text:
What's the chemistry behind only charging a Lithium-ion battery to 80% capacity at most, to increase its lifespan?
What's the chemistry behind only charging a Lithium-ion battery to 80% capacity at most, to increase its lifespan?
Over on the Skeptics StackExchange, a question has come up about charging Lithium-ion cells to just 80% of their maximum, in order to extend their lifetime.
Is there a sound basis in chemistry for such a recommendation, and if so, what is it?

Actually, the chemistry is just the same as chemistry of charging, it's the physics - mechanics of the microstructures collapsing.

Most commercially-available lithium-ion batteries use graphite as the anode (negative electrode, where the lithium ions go when the battery is charged) and $Li Co O_2$ as the cathode (positive electrode, where the lithium ions go when the battery is discharged). Both graphite and lithium cobalt oxide have a structure composed of layers stacked on top of one another; the lithium ions fit in between these layers in a process called intercalation.
The lithium sitting between the layers props the layers apart. If too much of the lithium is removed from either the cathode (over-charging) or the anode (over-discharging) then the layers can collapse back together and it becomes hard to put the lithium back in. Normally batteries have control circuitry to prevent the cells from over-charging or over-discharging but this circuitry is not always perfect, particularly for smaller, simpler battery packs.
That said, AFAIK the circuitry is already set to limit charging and discharging to well below the theoretical maximum capacity, so I would not recommend limiting the charge any further. Perhaps unscrupulous vendors will cut the amount of active material inside their batteries while keeping the rated capacity the same and therefore run the cells closer to their theoretical maximum, in which case keeping charge and discharge within their rated limits would be a good idea, but it should not be necessary for high-quality batteries.


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How do you calculate molar specific heat?
I know how to calculate the specific heat, but not molar specific heat. What is molar specific heat and how do you calculate it?

Specific heat has the units of $\mathrm{J/(K\cdot kg)}$.
Molar specific heat is in units of $\mathrm{J/(K\cdot mol)}$, and is the amount of heat needed (in joules) to raise the temperature of $1$ mole of something, by $1$ kelvin (assuming no phase changes).
So, the conversion factor you need, from dimension analysis, will have unit $\pu{kg/mol}$.
$\pu{kg/mol}$ is the SI unit for molar mass.
Multiply the specific heat by the molar mass to get the molar specific heat.
For example, the molar mass of water is $\approx \pu{0.018 kg/mol}$.
The specific heat of water is $\approx 4186\ \mathrm{J/(K\cdot kg)}$.
So the molar specific heat of water is $4186\ \mathrm{J/(K\cdot kg)} \times \pu{0.018 kg/mol} \approx 75\ \mathrm{J/(K\cdot mol)}$

In this case, exactly what is this "molar specific heat" that we obtain in the end? At constant volume/pressure?

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text:
Why does milk flake?
While drinking milk (or better 'while seeing the milk I'm gonna drink') a question came up to my mind:
Why does the milk sometimes flake, even if not in contact with some other substances?
I guess the correct term to use is flocculation, at least this is the word that raises up when talking about cheese production. I just learned the word casein. However, my question is not about cheese production.
My question is also not about flaking milk when poured into a cup of coffee which seems to be dependent on the acidity of the coffee. But casein is again the keyword.
I think the answer to my actual question will also contain the word casein, but here's the scenario:
I open a milk carton for the first time. This package doesn't have any hole and the shelf life isn't reached yet. I pour the milk into a clean cup and then observing some flakes at top of the milk.
What I was told as a child, when curiously asking, flaking is an indicator that the milk is getting acid. And in that case the keyword casein would fit again, but there's still one thing why I'm not persuaded: the milk was quite tasty and my stomach also didn't complain about it.
So, what exactly has happened to my milk when I see some flakes on top of it? Is the milk indeed spoiled and I actually shouldn't have drunk it (and just was lucky) or is there anything else which can lead to flakes?

I think the right word for it is curdle. Have you read, for example, the Wikipedia article on curd? It says:

“Milk that has been left to sour (raw milk alone or pasteurized milk with added lactic acid bacteria or yeast) will also naturally produce curds, and sour milk cheese is produced this way”

or this excerpt of the “Milk” page:

When raw milk is left standing for a while, it turns "sour". This is the result of fermentation, where lactic acid bacteria ferment the lactose in the milk into lactic acid.

So, milk naturally contains bacteria which turn it acid, and hence it curdles spontaneously. You see the beginning of this process.

An invaluable source of information on this topic is McGee’s On food and cooking. Great reading if you like chemistry and food.

As said before, fermentation of milk by microorganisms, e.g. Lactobacillus, produces lactic acid. This latter denatures the proteins in the milk, changing their "solubility", making them to agglutinate and forming curd.

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text:
How can I determine the pH of solution?
Could anyone tell me how I would go about calculating the pH of 100.0 mL of 0.100 M $\ce{H2NNH2}$ ($K_\text{b} = 3.0 \times 10^{-6}$).
Furthermore, how would this pH change when 20.0 mL of 0.200 M $\ce{HNO3}$ is added to the solution?
I’m not looking for only an answer, but rather the general method to solve these types of problems

Just to include what I’ve tried:
I first thought that (for the initial solution) $\mathrm{pH} = 14 - \log(\sqrt{K_\text{b} [\ce{H2NNH2}]})$, but that doesn’t seem to be the case.

That's what I got too (granted, I'm fairly rusty), why do you say that's incorrect?

Concerning the first part of your problem: In the general case, which you requested, we start with the mass action law:
$$
K_\text{b} = \frac{[\ce{HB+}][\ce{OH-}]}{[\ce{B}]} = \frac{[\ce{OH-}]^2}{[\ce{B}]}
$$
That’s essentially what you just reformulated to obtain your expression for $\mathrm{pH}$ above. At this point, you made an error in the sign of the second term on the right-hand side – it should be added, not subtracted, i.e. $\mathrm{pH} = 14 + \frac{1}{2} \lg \left( [\ce{B}]K_\text{b} \right)$.
Now comes the important thing: The mass action law uses the concentrations of our components in the equilibrium! What is given as 0.1 M is, however, the starting concentration $[\ce{B}]_0$ before equilibrium, which will of course decrease; $[\ce{HB+}]$ and $[\ce{OH-}]$ will increase by the same amount, which we will denote $x$. Thus, we rewrite:
$$
K_\text{b} = \frac{[\ce{OH-}]^2}{[\ce{B}]} = \frac{x^2}{[\ce{B}]_0 - x}
$$
Now, you just solve for $x$, which is the concentration $[\ce{OH-}]$.
One more thing. You will see that $x$ is actually pretty small in this case, so $[\ce{B}]_0 - x \approx [\ce{B}]_0$. By using $K_\text{b} = \frac{x^2}{[\ce{B}]_0}$, we essentially obtain the same result. You can’t always rely on this, however, and if you have stronger acids or bases, this will get important.


text:
Is this the correct relative Brønsted acidity in these four acids?
We are asked to sort these four acids by increasing Brønsted acidity:

The problem is, I am confused why $\ce{H2F+}$ is the strongest acid of them all. I understand that stronger acids will have weaker bonds, and the $\ce{F-H}$ bonds should be the most electronegative of them all, leading me to believe that they are the strongest bonds. So then why are these bonds the most easily broken? What have I forgotten here? Is the correct order really from weakest to strongest acid as follows? $\ce{CH3OH}$, $\ce{(CH3)2OH+}$, $\ce{CH3SH2+}$, $\ce{H2F+}$

Fluorine being the most electronegative element in the periodic table, it doesn’t like bearing a positive charge. Thus $\ce{H2F+}$, which has the positive charge on the fluorine nucleus, is not very stable and willing to expel the extra proton. So, it is indeed a strong acid.
$\ce{H2F+}$ is formed, e.g., by autoionization of liquid $\ce{HF}$, but only because that reaction also involves creating a $\ce{F-}$ ion (associated with another molecule of $\ce{HF}$ to form $\ce{HF2-}$), which is very favorable:
$$\ce{3 HF <=> H2F+ + HF2-}$$


text:
What's the energy density of liquid nitrogen?
Liquid nitrogen seems to be attracting a bit of attention at the moment as a medium of energy storage, both for electricity grid applications and for transport.
For example, Highview (via the Internet Archive) are doing round-trip electricity storage via liquid nitrogen.

The Dearman Engine Company (via the Internet Archive) are developing a "liquid-air" vehicle engine.
And the UK’s Institution of Mechanical Engineers has formed a Liquid Air working group (via the Internet Archive).
What is the energy density of liquid nitrogen, per unit volume and per unit weight, assuming we’ll be bringing it up to 20 °C, 1 bar? How does this compare to the energy densities of isooctane and of Li-ion batteries?

Just skimmed over the links:
Highview:

Using ambient heat to warm it, the process recovers around 50 per cent of the electricity that is fed in, says Highview's chief executive Gareth Brett. The efficiency rises to around 70 per cent if you harness waste heat from a nearby industrial or power plant to heat the cryogen to a higher than ambient temperature, which increases the turbine's force, he says.
Batteries under development .. have efficiencies of around 80 to 90 per cent, but cost around \$4000 per kilowatt of generating capacity. Cryogenic storage would cost just \$1000 per kilowatt because it requires fewer expensive materials, claims Brett.

Here's some evidence that people who know about gas and liquid air are interested in the technology: strategic partnership between Messer and Highview
Actually Highview has a presentation on their web-site that contains a lot of information. Note that they propose to run a 4-stage process against heat (ca. 100 °C), not ambient temp.
Here are some values for energy storage efficiency comparison found after a very quick search:
Batteries (Wiki)
German Wikipedia has lower efficiency values
Energy conversion table (Wikipedia)
Pump-stored hydroelectricity: 70/75 - 80% (Wiki + de.Wiki)
(Note the difference between technology that is actually in use, so efficiencies that are achieved in practical use are known and prototypes/scientific studies about achievable efficiency.)
Dearman claims 70% efficiency for the engine.

Unfortunately they don't say % of what (remember the low-temp heater manufacturers claiming > 100% efficiency because they used achievable efficiency with high temp exhaust gas as their denominator?).
Neither do they say something about the power at which they have this efficiency (ideal thermodynamic efficiency is at P $\rightarrow$ 0 ($t_{cycle} \rightarrow \infty$))
Neither can we assume that they start their calculation taking into account the efficiency of producing liquid nitrogen.
Nor storage loss.

