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Is there a way to contain fluorine gas for long term so that it can be visually observed?
Bromine, chlorine and iodine can all be sealed in a glass container for display without the elements reacting with the glass.
But if you try to seal fluorine in glass I believe it will react and fog the surface making it harder to see the gas, right?
What's the best way, if at all possible for long term containment of fluorine for display?

Alumina and several other ceramics are pretty resistant to fluorine gas at room temperatures.  Not sure what optical clarity you can get easily - I doubt you want to try and drill out a sapphire.  You could still seal the top with glass.

Sealing fluorine in regular glass will do more than fog the surface. It will fairly quickly etch the glass and escape.

Related: How is fluorine gas contained and transported, why aren't the metal cylinder and tube in this video attacked?

@uhoh I assume they select a proper material for the pressure vessel that does react with the fluorine in a manner to create a thin layer of metal fluoride that eventually becomes impermeable to the elemental gas. So it self-passivates the surface. The answers I'm getting here don't look to promising for any clear material that won't fog and won't disintegrate over time.

Thanks! Here's a quite exotic solution: https://youtu.be/TQDjILTly3s

This is not something to speculate on...it has been done.
Fluorine gas was observed by Moissan, actually, (i.e. over a hundred years ago and by the man who first isolated the element).  He used a platinum tube with CaF2 windows at the ends.
He actually published a color image--if you have access to a first class academic library or a good file server, may be able to see it there.  But here is an image from Wikipedia Commons (fluorine in the middle):

Although perhaps looking in the journal hard copy would be better...then again, reproduction of observed color is a tricky deal (even nowadays, don't get me started on three wavelength color matching for car paint!).
Journal cite:       H. Moissan, Ann. Chim. Phys. 25 (1892) 125
Also, I'm not clear if the Moissan image was a color photo or a drawing.  Image on wiki is not clear enough to tell.
But in any case, it was a tough experiment then and still would be now.  But if you have a good machine shop and the money...it's possible to do it again.  I would advise a light source to shine through the tube.  But the Pt cost and the seal of the flourite stoppers will be the hardest part.
Safety note:  Do it in a fume hood, please, and with proper safety precautions (limited amounts of gas, manual transfers).  Fluorine gas is no joke.  Don't be a "martyr".

As @JonCuster mentions, some materials are pretty resistant to fluorine gas at room temperature.
But out of curiosity, I checked Theodore Gray's website. He's made an effort to have a "periodic table" table with actual elements. He also has a beautiful book, called (naturally) The Elements.

Almost anything placed in the path of a stream of fluorine gas will spontaneously burst into flame. This includes things like, oh, say glass, steel, and other things not normally thought of as flammable. It is, therefore, fairly difficult to have a sample of it in an element collection. I used to have a statement here that there was no transparent container that could hold fluorine for any length of time, but I was corrected by a man who has figured out how to do it.

There's more here but I'll summarize. Basically, the problem with glass is the $\ce{Si-O-H}$ bonds, which will auto-catalytically react with $\ce{F2}$ to give $\ce{HF}$, and that etches the glass. So with a high quality quartz tube, a lot of annealing, and a fluorocarbon grease to coat the glass and minimize reactivity, it works.
Interestingly, I always had the idea that fluorine gas was a light blue-green. Instead, it's brownish.

Here's a slightly exotic, expensive solution: solid, crystaline $\ce{CaF2}$ tubing will do the job.
How to get the gas inside, then make a glass seal, is a bit of a challenge though. Surfaces can be optically polished ultra-flat, mated, and locally melted. Ultrasonic friction melting perhaps, or flame or IR. This is an expensive but viable option and would need some development.
According to the Periodic Table of Videos Fluorine Gas found in nature (NEWS) in 2012 existence of naturally occurring fluorine was found in "fetid fluorite" a form of fluorite where natural radioactivity has caused free fluorine to form and migrate within the crystal. The remaining Calcium has already fully reacted with fluorine and is therefore protected from attack.
A tube manufactured from polycrystalline $\ce{CaF2}$ might do the job nicely. It is a common optical material used in certain applications. For example, the UV transparency and resistance to UV damage and darkening (solarization) has made lenses of polycrystalline $\ce{CaF2}$  essential in the smallest dimension micro lithography for fabrication of computer chips.

above: Photonics.com back in 2003: In Search of Calcium Fluoride: Manufacturers face new production challenges as they attempt to meet rising need.

above: From optik-photonic.de Calcium Fluoride Crystals Blanks Offer Highest Transmission Rates at 193 nm and Below
Many sources for polycrystalline calcium fluoride or "$\ce{CaF2}$ glass" exist. Corning is one example, and it is possible that magnesium fluoride would work as well:

Corning® Calcium Fluoride (CaF2) - Code 9575

Corning® Magnesium Fluoride (MgF2)

Angewandte Chemie International Edition: Secret of “Fetid Fluorite” Aired: Elemental fluorine F2 detected for the first time in a natural mineral:

Florian Kraus of the TU Munich, as well as Jörn Schmedt auf der Günne and Martin Mangstl at the Ludwig Maximilians University in Munich have now obtained direct proof: Elemental fluorine is the guilty party that causes the unpleasant odor. By using 19F nuclear magnetic resonance spectroscopy (NMR spectroscopy), they were able to show for the first time that elemental fluorine is contained in “antozonite”.
How is this possible for such a reactive gas? The researchers explain that “antozonite” contains a tiny amount of uranium that, together with its radioactive daughter nuclides, constantly releases radiation into the surrounding mineral. This causes fluorite to split into calcium and elemental fluorine, forming the calcium clusters that give “antozonite” its dark purple color. The fluorine is contained in tiny enclaves surrounded by nonreactive fluorite, which shields it from the calcium, allowing it to maintain its elemental form.

I wonder if it would be cheaper to make your own fluoride glass from powdered CaF2 if you already had the appropriate equipment. One thing about fluoride glass is that it seems like the industry is completely centered around optics, which, in addition to meaning it's probably hard to find containers made out of it, means even the ones that use just CaF2 instead of rare-earth, etc, fluorides probably have the price driven up by purity requirements or maybe other requirements for optics that aren't necessary for containing highly fluorinating chemicals like F2.

Nice answer I upvoted earlier today! So all that is needed is a passivated metal cylinder, two optical grade calcium fluoride end windows, some fluorine-proof sealant or grease, and perhaps a mechanical clamping mechanism to securely fasten the windows to the cylinder.

@EdV great! I'm looking forward to hearing about the fluorine-proof sealant or grease!

Haloform 25-5S grease. Here is the spec sheet: https://chemistry.meta.stackexchange.com/a/4757/79678. I used this on the O-rings on my excimer laser's window seals. Pricey stuff, but worth it. I do not know if it could hold up indefinitely, but you could also use pure gold seals between the fluorite windows and the, e.g., nickel, cylinder ends. Smoother and flatter mating surfaces are obviously  better!

Halocarbon, not “Haloform”.

I checked and antozonite, in sealed ampoules, is available on ebay. Very dark color, almost black, and darker than I expected.

In addition to the suggestion of using metal fluorides like CaF2, I want to mention the possibility of using carbon-based fluoropolymers like polytetrafluoroethylene (PTFE), perfluoroalkoxy alkanes (PFA), and fluorinated ethylene propylene (FEP), all of which may be sold under the brand name "Teflon", which has become synonymous with PTFE for many people. These plastics are some of the most commonly used materials for dealing with highly reactive fluorine compounds like HF, UF6, and HSbF6, as well as other reactive compounds. All these plastics are colorless, but PTFE is more just translucent, and generally looks white, while FEP and PFA are the properly transparent ones.
You can see on this chemical resistance chart that PTFE is considered "resistant" to dry F2 up to at least 60°C, as well as everything else they have data for. PFA is apparently similar. FEP is also apparently compatible with F2 up to at least 60°C: https://www.polyfluor.nl/en/chemical-resistance/fep/ That page also mentions "better gas and vapor permeability", though, which sounds somewhat worrying if you want to hold gaseous fluorine in it. However, I think "better" might actually mean lower, since this pdf of general info on FEP films from Teflon.com mentions it having exceptionally low permeability to gases and non-gases.
Some rather vague comparisons of these 3 substances are given here.
These polymers may be cheaper than metal fluorides (though I'm not certain of that), but they are not fully fluorinated, and thus can still react with F2, e.g.:
(CF2)n(s) + nF2(g) → nCF4(g)
As this page by a company selling them puts it, "[t]he extremely potent oxidisers, fluorine (F2) and related compounds (e.g., chlorine trifluoride, ClF3 ), can be handled by PTFE/PFA only with great care and recognition of potential hazards.  Fluorine is absorbed into the resins, and with such intimate contact the mixture becomes sensitive to a source of ignition such as impact." That quote implies that teflon can burn in fluorine (as can normal oxide-based glass). This makes sense because I know it can burn in oxygen, although it does not normally burn in air, because "teflon" fires in the liquid oxygen tanks are believed to have been part of the cause of the Apollo 13 disaster. Also, those chemical resistance sheets I linked to list FEP, PTFE, and PFA all as "not recommended" for use with F2 at 100°C, and that Teflon.com pdf specifically mentions that FEP reacts with "molten alkali metals, fluorine at elevated temperatures, and certain complex halogenated compounds, such as chlorine trifluoride, at elevated temperatures and pressures." When combined with the fact that these fluoropolymers melt and chemically decompose into (somewhat toxic) gases at temperatures fairly typical for plastics, this means they are not very resistant to fire-like temperatures. (Thus, for example, you probably should avoid trying to close F2 in an FEP ampule by melting the tip, though maybe it would work if the F2 was in the form of a liquid far from the part you were melting. I've seen specific references to FEP being good for cryogenic temperatures and not becoming brittle, so it probably is the best for liquid F2.)
(There is a question on this stack-exchange specifically about what can attack PTFE: Is there ANY chemical that can destroy PTFE, or Teflon? .)
Metal fluorides, on the other hand, are much more refractory (though not as refractory as oxides). CaF2, in particular, has one of the highest melting points (my best number for the melting point is 1423°C from my CRC Handbook), though ScF3, LaF3, and CeF3 at least have higher melting points (but much lower boiling points). There is also no way for CaF2 and many other metal fluorides to be further fluorinated or burned in any way. (Some can react to form higher fluorides, which may even be gases, but there's no reason to use those.) One potential problem with CaF2 and many other metal fluorides is that they have non-trivial solubilities in water. The solubility of CaF2 in water is still very low, though, (similar to CaCO3) so I'm guessing it won't dissolve unless you leave it in a large body of water or use it as plumbing for a really long time, let it be rained on for many years, or dump acid on it. (The latter is how HF is commercially produced.) CaF2 and many other metal fluorides can be very transparent if pure (fluorides and fluoride mixtures like ZBLAN are used for fiberoptics, lenses, and windows on lab equipment), and I'm not sure if that's true of fluoropolymers.
SiO2, Al2O3, and some other other non-fluorine compounds are highly transparent, refractory, and insoluble in water, and also quite resistant to oxidation by F2. They can, however, still theoretically be corroded by F2:
SiO2(s) + 2F2(g) → SiF4(g) + O2(g)
ΔG°298K,1atm = -172.55 kcal/mol for glass or -171.13 for quartz (reaction is spontaneous)
2Al2O3(s) + 6F2(g) → 4AlF3(s) + 3O2(g)
ΔG°298K,1atm = -303.0 kcal/mol for corrundum to crystalline AlF3 (reaction is spontaneous)
Note that SiF4(g) is a gas (boiling point somewhere around -86°C or -90.3°C), while AlF3 is a colorless refractory solid (sublimation point 1290~1291°C) that might form a protective layer on the inside of a container, similar the NiF2 layer that forms on the inside of the nickel containers commonly used to store F2 and other highly fluorinating compounds. Thus, I suspect that Al2O3 is a better option than SiO2. It's possible this advantage could also be gained with "aluminosilicate" glasses (made of just Al, Si, and O), and, in any case, these probably work at least as well as fused quartz (pure SiO2 glass).
As I mentioned, melting an ampule containing F2 closed is not a very good idea, unless it's made of metal fluorides (and some metal fluoride glasses do have very low glass-transition temperatures), because both fluoropolymers and oxide-glasses react with F2 at high temperature, and may ignite under such conditions. Sealed ampules are also quite difficult to put things into or take things out of without breaking the seal. Other methods of making air-tight seals I know of tend to involve some kind of rubbery material ("elastomer", at least if a polymer). This may be used as an O-ring underneath some kind of cap or tube attatchment, or could be used to make a self-sealing membrane. I've often seen that silicones are used as these elastomers, especially for self-sealing membranes. There are fluorosilicones that may be quite resistant to F2, but normal silicone should generally be attacked by F2. Silicones all involve Si-O-Si chains, and at least almost always contain Si-C or O-C bonds, and these bonds can be attacked by fluorine. Some websurfing shows that apparently there are quite rubbery fluorocarbon elastomers, too, though. If you look at page 19 of this pdf, you can see that 3 elastomers they consider ("PERLAST®", "FKM (fluorocarbon)", and "FVMQ (fluorosilicone)") are listed as having "good" (but not "excellent") compatibility with fluorine. ("Good" apparently means they swell 10~20% when used with F2, indicating something bad might be happening.) I think the first two are both fluorocarbon-based, with PERLAST® being "FFKM", which is specifically perfluorocarbon-based, like PTFE or FEP, based on the website selling it. In any case, whoever makes the normal nickel or whatever containers F2 is sold in must have already solved this problem. (Actually, the comments on uhoh's answer point towards Halocarbon 25-5s grease as being fluorine-proof and in-fact specifically recommended for use with gaseous F2. I don't know if that's good enough by itself for a long-term seal, though.)

