}][\ce{OH-}]=10^{-14}$, give $\mathrm{p}K_\text{a}$ and $\mathrm{p}K_\text{b}$ of −1.74 for $\ce{H+}$ and $\ce{OH-}$ respectively. But those demonstrations seem to make use of the concentration $[\ce{H2O}]=55.56~\text{mol/L}$ instead of the activity $a_{\ce{H2O}}=1$ (for a solvent) in the equilibrium constant of the reactions.
So it seems to me that the $\mathrm{p}K_\text{a}$ of $\ce{H3O+}$ must be equal to $0$.
Is this question still debated?
PS1:
To add details about the question, the debate in the references mentioned in the discussion on the wikipedia page as well as in the link provided by Nicolau Saker Neto bellow seems to boil down to the coexistence of two different definitions for the equilibrium constant:
- one based on the activities of the compounds
- the other derived from the 'mass action law' and which thereby uses the molar fractions of the compounds
Reading through the Callen (Thermodynamics and an Introduction to Thermostatistics) gives me the impression that the derivation from the mass action law of an equilibrium constant involving molar fractions is based on the assumption of an ideal fluid. However, it is precisely the departure from ideality that seems to justify the use of activities instead of molar fractions. Is that correct?
PS2:
Another point, which I have not looked in details yet, seems to be a debate about the possibility to apply the definition for $K_\text{a}$ to the water molecules $\ce{H2O}$ considering that they are not a highly diluted solute since they constitute the solvent. Is this a valid issue and should then the acidity of $\ce{H3O+}/\ce{H2O}$ be determined in another solvent than $\ce{H2O}$ to be compared with the other acids?

There is the problem of having 55.56 molar hydronium re its counterion being included.  Consider water azeotropes of strong acids not by wt-% but by mole ratio.  Is there anything interesting to be had there?

The footnotes in this link talk about the calculation and validity of the $K_a$ of hydrated protons in water, with several references which may be of interest.

@Uncle Al, I'm sorry but I don't understand your sentence.

@NicolauSakerNeto, those references indeed overlap with the ones sited in the discussion of the wikipedia page. I looked at them and for now it seems to me that the issue boils down to the existence of two different definitions for the equilibrium constant:

@NicolauSakerNeto: (1) Sorry for the typos in my previous comment (I couldn't edit it anymore)... (2) I edited the question to resume the debate in the references as I understand it.

Water is 55.56 molar because every molecule within it is water, with low MW and high material density.  Hydronium must have a counterion,  Do better than hydronium fluoride.  I doubt you can stuff 55.56 moles of that into a one liter bag.

Two errors here. Firstly, Keq expressions use activities, not molar concentrations, and the activity of water is 1 in dilute solutions. Secondly, Ka = Keq = 1. Your statement that Ka = Keq x [H2O] is incorrect.

@ToddSilverstein First, I did stress in the question that activities should be used instead of concentrations and that the activity of water as a solvent should be 1. Second, I never stated that Ka = Keq x [H2O] (since Ka is an equilibrium constant, the equilibrium constant of the dissociation reaction of an acid).

The controversy surrounding the $\mathrm{p}K_\mathrm{a}$ of hydronium mostly arises from the definition of $K_\mathrm{a}$ or lack thereof.
There is no IUPAC definition of $\mathrm{p}K_\mathrm{a}$ or $K_\mathrm{a}$.
The closest IUPAC defined term is the standard equilibrium constant, which can be denoted as $K^\circ$ or just $K$.
Physical chemistry texts such as Levine and respected works such Bates's "Determination of pH — Theory and Practice" define $K^\circ_\mathrm{a}$ of an acid in water as:
$$\frac{a(\ce{A-})a(\ce{H3O+})}{a(\ce{HA})a(\ce{H2O})} \tag{1}$$
Where $a$ is activity.
Substituting that the acid is $\ce{H3O+}$:
$$\frac{a(\ce{H2O})a(\ce{H3O+})}{a(\ce{H3O+})a(\ce{H2O})} = 1 \tag{2}$$
and of course $-\log(1) = 0$.
The number $-1.74$ that some quote for the hydronium $\mathrm{p}K_\mathrm{a}$ comes from:

omitting the activity of water from the denominator of the $K^\circ_\mathrm{a}$ definition (equation $(1)$); and
taking the concentration of water (about $55.5~\mathrm{M}$) as the $K_\mathrm{a}$ of $\ce{H3O+}$.