=> so far, Highview's stationary plant concept looks more sound.

Let's have a look what thermodynamics says:

ideal efficiency
boiling T of liquid air: 77 K
ambient T: assume 290 K => $\eta_{ideal} = 1 - \frac{77}{290}$ = 73%
waste heat: assume 370 K => $\eta_{ideal} = 1 - \frac{77}{370}$ = 79%
(in other words, Dearman claims close-to-ideal efficiency)
energy density of liquid $\ce{N2}$: 620 kJ/l = 712 kJ/kg (from presentation at Highview)
The Linde process is said to need approx. 500 kWh / t = 1750 kJ/kg of air
(that would be ≈40% efficiency; no real source, just something someone answered in a German-language chemistry forum)
Here's short report that gives ≈20% efficiency for producing liquid air in Egypt. They also give storage loss (ca. 0.17 \%/d)

Thus, the 50% (achieved by Highview) to 70/80% (short term goal Highview against waste heat, quote for Dearman's engine) mean that

either the liquefaction was not part of the efficiency calculation,
or they have a super efficient air liquefaction that is about 2 - 4 times as efficient as the two numbers I found.
(Anyone around here who knows? Linde, Messer, AL?)

Update: just had a closer look at the Highview presentation. Their 50 - 80% are cycle efficiency based on "Available enthalpy for cold cycle ~300kJ/kg", so less than half of the energy density they give. Calculating with 1750 kJ/kg to produce the kg liquid air, we're at 17% (close to the "Egyptian" 20% efficiency for producing liquid air), of which they recover 0.5 - 0.8.
Conclusion: electric -> liquid air -> electric cycle with an overall efficiency in the order of magnitude of 10%

Given the track record of the question (4/5 dead links) I am somewhat concerned with the answer. Unfortunately I don't have the time right now to check.

According to Energy and Materials Flows in the Production of Liquid and Gaseous Oxygen (Shen and Wolsky, Argonne National Lab, 1980), the ideal (thermodynamically reversible) amount of work to liquify nitrogen is $\pu{21 400 J/mol}$, or $\pu{0.21 kWh/kg}$ (Table 12, page 17).
Using a simple Hampson-Linde liquefier at a high pressure of $\pu{200 atm}$ ($\pu{2940 psi}$) and a low pressure of $\pu{1 atm}$ ($\pu{14.7 psi}$), it would take $\pu{1.716 kWh/kg}$ to liquify nitrogen at $12~\%$ Carnot efficiency.
I would agree that the Carnot cycle would be appropriate for estimating how much energy it takes to liquify nitrogen, but in terms of expanding it by heating to perform useful work, wouldn't that be Charles' Law?

This is a nice fact, but I don't see it answering the question. For now, I am hesitant to clear the pending flag with either deleting or dismissing it. I recommend expanding this answer as it can be seen as more of a comment right now.

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text:
Can we separate positive ions from negative ions in a solution?
Can we make a solution containing only positive ions by using electrodialysis but instead of using 2 kinds of membrane (one blocks the positive ions and one blocks the negative ions) we can use only one kind (the one which blocks the positive ions) so the positive ions wont reach the electrode and wont be electrolysed while the negative ions will be electrolysed leaving the positive ions in the solution

While it is true that solutions are not always locally electroneutral, that non-neutrality is small, and does not extend to larger scale. Thus, it does not impact chemical concentrations: you cannot separate positive and negative charges in a solution at a macroscopic level. See for example this excerpt from Ionic Transport Processes:in Electrochemistry and Membrane Science:

For water and on macroscopic level we can't, as there is non-zero water dissociation constant.
Dropping solvent dissociation from consideration, we should consider molar charge. Faraday's number is molar charge and it is ~10^5 Coulomb/mol. So, if you can store really much charge (at least some Coulombs) in your solution, you can. But can anybody? It is really hard.

It's hard to understand the grammar here, and as I learner I do not know enough to clarify with an edit.


text:
Why is bread warm at room temperature?
If I buy a loaf of freshly-made bread and leave it out at room temperature for a day or two, when I reach into the bag the air inside is much warmer than the rest of the room. Why?

Maybe there is some crystallization of organic molecules going on while the bread turns hard. This might be a chemical question after all.

After reading some more stuff I actually think that crystallization of starch might be the answer to this question. There is a lot of literature about this and it might very well be that it can be sufficiently exothermic to warm the small volume of air in a bag.

To the above point i also want to include possible exothermic enzymatic reactions by microorganisms also

It could also be that inside the bag it's humid, and purely through suppressing the slow evaporative cooling of your skin it "feels" warmer. You could put on a thin nitrile glove to control for that.

(It turned out to be completely bogus, oops)

Is it actually warmer or does it just feel warmer? Have you measured the temperature?

@matt_black did you even read the comment above yours

After reading some more stuff I actually think that crystallization of starch might be the answer to this question. There is a lot of literature about this and it might very well be that it can be sufficiently exothermic to warm the small volume of air in a bag.

I call for an experiment. Bread, digital thermometer with recording device...

We need a controlled experiment before speculating. What sort of bread, bag, room? Moreover measure the temperature and don't rely on subjective feeling.


text:
How do surfactants spread in the case of multiple materials?
For example, consider a pot of water with a drop of oil floating on it, with surfactants present in the water.
Will the surfactants in the water spread evenly across all three interfaces (water/metal, water/air, and water/oil)?
Are there material-specific surfactants, such that each of the three interfaces might attract different surfactants?
If so, are there also surfactants that would tend to spread across all three interfaces evenly?

A surfactant is a material that has two regions, one that tends to be attracted to its environment, and another that is repelled by the environment. The amount to which they will be attracted to one interface or another depends upon whatever would minimize the total free energy of the system. Thus, unless the interactions for water/oil, water/air, and water/surface were equivalent, you should see that the surfactant preferentially moves to the interface where it will have the biggest impact on the free energy.


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Cleaning a round-bottom flask, especially after distillation
I was wondering, how can I clean my RBF after doing a distillation? Cleaning it with a simple pipe-cleaner or similar cleaner is a headache, due to the roundness of the flask.
Is there a more effective method?

@Permeakra suggests some good cleaning solutions. I have rearranged them (adding a few) in a suggested order from mildest to most evil. The specific thing you want to use will vary based on what you want to clean. If an initial rinse isn't good enough, you may want to let it soak for a few hours. Always take safety precautions (don't add acid washes to basic compounds, for example) and where the appropriate personal protection equipment (PPE).

Detergent + water + "elbow grease" - there is a lab grade soap out there called "microwash" that is pretty good, and I also tell my students that things are more soluble in elbow grease than water.
Acetone - Acetone will dissolve most organics, and it is less harmful to you and the environment than other organics.
Dilute acid (HCl or H2SO4) - good for some "baked"-on stuff or for inorganic compounds. Watch out for vigorous acid-base reactions.
Dilute nitric acid (HNO3) - since most nitrate salts are soluble, nitric acid does a better job with resistant inorganics. It is also an oxidant and will convert resistance organics to more soluble compounds. Also watch for acid-base reactions.
Other organic solvents - Use whatever will dissolve your crud, but use proper personal protective equipment (PPE), fume hoods, and waste disposal.
"Base bath" (saturated KOH in isopropanol) - This solution eats a fine layer off the glass, which also removes the crud stuck to it. Rinse first with water or neutralize any acid present to avoid vigorous reaction with the bath. Immerse the flask into the bath for a few hours to overnight and it should come out clean. Base bath will dissolve silicone grease. Leave it in too long and it will come out etched. Use heavy butyl gloves.
Concentrated HCl - more powerful than dilute acids, but more dangerous. Neutralize any bases present to avoid vigorous reaction. Use proper PPE.
KOH + KMnO4 + H2O - this solution is basic (eats a fine layer of glass) and oxidative (converts many things, like organics, to a more soluble form). Neutralize any acids present and wear appropriate PPE.
Concentrated H2SO4 - has the added bonus of reacting with some insoluble organics, like polyaromatics, to form soluble sulfonates. H2SO4 is more dangerous than HCl. Follow the same precautions, and be more careful. I would use use butyl gloves instead of nitrile or latex. It will cause serious chemical burns.
"No-Chromix" - a non-chromium alternative to chromerge (below), no-chromix is a proprietary cleaner that is added to sufluric acid. The main ingredient is ammonium persulfate (NH4)2S2O8. You can make a reasonable knock-off by dissolving 2 g of (NH4)2S2O8 in 100 mL of water and adding to 2.5 L of sulfuric acid. Feel free to scale down, as this stuff has no shelf-life to speak of.

The following five items have dangerous reputations, and for good reasons. Use them as last resorts.

Concentrated HNO3 - strongly acidic and powerfully oxidizing. Dissolves many resistant things, including burned-on crud, but may ignite low molecular weight organics (like solvents). Rinse the flask first with water to remove bases, then with acetone to remove organics, and then with water to remove acetone before adding HNO3. Use butyl gloves.
Dilute HF (no more than 5%) - HF eats glass also, and it does so more quickly than base baths. However, the the hazards associated with HF are severe. HF burns are not like other acid burns. The HF continues to bore into you and the damage might not be evident until later. DO NOT USE UNLESS YOU HAVE THE BURN ANTIDOTE HANDY. The burn antidote
is calcium gluconate, which neutralizes the acid and more important sequesters the fluoride ions. It is available as a gel or cream. You should still go overboard on PPE.
Aqua Regia: 1 part Concentrated HNO3 to 3 parts concentrated HCl. This is more powerful than just concentrated HNO3, since HCl is a stronger acid. It also decomposes to form NO2 and Cl2, both more powerful oxidants than Concentrated HNO3 alone. Aqua regia is famous as the only substance that can dissolve gold. It will also dissolve all manner of inorganics and organics, including silicone and hydrocarbon grease. Treat aqua regia with the respect deserved of its two components and keep it in the fume hood. Do not seal the vessels since it gives off gas. Aqua regia has no shelf life, but it remains a strong acid and will need neutralized before disposal. Note that aqua regia will not dissolve metallic rhodium, iridium, osmium, and ruthenium (all less active than gold).
Piranha solution: H2SO4 + H2O2 - this strongly acidic strongly oxidizing solution will eat off a lot of metal, organic, and inorganic residues by oxidizing then to more soluble forms. As a strong acid oxidant, it shares the same incompatibility warnings that concentrated nitric acid has. Piranha solution is good for glassware that needs to remain calibrated (volumetrics) or uniform (NMR tubes) since it does not eat glass. Piranha solution does not have a long shelf life.
Chromerge H2SO4 + K2Cr2O7 - forms chromic acid, a powerful oxidant. This stuff is considered toxic waste in most jurisdictions, meaning it will be expensive to get rid of. You can store it, however, if you store it in teflon. Do not use on NMR tubes, since chromium is paramagnetic. Treat chromerge with a respect due to the sum of everything above it on this list.