But are there any fluoropolymers that are transparent so that the gas can be observed?

This is quite a nice answer and I upvoted it earlier. i upvoted another of your answers as well, so thanks!

I think PTFE and PFA are both transparent, but they usually look white or at least a little bit cloudy whenever I've seen them. Looking up "transparent Teflon", I now see that fluorinated ethylene propylene (FEP) is apparently the usual choice for a highly transparent fluoropolymer. I'll add it to the answer.

Actually, it looks like both FEP and PFA are transparent, but PTFE is only white. https://adtech.co.uk/about/news/ptfe-fep-pfa-difference I have a tiny amount of reason to think FEP might be more inert to fluorine than PFA, though.

Actually that page makes it sound like PFA may be more resistant, at least in practice. In any case, they're all resistant, and FEP is still probably the best because it looks like it may be the best for cryogenic temperatures, and you may want to use liquid F2 at some point if you're handling F2. FEP seems to have worse thermal resistance than PFA or PTFE, though.

By "worse thermal resistance" I mean not as good for as high temperatures.

I think the tube they make the liquid F2 in in the Periodic Videos video must be made of a fluoropolymer, probably FEP, though maybe PFA or something else.

From some more websurfing and looking at images, I'm starting to become convinced that exactly how transparent these plastics are depends on more than just what material it is made of, except that purity is probably a big factor.

I think ultra-pure PFA or FEP is ultra-clear (or rather, just quite clear).

Here's a very clear FEP bottle: https://www.ahf.de/en/products/ultra-trace-analysis/equipment-ultra-trace-analysis/liquid-dispensing/seastar-dispenser/3987/baseline-bottle-fep-500ml I'm still wondering about how you would seal a gaseous sample in FEP or PFA, though.

Here's a very clear FEP "gas sampling bag" https://www.samplebags.eu/product/teflon-gas-sampling-bag-fluropolymer-fep/ It looks like the nozze/seal/whatever is PTFE-based. Actually "the fitting is made of polytetrafluoroethylene (PTFE)" but "The sample cap has a silicone septum that can be used for needle sampling". Th self-sealing mechanism is silicone-based, and the F2 would likely attack the silicone. I thought something like that might happen. There are such things as perfluorosilicones, though, so you probably could make one that could hold F2

I have sealed fluorine into glass ampoules routinely when I used it as an initiator for the polymerization of tetrafluoroethylene.
The borosilicate glass ampoules should be annealed before use and kept in an oven at 100°C
One must just know how to measure what is "condensed" in the glass ampoule.  Fluorine never liquefies or solidifies when cooled in liquid nitrogen. As long as the pressure in the glass tube is sufficiently low the glass will collapse and seal.
A friend knew u could do this.  He asked me to try fluorinating graphene in perfluoroheptane. We tried this and it worked. This led to a publication in the journal Carbon.

I upvoted this because I approve of you citing personal experience, which I, for example, and probably multiple of the other answerers, don't have, and also because borosilicate glas is an interesting new data point, though I feel a little doubtful about it, because the question did specifically ask about long-term storage, and I'm not sure if your answer addresses that or not.


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How much heat is needed to break CO2 down to CO?
I have found that because the oxygen is double bonded to carbon, it takes about $\mathrm{1500\ ^\circ C}$ to break it down from $\ce{CO2}$ to $\ce{C}$ and $\ce{O}$. But, I am curious, would it take less heat energy to break $\ce{CO2}$ down to $\ce{CO}$ and $\ce{O}$?

Heating CO2 at atmospheric pressure:
At 1000K it is still essentially all CO2.
At 2000K: 98% CO2, 1.4% CO, 0.7% O2
At 3000K: 44% CO2, 36% CO, 16% O2, 4% O
At 5000K: 50% CO, 50% O
only at even higher temperature does significant atomic C appear (9% at 6000K).
See Thermal decomposition of carbon dioxide in an argon plasma jet for more information.

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Buffer Solution - Sodium Benzoate

A buffer with a pH of $\mathrm{4.79}$ contains $\mathrm{ 0.39\ M}$ of sodium benzoate and $\mathrm{0.10\ M}$ of benzoic acid. What is the concentration of $\mathrm{[\ce{H} ]}$ in the solution after the addition of $\mathrm{0.058\ mol}$ of $\ce{HCl}$ to a final volume of $\mathrm{1.6\ L}$?

My thought is that
$$\mathrm{pH=pK_a+Log_{10}(\frac{[base]}{[acid]})}$$
When we added $\mathrm{0.058\ mol}$ of $\ce{HCl}$, the concentration of both acid and base decreased by $\frac{.058}{1.6}$, so I take that ratio and apply it.  I found $\mathrm{pK_a}$ by just applying what's given.  Not sure where I am going wrong.

In that equation, the acid and base are a conjugate acid/base pair.  We assume HCl quantitatively dissociates and protonates benzoates.  The concentration of the acid increases, and the base decreases.
Empirically, the pH should go down in this case.  Without seeing your specific calculation, we can't comment where you are going wrong, but the proper flow should look something like:
$$\mathrm{pH=pK_a+Log_{10}(\frac{[base]-[Added H]}{[acid]+[AddedH]})}$$
$$\mathrm{pH=pK_a+Log_{10}(\frac{0.39M - 0.036M}{0.10M + 0.036M})}$$
$$\mathrm{pH=pK_a+Log_{10}(\frac{0.354M}{0.136M})}$$
From your calculated $\mathrm{pKa = 4.20}$, you should get there pretty quickly.
Just applying the difference in the log values of the changed ratios, I get  $\mathrm{pH = 4.61}$

The concentration of [H] ion will be $2.45x10^{-5}$ if the pH was 4.61.

Generally it is favored to show how you arrive to the specific answer, so that we can assist in the understanding of how to solve the problem.

pH=-log[H+], then [H+]=antilog(-pH)   [H+]=antilog(-4.61)=2.45x10^-5

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How to calculate the temperature from heat of neutralisation?

The heat of neutralization of $\ce{HCl(aq)}$ by $\ce{NaOH}$ is $\pu{-55.9 kJ/mol}$ $\ce{H2O}$ produced. If $\pu{50ml}$ of $\pu{1.6M}$ $\ce{NaOH}$ at $\pu{25.15^\circ C}$ is added to $\pu{25ml}$ of $\pu{1.79M}$ $\ce{HCl}$ at $\pu{26.34^\circ C}$ in a plastic foam cup calorimeter, what will be solution temperature be immediately after the neutralization reaction has occurred?

I'm always overwhelmed by the question if it provided me too much information. I don't know what equation and concept to use first. (But I know all related-equations in this unit.)
I need a detailed explanation to understand! I'm not sure which one is the initial temperature? $\pu{25.15^\circ C}$ of $\ce{NaOH}$ or $\pu{26.34^\circ C}~\ce{HCl}$?

First, you need to recognize what the limiting reagent is.  Because NaOH and HCl react in a 1:1 ratio, and there is less HCl ($\pu{25mL}$ of $\pu{1.79M}$ is fewer moles than $\pu{50 mL}$ of $\pu{1.6M}$), HCl is the limiting reagent.
Second, given the amount of limiting reagent, how much heat will be released (or absorbed).  $\pu{0.025L} \times \pu{1.79M} \times \pu{55.9 kJ/mol}$
Third, you need to approximate that the solution has the heat capacity of water, which is $\pu{4.18 kJ/K L}$.
If you mix two volumes of the same substance at different temperatures, the temperature of combined volumes will be approximately the volume-weighted average.  So here: $$(50 \times 25.15 + 25 \times 26.35)/75 = 25.55$$
So finally calculate the amount the temperature of $\pu{75 ml}$ of "water" will increase when $\pu{0.025L} \times \pu{1.79M} \times \pu{55.9 kJ/mol}$ of heat is added, and add this to $\pu{25.55 ^\circ C}$.

If you mix two volumes of the same substance at different temperatures, the temperature of combined volumes will be approximately the volume-weighted average. As long as there is no phase changes.