With that, one obtains the $\mathrm{p}K_\mathrm{a}$ as $-\log (55.5) = -1.74$.  For example, Levine even has the $-1.74$ value in a figure comparing the $\mathrm{p}K_\mathrm{a}$'s of various acids, but has a footnote explaining that the value for $\ce{H3O+}$ is based upon the alternative $\mathrm{p}K_\mathrm{a}$ definition.
However, revisiting the analysis that $K^\circ_\mathrm{a} =1$ is probably the most authoritative paper on this issue:
New point of view on the meaning and on the values of Ka(H3O+, H2O) and Kb(H2O, OH-) pairs in water, which insists that
$$\ce{H2O + H3O+ <=> H3O+ + H2O}$$
"does not correspond to an actual chemical process" and therefore "it is not legitimate" to extend the concept of $K^\circ_\mathrm{a}$ to $\ce{H3O+}$ in water.
The article goes on to say that only by studying the $K_\mathrm{a}$ of $\ce{H3O+}$ in another solvent such as ethanol can $\ce{H3O+}$ be compared to other acids.
The $\mathrm{p}K_\mathrm{a}$ of $\ce{H3O+}$ in ethanol is $0.3$ and $\mathrm{p}K_\mathrm{a}$ values are $1.0 \pm 0.3$ units lower in water than in ethanol, so the article suggests a $\mathrm{p}K_\mathrm{a}$ of $-0.7$ for $\ce{H3O+}$ in water, for the purpose of comparison to other acids.

So it sounds like the origin of the -1.76 value arises from a simplified definition of pKa often used in introductory courses. In any case, what implications does this have, if the pKa is really 1.0 instead of -1.76? Does the pKa of protonated methanol differ significantly?

Also I found that the pKa of protonated methanol to be -2.2 Would H3O+ really be that much different from protonated methanol?

@Dissenter "what implications does this have, if the pKa is really 1.0 instead of -1.76" Nobody is saying 1.0, the only values mentioned are 0.0, -0.7 and -1.74.  But I don't think there is any consequences because, as the article explains, the only real equilibrium with H2O and H3O+ is Kw.  I can't think of any situation where I've ever used a value for Ka of H3O+ in water.

@Dissenter "I found that the pKa of protonated methanol to be -2.2 Would H3O+ really be that much different from protonated methanol?" In DMSO, water pKa is 2.4 units higher than the methanol pKa http://www.chem.wisc.edu/areas/reich/pkatable/

I think there is a misunderstanding between Keq and Ka:
for $\ce{H3O+ + H2O <=> H3O+ + H2O}$:
$$K_{eq} = \frac{[\ce{H3O+}]*[\ce{H2O}]}{\ce{[H3O+]*[H2O]}} = 1$$
however $K_\mathrm{a} = K_{eq}*[\ce{H2O}] = \ce{[H2O]}$
therefore $K_\mathrm{a}(\ce{H3O+}) = [\ce{H2O}]$
p$K_\mathrm{a}(\ce{H3O+}) = -1.74$


text:
Heat when dissolving solutes in water
Why do certain solutes dissolved in water release heat and other solutes absorb heat?

To understand this phenomenon first we must point out that there are very different solid substances and solvents. Different combinations of them lead to different interactions and properties.
Anyway for the regular case - ionic or soluble compound in water - there is at least two steps to realize the solution:
Edited Answer:

Break the solid structure (eg. the crystal lattice). This step generally absorbs energy (endothermic, see lattice energy);
Solvate the compound. Because there are interactions between matter, the solvent might organize itself around the compound (solvation shell). Those molecules involved are not any more available, they are tied to the compound, they are not any more free water. This step generally releases energy (exothermic, see solvation energy). You can think about this step as a reaction of the solvent to recover form the intrusion by coating the intruder with one or several layers of its molecules. Then it 'feels' if there were only water. You will find details of the complex solvation process in good Physical Chemistry book.

Succeeding these two unit steps leads to dissolution. Solubility is the thermodynamic limit beyond that you cannot dissolve further molecule of a substance. Off course this is related with every substance present in the solution and available free water. Solubility are generally expressed for pure substance only.
When you realize your solution, speaking of heat exchanged with the environment, you experienced the balance between those two reactions. This is why the heat balance can lead to a global endothermic or exothermic reaction.

Breaking a crystal lattice is endothermic. Solvation is exothermic - talking about enthalpy. Entropy terms then play a role as well. Your answer is good, but it says that breaking the crystal lattice is exothermic, and solvating is  endothermic. Isn't it the other way round??

Thanks to point out this mistake. I double checked it in Atkins Physical Chemistry. Answer has been edited.

Enthalpy of hydration is exothermic, Lewis acid-base neutralization.  Entropy of dissolution can be either positive or negative. Negative - ordering effect of ion on solvent is greater than the entropy increase of the crystal (highly ordered) lattice breaking down. Positive - increase in entropy because the solvent hydrogen bonding is disrupted.  Dissolving KOH is a very large exotherm,  Dissolving urea in water is a very large endotherm.  This is descriptive and qualitative,
http://www1.lsbu.ac.uk/water/kosmos.html
We invite the physical chemists to put a quantitative mathematical model on it.