Note for Piranha solution: it gives fumes with clear oxidative smell, beware. Note for $KOH + KMnO_4 + H_2O$ do not keep for long or have some $H_2C_2O_4$ to clean $MnO_2$ afterwards. General note: despite $KMnO_4$ being effective oxidizer in acidic solutions never, never make $KMnO_4 + H_2SO_4(cons)$ mixture. The resulting green $Mn_2O_7$ is really dangerous staff.

note for $H_2O_2$ based staff: it degrades quickly in presence of some d-element ions, like $Fe$ or $Mn$.

Another one for the "dangerous" list: Bromium + methanol. Don't ask me about specifics, i've prefered to never use it.

Make up a basebath out of conc KOH in isopropyl alcohol (Or whatever recipe you prefer). Leave it in there for an hour or two, then rinse with distilled water. Most of the time it will come out sparkling clean.
Be careful: Saturated base baths should be handled with care, rubber gloves (dish gloves) over lab gloves are a very, very good idea.

The simplest way is using some cleaning solution. The nature of solution depends on nature of remnants. In my practice, I used

$H_2SO_4(conc) + K_2Cr_2O_7$
$H_2SO_4(conc) + H_2O_2$ (porous glass filters mostly)
$KOH + iPrOH$
$KOH + KMnO_4 + H_2O$
$H_2C_2O_4 + H_2O$ (cleaning $MnO_2$ and $FeOOH$)
$HCl$(2%).
$HCl$ (conc.)
Organic solvents (toluene, alcohol, hexane, dichloromethane)

Again, the nature of cleaning solution depends on what you are cleaning after.


text:
Why is dialysis tubing impermeable to sucrose when it should retain >12,000 Da?
While writing up a lab for biology, I realized that the dialysis tubing for sale online seemed to retain large molecules (12-14,000 Da) but didn't specify an upper bound. I learned that the tubing is impermeable to Sucrose, molecular weight 342 Da, because it's too big. This doesn't make sense to me; the site says the tubing "retains >12,000 Da molecules", but Sucrose is clearly smaller than that.
So either A) Sucrose does indeed pass through dialysis tubing and I misheard/mis-learned or B) my class used tubing with much smaller pores.

So it seems that we used tubing with smaller pores and there is a discrepancy between the metric I was using to compare and the size of the molecules. Thank you to both answers.

There exist many types of semi-permeable membranes (the ones used for dialysis tubing), with various pore sizes. One of the very common lab experiment on the topic of osmosis is using a sucrose solution (sucrose is cheap) and small-pores membrane, such that water and small ions (typically Na+ and Cl–) can pass, but not sucrose. The one which you link to simply happens to be a variety with much wider pores.

Dialysis tubing (and most semi-permeable membranes) operate based on differences of size and not molecular weight. While the size of a molecule or ion does increase as its molecular weight increases, there is some nonintuitive size-mass discrepancies that can occur.
Many of these discrepancies are well known to polymer chemists. Most of the ways that we attempt to measure the average molecular weight of a polymer sample are actually measures of average molecular size. The shape that a molecule adopts in solution greatly impacts its apparent size. A polymer of MW in the 40 kDa range that adopts a rigid-rod conformation in solution will usually have a larger apparent size than a polymer of MW in the 100 kDa range that folds up into a tight globular conformation (spheres have a smaller surface area to volume ratio than cylinders).
Another factor in the "size" of a molecule is the number of solvent molecules that get dragged along with it. If the molecule is dissolved in a good solvent (for example sucrose in water), it will always have a number os solvent molecules associated with it. The solvodynamic radius of a molecule may be much greater that the molecule's actual size.
If indeed you had dialysis tubing rated to retain only greater than 12,000 Da (and like @F'x, I'm not sure that you did) it is still possible for sucrose to not pass through if the sucrose is trying to drag too many water molecules (from hydrogen bonding) with it.

It's a very valid point… but in the case presented, it would be one or two orders of magnitude difference between weight and size.


text:
Why does the passivation layer in DRIE protect just the mask and not the substrate?
Here's a good link as an introduction to DRIE:
BSAC: DRIE Process Development
The $\ce{C4F8}$ (refer Pg. 9 on the link) passivates the entire top layer. But when $\ce{SF6}$ ion bombardment occurs, only the region not covered by the mask is etched. Why does the passivation layer protect just the mask and not the substrate?
Additional reference: Surface Micromachining Techniques and Materials.


text:
Acid-base reaction is screwing up my synthesis
I'm gonna get kicked all the way to the Moon, but whatever.
I am doing some nitroglycerin. I know the process, since the reaction itself is pretty basic. I know that it is an electrophilic substitution and that it releases a hell lot of energy. I've handled that with an ice bath and by doing the respective stoichiometry. I add around 5 ml of glycerol every 10 minutes.
All of that works great, until I have to do the wash. I've learned how to do washing and extractions from MIT's OCW. Nitroglycerin is insoluble in water, so I require water to wash the mixture and pull out the acid. <- There's the problem. When I'm doing the wash, the temperature rises and a runway reaction begins.
Any advice, besides the usual: "Just… don't"?

It's one thing if you take the risk for yourself of synthesizing nitroglycerin. There's nothing we can do about that here, and that sorta covers the "Just ... don't" part you mention yourself. But don't expect anyone to step in and actively aid you in this. Especially given that you don't appear to be a trained chemist, but rather someone who does some whacky home brewing ("I learned how to do ..."). That's like telling a drunk lunatic who wields a gun how to turn the safety off because he couldn't figure that out himself.

Just… don’t, unless you have adequate mentoring and equipment.

Did you think ahead on how you could safely store such amounts of nitroglycerine? It is very shock-sensitive and becomes even more unstable over time. Do you have a container that would survive the explosion of the amount you want to produce? Can you guarantee that nobody else can accidentally get to the stuff and blow themselves up?

Just do a decantation. Simply discard the acid layer and then do the wash.

How did you determine that? More details would be useful…

The nitroglycerine is insoluble in water (even though it is hygroscopic) and it's density is higher than water's and lower than the $H_2SO_4$ and the $HNO_3$. It forms a layer which can be decanted before adding water to pull out the acid.


text:
If gases turn into liquids under pressure, what does carbon dioxide turn into?
Since there is no liquid state of carbon dioxide, what will carbon dioxide turn into when exposed to high pressure?

There is a liquid state for carbon dioxide. Borrowing the $\ce{CO2}$ phase diagram from Wikipedia, we can see that $\ce{CO2}$ will condense at a few atmospheres, dependent on temperature. At still higher pressures, the liquid will solidify. Below the triple point temperature, the gas will transition directly to solid.

just to add a little bit of cultural reference to @Richard Terret's already great answer, solid carbon dioxide is what most of us know more commonly as dry ice.

Carbon dioxide phase diagram,

This is extremely important.  $Trillion cumulative carbon sequestration demands deep "burial" of captured power plant carbon dioxide.  All proposed reservoirs are above 31 C.  Above its critical temperature, CO2 is an astoundingly effective penetrant (right through thick solid concrete for a one-day accelerated cure).  You can pump it down there, but it will come back up.  (Also note, PV = energy, 101.325 J/liter-atm.  Capturing, compressing, and pumping emitted carbon dioxide must consume at least 30% of a fossil fuel power plant's output.)


text:
The effect of structure on the boiling points of organic molecules
Why do molecules like methoxyethane, ethoxyethane and so on have lower boiling points than isomeric alcohols?
I think it has something to do with the structure, but I don't exactly know what it is.

Yes, structure is important in this respect. Most importantly, the alcohols are able to form hydrogen bonds among each other, which is not possible for the ethers. Since evaporation always means breaking intermolecular interactions in the liquid phase, any additional stabilizing interaction raises the boiling point.


text:
Why does hand sanitizer turn into a liquid after coming out of the bottle?
Hand sanitizer is a gel in its bottle, but when I pour it on my palm, it turns into a free flowing liquid. Why does that happen?
It doesn't turn to a liquid when I pour it on my floor or my table. It doesn't seem to be related to heat, as I tried it also with a hot surface.

have you tried wetting it on fruit or a similar organic surface?

Here is my experiment that I performed in 2017 when I was 16.
https://youtu.be/PC3Lv1KuopM
The answer to OP's question is that salt on your hands. If you want to go through the whole process of finding out why salt does this too hand sanitizer is up to you, but now we know it's the salt on your hands from sweating.

A little research and perhaps a small experiment might help you figure something like this out on your own. First, have you tried wikipedia? A small mention of research can go a long way. Describing what product you are interested in never hurts either. Is the polyacrylic acid, a dispersant in this substance? That material can absorb water effectively, so here is what I would do:
Take a very small amount of water and smear it on the same surface that it does not "turn into a free flowing liquid" and see if it does with the addition of water. If it works, you might conclude that your alcohol-based rubs use a suspension material that when contacted with a little water, such as from your palms, will cause the gel to have just enough water absorbed inside to decrease its viscosity.
Perhaps you will like to share the results of your experiment.
EDIT: OP seems to be observing a gel demonstrating Thixotropy, this I have determined from a quick google search

I tried, and it remains a gel even with water smeared on the surface. But I think polyacrylic acid would make the sanitizer more gel-like rather than liquid, because it is a powder and absorbs water to become a gel.

@duci9y Okay well, the only other explanation is thixotropy. I was under the assumption that you were simply putting it on your hand and doing nothing else. Simply rubbing the gel into your hand or allowing the oxygen or gas inside to escape (if any) will cause it to change significantly. The gas inside is used to suspend the materials, such as you would notice in your soap.

@duci9y It is due to Thixotropy. I have attended a workshop in which they talked about thixotropy of commercial products (such as tootpaste) and how to measure it

Thanks a lot an sorry for responding so late. Thixotropy seems the most plausible theory. It is incredible that there is so little research on these fluids.