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Difference between baking soda and baking powder
I learnt that the chemical symbol for baking soda and baking powder is NaHCO3. However, when I do a simple experiment which is put vinegar into both substances. I can get different observations.
When I put vinegar into baking soda, I get a fast and fuzzy reaction. But when I put vinegar into baking powder, instead of a fast and fuzzy reaction, I get a slow and bubbly reaction.
So, my question is what properties do each of the both substances have to make different reactions with vinegar although their chemical symbols are the same?

As LDC3 says, Baking soda is simply sodium bicarbonate, $\ce{NaHCO3}$. When one adds an acid, the following reaction occurs: $\ce{HCO3- + HA -> H2CO3 + A-}$ and the carbonic acid releases carbon dioxide: $\ce{H2CO3 -> CO2 + H2O}$ (these are both equilibrium reactions, but as you can observe when you add vinegar, a lot of gas is produced and the gas is generally lost, pushing the equilibrium towards making carbon dioxide).
Baking powder is a mixture of sodium bicarbonate and one or more weak acids. When dry, no reactions occur, but when wet, the components dissolve and can react. The observed differences when mixing vinegar with baking powder probably result from the fact that there is less sodium bicarbonate in a given volume of baking powder than baking soda, and baking powder often contains starch, which can influence the properties of the added liquid, giving the bubbles a different appearance.

Baking soda and baking powder are not the same.
Baking soda is (almost 100%) sodium bicarbonate ($\ce {NaHCO3}$).
Baking powder is a mixture of sodium bicarbonate, and some acids.
From Wikipedia: Typical formulations (by weight) call for 30% sodium bicarbonate, 5-12% monocalcium phosphate, and 21-26% sodium aluminium sulfate.
When you mix baking powder with water (in a dough), it will cause bubbles to form. Nothing will happen with baking soda.


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SN1 Reaction Involving Carboxylic Acids
Which oxygen atom is more nucleophilic in carboxylic acids?
If we look at this page, it says the carbonyl oxygen is more nucleophilic:
However, I also learned in class that $\mathrm{sp^3}$ atoms are more reactive than $\mathrm{sp^2}$ atoms.
So, in an $\mathrm{S_n1}$ reaction where the solvent is acetic acid, which oxygen attacks?

"However, I also learned in class that $\mathrm{sp^3}$ atoms are more reactive than $\mathrm{sp^2}$ atoms.", that might apply (if so) for $C$ atoms, not for others. Anyway, I would avoid this kind of reasoning. Using the concept of hybridization is not appropriate when it comes to try to justify or predict reactivity.

The carbonyl oxygen is more nucleophilic. You can draw a resonance structure of the carboxylic acid that places a formal negative charge on that oxygen which rationalizes its greater nucleophilicity. Also, after it adds to the carbocation, the resulting intermediate will be resonance stabilized. If the other oxygen attacked, no resonance structures would be possible.


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Yttrium -- rare earth or transition metal?
The element yttrium is called a rare earth element, yet periodic tables label it as a transition metal.
Why is that?
Thanks

Do you perhaps mean to ask why Y is counted amongst the rare earths even though it isn't an f-element?

Uh, why not both?

Comparing "transition elements" to "rare earth elements" (REE) is a bit like comparing apples and oranges. It all boils down to the different ways in which the elements can be classified.
Let's take a look at the periodic table:

(modified from this)
The rare earth elements are marked with a red outline.
Why are all rare earth elements are technically transition elements?
Strictly speaking, transition elements are the elements in groups 3 to 12 of the periodic table. The rare earth elements, in their widest sense, are Sc, Y and the lanthanides, La to Lu. You can see that if you group the lanthanides as one element, they fit nicely below Y in group 3 of the periodic table. The lanthanides are occasionally called the "inner transition elements".
Why is yttrium a rare earth element?
There are chemical and historical to the inclusion of yttrium in the rare earth elements.
Historically, yttrium was among the first elements to be isolated from the REE-bearing minerals of the Ytterby mine in Sweden. What was though as first to be "yttria" was later shown to contain also the rest of the heavy REE. It just took a longer time to purify them.
Chemically, yttrium commonly occurs with the other REE in the same mineral deposits around the world. You probably know the the chemical properties of the lanthanide series vary continuously from La to Lu. Yttrium also follows that pattern, and in terms of chemical properties it sits very nicely somewhere between Dy and Ho (that's why it's associated in mineral deposits with the heavy REE).
Yttrium is commonly used in the same high-tech applications as the other REE: it is commonly alloyed with other REE for speciality materials. It is also commonly used in combination with the other REE in lasers and various detectors in electron microscopy and analytical chemistry.
A different look
If you take a look at this figure, taken from here: https://earthscience.stackexchange.com/a/2951/725

You can see that most transition elements, as they appear in nature, usually have a cationic charge between 60 and 80 pm. The REE (including Y) are somewhat larger than that, resulting in slightly different properties.
You may also be interested in reading the answers to the following questions, also here on Stack Exchange:
What are rare earths and why do they cluster near alkaline magmatism?
Why are the rare earths erbium (Er), terbium (Tb), yttrium (Y) and ytterbium (Yb) named like that?


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Finding pH of a buffer solution on addition of hydrogen iodide to it
Question:

A buffer with $\mathrm{pH} = 4.88$ was prepared by dissolving $\pu{0.10 mol}$ of benzoic acid ($K_\mathrm{a} = \pu{6.3E-5}$) and $\pu{0.50 mol}$ of sodium benzoate in sufficient pure water to form $\pu{1.00 L}$ solution. To a $\pu{70.0  mL}$ aliquot of this solution was added $\pu{2.00 mL}$ of $\pu{2.00 M}$ $\ce{HI}$ solution. What was the $\mathrm{pH}$ of the new $\pu{72.0 mL}$ solution?

My solution:
Using the Henderson–Hasselbalch equation
$$4.88 = -\log(6.3 \cdot 10^{-5}) + \log \left( \frac{0.10-x}{x} \right),$$
$$x = \pu{0.0173 mol}$$
Amount of conjugate base in $\pu{70.0 mL}$ is
$$0.0173 \times 70/1000 = \pu{1.211e-3 mol}$$
Amount of $\ce{HI}$ added is $\pu{4e-3 mol}$.
I'm stuck here having more $\ce{H+}$ than my base.

Amount of acid added:
$$0.002\ \mathrm L \times 0.2\ \mathrm{mol/L} = 0.004\ \mathrm{mol}$$
This will neutralize $0.004\ \mathrm{mol}$ of the benzoate: $$0.50\ \mathrm{mol} - 0.004\ \mathrm{mol} = 0.496\ \mathrm{mol}$$ of benzoate
This will increase benzoic acid by $0.004\ \mathrm{mol}$: $$0.10\ \mathrm{mol} + 0.004\ \mathrm{mol} = 0.104\ \mathrm{mol}$$ of benzoic acid
$$[\ce{H+}] = K_\mathrm a\left(\frac{n(\ce{HA})}{n(\ce{A-})}\right)$$ (yes, it's supposed to be $[\ce{HA}]/[\ce{A-}]$, but volume cancels out leaving amount of substance so long as your numbers leave you inside the titration curve between the starting $\mathrm{pH}$ and the equivalence point $\mathrm{pH}$)
$$[\ce{H+}] = 6.3 \times 10^{-5} \left(\frac{0.104\ \mathrm{mol}}{0.496\ \mathrm{mol}}\right) = 1.32 \times 10^{-5}$$
$$-\log\left(1.32 \times 10^{-5}\right) = \mathrm{pH} = 4.88$$
(actually $4.8794$, so you have a small change in $\mathrm{pH}$, as you would expect with a buffer with so much of the conjugate base in solution, but not enough to make a significant difference.)
I'm not at all clear why people are married to the Henderson–Hasselbalch equation for buffer problems. The equilibrium expression covers just about any situation I can think of, there's no formula to remember if you know how to do an equilibrium expression.


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Why the salts in a salt bridge?
Take the example of a copper and zinc galvanic cell, connected by a salt bridge of $\ce{KNO3}$. I understand how the reactions will result in positive and negative charges, and that the ions of the salt will migrate to maintain neutrality. But why is the salt needed in the first place?
In one beaker we will have an excess of zinc ions (resulting in positive charge) and the other with an excess of copper ions (resulting in negative charge). If we just have a bridge, perhaps just soaked in water, wouldn't the ions, if not purposefully then randomly, travel across the bridge and maintain neutrality? What is the purpose of the salt then, if it works without a salt?

The purpose of the salt bridge is to prevent the two half-cell solutions from mixing. It is possible to make a really bad galvanic cell by putting both half-cells in a single solution, but they rapidly self discharge as the oxidizing agent ($\ce{Cu^+}$ in your example) can diffuse through the solution and react directly with the other electrode (Zn in the example), with the electrons being transferred directly instead of going through the external circuit.
Having a "bridge" of just water soaked paper is similar to having both half-cells in the same container but poses some issues. First, the diffusion of ions through the bridge is going to be very slow, increasing the cell resistance, second, the self-discharge problem above will occur, and third, because different ions diffuse at different rates, a junction potential develops as the two containers diffuse into one another.
So a salt bridge's job is to not only maintain electroneutrality between the two containers, but to do so without allowing the solutions to mix. An inert salt is used to prevent any reactions from occurring between the salt and the solutions, and things like $\ce{KCl}$ are commonly used for analytical cells as potassium and chloride ions have very similar diffusion coefficients, minimizing junction potential.

But even with a salt bridge don't the two solutions mix anyway?(at a small rate) does the salt impede this mixing?

Sure, if you leave it long enough. Often a salt solution is turned into an agarose gel to fill the bridge or glass frits placed at either end of the tube to limit the rate of diffusion.

Why do the ions in the salt bridge have to diffuse into the half cell? Could we plug up the bridge completely and just let the ions electrically neutralize the cell by moving into the cell volume while staying inside the salt bridge tube without diffusing into the cell solution?

This is an interesting question with two possible answers that I can see. If you are talking in terms of a galvanic cell used as a source of electrical energy, the current supplied by the cell has to be matched by the current flowing between the two half cells. As pure water is very nearly a perfect insulator due to its extremely low degree of ionisation, the internal resistance of the cell would make it impossible to supply any useful quantity of electical power. The use of an ionic solution overcomes this problem and the higher the concentration of ions the better, hence saturated solutions. However, if your question is referring to cells for measuring electrode potentials, I am not so sure of the answer. In principle pure water should do, as you suggested because its slight degree of ionisation means that it does conduct electricity very slightly and this should be all that is needed since electrod potentials should be measured when no current is flowing. I am only sorry that I did not think of this question when I had access to the apparatus needed to try it out.