"Entropy of dissolution can be endothermic" is quite a poor way to word it. Endo and exothermicity are related to enthalpy only, so perhaps you could clarify your sentence? And I'd like to reinforce that the linked site is exceptional. It is definitely one of the most complete descriptions of a single substance I have ever seen.

Entropy at J/K-mole is ~1000-fold smaller than enthalpy at J/mole (KJ/mole typical), but temperature takes up a lot of slack.  Water is extremely ordered re its large specific heat via hydrogen bonding, offering a large free energy sink and temperature drop when disordered other than by adding energy as heat.

Entropy cannot be qualified as endo or exothermic, those terms are for Enthalpy only. Please rewords for the sake of your answer and the exactness of thermodynamic.


text:
how to determine the relative differences in activity between metals
I am doing a lab for school and I do not know what this question is asking. In this lab we are making electrochemical cells from different metals.

Welcome to Chemistry.SE! Could you provide a little more detail in your question? What exactly are you supposed to do?

You have a standard pot of liquid electrolyte into which are dipped non-touching same-spaced paired strips of different clean metals.  The external circuit between the strips is a sensitive voltmeter (with extremely high electrical resistance - hard by zero current flow).  You measure the potential (noting direction!) between all combinations of paired strips, (n)(n-1)/2, and then draw certain conclusions.
Why are amalgams saturated with respect to excess solid metal interesting substrates for this experiment (e.g., aluminum)? How do you construct a standard hydrogen electrode?


text:
Flame test: Is the metal atom or the metal ion responsible for the flame colour?

A sample of $\ce{CaCl2}$ is placed in the flame of a Bunsen burner. The component of the substance which is responsible for the production of the flame colour is:

$\ce{Ca^2+}$
$\ce{Ca}$
$\ce{Cl-}$
$\ce{Cl}$

Why is the textbook answer $\ce{Ca}$ and not $\ce{Ca^2+}$ ?

What are your own thoughts on the matter? Why do you think it should be $\ce{Ca^{2+}}$ rather than $\ce{Ca}$? Do you know which process is responsible for the flame colour? Please, have a look at our homework policy.

isnt it the presence of the cation that causes the change in colour in the flame test?

The phenomenon that is witnessed during a flame test is an "atomic emission".
This statement may seem inappropriate, since it is a solution of metal ions (and not atoms) that is tested. The reason for calling it atomic emission lies in the process occurring in the flame. One of the steps of the process is an atomization step. That is, the flame converts the metal ions into atoms. When a solution of sodium chloride is placed in a flame, for example, the solvent evaporates, leaving behind solid crystalline sodium chloride. This evaporation is then followed by the dissociation of the sodium chloride crystals into individual ground state atoms - a process that is termed atomization. Thus sodium atoms are actually present in the flame at this point rather than sodium ions, and the process of light emission actually involves these atoms rather than the ions.

@confused Glad, I could help.

While it is an atomic emission in the case of alkali metal salts, for Ca-Ba monochloride molecules are also of consideration, and, in fact, for blueish copper flame CuCl molecules are of essence (greenish of yellowish flame is formed otherwise)

Actually, screw it. Here is an article that discuss the question of Ca(NO3) and CaCl2 solutions in flame test. The majority of the visible spectra is derived from molecular particles www.jpyro.com/wp-content/uploads/2012/08/Kos-710-731.pdf

I got a question in "That is, the flame converts the metal ions into atoms"... so is this considered a reducing flame for the metal ion?

I got another question as to the word "atomisation" in "One of the steps of the process is an atomization step,  That is, the flame converts the metal ions into atoms." isn't atomisation other way round- solid substances broken down into their atoms? is it applicable for ions converted to atoms also? isn't it Ionisation enthalpy?

But alas it appears the answer appears to been plagiarised [link]http://faculty.sdmiramar.edu/fgarces/labmatters/instruments/aa/AAS_Theory/AASTheory.htm#:~:text=One%20of%20the%20steps%20of,the%20metal%20ions%20into%20atoms.&text=This%20evaporation%20is%20then%20followed,process%20that%20is%20termed%20atomization.

Or it could be that the faculty copied your answer, or maybe you being part of the faculty

@Philipp It would be nice if you could address the issue with possible plagiarism. This might be a serious issue. That PDF appears to be compiled way prior to your post, in 2007, and the author's name doesn't look like your nickname. You should either properly attribute cited work, or claim the authorship in convincing way, or delete the post altogether as a plagiarism.