If it is Thixotropy (shear thinning) then you could see if it is reversible. Shake the closed bottle vigorously, see if it turns into a liquid. The let the bottle stand for say an hour, and see if it has turned back into a gel. You could also compare to putting  a large blob gel on your hand, move it into a liquid, then scrape into a small dish cover it (with plastic wrap, to stop evaporation) and leave it and compare after an hr.

GermX thins when it hits your hands because of the sodium ions present in your hands. The polymer used in GermX is a polyelectrolyte - in this case it is poly(acyrlic acid) that has been neutralized with a base made with a large cation. The large cation is too large to effectively separate and quench the electrical repulsion between the neighboring negative charges on the acrylate monomers. When sodium ions are available, like when it hits your hands or when you add salt water, the smaller sodium ions may exchange with the large cations and collapse the polyelectrolyte. This collapse is thermodynamically favorable both in terms of enthalpy and entropy change. The answers using rheology to back up their claims are erroneous.
Source: PhD Polymer Science and Engineering, and my professor, who patented this technology.

The references you provide are flimsy at best. It would be better to link to or cite verifiable sources, maybe mention who is "your professor".

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


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How do I find change in Gibbs free energy of water formation?
How do I find the change in Gibbs free energy of the reaction
$$\ce{H2 + \frac{1}{2}O2 -> H2O}$$
when it is at $\pu{75^\circ C}$ and $\pu{1 atm}$? Do I calculate the Gibbs free energy of each species and then do
$$\sum\Delta G{\mathrm{(products)}} - \sum\Delta G{\mathrm{(reactants)}}?$$

You're using the correct equation, $\Delta G^0_r(T;P)=\sum \nu_i \mu_i^0(T;P)$,
where $\Delta G_r^0(T;P)$ is the standard free energy change for the reaction at $T$ and the pressure $P$.  The task is to calculate each $\mu_i^0$.  Provided the same convention for the standard state pressure $P^0$ and for the zero value of $\mu_i$ is used, then $\Delta G_r^0$ for any reaction will be the same.  Here are some ways to calculate $\mu_i^0(T;P)$ for each species to use in the equation and the data required to calculate them (available in tables). In the following, $P^0$ is the standard state pressure used in the tables and subscript $fi$ denotes the reaction forming the species from the elements in their standard states.  If the tabular values are available at a different pressure than the one you want ($P$), you must add $RT\ln(P/P^0)$ to the values of $\mu_i$ indicated below.

$\mu_i^0 = \Delta G^0_{fi}(T;P^0)$.
$\mu_i^0 = \Delta H^0_{fi}(T) - T\Delta S^0_{fi}$.
$\mu_i^0 = \Delta H^0_{fi}(298.15) + \int_{298.15}^Tc_{pi}^0(T) dT -Ts_i^0(T)$.
$\mu_i^0 = \Delta H^0_{fi}(T) - Ts_i^0(T)$.
$\mu_i^0 = \Delta H^0_{fi}(298.15) + \int_{298.15}^Tc_{pi}^0(T)dT -T\left[s_i^0(298.15) + \int_{298.15}^T \frac{c^0_{pi}}{T}dT\right]$

There are many other ways.  It's an interesting exercise to prove why all the above routes give the same value of $\Delta G^0_r$ for any reaction.

Warning: this is only a way to estimate and not necessarily good practice.
I never actually did one like this, but I believe that one must use the equation $\Delta G$° $= \Delta H$°$ - T\Delta S$° with the assumptions that the information you will need to look up for $\Delta H$° and $\Delta S$° are temperature independent. This will only allow you to estimate $\Delta G$° and remember to convert to kelvin when plugging in the temp.

This isn't quite right.  You'd think that $\Delta G^\circ$ would equal $\Delta H^\circ - T \Delta S^\circ$, but in fact $\Delta S^\circ$ is usually defined on a scale with a zero point defined by the third law, whereas $\Delta H^\circ$ and $\Delta G^\circ$ are defined to be zero for a pure substance at standard conditions. These incompatible scales mean that equation doesn't hold. Luckily, though, the values of $\Delta G^\circ$ are usually tabulated, so you can just look them up.

@Nathaniel You are probably very correct, I only was offering a way to estimate, unfortunately I am not an engineer so my answer should be more clearly stated as an estimate, not a good answer. I wonder though, how far off the estimate would be from the real answer.

Well, you can always give it a try. A good textbook will have all three values ($\Delta G^\circ$, $\Delta H^\circ$ and $\Delta S^\circ$) for various compounds in a table at the back, so you can try calculating $\Delta H^\circ - T\Delta S^\circ$ and see what you get. My guess is that it will usually be very far away from $\Delta G^\circ$ though - they're on different scales, so it's a bit like adding miles to kilometres.


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What components of safety should be included in a chemistry laboratory experiment conclusion?

The focus of my question here is this: In a laboratory there is a Bunsen burner, a hot plate, hydrochloric acid, and concentrated ammonia. What would you mention about safety precautions?

As students of chemistry and science, we often need to write detailed conclusions about our laboratory experiments. This eventually becomes second-nature, but to have an an idea of which of the safety measures taken to include are often useful to ensure ourselves that we have not left anything out.
I believe that the following are some of the the most important components of any well written conclusion in a lab entry, something that your lab instructor will read and grade you on.

Purpose: Explain the goal and purpose of the experiment in a clear and concise manner.
Findings: Presents a reasonable interpretation of, and logical explanation for all findings pertaining to the problem and stated purpose.
Discussion: Discusses possible sources of error in detail, including their effect on the results and ways of avoiding them in the future.
Referencing experimental findings and explaining the known/expected results we were looking for. Mentioning and discussing reasons for trends, if any.

In particular my instructors last year were often interested in:

Safety: This is often what I always got marked down for. I would explain that hydrochloric acid is a caustic substance and should be treated with care to not get on your tissue by wearing clothing that does not expose skin and closed toe shoes, and to always carry out the experiment with safety goggles securely fastened. It never seems to be enough for them, even if I mention eye flush and chemical shower in case of emergencies. Should I mention to not snort or freebase it? /end sarcasm

Have I missed any safety components that should be included in a "complete" conclusion for the average experiment? What else might I have left out that instructors tend to look for?

Perhaps your instructors want more specific information about the hazard itself, rather than generic safety advice.  Some examples:
1) HCl: Strong acid, highly corrosive (nitpick: caustic is usually reserved for strong bases, corrosive for acids), emits corrosive and toxic fumes (not all acids do)
2) 29% NH4OH: Caustic, emits strong fumes (lachrymator--makes your eyes water), work in a fume hood, outdoors or in a well-ventilated area
3) Bunsen burner: No open flames in the presence of flammable vapors or liquids with high vapor pressure (much bigger hazard than just burning yourself)
4) Hotplate: Hotplate looks the same whether hot or cold, so assume it is hot, take care with flammable liquids with low flash points
Again, these are just examples, but I tried to tailor the answer to the item in question.

Stumbled across this very old thread, but I'm a firm believer in safety. // As former teaching assistant who worked a lot of freshman labs, one of the real problems was safety goggles/glasses. They are like blinders on a horse. So you must turn and look before going off in a different direction.


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Determining the exact charge of ions that will be formed by an atom using the periodic table
I don't know how to obtain the "exact charge of the ions" that will be formed by each atom mentioned. I also have to figure out whether they are positive or negative.
Hydrogen, for example, can form the ion:
$$\ce{H -> H+ + e-}$$
Now if I were to try to get the exact charge for say, boron using only the periodic table, how would I do so?

Each atom tends to imitate the noble gases which have a full valence shell. Looking at your table you can see that boron is 3 spaces above helium, but 5 spaces behind neon. Boron would "prefer" to have a configuration like the noble gases, and because losing 3 electrons seems easier than gaining 5, boron, as boring as he is, decides to do the obvious thing and three electrons walk the plank... Now he has 5 protons and only 2 electrons therefore imitating helium's electron configuration. This means boron has a charge of $+3$.
Oxidation states are common on most periodic tables, which show you the result of less than obvious behavior of atoms when this trend does not always hold. In the case of hydrogen, it is common than it will bond with another hydrogen to imitate helium, otherwise it will tend to form a $+1$ ion.


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Phase diagram for first 10 elements
I've been looking on the net for a reference which shows a phase diagram for elements at least from $\ce{H}$ up to $\ce{Ne}$. Specifically, I'm looking which of these elements can be solid or liquid in a vacuum.
I haven't really found anything, except some anecdotical comments here and there that hydrogen and helium evaporate in a vacuum at any temperature.
Any suggestions as to where these data might be published?

For the most part, I would expect any of the elements that usually exist as diatomic molecules such as hydrogen, fluorine, oxygen, nitrogen. Would fly apart extremely fast since nothing holds them together. While a substance like carbon graphite or diamond is all connected, and the vacuum probably would not break the covalent bonds.

This may be what suffices you are looking for :

Periodic Table of Elements w. Selected Phase Transition Temperatures

This may get you started:

MatDL - The Materials Digital Library Synopsis
NIST SRD ( I would write them)
Java Applet: Phase Transition Temperatures (untested)

It is a very good question, which I came across as well. You could just google-image search phase diagram + ELEMENT to copy and past the most appealing charts into a table in your editor of choice.
If the intend is personal use, this might be the fastest route.
wolfram alpha should allow you to look up the state of the element at a given pressure as follows:
sodium phase at 0 bar

Helium itself is a particularly interesting subject.
I hope this helps for a start...

+1 for link to Wolfram alpha... at least I now have a partial phase diagram for ethanol (who knew it would be so rare?!)