In fact, it is possible if only water and some gelatin or agar-agar is present in the salt bridge. But for $\ce{SO4^{2-}}$ Or $\ce{Zn^{2+}}$ to travel through salt bridge it will have to face a lot of resistance resulting in low voltage. Instead, it is better to have strong electrolytic salt which can maintain neutrality of solution as soon as zinc atom is oxidized or $\ce{Cu}$ atom is reduced.


text:
Reason for inability to find the absolute value of internal energy
Today in my thermodynamics lecture, my teacher told me that it is not possible to find the absolute value of internal energy and so we have to calculate the change in internal energy.
So, my question is why isn't it possible to find the absolute value of internal energy? Is this statement true?

How would you find the internal energy of hydrogen gas? Even when it is cooled way down (to absolute zero), it still contains energy (the atoms never stop vibrating). There is no known way to determine the internal energy.

The internal energy of an ideal gas is the ensemble average determined by statistical mechanics and is $U=(3/2)NkT$, or in general $U=(N/Z)\Sigma_j\exp(-\epsilon_j/(kT))$ where Z is the partition function, k Boltzmann const, $\epsilon_j$ energy of level j, and T temperature.

In order to answer this question, one needs to define what the absolute energy of a system is. Energy can be trapped in a system in ways not yet discovered or fully understood. Think of the energy associated with mass ($E=mc^2$), which is a result that was not known to the early founders of thermodynamics.
We need to define reference points, which we can use to compare our products with. Usually this "zero" point is defined as the energy of the individual atoms in their most energetically favorable state (phosphorus is an exception, since red phosphorus is much more available), such as $\ce{H2}$ and $\ce{He}$.

I also heard that and it's not exactly true, you could find total value using mentioned earlier $E=mc^2$. It shouldn't be called absolute, though, as mass-energy depends on frame of reference - masses measured between inertial and non-inertial frames of reference are different. Also, as pointed in comments, you can't account for zero point energy of vacuum this way, although I wouldn't call it a problem, as this energy is the same when the object of measurement is absent.
Still, if you weighed something and multiplied this mass by $c^2$ you would get its total energy. It would be a very rough estimate, though, as one gram is equivalent to about 90 terajoules of energy, which is really large amount. Even if you would measure a very small mass with great precision, it would still be a rough estimate.

I don't think using mass-energy equivalence lets you decide on an absolute scale for energy.  Reference states still matter.  There are differences between the internal energy of a proton and electron infinitely separated at rest, the internal energy of a ground-state hydrogen atom at rest, and the internal energy of a ground-state hydrogen atom moving at the near the speed of light.

@CurtF. You're talking about frame of reference? That is indeed important, I'll edit.

"Frame of reference" not only including velocity but also geometry.  If you want to find the "absolute" energy of the hydrogen molecule $\cf{H2}$, relative to what is still a valid question.  The reference state could be infinitely separate ground state hydrogen atoms, or it could be infinitely separated pair of protons and also infinitely separated pair of electrons.  Or it could be an infinitely separated set of six quarks, ... etc.

@CurtF. Technically, "frame of reference" refers to your spacetime coordinate system, not the geometry of a particular object. Moreover, to the extent that general relativity can be ignored, the absolute internal energy measured at a given reference frame is indeed well-defined. Here, the absolute reference state of zero energy is the vacuum, and, for example, an electron with zero momentum has an absolute energy of $m_e c^2 = 511\mathrm{keV}$.

The absolute energy of a $H_2$ molecule of zero momentum is given by $E_{\mathrm{H}_2}=2m_e c^2 + 2 m_p c^2−2×(\mbox{binding energy of an H atom})−(\mbox{H-H bonding energy})$. This value divided by $c^2$ is the mass of this molecule, but notice that among the four terms, the last two is negligibly small compared to the first two. On the other hand, only the last two terms are of interest in chemistry.

"Absolute" relative to what?  An infinite separation of the quarks that comprise the two protons?

@CurtF. Relative to the vacuum state. Quark confinement means that it takes infinite energy to free the quarks from, e.g., a proton, while a proton has a finite energy ($m_p c^2 = 938 \mathrm{MeV}$) relative to the vacuum.

I think we disagree about the meaning of "absolute".  "Relative to the vacuum state" still sounds like you are defining energies relative to a reference state to me, but I can see how others might disagree.

@CurtF. As you point out, there should always be a reference state. But nothing can have a lower energy than vacuum, and "the energy relative to the vacuum state" has an absolute sense.

higgsss thanks for this thread...I learned a lot.  As a closing bit of pedanticism, let me just ask...shouldn't there be a gravitational term in there?  :-)

@CurtF. I don't know much about gravity, but it should be safe to ignore general relativity effect  in chemistry :P  From what I heard, (1) there is in general no globally conserved quantity (e.g., energy) in a curved spacetime. (2) if we define quantum theory in a curved spacetime background, the vacuum state is not globally defined (this leads to Hawking radiation).

http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html

In classical thermodynamics the change in internal energy is defined by the first law as
$$\Delta U = q + w$$
so that only the difference in $U$ is known; $q$ is the heat absorbed by the 'system' and $w$ the work done on the system.
For example in a closed system (no exchange of matter with environment) we can write for a reversible change
\begin{align}
 \mathrm{d}q &= T\mathrm{d}S \\
 \mathrm{d}w &= -p\mathrm{d}V
\end{align}
and then if the only form of work on a gas is volume change
$$ \mathrm{d}U = T\mathrm{d}S - p\mathrm{d}V$$
and this is the fundamental equation for a closed system. Thus only difference in internal energy are measurable from thermodynamics, and this follows from the first law. (Even if you integrate this equation from say state $a$ to $b$ the result will be $U(b)-U(a)=\ddot {}$ in other words $\Delta U$.)
Thermodynamics was developed before the nature of matter was known, i.e. it does not depends on matter being formed of atoms and molecules. However, if we use additional knowledge about the nature of molecules then the internal energy (and entropy) can be determined from statistical mechanics.
The internal energy ($U$ not $\Delta U$) of a perfect monoatomic gas is the ensemble average  and is
$$U=(3/2)NkT$$
or in general $U=(N/Z)\Sigma_ j\exp(−\epsilon_j/(kT))$ where $Z$ is the partition function, $k$ Boltzmann constant, $\epsilon_j$ energy of level $j$, and $T$ temperature and $N$ Avogadro's number. The absolute value of the entropy $S$ (for a perfect monoatomic gas) can also be determined and is given by the Sakur-Tetrode equation.

So what  is it? you say you can not, then say you can!  Instead of integrating from a to b; integrate from 0 to a and 0 to b and subtract.

The statement that it is impossible to calculate the absolute energy at an ambient T seems to be intuitively incorrect, difficult possibly, but not impossible. Since U is a state function the value of U for a given set of conditions is invariable and can be set as an arbitrary zero and changes in U calculated. So all one must do is work in reverse to absolute zero. This is done for entropies using the 3rd Law. The general approach in thermodynamics is that it is entirely accurate to work with the changes in state functions and there is NO NEED to calculate absolute internal energies from absolute zero.

Will the downvoter please give a reason for the downvote!

At absolute zero, internal energy would be almost the same as at any other (not extremely) high temp.

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


text:
Difference between internal energy of combustion and enthalpy of combustion?
I am having real trouble understanding the first law of thermodynamics and I think the question, below, highlights most of my issues:
1) (a) A bomb calorimeter was calibrated by igniting a $0.825\ \mathrm{g}$ sample of benzoic acid in the presence of excess oxygen. The heat given out made the temperature of the calorimeter rise by $1.940\ \mathrm{K}$ from $298\ \mathrm{K}$. The internal energy of combustion of benzoic acid is $-3251\ \mathrm{kJ\ mol}^{-1}$. Calculate the heat capacity of the calorimeter.
(b) In two separate experiments in the same apparatus, $0.498\ \mathrm{g}$ of fumaric and $0.509\ \mathrm{g}$ of maleic acid (both of RMM 116) were ignited and gave temperature rises of $0.507\ \mathrm{K}$ and $0.528\ \mathrm{K}$ respectively. Use the heat capacity of the calorimeter, calculated in (a), to calculate (i) the molar internal energy of combustion, (ii) the molar enthalpy of combustion, and (iii) the molar enthalpy of formation of fumaric and maleic acids. Comment on the difference. The standard enthalpy of formation of water is $285.8\ \mathrm{kJ\ mol}^{-1}$ and of $\ce{CO2}$: $-393.5\ \mathrm{kJ\ mol}^{-1}$.
Part (a) was fine - the heat capacity at a constant pressure = change in internal energy/temperature change. Then divide by number of moles.
Part (b) is where I struggle conceptually. I got the answer correct but only by following lecture notes. Sticking with just fumaric acid because there's no point talking about both. It's easy to find the molar change in internal energy using the constant volume heat capacity ($-1338\ \mathrm{kJ\ mol}^{-1}$).
However, to find the molar enthalpy change, the change in $H = U + pV$. but since $\ce{pV=nRT; H=U+nRT}$ where n depends on the change in the number of gaseous moles from products to reactants - in this case +1. So $H = U + RT$ giving $-1336\ \mathrm{kJ\ mol}^{-1}$ (less exothermic).
Firstly: Why are the values negative. If the system is the fumaric acid/products of reaction then surely there is an increase in internal energy because the particles will be moving faster after the bonds have broken. (Am I right in thinking that the system is the chemical)?
Secondly: Why is the enthalpy value less exothermic? (I have a very poor grasp for what enthalpy is especially in a situation like this).
Any help with answering my questions or helping me with this concept would be greatly appreciated. Also, do you know any good books/videos I could watch to help me undertand?

Combustion releases energy from the system usually as heat and/or work to the surroundings, that's why the standard energy of combustion has a negative sign. The bomb is the system in this case (alternatively you can see the chemical as the system, but you will need to include the gaseous products in your system too, changing the volume of the system, which I find a less preferable viewpoint).
The value for the enthalpy will be lower since energy is trapped in the form of increased pressure introduced by the reactions products (mainly by $\ce{CO_2}$) in the bomb. This pressure can be used to do work on the surroundings (in an open system this will be usually on the atmosphere). If you determine the final pressure, you can calculate your $\Delta U$ back ($\Delta U = \Delta H - (p_{end}-p_{atm})V_{bomb}$).
What book do you use for the lectures?

Physical Chemistry by Peter Atkins. He used to be a lecturer here so they strongly advise using it.

@user3764899 That book is fine, although I find the newer versions sometimes so focused on "pushing the material" that it becomes confusing (all the different colors and frames). Did my answer help you?

Yes, the answer was great. However, what do you mean by "you will have your delta U back"? Also, surely the bomb is the surroundings because as the chemical gives off energy, surely the internal energy of the bomb would increase not decrease?

Also, do you know a better book? Perhaps even a good alternative? We have really good libraries here so I will definitely be able to find one.