Apart from Philipp’s practical answer, we can also approach this problem theoretically. Both the $\ce{Li+}$ ion and the $\ce{Li}$ atom have electrons in lower energy levels that can be excited to higher levels given adequate heat and that can re-relaxe under photon emission to their original energy level.
In the case of a $\ce{Li}$ atom, we see an emission at $670~\mathrm{nm}$ corresponding to an energy of $1.85~\mathrm{eV}$. This relaxation corresponds to a $\ce{2s<-2p}$ transition; i.e. two quantum states with the same principal quantum number $n$. We need to add at least this transition’s energy if we want to excite an electron, since it is the lowest possible excitation energy.
In the case of a $\ce{Li+}$ ion, we can at first approximation assume the energy levels of a helium atom. Thus, the lowest excitation should be $\ce{1s->2p}$ ($\ce{1s->2s}$ is forbidden), including a change in the principal quantum number $n$. This corresponds to an energy of $\approx 20~\mathrm{eV}$. That corresponds to a wavelength of $62~\mathrm{nm}$ — well in the ultraviolet range, already approaching x-ray wavelengths. That would mean two things:

Undergraduate labs would likely disallow this experiment due to the very harmful wavelength (the longest-wavelength line of the Lyman series of hydrogen is $121~\mathrm{nm}$, just for comparison).
You would need much more energy to excite the electrons than a simple lab flame would make available. Remember that a bunsen burner flame has a temperature of $\approx 1300~\mathrm{K}$; according to this image on Wikipedia that is not really hot enough to truely emit ultraviolet photons in any reasonable concentration let alone come close to $62~\mathrm{nm}$

By comparing this to the experimental result (‘we can colour the relatively cold flame with lithium salts’), we see that the assumption ‘$\ce{Li+}$ ions are responsible for the flame colouring’ must be rejected.
Note: All calculations and predictions presented herein are based on some kind of layman’s simplification and may be physically inaccurate.


text:
Elution Rate - Alkanes VS Alkenes in Gas Chromatography
Some references say that alkanes elute faster than alkenes, whereas in my textbook the higher volatility of alkenes means they elute faster than alkanes. This is for gas chromatography.

Alkenes are more polar than alkanes due to more electron-withdrawing
  $\text{sp}^2$-hybrid orbitals (more s-character draws electrons closer to
  nucleus).

Can someone please clarify - bit confused.

Chromatography in general is a separation method in which the partition equilibrium of a compound between two phases is permanently disturbed since one of the phases is changing/exchanged (mobile phase).
Apparently, the mobile phase in gas chromatography is the carrier gas.
The stationary phase isn't just a hollow tube but a capillary covered with various differently functionalized materials, that may be rather unpolar (squalane), more polar (polyethylene glycol) or polar and shape-selective (cyclodextrines). Other very common materials are alkyl-substituted polysiloxanes, such as polydimethylsiloxane.
Using the latter stationary phase, Alan Katritzky and coworkers published QSPR: the correlation and quantitative prediction of chemical and physical properties from structure, Chem. Soc. Rev., 1995, 24, 279-287 (DOI, PDF). Here, the retention times for dodecane and 1-dodecene were given as 4.59 and 4.45 min, respectively.
Ladislav Sojak published a freely available 35-page review Separation and identification of isomeric hydrocarbons by capillary gas chromatography and hyphenated spectrometric techniques in Petroleum & Coal, 2004, 4(3), 1-35 (PDF). You might want to have a look at this article on the effect of having non-terminal $\ce{C=C}$ double bonds and cis-trans isomers of longer alkenes, which therefore have significantly different shapes.

ah interesting... so it really depends on the molecule itself then... no general rule of one before the other.

@confused It's apparently a complex matter of interaction of the compounds with the stationary phase. Have a look at Fig. 9 in the article by Sojak. There, the order of retention times for trans-6-pentadecene and cis-5-pentadecene is reversed when a column with a thicker coating of the same stationary phase is used.


text:
How to memorize 7 crystal systems?
Is there any trick to memorize the angles and sides of 7 crystal systems of crystalline solids?
1. Cubic
2. Tetragonal
3. Orthorhombic
4. Monoclinic
5. Hexagonal
6. Rhombohedral or Trigonal
7. Triclinic

The first step is understanding what you have.  The seven crystal systems are derived from decreasing symmetry (lifting degeneracies) of the cubic system by altering lengths and angles.  Example: Cube volume is $\ce{a^3}$.  Parallelepiped volume with all symmetries explicit is
http://en.wikipedia.org/wiki/Parallelepiped

http://www.rockhounds.com/rockshop/xtal/index.shtml
Good reading about crystal systems and their expression.
http://en.wikipedia.org/wiki/Space_group
As usual, nomenclature is a mess.
http://en.wikipedia.org/wiki/Hermann%E2%80%93Mauguin_notation
That's a winner.
Crystallographers often translate the unit cell within the lattice to make its contents  "nice." The rhombohedral system is particularly nasty for choosing the unit cell itself.  65 Sohnke space groups of 230 overall contain chiral formula units.  Racemic screw axes make that interesting, ${2_1}$, ${4_2}$, ${6_3}$.  Within the 65, 11 enantiomorphic space groups - themselves chiral without regard to unit cell contents - allow the same substance to crystallize in two mirror-image space groups.

For bonus points: There is the monoclinic crystal system (two right angles).  Why there is no biclinic crystal system (i.e. a crystal system with one right angle)?