If you want to know the temperature at atmospheric pressure below which an element is a solid, just look up the melting point of that element.
As for evaporation of solids, ALL materials evaporate.  That is, all materials exhibit a positive non-zero vapor pressure.  That vapor pressure will vary with temperature.  When the vapor pressure hits 1 atm, you are at the boiling point.
For phase diagrams of the elements simply Google for them.  One promising source is here

from thermodynamical point of view at absolute zero pressure each and every compound will evaporate, given enough time. So, it is better to stick with finite (i.e. small, but non-zero) pressure.
Then again, phase diagrams usually has 'triple point' (at which solid, liquid and gas coexists) with three areas - lower-right one, where gas is stable, left one, where solid is stable, and upper, where liquid is stable, with barrier between liquid and gas fading at some point. This means, that liquid is doubtful to exist in vacuum.
Given that, you can hope for slowly evaporating solids, which are ones with crystal structure hold by metallic or covalent bond. Evaporation speed should drop with boiling point raise, which drops in the row of $\ce{C}$ (3915 K), $\ce{B}$ (3927 K), $\ce{Be}$ (2725 K), $\ce{Li}$ (1342 K). $\ce{Be}$ is sometimes used as construction material in satellite construction, so it should work fine, but is very toxic. Carbon should work fine as well, as to my knowledge was used in thermonuclear fusion program (where both vacuum and hight temperature stability are required) , but is hard to work with. $\ce{Li}$ and $\ce{B}$ are out of consideration except VERY special application, as $\ce{Li}$ is highly corrosive and flammable and $\ce{B}$ is almost as hard to work with as diamond.

All elements exist in as a solid in at least some pressure and temperature conditions, i.e. in some region of their phase diagram. In fact, all elements present a solid form at zero pressure (or atmospheric pressure) if you cool them enough, i.e. when they reach 0 K. All…  except helium, which requires some pressure to solidify.

sounds promising, but i would like to know at what temperature does this happen for, say, hydrogen, at vacuum pressure

@lurscher in your question, does “vacuum pressure” mean “zero pressure”?

yeah, vacuum pressure means zero pressure


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What meaning do multiple real solutions of a truncated virial equation have, if any?
Let's say I truncate the virial equation after the third term for use as my equation of state:
$${P\over\rho RT}=1+B\rho+C\rho^2$$
I have tabulated values for $B(T)$ and $C(T)$.  I know $P$ and $T$ and want to solve for $\rho$.  The equation is cubic in $\rho$ meaning I could potentially have three real solutions.  I suspect that since the virial equation is intended to calculate the densities of vapors or supercritical gasses at relatively low pressures, that the lowest real root is the "real world" value.  Is this correct?  Do the other real values, if they exist, have any meaning?  Can this be generalized to higher order virial expansions?

I don’t have much practical experience with virial equations, though I teach it :)
My recommendations would be to identify the correct root based on physical knowledge of the system: first, the root has to be real and its value of $\rho$ has to be positive. This part is obvious. The second thing we know is that the virial equation represents a deviation from  the ideal gas law. So, I would take the root that is closest to ideal behavior.
I've tried to confirm it with an example, so I investigate water at 298 K (based on virial coefficients given in K.M. Benjamin et al., J. Phys. Chem. C 2007, 111, 16021-16027). Here's a plot hastily made:

At least in these conditions, this rule works.

OK, if I'm reading your plot correctly, the largest solution decreases with increasing pressure, not a physically realistic solution behavior.  I suppose this is enough to discredit it as a meaningful root.  You have answered my question regarding truncation after the 3rd term although I still wonder about higher order terms.  Since as @Paul J. Gans says below, practically speaking 4th and higher virial coefficients aren't available it is kind of an academic point.

The virial equation is set up for the gas phase only.  The virial coefficients $B$ and $C$ are found from gas phase measurements.
If you get three solutions, I suspect that it is the largest one that you want.  I'd do a trial solution with something like water vapor to see.
Higher order virial equations are problematic since it is exceptionally difficult to determine higher virial coefficients from experimental data.  Higher order virial equations do have theoretical interest, but that's another story.

I think you meant the smallest solution is the one I want since I wrote the virial equation in terms of molar density rather than molar volume.  As an aside, you obtain a quadratic equation in $\rho$ if you truncate after the 2nd term and thus two solutions.  The behavior of this quadratic system is that one solution is near zero (the physically correct one) and increasing with pressure whereas the other is positive and decreasing (the extraneous solution).  So, the quadratic case can be understood rather easily.

You are quite right. It is the smallest value of the density. My bad. Apologies to all.


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How much of the anion species is left over after the first precipitation?

Some barium chloride is added to a solution that contains both $\ce{K2SO4}$ ($\pu{0.050 M}$) and $\ce{Na3PO4}$ ($\pu{0.020 M}$).
  (a) Which begins to precipitate first: the barium sulfate or the barium phosphate?
  (b) The concentration of the first anion species to precipitate, either the sulfate or phosphate, decreases as the precipitate forms. What is the concentration of the first species when the second begins to precipitate?

I've figured out the first part of this problem, that barium phosphate is the first to precipitate (due to lower $K_\mathrm{sp}$), but I can't seem to figure out the other part. I even read about selective precipitation and I can't get the answer that was in the back of the book (namely, $\pu{1.1 \times 10^{-6} M}$).
Granted, sometimes I'm pretty there are errors in the books I use to teach myself, but this is not one of those instances where I am so sure.


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Maintaining a 40% CO₂ aqueous environment?
I wish to grow algae in a 40% $\ce{CO2}$ (water under air with 40% $\ce{CO2}$ content) aqueous environment.
Simple DIY methods are available, but they are generally unreliable (For more info on DIY read here) because they rely on a pH/KH chart.
Atmospheric $\ce{CO2}$ levels are measured based on IR absorbency. However, such measurements are difficult and "complicated by the temperature signal generated by carbon dioxide as well as by the signatures of ozone and water vapor."
Is there a reliable way to measure the $\ce{CO2}$ concentration that is relatively inexpensive (under $1,000)? Is the drop checker method reliable?

ergh. Please, clarify: do you need 40 weight % water solution of $CO_2$, water under air with 40% of $CO_2$ with atmospheric pressure or something else?

there are companies, providing industrial-level solutions. Home-brew solution should include arduino board, $CO_2$ gas cylinder, $CO_2$ sensor and electronically controlled valve.

Unfortunately, I feel that the answer you desire could be a hair too involved to set up a home-made system which appears to be your desire in this case. I can guarantee you a wide family of correct answers to your problem await via a quick google search, but, and please understand that I only wish to help here, I fear that two sentences and a bounty are not a significant step towards what is likely an in-depth solution of the many out there. I wish you the best of luck however, and hope you revise your question soon.


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Why do H₂ and O₂ create H₂O?
Both $\ce{H2}$ or $\ce{O2}$ are stable molecules. However when given energy they heat up and form $\ce{H2O}$. My teacher explained that the increased energy breaks the bonds, allowing new ones to form. However, why don't the old forms reform instead? Why are new bonds created?

Are you familiar with activation energy?  Consider a hole at the top of a large hill.  If a ball or some water was in that hole it would seek the lowest point it could find.  Clearly, the bottom of the hill would be preferable to the bottom of the hole at the top of the hill, however, in order to get there it must first be brought up over the top of the hole.

In this picture, the $H_2$ and $O_2$ are on the right side of the curve.  Just like water in the above analogy, they'd like to move to the lower area, but first must be brought up over the bump.  So some energy must be added to accomplish this, and then they end up in the much lower energy position on the left.
Once over the bump, they stay as $H_2O$ because that is the state they'd like to be in at any time, but were prevented from being in before by the activation energy bump.

Thank you. One question: why is $H_2O$ considered to be the lower part of the hill? In other words, why does $H_2O$ have less potential energy?

The hand waving answer is that the total energy in all the bonds of $2 H_2O$ is more than the total energy in $2 H_2 + 2 O_2$.      http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html

That is not hand waving at all, it is thermodynamics.  Enthalphy gains offset entropy losses.

Adding on to DaleSwanson's correct answer: The ball-and-hill analogy is just saying that a ball will roll down a hill. This is because if the path is available it will find a way to the bottom of the hill due to there being less potential energy in the ball down at the bottom than up at the peak. Think of the activation energy as the little 'push' needed to get the ball moving off the edge of the cliff, once it is nudged off the side it will release almost all the potential energy that was released through kinetic energy when it hits the ground (small amounts of potential energy will also be released in other forms, i.e: friction of the air and ground).

To answer your question, I ask you this: you accidently knock a pen off your desk, it required so little energy it was almost unnoticeable. Would it require more energy for you to reach down and put the pen back on your desk? What position is "more stable" for the pen?

Hydrogen and oxygen are stable, yes, they have full valence shells and are satisfied. Just like a ball sitting on the top of the hill is stable, until you give it a nudge. Now think of the top of the hill as the hydrogen and oxygen molecules. Also, think of the bottom of the hill as the water molecule. Take a look at the balanced chemical equation now:
$\ce{2 H2 + O_2 -> 2H_2_O}$
There is less energy at the bottom of the hill, the same way that there is less energy in the covalent bonds that make up the products. There is more energy stored in the reactants.
It is very possible to calculate the energies of the bonds and compare the energies of the reactants to the products in the balanced chemical equation by looking up heat of formation's for these species, and calculate the change in energy from going to reactants to products. That is, we can calculate how much energy is released when two moles of $\ce{H_2O}$ are created from this reaction.
Similarly we can calculate how much energy a ball will release when falling from the cliff. The release in energy to the surroundings during a chemical reaction is said to be an exothermic process, whereas an endothermic process would take away energy from the surroundings, such as vaporizing water molecules into steam.
Now using the analogy you have learned, would it be reasonable to say that an equally small nudge that caused the ball to fall would be enough to cause the ball to rise up back onto the peak of the hill? You tell me.
A better picture of the path from reactants to products would be different than a smooth 2d hill picture. Sure, the change in energy is not as outrageous as I portray it, and perhaps in some miraculously rare situation some invaders from another country find the ball at the bottom of the hill, they put it in a catapult and launch it at their enemies at the top of the hill. This is extremely rare and very situational. In other words there is almost no chance it will happen. As it turns out, you are thinking very rationally when you ask why the reverse process does not happen! The problem is you have to think on the molecular scale where molecules are colliding into each other. It is very likely that a $H_2$ will collide with the $O_2$ and one side of the oxygen breaks into the bond of the hydrogen and you now have $H_2O + O$. But how would seperate $H_2 + O_2$ form from two $H_2O$ colliding, the answer is that it is just extremely unlikely, it would more likely create $H_2O_2 + H_2$ if at all! There are other steps in-between that would need to account for going backwards in the reaction. You can think of this as a separate path around the hill. There are more ways back up the hill in other words, but they may lead to dead-ends!

Thank you. One question: why is $H_2O$ considered to be the lower part of the hill? In other words, why does $H_2O$ have less potential energy?