@user3764899 See edit. I believe the bomb is only the internal chamber of the bomb calorimeter, which has the diathermic wall (see e.g. http://www.precisionnutrition.com/wordpress/wp-content/uploads/2010/02/bomb-calorimeter.jpg). The bomb is usually made of a material that can withstand high pressure. Atkins is perfectly fine! It is regarded as an authoritative book within its field.


text:
Potassium permanganate (VII) + glucose + sodium hydroxide =?
I saw this experiment many times but still can't explain why.
The experiment is first prepare two solutions. The first one is to dissolve potassium permanganate crystal into distilled water to form potassium permanganate (VII) while the second one is to put sodium hydroxide and glucose (sugar) into distilled water.
Secondly, I pour the second solution into the first one. My observation is the purple solution will slowly turn into blue, then slowly turn into green and lastly turn into yellowish orange.
So, what really happen in the solution?

This is a redox reaction in which the permanganate ion is reduced and the glucose is oxidised. Potassium permanganate is usually used in acid solution and under these conditions is a very powerful oxidising agent in which the manganese is reduced from the oxidation state of +7 to +2 with a colour change from purple to colourless (actually extremely pale pink). In alkaline solution manganese is only reduced from +7 to +6 changing colour from purple to green. (The blue colour you mention is due to a mixture of purple and green). The green manganate(VI) ion is unstable and slowly disproportionates to manganate(VII) (purple) and manganese(IV) oxide which is brown. The manganate(VII) goes on to react as before until the brown manganese(IV) oxide is all that remains. In alkaline solution this tends to form a colloidal suspension which (if fairly dilute) can appear orange. The oxidation of the glucose forms no coloured productss.

Alternatively the blue colour might be Manganate(V) ( https://en.wikipedia.org/wiki/Manganate#Manganate(V),_manganate(IV),and%22manganate(III)%22 ), which disproportionates to Maganate(VI) and MnO2, and then as in the answer


text:
What would be the reaction if melting iron is put in normal water?
What would be the reaction if melting iron is put in normal water? Will water be chemically changed?

Unlike the other metals we usually see in the reactivity series, such as calcium, magnesium, zinc, etc., hot iron does not form iron hydroxide, instead it forms iron oxide. The reaction is:
$\ce{3Fe + 4H2O <=> FeO.Fe2O3 + 4H2}$
Iron is similar to magnesium and zinc as they react only with hot water (not cold water). But the reaction of iron with hot water is less vigorous than that of magnesium and zinc.


text:
Why are double bonds at more substituted carbon more stable?
That $\ce{C=C}$ bonds are more stable at more substituted carbon  atoms is a common argument used throughout Clayden et al., but is never actually explained (at least not in the first 21 chapters). They do explain however, why the HOMO is increased in energy for more substituted double bonds, through the argument of electron donation by means of hyperconjugation of neighboring groups, increasing the electrophilicity of the $\ce{C=C}$ bond.
I find this somewhat counterintuitive. Usually when something is more reactive, it is less stable (I understand that of course by kinetic pathways this does not always hold, e.g. with $\ce{I-}$ as nucleophile).
How can both observations be incorporated in a consistent explanation in this case?

Let's start by providing some useful background information that will help answer your question.
1,3-Butadiene is more stable than 2 separate ethylene molecules due to the extended conjugation possible in the diene.  We can illustrate this by drawing simple resonance structures for the diene that can't be drawn for ethylene.

Huckel calculations show the same thing, 1,3-butadiene is more stable than 2 individual ethylenes.  Here is the Huckel MO diagram for ethylene.  The two electrons in ethylene have a net stabilization of 2$\beta$ compared to an electron in an isolated p-orbital.  4 electrons in 2 ethylenes would have an overall stabilization energy of 4$\beta$

Here is the Huckel diagram for 1,3-butadiene.  The 4 electrons have a net stabilization of [(2 x 0.62) + (2 x 1.62)] = 4.48$\beta$.

Even though the HOMO in 1,3-butadiene is higher in energy then the HOMO for ethylene, overall 1,3-butadiene is more stable than 2 isolated ethylenes by 0.48$\beta$.  This is because the MO below the HOMO is stabilized much more (0.62 $\beta$ compared to ethylene).
Now let's return to your question.  Hyperconjugation is a reasonable way to explain why more highly substituted double bonds are more stable. Just like in 1,3-butadiene, when we delocalize electrons over a larger number of atoms, which is what hyperconjugation does, an overall stabilization of the molecule should result.  And, as the hyperconjugated resonance structure shows, electron density will shift from the alkyl group to the double bond.

Even though the HOMO may be raised in energy, (at least some of) the other occupied MO's below the HOMO will be lowered in energy (just like in 1,3-butadiene) and overall a net stabilization of the molecule will result.


text:
Limiting reactant and percent yield
If 2.30 mol of sodium reacts with 1.95 mol of water, how many moles of sodium hydroxide are produced?
Would it be 2.3 mol NaOH?

The first step is to figure out a reaction equation.
[I'm assuming elemental] sodium reacts with water violently because it releases a lot of energy when its only valence electron gets ripped apart. It creates \ce{NaOH} and \ce{H2}.
$\ce{Na + H2O -> NaOH + H2}$
We need to balance this.
$\ce{2Na + 2H2O -> 2NaOH + H2}$
$\ce{Na}$ and $\ce{H2O}$ are consumed at equal rates, so the $1.95$ moles of $\ce{H2O}$ are going to be used up completely (leaving $2.30 - 1.95 = 0.35$ moles of $\ce{Na}$ not reacted).
$\ce{NaOH}$ is formed at the same rate of consumption of the other two, so it's going to have $1.95$ moles produced.
--
Mathematically...
$\frac{2.30mol_\ce{Na}}{2mol_\ce{Na}} = 1.15$
and
$\frac{1.95mol_\ce{water}}{2mol_\ce{water}} = 0.975$
Effectively, water burns out faster in the reaction than sodium. So, we have to base our calculations off of moles of water.
For every 2 moles of water used, 2 moles of sodium hydroxide are formed.
$1.95mol_\ce{water} * \frac{2mol_\ce{NaOH}}{2mol_\ce{water}} = 1.95mol_\ce{NaOH}$

$$\ce{Na + H2O->NaOH + \frac12H2}$$

If 2.30 mol of sodium reacts with 1.95 mol of water, how many moles of sodium hydroxide are produced?
  Would it be 2.3 mol NaOH?

No cause the limiting reagent here is water which is in lesser amount, since both reactants have same stoichiometric coefficients, so the product will be formed in accordance with amount of water, i.e. 1.95 moles of $\ce{NaOH}$


text:
How can we confirm the number of protons in an atom?
The periodic table tells us that there are 6 protons in a carbon atom. Is there a way to verify this first-hand? Or are we just expected to believe it unquestioned?

Carbon is defined as an atom with six protons (and 6 electrons). That is the nature of the periodic table. It is so per definition. I suppose your real question is: how can we determine if something is really carbon. Right?

My real question would be 'how do we really know that what they've been telling us in school about atomic structure is true?'

That what about the atomic structure is true? The atomic structure is a model. It is not known if it represents reality completely (probably not). This can never be known.

@Jori then why claim that carbon has x protons as if it were the truth ?

Excellent question : what has been affirmed without proof can also be denied without proof

1) While not easy, it is possible to obtain mass-spectrum of full range of ions, including fully ionized. By number of peaks with Z/E corresponding as X, X/2, X/3 ... X/n it is possible to ensure that an element has exactly n electrons and protons. It is still difficult to ensure full ionization of heavier atoms, so the method is not applicable for heavier elements. This, however, is the only direct method I can think of. Should work for carbon, though.
2) Various X-ray-derived spectral methods. While not providing direct evidence, they do provide information on electronic shell structure, that can be compared with theoretical calculations.

I'm still not convinced


text:
Why are the ozonides of heavier elements more stable?
The ozonides of $\ce{Cs}$, $\ce{Rb}$, $\ce{K}$ are well known and relatively stable, but there is little mention of the ozonides of $\ce{Na}$, $\ce{Li}$. Why is this?

They are extremely unstable according to Wikipedia. You can sort of guess why- the ozonide ion is big for the tiny Sodium and Lithium ions so it is stretched around it due the distributed charge on the ozonide ion- puts it under large internal stresses.

Potassium and heavier alkali metals ozonides are formed by treating with ozone, or by treating the alkali metal hydroxide with ozone. They are very sensitive explosives that have to be handled at low temperatures in an atmosphere consisting of an inert gas.
Lithium and sodium ozonide are extremely unstable and must be prepared by low-temperature ion exchange starting from $\ce{CsO3}$. Sodium ozonide, $\ce{NaO3}$, which is prone to decomposition into $\ce{NaOH}$ and $\ce{NaO2}$, was previously thought to be impossible to obtain in pure form. However, with the help of cryptands and methylamine, pure $\ce{NaO3}$ may be obtained as red crystals isostructural to $\ce{NaNO2}$.
As pointed out by user2617804, lithium and sodium ozonide are extremely unstable because the ozonide ion is big for the tiny sodium and lithium ions so it is stretched around it due the distributed charge on the ozonide ion- puts it under large internal stresses.
See wikipedia for more details.

We've actually gotten ozonide ions around a magnesium cation center. But it requires an argon matrix. See the bisozonide section here. (The monozonide may have $\ce{O3^{2-}}$ instead of $\ce{O3^-}$; we can't really tell that from the limited information in the article.)


text:
How can compounds like XeO4 form tetrahedral structures despite having double bonds?
Xenon tetroxide is tetrahedral despite having double bonds. Is it something to do with $d$ orbitals? Also things like sulfur trioxide (trigonal planar with three double bonds but the sulfur is hypervalent). Is it possible for the $\pi$ bond to be formed from overlap of $d$ orbitals from the central atom with $p$ orbitals of the oxygen?

$\ce{XeO4}$ does not have double bonds. Each $\ce{Xe-O}$ bond has order 1.

$\ce{XeO4}$ has a tetrahedral structure as you noted, but $d$-orbitals are not involved in explaining its geometry or hybridization. The central xenon atom in $\ce{XeO4}$ is best described as simply $\ce{sp^3}$ hybridized.  We have 8 xenon electrons so we need 4 orbitals to put them in.  The 4 equivalent orbitals are formed by mixing the $s$- and $p$-orbitals (just as we do with carbon) to form the equivalent $\ce{sp^3}$ orbitals pointing towards the vertices of a tetrahedron (which minimizes electron-electron repulsions between the 4 bonds).
The molecule can be represented by several resonance structures and I've drawn two of them (A and B) below.  I've only drawn the lone pairs on one of the oxygen atoms for each resonance structure, but they are the same for all oxygens in each resonance structure.  So the resonance structures tell us that the xenon-oxygen bonds in $\ce{XeO4}$ are some mix of single and double bond character.  In resonance structure A, the xenon and oxygens are neutral; while in resonance structure B, the xenon has a charge of +4 and each oxygen has a charge of -1.