A triclinic system could have one right angle, but that wouldn't make it more symmetric. Right?


text:
Are multiple transition states possible? Criteria to choose the correct one?
I am studying a reaction using electronic structure methods using various software packages (ORCA,GAMESS, G09) and have found two possible transition structures between product and reactant. Both have imaginary frequencies, but each has a different energy value, frequencies and geometries. Some software coincide in the found transition state while other differs, adding to the confusion.
So my questions are:

Is it possible to have multiple transition states, if so what is a good criteria to choose between candidate states?
Should I expect the frequency to be within certain range? (for example if the reaction is the movement of a H atom within the molecule)
Is there some quantitative value(s) (energy, vibration amplitude/frequency, vibration 3d direction vectors, 3d geometry, etc.) that can be used as a reasonable measure to choose the "correct" state? (That is to avoid visualizing the imaginary mode of vibration or computing the IRC).

Is it possible to have multiple transition states connecting the same two points on a potential energy surface (e.g. reaction pathway)?  Sure!  For a hydrogen transfer, you would expect to have a an imaginary frequency with a large magnitude associated with that motion.  Not sure what you mean by question number 3.  Also, have you tried performing an IRC using the transition states you characterized?  They should connect the TS to your reactants and products.

Edited: question 3 to clarify.

WRT to #3: Not that I'm aware of. The only quantitative (and valid) procedure to ensure a TS connects the poitns on a reaction pathway that you are interested in is to do something like an IRC.

You could look at the frequencies and figure out the displacements (in the X, Y, or Z directions) of each particular atom, but that still doesn't really tell you what you're looking for (plus it takes more effort to do this over visualizing the mode in a program).

Can you be more concrete with your question? Different codes use different implementations and can therefore vary slightly. If you see significant difference, can you find the same TS in other program? Geometry convergers are quite tricky. And do you use appropriate level of theory? If you reach the limits of validity, every program can fail in its specific way.

One comment: Many TS search algorithms exist, and sometimes they may lead to different points along the PES. I generally use the Freezing String Method (FSM), and in most cases it has helped me successfully locate the actual (physically more meaningful) TS structure. You might want to read more here: http://scitation.aip.org/content/aip/journal/jcp/135/22/10.1063/1.3664901


text:
Identifying 5 unknown substances
I need help identifying the following 5 substances; G, Y, T, L and J

G is a white solid which is insoluble in water. Consider the following reactions carried out on solid G.
Step 1:
Excess dilute nitric acid is added to a small amount of G in a test-tube. A colourless and odourless gas Y is given off and it is noticed that Y extinguishes a lighted splint.
The solution remaining in the test tube is divided into two different test tubes, sample A and sample B.
Step 2:
A few drops of potassium iodide solution are added to a sample A and a bright yellow precipitate L is observed.
Step 3:
Sodium hydroxide solution is added drop wise to sample B. A dense white precipitate, T, is immediately observed. Precipitate T dissolves when excess sodium hydroxide solution is added to it.
Step 4:
When some G is heated in a hard glass tube, gas Y which is colourless and odourless is given off. The solid J remaining in the hard glass tube changes colour on cooling from red-brown to yellow.

These are my solutions:

G = $\ce{PbCO3}$ because
Y = $\ce{CO2}$ - colourless and odourless gas
L = $\ce{PbI2}$
T = $\ce{Pb(OH)2}$
J = $\ce{PbO2}$

Are they correct or not?

Your solutions sound good so far. But why do you think product I is $\ce{PbO2}$ and not, for example, $\ce{PbO}$? How would you explain the color change during cooling of I?

If you don't have any idea:

Start with the hypothesis that your starting material is the salt of some metal.
Some of the test are apparently useful to identify the cation, others tackle the anion.
Is it likely that the flame-extinguishing gas comes from the cation?
If it is not the cation, it must be the anion from which the gas is formed.
Think in the main property of diluted nitric acid. Is it an oxidant in diluted form, or just something else?

I almost completely agree with the solution given in the question, except for the last step.
The three known lead oxides are

$\ce{PbO2}$: dark brown to black, strong oxidant
$\ce{Pb3O4}$: red-orange, used as a rust-proof primer (in German: Bleimenninge)
$\ce{PbO}$: (normally) yellow

When dark $\ce{PbO2}$ is heated, it splits off oxygen to form red $\ce{Pb3O4}$, when it's heated up more, more oxygen is released and yellow $\ce{PbO}$ is formed. Properly done in a quartz tube the colours of the German flag can be produced this way.
In the puzzle, we are not starting with $\ce{PbO2}$ but with $\ce{PbCO3}$.
When $\ce{PbCO3}$(= G) is heated, $\ce{CO2}$ (= Y) is released. The questions mentions a hard glass tube, which suggests that the tube has to withstand high temperatures.
Let's assume that the reaction simply is $\ce{PbCO3 ->[\Delta] PbO + CO2 ^}$
A possible explanation for the colour change would be that at high temperatures, red $\ce{\alpha-PbO}$ is initially formed, which converts to yellow $\ce{\beta-PbO}$ upon cooling.


text:
How do I prepare a 5N HCl solution?
I am trying to make  5N $\ce{HCl}$ solution. So far I understood that you need to get the molecular weight (36.5) and divide it by the number of $\ce{H+}$ that react in the acid-base reactions. The problem is that my $\ce{HCl}$ stock is 37 %. How should I take this into account?