@Farhad Yusufali The strength of the bonds in the products are stronger than in the reactants. The heat of formation for water is negative (the system, or chemical process, lost/gave energy to the surroundings). For elemental hydrogen and oxygen gas it is a zero value (kJ/mol). Engineers think of energy change as positive when the rest of the universe gains energy, we think of it as the other way around (positive when the system gains energy).


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Are there ways to predict which chemicals will kill certain bacteria?
Are there ways to compute what chemicals kill a give bacteria but not another given bacteria ?
Does that help much to make medication ?

The basic idea is sound, and is used to some degree; however, there are significant difficulties inherent in "tailoring" a chemical to selectively target certain organisms and not others.
There are three things that a chemical can do that make it toxic to cellular life:

Rip some vital chemical to shreds. Chlorine and other light halogens (fluorine, iodine) are huge here, because they have such high electron affinity that they can tear apart the otherwise relatively stable bonds within most organics, destroying that molecule. The most effective ones don't form any stable compounds; fluorine, for instance, is very effective at shredding organic compounds because it isn't happy just barging in on any one organic molecule and making itself at home; it'll deprotonate this one, split that one in half, before finally binding tightly and insolubly to calcium (robbing your body of that essential mineral).
Bind to a vital chemical and not let go. Carbon monoxide is a deadly danger to most aerobic organisms because the hemoglobin (or related compound) that loosely binds oxygen for transport to cells will bind more stably with CO, so the chemical can't transport oxygen and can't get rid of the CO. A lot of more selective poisons attach themselves to enzymes, not necessarily destroying them, but preventing the enzyme doing the work it's supposed to. Cyanide is poisonous specifically because it works like (well, it is) a nitrile group, of which there are very many in a lot of inorganic biochemicals, and specifically it targets a key enzyme involved in cell respiration and programmed cell death.
Mimic a chemical's interaction with a key receptor, causing it to go haywire. Many other "selective" poisons and antibiotics work this way; they fit into the receptor sites that a neurotransmitter or enzyme would, triggering some behavior of the cell, and by flooding the cellular environment with this chemical you cause the underlying programmed reaction to go haywire (or stop completely). A lot of insecticides and pesticides do this; they increase or shut down cell metabolism, or in creatures with a nervous system they mimic or trigger the release of massive amounts of neurotransmitters, basically causing cardiac or respiratory arrest because any normal message to control these actions is completely overridden.

The problem with all this comes in identifying which bacteria have a particular enzyme, receptor or other targeted chemical process, and thus would be susceptible to poisoning by the chemical. In addition, the chemical must not be toxic to human biochemistry (otherwise, no matter how beneficial it might be as a disinfecting agent, it will never see approval as an antibiotic medicine).
Antibiotics must have selective toxicity; while they do not affect eukaryotic cells (such as the ones that make up most plants and animals, thus allowing the chemical to be taken internally or directly applied), they have a much more pronounced effect on bacterial and archaeal cells. Usually, the chemical targets something common to a large class of bacteria (such as the thick peptidoglycan mesh forming the cell wall of Gram-positive bacteria), or most/all bacterial cells (such as the S30 robosomal subunit, key to bacterial protein synthesis and thus found in nearly all bacteria, but completely absent from eukaryote cells which use a different mechanism). However, even within an antibiotic's "target class", there are bacteria that are naturally resistant (perhaps they don't accept the chemical through the cell membrane due to some variation unique to the genus) or develop genetic resistance through overuse or misuse.
Also, bacteria are fickle; most are at least harmless if not beneficial in certain areas (E. coli, for instance, produces the essential vitamin K2 in your intestine), but lethal in others (ingestion of e-coli or introduction through an open cut can cause a fatal infection). Staph is relatively harmless on the skin, but fatal in the spine. Antibiotics don't stop and question bacteria; they kill any bacteria that comes across them, and so overuse of antibiotics to kill one harmful bacteria will kill the bacteria in beneficial places as well.
Lastly, not every infection is bacterial. Fungi and protozoa produce a significant portion of food-borne illnesses and poisonings, as well as other types of infection. These life forms are eukaryotes and unaffected by most antibacterial agents (in fact, penicillin was discovered when a fungus contaminated a bacterial culture; the fungus produces the antibiotic chemical as a natural defense).


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How useful is the assumption that water doesn't auto-ionize in the following?

Now in the first row of the iCe-table, $[\ce{H3O+}]$ is said to be 0. But water autoionizes to some extent, so it shouldn't really be 0. I accept that, because the pH of this solution is 2.5, the autoionization effect is negligible but are there instances where this can't be ignored (e.g. where the pH is close to 7?)? Academic, industrial (or hobby) examples would all be fine.

First, while it is useful to know approximate methods of solving aqueous solution equilibrium problems, you have to realize that in a professional context, these methods are not used other than for back-of-the-envelope calculations. There exist software that perform that kind of calculations without introducing approximations, and they are widely used. So, to answer the question in your title: such approximations are useful for simple calculations, and for those only.

Now, let’s try to evaluate the effect of neglecting auto-ionization of water on your problem.
You can write the full system of equations describing the aqueous equilibrium:

conservation of mass: $[\ce{HA}] + [\ce{A-}]=c_\ce{A}$, where $c_\ce{A}$ is the analytical concentration in chlorobenzoic acid (0.1 mol/L).
electroneutrality: $[\ce{H+}]=[\ce{A-}]+[\ce{HO-}]$
water autoionization equilibrium: $[\ce{H+}][\ce{HO-}]=K_\text{e}$
acid-base equilibrium: $\displaystyle K_\text{a}=\frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}$

You can solve this system of equations, as you have four equations and four unknowns ($K_\text{a}$, $[\ce{A-}]$, $[\ce{HA}]$ and $[\ce{HO-}]$). The answer gives $\mathrm{p}K_\text{a} = 9.19$ (given the precision given in experimental measurements).
You don’t say what figure you obtain, but I can take the same system above and simulate “neglecting auto-ionization” by setting $K_\text{e}=0$. Doing this, I obtain $\mathrm{p}K_\text{a} = 9.18$, which means the error involved in your approximation is 0.01 on the $\mathrm{p}K_\text{a}$. So, this is indeed a good approximation.

Oddly enough, I just reached that sort of system of equations in my reading. Mea culpa.


text:
Does this set of quantum numbers have a unique solution?
I received the following question:

Complete the missing values for the four quantum numbers:
  $n=?$, $\ell = 2 , m_\ell = 0$ , $m_s= ? $ .

The problem is that I think $n$ can be any number greater then 2, and $m_s$ can be $\frac{1}{2} $ or $-\frac{1}{2}$, so there seems to be no unique solution to this problem.
Is this correct?

You haven't gotten anything wrong, you are entirely correct!
A set of quantum numbers can be used to identify the specific electron in the sub-atomic world of an atom, and what path it occupies in three dimension.
$\ell = 2$ refers to the $d$ orbital. Now how many electrons can a d orbital contain? It has 5 "slots" because we can go from -2, -1, 0, 1, 2. These represent the orientations, each one holding up to two electrons at different spins (up or down, -1/2 or +1/2). We therefore know that $\ell = 2$ refers to a possible 10 different unique sets of quantum numbers, because each set could represent a specific electron, of which there could be up to 10 in the d orbitals, only 6 for p orbitals, etc.

The principal quantum number ($n$): Has only integral values (1, 2, 3, etc) starting at 1. Think of this as the power level. An electron in a 3s, 3p, 3d, etc orbital is in the third power level. It represents the numbers before the letter in the electron configuration such as 1s$^2$2s$^2$2p$^6$ so in other words, it represents what level the electron is on.
The angular momentum quantum number ($\ell$): goes only from 0 to $n-1$, also uses integral values like $n$. Think of this as the letter after the number in the electron configuration such has 1s$^2$2s$^2$2p$^6$ so in other words it represents which orbital we are talking about on that power level. 0 represents $s$, 1 for $p$, 2 for $d$, 3 for $f$ etc.
The magnetic quantum number ($m_\ell$): Goes from $-\ell$ to $\ell$ including 0. Think of this value as representing a particular orientation of an orbital. The number of orientations will vary depending on the orbital. s orbitals are spherical and can only have one orientation, s orbitals are referred to by saying $\ell = 0$. If we said $\ell = 1$ we would be talking about the p orbital, which has -1, 0, and 1 orientations (3 total). This can be thought of as the x, y, and z axis. $d$ orbitals have 5 possibilities, $f$ has 7, etc. I like to refer to each of these orientations as a "slot" which can hold two electrons.
The value for $m_s$ is used to identify the spin either up or down, but not both.


text:
Lewis structures of molecules containing oxygen
As far as I know, when writing Lewis structures for molecules, we need to consider the octet rule and the valence electrons each atom has.
But, if we take for example, the molecule $ Cl O_2^- $ , we have the following Lewis structure: (as it appears here )
O - Cl - O with 6 valence electrons surrounding each oxygen atom, and 4 valence electrons surrounding the chlorine atom.
My problem is:
The oxygen atom has 6 valence electrons. By forming a chemical bond with the chlorine atom, it shares one electron, which leaves it with a total of 5 non-bonded electrons. So, how can it be that in the Lewis structure of this molecule, it has 6 non-bonded electrons and one shared electron? (or a total of 7 electrons! )

If you consider the $\ce{ClO2-}$ as coming together from neutral atoms (plus the extra electron), all the bonding electrons "come from" the chlorine. This is commonly called a 'dative covalent bond' (or just 'dative bond' for short), and is sometimes drawn like this, instead of just a line:

As this image suggests, you can think of it as an atom with a lone pair bonding with an atom with no spare valence electron.


text:
What is the correct lewis structure of SCH⁻?
I need to find the Lewis structure of the ion $\ce{SCH^-}$.
I think the Lewis structure should be:
$\ce{S=C-H}$, with one lone pair on the carbon and two lone pairs surrounding the sulfur.
Is this correct?

This model seems reasonable, as you can see from the formal charges carbon has the -1 value while the rest are 0. One thing that you have forgot is to add brackets around the structure to show that it is a ion, include its overall charge.
$$\ce{[\overset{..}{\underset{..}{S}}=\overset{..}{C}-H]-}$$
Although, I suppose a structure of $\ce{[\overset{..}{C}#\overset{..}{S}-H]-}$ would work with the same formal charge, except that sulfur has the expanded octet, which is possible but I am not sure if that is more reasonable or not, it probably is not.
It is a very odd ion, to say the least, probably very unstable. Perhaps you have mistaken the hydrogen with nitrogen? It would be more reasonable to be asked to draw the lewis structure for $\ce{[SCN]-}$, thiocyanate ion. I am not even know how to pronounce or name the $\ce{[SCH]-}$ion you are required to draw, or if it even exists. Maybe that is why you need to draw it though.