The $\ce{Xe-O}$ bonds in $\ce{XeO4}$ are very weak. In resonance structure A the pi overlap is very poor due to the different atomic sizes of xenon and oxygen. In resonance structure B we just have a simple single bond connecting the oxygen and xenon and it contains 2 electrons, both from xenon and shared by oxygen, oxygen does not contribute any electrons to this bond. The single bond resonance structure B probably contributes more to the description of this molecule than resonance structure A. The bonds are so weak that $\ce{XeO4}$ is unstable and explosively decomposes around -36 C.
In summary we can say that

the xenon in $\ce{XeO4}$ is $\ce{sp^3}$ hybridized
the $\ce{Xe-O}$ bond is a partial single-double bond with the single bond probably being a more accurate description, and
the $\ce{Xe-O}$ bonds are extremely weak because of the poor pi overlap in resonance contributor A; and oxygen does not contribute any electrons to the xenon-oxygen bond in resonance contributor B.

Xenon is $[\ce{Kr}]4d^{10}5s^25p^6$ which, in its excited state, becomes $[\ce{Kr}]4d^{10}5s^15p^35d^4$. Now the $s$ and $p$ mix to form the hypothetical $sp^3$ hybridised orbitals which form $\sigma$-bonds with each electron in $p$ orbital of oxygen. The rest $4$ electrons in $d$ orbital form $\pi$-bonds with each electron in $p$ orbital of oxygen. (Oxygen has two unpaired electrons in $p$ orbital.)
In the case of $\ce{SO3}$, sulfur excites to $[Ne]3s^13p^33d^2$ where $sp^2$ forms $\sigma$-bonds and the remaining three orbitals, the $\pi$-bonds.
The $\pi$-bonds are ignored in determining the shape of the molecule. The $\pi$-bonds shorten the bond lengths but do not affect the shape. This approach explains the $\sigma$-bonding and the shape of the molecule, but the explanation of $\pi$-bonding is unsatisfactory. This is better explained by MOT (Molecular Orbital Theory.)

This is incorrect.


text:
Compressible Factor
I 've a soft question on the terminology of compressibility factor. It is termed compressible factor because, its value determines how any reals gas is compressible with respect to an ideal gas according as its value-less than or greater than 1. If $Z>1$ then $\frac{pV}{nRT}>1$ and so $p>\frac{nRT}{V}$ which means it is more compressible than ideal gas. Because for ideal gas it is $p = \frac{nRT}{V}$. Simillar for negative deviation. But the question arises why we have not taken $V$ on LHS in these equations and inequations. I mean why does scientists doesn't named this as something like "volumetric factor" (actually can't find any specific word, but it is not important, hope readers can understand my question.) I mean something like Volume is greater than of ideal gases.
And another one is we always have graphs of $Z \rightarrow p$ to understand the deviation in real gases from ideal gases. (I'm posting a graph for instance.)

Why we don't take $T$ or $V$ on the X-axis to understand this concept.

The compressibility factor measures the ratio of actual volume of the gas (under the given conditions of temperature and pressure) to the volume the gas must have occupied if it were an ideal gas (under the same temperature and pressure). This ratio tells us how much compressible is a gas (more compressible if the volume is lesser) by a given pressure applied on it. At this point, we tend to take the temperature as constant and assume the volume is a function of pressure alone. Then, suppose changing the pressure and observe how the volume changes for the gas. This is what the graph in your question shows. It illustrates the variation in compressibilities of different real gases with pressure. The term as well as the graph makes sense in that they show the willingness of a given real gas to get compressed upon subjecting them to various pressures.

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I look at it in a different way; not about volume but rather the relationship between molar volume and 'ideal' molar volume, $\frac{V}{n}$  and $\frac{RT}{P}$.
From $PV = ZnRT$ write
$$\frac{V}{n} = Z\frac{RT}{P}$$
The reason the compressibility factor Z is called 'compressibility' factor is that it relates these two quantities like a 'spring' constant. It relates how many molecules you can actually 'squeeze' or rather  'compress' into a volume at some particular state of temperature and pressure.      For Z > 1 the number of molecules, n is reduced for the same volume, temperature and pressure than when Z = 1.

authors: Богдан Войтенко


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Why can't the conjugate base of benzoic acid be stabilized by conjugation with the aromatic ring?
All the carbon and oxygen atoms are sp2 hybridised and can have a p orbital in the correct plane yet only conjugation of the COO- group occurs. I can't think why. I can't draw resonance structures for the negative charge being delocalised around the ring but if you look at it from the orbital perspective (sp2 hybridisation) it seems clear that this conjugation should occur.

Check the following, which might answer your question. Be sure to read the answers, since the structures in the question are incorrect: http://chemistry.stackexchange.com/questions/16372/carboxylate-cage-concept

Here are some of the resonance structures we can draw for the carboxylate anion of benzoic acid.

You're correct that we can't delocalize the negative charge from the oxygen into the benzene ring.  Even though there is continuous p-orbital overlap between the oxygen and benzene carbons, there is just no way to push the electrons so that a negative charge winds up in the benzene ring. However, as the resonance structures show, we can move electrons the other way and place a second negative charge on the oxygens and leave a positive charge delocalized around the ortho and para positions in the benzene ring.  This is why the carboxylate anion of benzoic acid will not react with electrophiles (electrophilic aromatic substitution) at those positions in the aromatic ring.

Why can't we have a resonance structure in which the Oxygen donates 2 of its electrons to the carbon and thus is slightly positive with only 6 electrons, this will push electrons into the ring.

Oxygen doesn't like to be positive because of its electronegativity and in the benzoate anion there is actually a formal negative charge on oxygen.  But maybe I am not understanding your question.  If so, please ask it as a new question with a drawing of your proposed resonance structure.


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Trouble to predict the effect of pressure and temperature for a reaction in equilibrium
I have trouble to predict the effect of decreasing pressure and increasing temperature on the reaction $\ce{N2 + 2O2 <=> 2NO2}$. Increasing temperature caused this reaction to shift in the forward direction and decreasing pressure brought no change. I know increasing temperature favour endothermic reaction and decreasing temperature favour exothermic reaction. I also know on increasing pressure the system will shift to the direction where less moles are present. Please tell me how to use this information to predict the effect for increasing temperature and decreasing pressure.

When I posted this question I had no idea how changing each factor bring change in equilibrium. Then I bombarded my brain with everything I know about equilibrium and I found a fairly easy way to understand this.
Change in concentration
Imagine two vessels A and B connected to each other through a thin hollow pipe through the middle of the vessel. Let vessel A be reactant and vessel B be product. The vessels are filled with water a little above the pipe. If I add water to vessel A  water flow from A to B. Thus increasing concentrated of A caused the water in vessel A to flow to B; which in other words is the forward reaction.

Change in temperature
Think of a reaction which is in equilibrium.
1)Decreasing the temperature of reaction in equilibrium.
We know that decreasing the temperature favour exothermic reaction
 and exothermic reaction are favoured by low temperature. Exothermic mean giving out heat which remove the deficiency of temperature to bring it back to normal in equilibrium.
2) Increasing the the temperature of reaction in equilibrium
We also know that increasing the temperature favour endothermic reaction or endothermic reaction are favoured by high temperature. Endothermic means to absorb heat energy which causes the excess heat energy ( from high temperature) to bring it back to normal in equilibrium
Effect of pressure
Changing pressure play a significant role in gaseous reaction which are accompanied by change in moles. On increasing pressure of the system equilibrium will shift to the direction where number of moles decrease or where pressure decrease (because pressure is directly proportional to the number of moles). If pressure on the system is increased equilibrium shift in the direction in which decrease in number of moles take place.

This is the second time I answered on stack exchange so please let me know my mistake if any. If I had known how to create the diagrams I usually find in other answers I would have added them.


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Fermentation in the presence of UV light
Does the presence of UV light, or any radiation for that matter, affect the process of fermentation? I'm not sure, but I seem to recall somebody telling me before that yeast will not function properly, but I can't seem to find it anywhere. Thanks for any help.

Depends on the wavelength. Fermentation is basically metabolic activity by yeast and bacteria. UV-C (<280 nm) is pretty good at killing a lot of micro-organisms. Some species are more resistant than others, but it's safe to say that exposure to this kind of light will probably interfere with fermentation. Depending on the medium and how the light is applied (thickness of material, power, etc.), it may have more or less of an effect depending on how much light reaches the yeast or bacteria.


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Chemical composition of petrol
What is the chemical composition of petrol? What are the compounds available in petrol? Does octane number have something to do with actual octane  concentration?

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The composition of petrol, also called 'gasoline' or 'gas' in the USA, despite it being a liquid at room temperature, contains a complex mixture of hydrocarbon compounds selected for use in internal combustion engines.
The term "95% octane" refers to the 'octane rating', 'octane number' or 'anti-knock index' of the fuel, not the amount of 'octane' in the fuel (since 'octane' is a specific organic compound in the group of alkanes).  The octane number / rating is an 'anti-knock index', based on an arbitrary scale indexed to a liquid mixture of iso-octane (2,2,4-trimethylpentane and n-heptane). Whilst different countries have different standards for anti-knock index, in the USA, the American Society for Testing and Materials (ASTM) defines two popular indices called the 'research octane number' (RON) and motor octane number (MON). Regardless if the specific stanard, the general principle is that the higher the octane number, the more the fuel can be 'compressed' before 'auto-igniting'. Auto-ignition is not desirable in petrol engines as it causes 'knocking' of the pistons which greatly shortens their life.
What petrol (gasoline) actually contains is a blend of refinery products, including reformate (high in aromatic hydrocarbons, low in alkenes), olefinic hydrocarbons, alkylate (high in paraffinic hydrocarbons) and isomerate (high in branched isomers of pentane and hexane). Specific blends vary from manufacturer. Manufacturerers even offer a range of blends at the bowser (gas station) including 10% ethanol and 'premium' blends for high performance engines.
Refer to 'Development of a Detailed Gasoline Composition-Based Octane Model' for more information.


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Which vibrational modes of CO2 and CSO do not lead to IR absorption bands?
I recently came across this question, I have no idea about IR:

Which of the following vibrational modes show no IR absorption bands?
A) Symmetric CO2 stretch
B) Antisymmetric CO2 stretch
C) Symmetric S=C=O stretch
D) Antisymmetric S=C=O stretch

related question: http://chemistry.stackexchange.com/questions/19119/is-carbon-dioxide-ir-inactive?rq=1

A) Symmetric $\ce{CO2}$ stretch.
A particular vibrational mode is considered IR active, if there is a change of transition dipole moment. $\ce{CO2}$ is linear, and the symmetric stretching mode means both oxygens move in/outward concurrently.
You can see an illustration of the symmetric and antisymmetric stretching modes of $\ce{CO2}$ here: https://www.youtube.com/watch?v=TMLnUmbLwUI&t=1m50s

For the symmetric mode, the overall center of charge stays the same, thus no change in dipole moment, thus the mode is not IR active.
For the antisymmetric mode, the center of charge shifts, thus the dipole moment changes and the mode is IR active. This is also true of the $\ce{S=C=O}$ antisymmetric stretch.
The $\ce{S=C=O}$ symmetric stretching mode is also IR active, since $\ce{S}$ and $\ce{O}$ are two different atoms, with different partial charges. As they vibrate, the total center of charge moves.