Hydrogen chloride in its standard state is a gas. We usually do not make solutions of $\ce{HCl}$ by the method of dissolving a known mass of the the solute in enough solvent to make the appropriate volume of the specific concentration since gaseous substances are difficult to weigh (and weight is one of the most direct ways we have to determine mass in the lab).
Hydrochloric acid (aqueous solutions of $\ce{HCl}$) is commercially available in several concentrations, usually $1\ \text{N}$, $0.1\ \text{N}$. and "concentrated".
"Concentrated" simply means that you have in that bottle the highest concentration of $\ce{HCl}$ water that is stable. For $\ce{HCl}$, a concentrated solution is 36-38% by mass. If you know the density of the solution (should be on the bottle), you can convert % by mass into moles of $\ce{HCl}$ per liter of solution (you also need the molar mass of $\ce{HCl}$).
For example, if you had a solution that was 20% $\ce{HCl}$ by mass with a density of $1.098 \ \text{g/mL}$, you can determine molarity by converting mass of $\ce{HCl}$ into moles and mass of solution into liters:
$$20\% \ \ce{HCl} \ \text{by mass}=\frac{20 \text{ g }\ce{HCl}}{100\ \text{g solution}}$$
$$20 \text{ g }\ce{HCl} \times \frac{1 \text{ mol }\ce{HCl}}{36.46 \ \text{g }\ce{HCl}}=0.5485\ \text{mol }\ce{HCl}$$
$$100\ \text{g solution}\times \frac{1\ \text{mL solution}}{1.098\ \text{g solution}}\times \frac{1\text{ L}}{1000\ \text{mL}}=0.09107\ \text{L solution}$$
$$\frac{0.5485\ \text{mol }\ce{HCl}}{0.09107\ \text{L solution}}=6.023 \ \text{M}$$
However, other folks have the done the math for you. The Wikipedia article of hydrochloric acid lists the concentrations (and densities) for several $\ce{HCl}$ solutions as a function of % by mass. This listing of concentrated reagents lists $37.2\% \ \ce{HCl}$.
With a concentration for $37.2\% \ \ce{HCl}$, you can proceed to work through the dilution process using $c_1 v_1 = c_2 v_2$.


text:
What is going on when pen stops working on a section of paper?
Ever had a ballpoint pen that would all of a sudden hit a 'dead spot' on a sheet of paper and refuse to write? It's clearly not the pen, because when moved off to the corner it would draw perfect circles instantly. Move it back to the spot, and it stops working again. Usually I can get something blotchy if I write out the word a dozen or so times while pressing hard.  What's going on with the paper and ink here?
Some pages of my notebook are like this and it's very annoying. My best guess is that it got splashed and something about the paper changed, but my guess probably isn't a very good one since it doesn't have the signature crinkly waviness of wetted paper.

Perhaps the paper is really smooth at that spot to the point where there isn't enough friction to 'turn the ball point'.

If in doubt, try it out.
I just spilled some olive oil on a sheet of paper and wiped it off again.
Result: The ballpoint pen does not write on the oily spot.
Did you fat-finger the paper?

EDIT
The experiment above confirmed the finding in the question but didn't give an explanation.
Actually, two hours later it is still not possible to write on the oil-soaked and more transparent parts of the paper with a ballpoint pen, while a felt-tip pen (here: an Edding, ther German equivalent to a Sharpie) leaves perfect lines on every part of the paper.
Paper is fibrous material with a porous structure - without that and the capillary effects coming with it, fountain pens wouldn't be able to write on it.
For ballpoint pens, this probably isn't so important - the ink is very viscous.
But paper also has a rough surface - and this might be crucial, as already suggested by Lord Stryker in his comment. The rough paper surface provides friction, allows the tungsten carbide ball to roll, take up ink and spread it on the paper.
On a smooth or slippery surface - and this seems to be the case for the oily regions on the paper - the tip of a ballpoint pen only slides but does not roll. As a result, no ink is transfered to the paper.

I suggest another experiment: usually when you write a lot, your fingers and the side of you palm sweats and becomes fatty. Just try it on  a piece of paper with a pen that has such a problem, that you strongly rub your finger against the paper, and then you can't write on it with the pen. The only solution is found is that I put a piece of anouther paper under my hand while writing. Not every pen has such a problem however, there are pens that can quite well write on fatty/oily paper, I wonder what the reason to that might be. Probably the ball is more loose and turns more easily.