That second structure of $\ce{CSH-}$ you propose should be purged into hell and it is about as unlikely.

The ion in question is the deprotonated form of thioformaldehyde. While Lewis structures are in general not a very accurate way to depict the molecular structure, it should be made clear, that the molecule is not linear.* It is also necessary, that the formal charges within the molecule add up to the overall charge of the ion. Hence the probably best Lewis structure is the following:

Please note that the expanded octet around sulfur, which would involve the d orbitals of sulfur, has been disproved, therefore an alternative structure as proposed in another answer is not correct.

Footnote: Just to make sure, I have run a conformational analysis using CREST and GFN2-xTB, see GitHub. It yielded one bent structure, which I have further optimised with Orca 4.2.0 on the RI-BP86-D3(BJ)/def2-SVP level of theory. Normal coordinate analysis confirmed no imaginary frequencies. If you would like to visualise the ion, e.g. with Avogadro, below is a simple Xmol file, including the absolute energy on the given level of theory, and a rendered image.
3
FINAL SINGLE POINT ENERGY      -436.706625474783
S        0.024944000      0.000000000      0.050790000
C       -0.062532000      0.000000000      1.757406000
H        1.046395000      0.000000000      2.108467000

I had a feeling but could never be sure. Do we have any references here, provided this is a bit rare to come across? And do we have some basic sites that are mostly accepted on Chem SE? Many high-rep users include reliable sources, but are mostly out of our search results, but would be nice if one could cross-check facts with those references.

@SirArthur7 I'm not entirely sure what you mean, i.e. what kind of references would you like. For this kind of homework exercise, I really do not want to go into too much detail about the electronic structure.

I'm sorry. I just kinda wanted to know if this ion finds a mention in some article or so. But you went far more than what I could ask for. Thanks for your time :) this got a solid response now. I was actually asking something else too, maybe kind of a meta post, where we could list down some sites that are mostly considered reliable sources here on this site. I saw some meta discussion going on about this, we have one for books and resources for learning chem at different levels, another could be useful too.


text:
What does it mean when "a gas is dilute in CO₂"?
I'm reading an article about carbon sequestration, and I came across a strange phrase:  "The flue gas is dilute in $\ce{CO2}$".  I can't quite ascertain its meaning from the context.  I think it means that the $\ce{CO2}$ is in low concentration in the flue gas, but I'm no chemistry expert.
Does this have a technical meaning in chemstry when a gas is dilute in another substance?

Thank you for including the article, my guess is that they are saying there is "not enough" $CO_2$ from what I can read, judging by the line following which reads "...resulting in... Low $CO_2$ partial pressure." The context is "barriers to implementation" which I assume... means not helping us scrub it.

@Leonardo: I added the link to the article.  The phrase is included at the top of table 1, under "Barriers to implementation".

Yes, kind of counter-intuitive that there is more $CO_2$ before combustion rather than after. Looking at the chart it seems to have all the answers but I have never looked into the process which is why I am only commenting. I think a good question would be about this process itself and why is there more carbon dioxide before combustion, and this is coal burning right? This is looking at the middle flow-chart for pre-combustion.

In this case, it's comparing post-air-combustion $\ce{CO2}$ capture with pre-combustion capture, and with post oxy-combustion capture.
In the two latter cases, the  $\ce{CO2}$ stream is very highly concentrated.  However, with air-combustion, the flue gases contain a lot of nitrogen and NOx; hence the $\ce{CO2}$ is diluted in the flue gas, compared to the other two methods, where it's concentrated.
And as to Leonardo's supplementary question in the comments, about why the pre-combustion  stream has higher concentrations of $\ce{CO2}$ than the post-air combustion stream:
the comparison here isn't between two points in the same process. It's between analagous points in two completely different processes.
With post-combustion capture, the $\ce{CO2}$ is stripped out after the hydrocarbons (coal, gas, oil) have been burnt in air.
Whereas with pre-combustion capture, the hydrocarbon goes through a gasifier/shift process, resulting in a stream of $\ce{H2}$ mixed with $\ce{CO2}$. And at this point, the $\ce{CO2}$ is stripped out, and the $\ce{H2}$ is burnt.
And that's why the $\ce{CO2}$ concentration is higher in the pre-combustion capture process, than in the post air-combustion capture process.


text:
How do amines react with flue gases to absorb CO₂?
I'm reading this article about carbon capture from the combustion of fossil fuels.  In this article, it says that amines are used as a chemical absorbent to capture the $\ce{CO2}$ from the exhaust of the combustion.
In particular, I have read that N-Methyl diethanolamine (MDEA) is used (a  reference  lists the formula as $\ce{CH3N (C2H4OH)2}$).  I am curious if anyone can explain what happens during the reaction with $\ce{CO2}$ and how it absorbs the $\ce{CO2}$ from the flue gas.

In addition to the reaction in permeakra's answer, some amines may react with $\ce{CO2}$ in absence of water also. This type of reaction is a Lewis acid-base reaction forming a zwitterionic complex: $\ce{R3N + CO2 -> R3NCO2}$.

MDEA adds some stability to this complex. The alcohol groups can stabilize the carboxylate group through hydrogen bonding.

All amines are proton bases. $\ce{CO2}$ in water containing solutions acts as a weak acid. An overall reaction is $\ce{NR3 + H2O + CO2 -> [NHR3][HCO3]}$ and is possible for all amines. Aminoalcohols are cheap as they are produced from easily available oxirane (ethylene oxide) and have a high boiling point, so for them this reaction can be reversed by heating, with the regeneration of working solution.

I don't think this is clear enough for someone who obviously does not have much chemical knowledge. Also, your reaction is wrong, and should contain charged species.

I see nothing wrong with permeakra's reaction. The reaction does contain charged species, $\ce{NHR3+}$ and $\ce{HCO3-}$. However, like all ionic formulae, when you write the two ions together, we often choose not to write the charges. For simplicity, we often write $\ce{HCl + NaOH -> H2O + NaCl}$ and not the full ionic equation.


text:
Understanding dissolution of CO₂ in Selexol and Rectisol
I'm trying to understand pre-combusiton carbon capture processes as described in this presentation.
I'm told that solvents such as chilled methanol ($\ce{CH3OH}$) and dimethyl ethers of polyethylene glycol ($\ce{C_{2n}H_{4n + 2}O_{n + 1}}$) can dissolve $\ce{CO2}$ from a gasified coal fuel (before combustion) under the action of increased pressure.  I'm struggling in understanding how the increased pressure induces the dissolution of $\ce{CO2}$ from the gasified coal.  I'm also a bit rusty with my chemistry, by saying $\ce{CO2}$ is dissolved by the solvents, what is really going on, chemically?

dissolving something is really something simple that you can find on wikipedia. When water dissolves a salt the ions are seperated and surrounded by water molecules. Anything can dissolved with a proper solvent. The molecules in the air we are breathing is dissolved in nitrogen if I recall correctly, although I do not think bonds other than ionic bonds are broken in dissolutions. Think of the solvents you mention as the policemen tackle the bad guy, more pressure means more cops nearby I suppose.

@Leonardo: is it proper to say that the CO2 chemically bonds to the solvent?

Honestly I can not be sure and say that no chemical reaction happens. For the most part, when a solvent dissolves a solute, ionic bonds tend to be broken (CO2 is covalent not ionic) and what is usually left are the intermolecular attractions, but in the case of the CO2 I would not like to mislead you by saying that no chemical bonds are formed with the CO2, although that is my guess due to pressure being one of the ingredients.

Simply, $\ce{CO_{2}}$ is dissolved in those solvents in a similar way to gases dissolved in water. The molecules do not react with the solvent but mix. The higher the pressure of the gas you pump into the liquid, the more that dissolves in it (rather than turning into bubbles and escaping). In fact, amount of gases dissolved in liquids can be measured both as a concentration (mol/L) or as a partial pressure (Pa).
Henry's law states that:

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

The formula is:
$p = k_{H}c$
where $p$ is partial pressure of the gas in solution, $c$ is its concentration and $k_{H}$ is a constant based on what the solute is (in this case $\ce{CO_{2}}$), what the solvent is (eg methanol) and the temperature.
The harder you push, the tighter you can pack it in.
Here is a picture:

In contrast, with the amine-based solvents, the $\ce{CO_{2}}$ actually reacts, which is why a degasification process needs to happen to release it.


text:
Lewis structures of H₂(C)xCH₂ molecules
I need to calculate the angles between CCC, CCH, HCH in the following molecules:

ethene ($\ce{H2CCH2}$),
propa-1,2-diene ($\ce{H2CCCH2}$), and
buta-1,2,3-triene ($\ce{H2CCCCH2}$).

If I'm not mistaken, the Lewis structures of all of these molecules have the carbons in the middle, connected by double bonds to each other, and the hydrogen atoms connected to the carbons in the edge. For example:
H           H
 \         /
  C = C = C
 /         \
H           H

The carbons at the edge are $\mathrm{sp^2}$ and the middle carbon is always $\mathrm{sp}$ .
This means that the angle CCC is always 180°. But how can I determine the angles CCH, HCH?
In addition, how can I determine if the hydrogen atoms in the three molecules lie on the same plane?

What are the bond angles in a normal trigonal planar arrangement? Treat the center atom as in H C H as a trigonal planar arrangement with the middle (carbon) atom in the center with a third atom attached on the left over position (another carbon). What angle maximizes their separation?
Your other question about the plane of the molecule is a bit more interesting. The pi bonds between the carbons ensure that the molecule is not rotating in space. To maximize the distance between the terminal hydrogen atoms, one set should be aligned up and down, while the other two on the opposite side should be rotated 90° in space, otherwise they would ellipse one another. In other words, one pair is horizontal while the opposite is vertical for example, depending how you want to imagine it.

Does the answer to your first question is 120 degrees?
As for the second explanation, is there any picture or a diagram that can be explain this structure? I can't really understand it completely...
Hope you'll be able to help
Thanks !

Yes, I get the same answer as you.. just out of curiosity though, why not google "trigonal planar" and the answer is literally the first thing your eye should notice. As for newman projections, I am sure you will enjoy some khan material. In your case, I recommend you keep watching every video until confusion is eventually obliterated. Good luck!