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How can we measure electronic excitation in some "nm"(nanometres)?
I have no idea to which concept this belongs:

The widest range over which electronic excitations in organic compounds occur, is
A) 200 nm – 780 nm
B) 220 nm– 500nm
C) 250 nm– 700 nm
D) 290 nm – 1000nm

Don't make junky tags, please.

@jonsca i didn't knew any related to it

So try to get as close as you can and if they are not pertinent, someone can put the right ones on.

The question is asking what wavelengths of light can cause electronic excitation in organic compounds, i.e. what energy photons correspond to the energy difference between the ground and excited states. They're not the same in all compounds (which is why different compounds have different absorbance spectra), but there is a range of energies in which most of the electronic transitions in organic compounds occur.

I think an important component of the question is whether, for example, there are compounds that will have electronic excitations  at 200 nm, 220, 250, 290nm, etc. and the same for the long wavelengths  (i.e., will anything have an electronic excitation at 1000 nm?)


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Why are square planar coordination compounds with four different ligands optically inactive?
Why do square planar coordination compounds of type $\ce{[Mabcd]}$ not show optical activity, although they contain 4 different ligands (i.e. chiral central metal atom)?

Even though the central atom has 4 different ligands this does not necessarily mean that the compound is optically active. The condition for that rather the following (from Wikipedia):

A molecule is achiral (not chiral) when an improper rotation, that is a combination of a rotation and a reflection in a plane, perpendicular to the axis of rotation, results in the same molecule[...] For tetrahedral molecules, the molecule is chiral if all four substituents are different.

So for 4-coordinated carbon atoms which are tetrahedral having four "ligands" actually establishes optical activity. But in the case of a square planar metal complex you always have a mirror plane, namely the one made up by the central atom and the ligands:

And as a result you don't get optical activity. But it might be possible that some vibrational modes destroy the mirror plane symmetry element of the metal complex and this might lead to some optical activity depending on how well this mode excited.

"it might be possible that some vibrational modes destroy the mirror plane symmetry element of the metal complex and this might lead to some optical activity depending on how well this mode excited." Hello, thanks for your answer! Could you please give an example of this "vibrational optical activity"? Also, is it possible to predict it theoretically?


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Electrolysis on aqeuous solution
I know when electrolysis on a concentration aqeuous solution which contains halide ions($Cl^-$ , $Br^-$ , $I^-$). The halide ions is discharged although $OH^-$ is more easily to discharged.
For example, electrolysis of concentration aqueous sodium chloride. The cation $Cl^-$ and $OH^-$ will attract to the anode. The $Cl^-$ ions will be discharged although according to the electrochemical series that the $OH^-$ will be more easily discharged.
Why do that happen?

What do you mean by higher position? You question as a whole is rather unclear to me. Furthermore, could you try to formulate a more descriptive title and include some of your own thinking about the problem in the question.

I just edit the question

Let's explain this using an example. Consider a cell in which an aqueous solution of sodium chloride is electrolyzed. The $\ce{Na^+}$ ions migrate toward the negative electrode and the $\ce{Cl^-}$ ions migrate toward the positive electrode. But indeed at the anode, there are  two substances that can be oxidized at the anode: $\ce{Cl^-}$ ions and water molecules.
$$\ce{2Cl^-   \rightarrow Cl_2 + 2e^-}$$ $$\ce{2H_2O   \rightarrow O_2 + 4H^+ + 4e^-}$$
The standard-state potentials for these half-reactions are $E^0_{ox}= -1.36 V$ for the first couple and  $E^0_{ox}= -1.23 V$ for the second couple.
At first glance, it would seem easier to oxidize water ($E^0_{ox}= -1.23 V$) than $\ce{Cl^-}$ ions ($E^0_{ox}= -1.36 V$).
It is worth noting, however, that the cell is never allowed to reach standard-state conditions. The solution is typically 25% $\ce{NaCl}$ by mass, which significantly decreases the potential required to oxidize the $\ce{Cl^-}$ ion. The pH of the cell is also kept very high, which decreases the oxidation potential for water. The deciding factor is a phenomenon known as overvoltage, which is the extra voltage that must be applied to a reaction to get it to occur at the rate at which it would occur in an ideal system.
Under ideal conditions, a potential of 1.23 volts is large enough to oxidize water to $\ce{O_2}$ gas. Under real conditions, however, it can take a much larger voltage to initiate this reaction. (The overvoltage for the oxidation of water can be as large as 1 volt.) By carefully choosing the electrode to maximize the overvoltage for the oxidation of water and then carefully controlling the potential at which the cell operates, we can ensure that only chlorine is produced in this reaction.


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Why do many pharmaceuticals contain fluorine?
What exactly does fluorination of some compound do in a biological sense?
I read the following things about what fluorine does for drugs:

Fluorine withdraws electron density and can make an acid stronger.
Fluorine substitution can increase lipophilicity - I don't understand this. Lipids are non-polar; I can't see why adding fluorine would generally decrease the polarity of some molecule ...
Carbon-fluorine bonds are exceptionally stable and resist change.

However, can anyone provide some concrete examples of fluorine influencing the behavior of some drug? I could only find very general points on the internet.

A definitive reference is:
"The Many Roles for Fluorine in Medicinal Chemistry" J. Med. Chem., 2008, 51 (15), 4359–4369.
In reference to the bullet points:

Fluorine obviously affects $\ce{p}{K_{a}}$ and can make an acid stronger but also a base weaker which could be advantageous when optimising a drug candidate (lead) molecule.
I guess here they refer to $\ce{-CF_3}$ which can increase lipophilicity
That's one of the most useful roles of fluorine. A lot of times you may have a very potent and selective molecule, but when you test it in vivo let's say in rats, it has no effect because it gets metabolised i.e. chemically modified to an inactive molecule by the rats liver (and other mechanisms). This can maybe prevented by changing to an $\ce{C-F}$ bond. For example electron rich aromatic sites are prone to metabolic oxidation. You can block this by fluorination of this site, exchanging the proton for a fluorine.

Have a look at the paper on top, you will find a new world of information.

how does the CF3 group increase lipophilicity?

@Dissenter The explanation is not as simple and there are different factors that will weigh in depending on the environment of $\ce{-CF_3}$. For example in $\ce{-XCF_3}$ you can claim that the EWG effect decreases the ability of X to form hydrogen bonds with water molecules. To make things more complicated $\ce{-F}$ substitution (as opposed to $\ce{CF_3}$) is less capable of increasing lipophilicity and in fact sometimes it lowers clogP. If you are interested, I suggest you can have a look in relevant literature or submit a new question

Most helpful I've seen is point #3. In many instances a simple fluorine substitution off a methyl or phenyl/aryl significantly reduced CYP450-based metabolism. The differences in the HPLC trace was sometimes dramatic.


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How is the octane percentage defined?
Normally octane percentage is mentioned on petrol (95% octane). How is this defined? Is it percentage of chemical or some other parameters?

@Qmechanic Always wanted to learn about Hydrocarbon's in physics class

See the Wikipedia article on octane rating. 95% octane is the same as an octane rating of 95, and it means the petrol has the same knocking characteristics as a mixture of 95% iso-octane and 5% heptane.

Thank you for the answer. So if the higher octane number  will it produce more energy? Or is it not even considered on octane indexing?

@sugunan: The amount of energy stored in the fuel is not related to the octane number. However a high octanc number allows the engine to use a higher compression ratio, and this increases the power output of the engine. So powerful engines tend to use high octance fuel, but only for its knock resistance, not because high octane fuels contain more energy.

Wow, only a 50% success rate in spelling octane :-)

I didn't get the point :S


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stone which turns to black on exposing to atmosphere?
My mom is using a stone like substance as a dye for hair recommended by a herbalist. That stone becomes black when exposed to air. Do you know which compound is this?

Could it be lead aceteate? This is a white crystalline substance which turns black when it reacts with sulfur. It is used in 'progressive' hair dyes, where it reacts with the sulphur in hair proteins to slowly turn grey hair black (over several applications, so 'no-one' will notice). Alternatively, it could be 'white lead' which turns to black lead sulphide when it reacts with hydrogen sulphide in the atmosphere.

@theo If the OP doesn't provide any additional information which might point in a different direction then I guess this comment of yours will make a fine answer.

@TuBaTariq Could you add some more information. For example what color is the stone initially? How is it used to dye the hair etc. In its current state your question is difficult to answer without guessing.

There is a compound called cerussite or 'natural white lead' which turns to black 'lead sulphide' when it reacts with hydrogen sulphide in the atmosphere. It was commonly used as make up by women in medieval times and is sometimes still used as a white pigment in 'lead paint'.
Of course, we all now know lead is toxic to many organs and tissues including the heart, bones, intestines, kidneys, and reproductive and nervous systems, as well as causing potentially permanent learning and behavior disorders, particularly in children.
For this reason, it's use has diminished over time, except perhaps amongst New Age crystal 'healers' who claim it can "infuse your aura with light" and "bring amazing change into your life", not only through a lowering of general intelligence, but, it can also help prevent bacterial infections.
Having said all this, you may be surprised to know that another toxic lead-based white crystalline substance called lead acetate is approved by the FDA for use in 'progressive' hair dyes, where it reacts with the sulphur in hair proteins to slowly turn grey hair black, over several applications of course, so 'no-one' will notice.

image of lead(II) acetate crystals, courtesy of Wikipedia
PS: In case you were wondering why lead-based products are still used, given that there are many alternatives out there, many of which are labelled 'non-toxic', 'natural' and 'organic'...
the National Cancer Institute reports that there are over 5000 different chemicals used in hair dyes, including toxic coal-tars, aromatic amines and phenols. Some of these are also carcinogenic (cancer-causing agents) and can be absorbed through the skin or inhaled. Whilst the industry has been working towards eliminating some of these chemicals since the late 1970's, people who regularly began using hair dyes before 1980 may be at increased risk of developing non-Hodgkin lymphoma (a form of blood cancer).
The International Agency for Research on Cancer (IARC) has said that "whilst workplace exposure as a hairdresser or barber is probably carcinogenic to humans", personal hair dye use is “not classifiable as to its carcinogenicity to humans,” based on "a lack of evidence from studies in people".

I'm wondering if anyone is getting grey hair worrying about the increase in cancers on rats and mice, caused by testing these hair dyes.


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When is the carbaldehyde suffix correct?
Is benzene carbaldehyde a correct name for Benzaldehyde? I'm (always) getting confused between common and IUPAC names. If the above name is correct then is it a common one or IUPAC? And why don't we use it more often?