The reason is, when you start writing on the top.. it is quite good, it does not happen but when you move at the bottom it starts, right?
It is due to your sweat and oils your body release due to humidity and heat and other reason would be keeping your book open for a long time under the influence of the atmosphere.
what you can do to prevent it, is..
use a piece of paper below your palm and don't leave the book open for a long time.


text:
Which units do I use in calculating heat capacity?
The specific heat capacity of liquid water is $4.18 \, \text{J} \, \text{K}^{-1} \, \text{g}^{-1}$. Calculate the energy required to heat $1.0 \, \text{mol}$ of water from $298 \, \text{K}$ to $363 \, \text{K}$.
My question is: Do I have to change the moles of water into grams or should I sub in $1.0$ into the formula for specific heat capacity?

General tip for any calculation: do the units. Literally write every number out along with the units (respectively placed in the numerator or denominator). Then go through the right side of the equation crossing everything out and make sure you get the units you would expect for your final quantity on the left.

Yes, you would have to change it to grams.
Units must be consistent. Specific heat is defined as the energy required to raise one gram of water by one kelvin. At atmospheric pressure in a closed container, the specific heat of water is 4.184 joules.
One mole of water is 18.016 grams. So:
$E =mc \Delta T $
Where $m$ is the mass of the water, $c$ is the specific heat capacity of water, $\Delta T $ is the change in temperature, and $E$ is the energy required. Substitute in your numbers, and you shall get the right answer:
$E = (18.016\ \mathrm{g})(\frac{4.184\ \mathrm{J}}{\mathrm{g\ K}})(363\ \mathrm{K} - 293\ \mathrm{K})$
$E = 5280\ \mathrm{J}$
If you wanted to use moles directly instead of converting to grams, you would have had to use the molar heat capacity. Generally, people prefer to use specific heat. Engineers sometimes use the volumetric heat capacity as well.

You have the operative value/g.  Have it act upon grams of sample.  Keep your units consistent.
Water boils at 373 K, so one presumes it is a reasonably closed container - or evaporation will skew the real world number, leveraged by surface area.  A mole of water is a little more than a tablespoon. Call it about 18 calories/K.


text:
Naming an ensemble of compounds
As part of intern work I am running simulations on a group of compounds, but as it is I have little in the way of O-Chem experience and I wanted to be able to at least get some direction on naming these compounds (maybe some examples).
I think I've got the names of the functional groups down right, and I've recognized a few of the reaction types, but after that what do I do?
What are the other 2 reaction types?
How should I take the names of the products/reactants and the functional group names and stick em together.
What about when the functional group listed is bigger than the main group. (example R8 and ring opening)?

I should have written F1-F8 but R1-R8 represent functional groups if I'm not mistakened.
Here are my best guesses
R1:alkane
R2:alkene
R3:alkyne
R4:amine
R5:alcohol
R6:ketone
R7:amide
R8:benzene

Pericyclic ring opening, cyclobutene to butadiene (Woodward-Hoffmann rules).
Claisen rearrangement, vinyl ether to ketone (re bullvalene).
Diels-Alder addition.
Sn2 displacement; AgF will move that along (not a soft Lewis base solvent).
Alkene to expoxide is not done with water.  Direct catalytic oxidation, base elimination from the cyanohydrin (first adding HOX), peracids.
Primary alcohol to aldehyde is Jones reagent (in water), Collins reagent (in anhydrous pyrdine, and much better. There is no product in the sludge), Swern oxidation (makes you unpopular downwind), fancier stuff.
Markovnikov addition of HCl to an alkene.
The Woodward-Hoffmann rules tell you what conditions are required (thermal vs. photochemical) and the associated stereochemistry (suprafacial vs. antarafacial) re conservation of orbital symmetry.

Consider Foldit, the minimizing protein conformation energy game. The very best players are not chemists. The best drone drivers are gamers not pilots. I'm rooting for the OP discovering something useful.


text:
What is a simple ratio in which atoms combine?
According to Modern Atomic Theory:

The ratio in which the different atoms combine may be fixed and integral but may not always be simple.

The example given is that in sugar molecule the ratio of $\ce{C,H}$ and $\ce{O}$ atoms is $12:22:11$ which is not simple.
Well, what I see here is that 12,22 and 11 are whole numbers, then why is this not a simple ratio?

The ratio of C, H and O atoms of 12:22:11 is likely to be sucrose. Glucose has a simpler ratio of 1:2:1 (even though it has 6 carbons). The usual context for presenting the theory is with inorganic compounds with no more than a 2:3 or 3:5 ratio. With this limitation, they use the words simple ratio since the numbers are small and can easily be found. The ratio of 12:22:11 could be confused with another ratio if your calculations are off by 20%.

It is often not true within inorganic chemistry (isomorphous substitution, fractional valences; e.g., ruby, sapphire).  Discrete organics are overall exact single compositions.  Polymers are not.  Hydrocarbons have simple empirical formulas.  Add heteroatoms and all bets are off.
Hemoglobin is $\ce{C2952H4664O832N812S8Fe4}$. Every high molecular weight Kraton thermoplastic elastomer triblock polymer contains a single atom of silicon at its exact middle.  Plasma-drawn Spectra ultra-high molecular weight gel-crystallized polyethylene, some 20 times by weight stronger than steel cable, is $\ce{CH2}$.
Dicoronylene: $\ce{C48H20}$
Bacitracin: $\ce{C66H103N17O16S}$


text:
1H-NMR and deducing structures

here is the data booklet we must refer too:

Here is the exam paper solution:

I dont understand how they conclude this.

Peak areas and the chemical shifts seem to be different for me. For example there is a peak area of 2 for the chemical shift (ppm) of 2.0. How can this be when the group is CH3-COOR... isn't the peak area 3 if there are 3 hydrogens? Can someone please help me clarify reading this table, and deducing structures. How do link the R groups and when they bold H, does that mean the H can be any number?

I found it quite hard to follow both the question and the answer, so I am going to state a few relevant things and paraphrase the answers of @LDC3 to add some clarity.
Chemical shift (ppm) and peak area (integral)
These are independent quantities. One tells you about the environment (shift) the protons are in, and the other tells you how many protons are in that environment, compared to the number of protons in all the other environments.
That the areas need to be considered as a ratio instead of absolute values is important: if the molecule was symmetric, with an ethyl group either side of a ketone e.g. C₅H₁₀O you would see a peak ratio of 3:2 (two environments), but there would actually be 6 protons in the first environment and 4 in the second. This is a bit like empirical vs molecular formulae.
Some tips for deducing structures
1) Start with 'characteristic' shifts. These are usually the lowest and highest shifts.
Here the 4.1 ppm shift is characteristic of ester protons, i.e. -(C=O)OC H₂R (R can also be a proton too!) where, much like the data sheet above, bold is to tell you which protons are being referred to in a more complicated structure. The shift at 0.9 ppm is characteristic of protons which are attached to a standard saturated alkyl chain. This tells you that those protons aren't very near any oxygen, nitrogen or other electron withdrawing or electron donating groups.
2) Try and explain the other shifts using functional groups that you have identified from your characteristic shifts
The shift at 2.0 ppm is likely to be alkyl chain protons which are next to an electron withdrawing group, which conveniently you have already identified as an ester group by looking at your characteristic shifts! In this case, that means you can put the 0.9 ppm shifted protons at one end of your structure, because there is nowhere else to put them.
3) Check your peak assignment makes sense when you look at your peak areas
The last thing to check is which end of the molecule the 0.9 ppm shifted protons go on. i.e. is the molecule methyl propanoate or ethyl ethanoate? To do this you can look at the peak areas. For this example it is quite easy to check because there are 3 protons at 4.1 ppm, so you know that you have a methyl ester group -(C=O)OC H₃. Our R group from earlier is an H after all! The resulting structure is the one shown in the example: methyl propanoate.

For the chemical shift of 2.0, the person used the example of a methyl group attached to the carboxyl group, he didn't mean to imply that there are 3 hydrogens. In the structure he presented in B, you will see that there are only 2 hydrogens, not 3.
Also, for the chemical shift of 4.1, he used ethyl (which has 2 hydrogens) instead of methyl (3 hydrogens).

So for 2.0 its CH2-CO-OR...? But how can you change the num of H from the reference chart, when it shows CH3-CO-OR... ?
and i dont get 4.1. You said he used ethyl, 2 H, but the peak area is 3, so shouldn't it be 3?
In the answer he says R-COO-CH2-CH3 (How is this 3 Hydrogens?)
In the diagram it is:
R-COO-CH3.
How can you just remove the ch2? where did that go?
dont you need the same Hydrogens as the reference?

@confused I can see why you are confused. The area for the hydrogens is independent of the shift. For different molecules you can have a different area (meaning a different number of hydrogens), but still have the same shift. Just because the table has 3 hydrogens, your molecule is different, so it might have 2 (or even 1 hydrogen).

oh... okay... i see.. ! So when reading the reference table, i am only looking at the position of the H, as in what its connected to?

@confused Yes, that's right.

Okay, one last question. If you look at the reference chart, for 0.9, it says R-CH3
but CH3 is NOT connected to an R group, alkyl, it is connected to CH2-COOCH3... ? uh, i thought we can only deduce chemical shifts when it is clearly connected to only an alkyl group

@confused Groups beyond 1 carbon have very little influence on the chemical shift. Only when the group is held close to the hydrogen, do you see an influence. In benzoic acid, the hydrogens on the ring should not be influenced by the carbonyl group (2 carbons away), but the closest hydrogens have a different shift than the others.