Both are IUPAC names, but used in different situations. For a 1 carbon aldehyde attached to a benzene, the one that must be used is benzaldehyde.
Aldehydes ($\ce{R-CHO}$) take the suffix "-al". If other functional groups are present, the chain is numbered such that the aldehyde carbon is in the "1" position, unless functional groups of higher precedence are present.
If a prefix form is required, "oxo-" is used (as for ketones), with the position number indicating the end of a chain: $\ce{CHOCH2COOH}$ is 3-oxopropanoic acid. If the carbon in the carbonyl group cannot be included in the attached chain (for instance in the case of cyclic aldehydes), the prefix "formyl-" or the suffix "-carbaldehyde" is used: $\ce{C6H11CHO}$ is cyclohexanecarbaldehyde. If an aldehyde is attached to a benzene and is the main functional group, the suffix becomes benzaldehyde.
For example, this guy would be cyclohex-2-ene-1-carbaldehyde:

Both names, ‘benzenecarbaldehyde’ and ‘benzaldehyde’, are unambiguous and describe the same compound. However, the preferred IUPAC name (PIN) is ‘benzaldehyde’.
Concerning aldehydes, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-66.6.1 Systematic names of aldehydes
Aldehydes are systematically named in three ways:
(1) substitutively, using the suffixes ‘al’ for $\ce{-(C)HO}$ and ‘carbaldehyde’ for $\ce{-CHO}$;
(2) by changing the ‘ic acid’ or ‘oic acid’ endings of retained names of carboxylic acids into ‘aldehyde’; the nomenclatural properties of acids are transferred to aldehydes; thus, preferred names of aldehydes correspond to preferred names of acids, and carboxylic acids that are not substitutable generate nonsubstitutable aldehydes;
(3) by using the prefixes ‘oxo’, denoting $\ce{=O}$, or ‘formyl-’, denoting the substituent group $\ce{-CHO}$.

In particular,

P-66.6.1.1.3 The suffix ‘carbaldehyde’ is used when the $\ce{-CHO}$ group is attached to a carbon atom of a ring or ring system, or to a heteroatom.

For example, ‘cyclohexanecarbaldehyde’ is a PIN.

However, according to P-66.6.1.2.1, the PIN for the compound given in the question is ‘benzaldehyde’. (This name is analogous to the preferred name of the corresponding acid ‘benzoic acid’, see P-66.6.1 (2) above.)

Nevertheless, according to P-66.6.1.2.2, the PINs for phthalaldehyde and terephthalaldehyde are the systematic names ‘benzene-1,2-dicarbaldehyde’ and ‘benzene-1,4-dicarbaldehyde’, respectively. (These names are analogous to the preferred names of the corresponding acids ‘benzene-1,2-dicarboxylic acid’ [not ‘phthalic acid’] and ‘benzene-1,4-dicarboxylic acid’ [not ‘terephthalic acid’], respectively.)


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How is ethanenitrile protonated?
I know that the proton will attack the HOMO, but I'm having difficulty imagining how this will happen.

Have you drawn out the structure of ethanenitrile? Have you thought about the molecular orbital diagram? These are things you need to do to identify the HOMO.

The structure of ethanenitrile is:
$$\ce{H3C-C#N:}$$
Qualitatively, we can draw some conclusions about the molecular orbitals in the structure. If covalent bonds result in in the involved electrons being lower in energy, then which pair of electrons in the structure has the highest energy?
However, if you are looking at nitriles, you probably already know that both lone pairs and pi bonds are good proton acceptors.
Let's consider what the difference is if the lone pair is the base or if a pi bond is the base:

Lone Pair

$$\ce{H3C-C#N + H+ -> H3C-C#N+-H}$$

Pi Bond

Which atom will the proton end up attached to? Which of the two following carbocations is more stable?
$$\ce{H3C-C#N + H+ -> H3C-C+=NH}\ \  or\ \  \ce{H3C-CH=N+}$$
Now... what is the relationship between the two conjugate acids drawn?
$$\ce{H3C-C#N+-H}\ \ and\ \ \ce{H3C-C+=NH}$$
Where is the HOMO?


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If dS ≥ dq/T for a spontaneous change is dS ≤ dq/T for a non-spontaneous change and is that even possible?
I am struggling with the concept of entropy but might be on the verge of understanding it. The Clausius inequality, $$\mathrm{d}S \geq \mathrm{d}q/T,$$ is true for all spontaneous changes with the equality for reversible changes. But what if a change is not spontaneous or is that impossible i.e a change cannot be forced. Any help with answering this question or to help conceptualize and understand the entropy would be greatly appreciated.

As you perfectly say, Clausius inequality states that $$\mathrm{d}s\geq \mathrm{d}q/T.$$ This has to be true for any process.
For reversible processes, what happens is that $\mathrm{d}s= \mathrm{d}q/T$, and for non-reversible processes, $\mathrm{d}s > \mathrm{d}q/T$. If for a given process $\mathrm{d}s< \mathrm{d}q/T$, it would not fulfill Clausius inequality.
Consider the situation where the system is isolated from its surroundings. In this case, $\mathrm{d}q_\mathrm{(sys)} = 0$ , and thus the Clausius relation implies $\mathrm{d}S\geq0$, so this tells us that the entropy of an isolated system cannot decrease in the course of a spontaneous change.
Since the universe is itself an isolated system, this result shows that we can use entropy as the signpost of spontaneous change. Processes are only spontaneous if they cause an increase in the total entropy of the universe.

One way of thinking about this is to consider what happens when you clean a messy room. Your room is an example of an open system (i.e. heat, work and mass can move across its boundary). If you put a lot of work into your room, in the form of good old fashioned elbow grease, you end up with a nice, clean room that has a lower entropy than what it started with. However, it is important to note that the entropy of the universe, which is an isolated system that happens to contain the open system of your room, will have increased as a result of your work. This is because you could not do the work in a reversible fashion. Your body heat, friction from moving things around, etc. all produced irreversible changes in the heat flow to your surroundings. This is where the Clausius equation holds. To describe the entropy decrease in your room you have to know the equation of state that can provide the initial and final entropies. Not really practical for an abstract concept for a room, but for a fluid such as a gas or liquid it's usually clear-cut.


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Exothermic reactions at high temperature and the chemical equilibrium
Why are many exothermic reactions, like Haber process, conducted at high temperature, thus  their unfavourable equilibrium position?

There are two competing problems in the Haber Bosch process: the equilibrium position and the reaction rate. The reaction: $\ce{N_2 + 3H_2 -> 2NH3}\ (\Delta H = -92.4 \mathrm{\ kJ/mol})$ is exothermic and the equilibrium is actually favourable at room temperature, but the rate is very slow due to the large energy barrier to breaking the triple bond in $\ce{N2}$. To increase the reaction rate, a catalyst is used at high temperature. As you say, the high temperature pushes the equilibrium towards the reactants (by the Van 't Hoff equation) so there is a competition between yield and rate.
In practice, high pressure is also used to increase the yield (since the products side of the equation has fewer molecules), and like many industrial processes, it's run as a recirculating process. At manageable pressures and temperatures, they may only get 10 or 15% yield after one pass, but the ammonia is removed and the unreacted gases passed back into the reactor, giving quite high overall yield.


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A way to calculate orbital energy of an electron
Maybe this is just not included in my course, but I still wonder how to calculate the binding energy of an electron .at a certain electron orbital.
Intuitively,The ionization energy of an element should vary depending on which orbital the valence electron is in.
PS: I only know how to calculate ionization energy using principle level.

I think we'd need to know what course you're taking. This can be determined from quantum mechanics, but in general the numbers are tabulated because the equations get extremely complicated.

High school general Chem

You are correct that the ionization energy of an electron depends on the orbital. There is no easy formula, because the electrons in atoms are governed by quantum mechanics and for multi-electron atoms, the equations are un-solvable algebraically.
That said, there are some reasonable tabulations, which have been verified by calculations and by spectroscopy, e.g., http://www.colby.edu/chemistry/PChem/notes/AOIE.pdf (from a quick Google search).
Both ultraviolet photoelectron spectroscopy and x-ray photoelectron spectroscopy can be used to eject electrons from various materials, and you can thus identify elemental composition (and other properties) from the known orbital energy levels

What is name of the name of the equation and why proposed it?

If you'd like to read more about quantum mechanics you would want to read about the Schrödinger model of the hydrogen atom. But it helps to have some familiarity with differential equations first.

I just stumbled across this question ... a little late....
A common approximation in XPS to solve this question is Koopmans theorem (frozen-orbital approximation). If the remaining (those were not photo-electron was removed) orbitals of the exited atom after photo-ionisation do not change between the initial and final state (no relaxation)  the experimental binding energies are than taken to be equal to the respective negative Hartree-Fock/KS orbital energy.
A good way to start reading is this Book:
Book
See also:
Link
Link2


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How does the presence of chlorine atoms in PVC make it flame resistant?
In a paper that I'm reading, it states that:

The presence of chlorine atoms in polyvinyl chloride make it flame resistant

and I'm just wondering why the presence of chlorine atoms will make it flame resistant. To the best of my knowledge, burning something is just providing a material with enough activation energy such that it can react with oxygen to produce products such as carbon dioxide and water. If so why does the chlorine prevent this process in any way?

It'd be helpful if you had also cited that paper, as a Scholar search yields no results relevant to your blockquote (link)

PVC is not truly flame resistant, it does decompose on relatively gentle heating (around 150 °C) However, the process produces significant volumes of hydrogen chloride, that is not flammable and isolates the plastic from the flame. This makes makes PVC self-extinguishing or fire-retardant.

Many of the hydrogen halides ($\ce{HCl}$ and $\ce{HBr}$) are able to inhibit the combustion process, as when they are added to a flame they alter the flame chemistry. What will happen is that they react with oxidizing radicals such as hydroxyl radicals to form halogen atoms. The halogen atoms are able to react further but their reaction with fuels such as methane will be less exothermic than the reaction of hyroxyl radicals.
Also, Chemosphere, 1991, 22 (1–2), 67-76 states that alkyl radicals in flames can react with $\ce{HCl}$ to form alkyl chlorides. This will leave behind hydrogen atoms. It has also been stated by M.J. Thomson, A.D. Lawrence and J. Bu (Combustion Science and Technology) that HCl inhibits the combustion of benzene, it increases the carbon monoxide yield and reduces the chain branching which occurs during the combustion.

Combustion reactions are radical reactions, and the chlorine atoms formed during decomposition of the polymer act as radical traps.

How does chlorine act as radical traps? Why might chlorine act as radical traps?

My intuitive answer is that Cl has a relatively high electronegativity, so it pulls electrons towards itself, making it harder for the oxygen to react with the molecule. Reactions generally have an oxidizer which attracts electrons, and a reducer that supplies electrons. Oxidation of an organic compound usually means the carbons supply electrons to incoming oxygen atoms, making polarized and thus stable bonds. The bonds with the Chlorine atoms are themselves polarized, making the chlorine somewhat negative and the rest of the molecule slightly positive, making it less eager to give up electrons to an oxygen atom, meaning a resulting bond with oxygen would be less polarized and less stable....