Be aware, that in $\ce{H2C=CH2}$ all hydrogens are in plane, and the same holds for $\ce{H2C=(C=C){$n$}=CH2}$ because of symmetry reasons. For any $\ce{H2C=(C){$n$}=CH2}$ the alignment of the hydrogens is perpendicular. You can explain this with the pi orbitals.


text:
Galvanic corrosion between nickel brass nut and stainless bolt
I am planning to install something on a boat that will be in salt water, and my question is, will the nickel plated brass fastener not corrode if I connect it to stainless steel?
I looked it up on the chart and it says that there should be less then 0.15 V difference in the "Anodic Index" for harsh environments and while brass and stainless is within the 0.15, nickel is just outside.
The environment is not entirely harsh as well, as these will be used to fasten just the windows, so while there could be water there, mostly it will be fresh water from rain and the occasional spray from salt water.

While I cannot comment in a technical level, I can speak from a background in aerospace experience (in my family) and lifetime of building things that live outdoors.
In time, dissimilar metal corrosion always sets in. Most stainless steels are inhibiting, not a 'way out' of the problem.
On top of that, I think the ambient salts will kills you on this in the marine environment.  If you have a regular maintenance plan, then perhaps you can re-service the bolts yearly, as needed.


text:
Will the current at the inversion potential of a CV trace asymptotically approach zero?
Reading up on the theory of cyclic voltammetry, I note that the current at the inversion potential (that is, the point at which the applied potential goes negative) in every reversible CV trace I have seen is nonzero. My intuition says that this is related to the finite speed of enlargement of the depletion zone around the working electrode and suspect that given sufficient time this current will drop away to zero as the bulk solution homogenises. Is this correct? If not, what accounts for the current?

Do you mean the point at which the $\Delta$ of the applied potential goes negative? Or do you mean the point at which $V_{\text{app}}$ goes back to $V_0$? I think you should add a graph to your question!

I'm re-reading Skoog and Holler, ch. 25. Extremely interesting stuff. From what I remember pre-reviewing, CV studies diffusion controlled redox processes, so yes, the current should tend to a minimum, but definitely not 0. There are basically two currents, faradaic and non-faradaic (or capacitive, because of the capacitor that forms at the electrode boundary), competing, the latter being impossible to eliminate. I can't offer a better answer, but you got me interested in Voltammetry again!


text:
Why are atoms with eight electrons in the outer shell extremely stable?
Atoms that have eight electrons in their outer shell are extremely stable. It can't be because both the $s$ and the $p$ orbitals are full, because then an atom with 13 or 18 valence electrons would be extremely stable. ($d$ has 10, and 5 is also stable).
Why is it that atoms with eight electrons in the outer shell are extremely stable?

Related question: http://chemistry.stackexchange.com/questions/444/can-an-atom-have-more-than-8-valence-electrons-if-not-why-is-8-the-limit

First, this isn't quite true.  It is true for the first row of the periodic chart (from lithium to neon). It is almost true for the second row (from sodium to argon.  But there are exceptions there.  Beyond that it really isn't true at all for the elements beyond the first two columns.
The reason for the increased stability for the first two rows lies in quantum mechanics.  Classically we can note that there are no $d$ electrons there.  Another ways of looking at it from a classical point of view is that the early elements are too small to allow too many other atoms or groups of atoms around them.  That tends to go away as you go down the periodic chart and the atoms get "fatter".   A typical example is chloroplatinic acid which has six chlorines around it.
Most transition metals also can have more than four groups around them as well.
I suspect that this isn't an exceptionally useful explanation.  As I said, the answer really lies in quantum mechanics.  In looking up "molecular orbital theory",  one reference can be found here.

This video nicely explains your question.
The essence is that paulis exclusion principle states that two electrons(or generally fermions) cannot occupy the same quantum state.
This is because electrons are fundamentally indistinguishable,but their wavefunctions are not. So there must be some quantum factors which allows to distinguish between their quantum states. In general we say that quantum properties of electrons can be conveyed by some numbers which we call the quantum numbers of that electron.
The distinguishablity in the first shell electrons is assured by spin quantum number. They can have $+1/2$ or $-1/2$ spin. So, there are 2 electrons in the first shell. But in the second shell,there are additional quantum numbers like azimuthal quantum number and magnetic quantum number, that can vary which allows for upto 8 electons in a single shell.
So,why are 8 electrons stable? because the protons and electrons attract each other.so,the atom in which all electrons  are maximum filled is the most stable system.

The essence of your answer is “because protons and electrons attract each other,” which I find wholly unsatisfactory. This doesn’t explain the favorable energy state of noble gases, the electron affinity of chlorine, etc. There must be some fundamental law governing the energy of various electron states that causes certain electron configurations (ie, a full octet) to be more stable than partially filled s and p orbitals.


text:
How is the yellow colour of nitric acid removed?
Nitric acid obtained in laboratory is slightly yellow due to the dissolution of nitrogen dioxide which is produced due to thermal decomposition of a portion of $\ce{HNO3}$.
$$ \ce {4HNO3  -> 2H2O + 4NO2 + O2}$$
Its says here that if $\ce{CO2}$ is bubbled through the acid, it turns colourless- because it drives out $\ce{NO2}$ from warm acid which is further oxidized to nitric acid.
What does it mean by drives out ? How can carbon dioxide drive out nitrogen dioxide that's dissolved in the acid? And also, the oxidation of 'what' is being talked about here?

Look up "Henrys law"

I think you can explain this using the equilibrium:
$\ce{NO_2(g) <=> NO_2(dissolved)}$
When you pass $CO_2$ through the warm acid, the partial pressure of $CO_2$ ($P_{CO_2}$) above the gas increases (Since partial pressure generally represents the amount of gas above the solution).
Since

total pressure $P = P_{NO_2} + P_{CO_2}$ (Dalton's Law)
And $P$ remains constant at atmospheric pressure (assuming that the system is not closed).

$P_{NO_2}$ will decrease.
According to Le Chatelier's Principle, the equilibrium proceeds backward to compensate the decrease in $P_{NO_2}$ and thus more $NO_2$ gas is released. The gas flows quickly along with the bubbled gas and this prevents the forward reaction.
This seems like quite a roundabout method but it works. I'm sure there could be a much simpler explanation though.
The oxidation mentioned is most probably the disproportion reaction of $NO_2$ when it comes in contact with water:
$\ce{3NO_2 + H_2O -> 2HNO_3 + NO}$


text:
Energy of the n-th level for an atom
Going through the Bohr's model and his assumptions, I came across with this formula to find the energy of the n-th level of any atom:
$$E = - \frac{Z k_e e^2}{2r_n} = -\frac{Z^2(k_e e^2)^2m_e}{2\hbar n^2} \approx \frac{-13.6Z^2}{n^2}~\pu{eV}$$
Now, let's say we take as an example the atom of hydrogen, the energy of the $n = 1$ and only level is ($\pu{-13.6 eV}$). Now, according to Wikipedia (not really a reliable source) this can be interpreted as

An electron in the lowest energy level of hydrogen ($n = 1$) therefore has about $\pu{13.6 eV}$ less energy than a motionless electron infinitely far from the nucleus.

I don't really understand why there is a "minus" in the result. Why does it say that it has less energy than a motionless electron when he, himself, suggested that the electron's acceleration does not result in radiation and energy loss?
I also understand this formula only worked for the atom of hydrogen (that's why I took it as an example) and due to this problem and that's why the Sommerfeld model was postulated.

It's like an asteroid falling toward Earth. Potential energy converts to kinetic, until it crashes, then the energy is found in whatever aftermath there is.  The total energy is still zero. The "falling" from a great distance is a change from high potential energy to low. By convention, that higher level is assigned zero, so the lower energy must therefore be negative.
For an electron and a bare nucleus far apart, and not counting any nuclear or non-chemical phenomena, we assign zero as the total energy.  When the electron is allowed to fall, it falls faster and faster, potential energy converting to kinetic, until "crashes" - it enters the 1s orbital, n=1, releasing a photon of 13.6eV. (In constrast, an asteroid makes a big crater, stirs up debris, and creates tremendous heat.)  The photon carries 13.6eV, and so the newly formed hydrogen atom is -13.6.  This is  called binding energy, and is negative by the convention physicists and chemists have followed since long ago.
BTW, that formula is only for any atom with exactly one electron, and Z not too big.  With two electrons it no longer applies, and for the heavier elements such as uranium, there are relativistic effects and finite nucleus size effects requiring more sophisticated math and at the very least, some sort of correction terms to that formula.

If it only applies to any atom with one electron, why would we be talking about the electron falling to a lower orbital (1s orbital), it it has always been there.?

Electrons can also exist in excited states, with n>1. This occurs frequently in spectroscopy, where we are measuring the absorption of electromagnetic radiation. That energy has to go somewhere - it promotes electrons to higher energy levels. The Bohr equation does a fine job of predicting/explaining the line spectrum of hydrogen. For 2 electrons, the three body problem prevents there from being an easy solution to the energy of an electron.

Oh!!! Let me tell you what actually happens.
Hopefully you might have been studying physics along with chemistry because the concept used is physics.
..........................................
The thing is this question is very important and often a point of confusion. First lets prove it mathematically then i will reason it out.
Potential energy is calculated by the formula kq1q2/r
So the charge on the nucleus as a whole is +ive and that on the electron is -ive so when you take the product you get a -ive. Hence the potential energy of this system is -ive with respect to the standard infinite distance condition which means that the electron is at an infinite distance from the  nucleus (unbound electron). Hence the energy is negative in comparison to an unbound electron.
Now in the just above formula if you imagine an electron to be a positively charged particle then you would get a positive potential energy which actually means that the electron will be repelled until it retches infinite separation.
From this we actually need to interpret that the -ive total energy implies a bound system, like that of an electron in an orbit. On the other hand +ive or zero energies represent an unbound system where particles stay or prefer to stay at infinite separation.
Remember The potential energy of any bound system is -ive!!!


text:
Why not "Oxygate" OO4(2-)?
So many atoms can be surrounded by oxygen atoms to make anions - chromate, phosphate, etc.   In particular there's Tellurate $\ce{TeO4^{-2}}$, Selenate $\ce{SeO4^{-2}}$ and Sulfate $\ce{SO4^{-2}}$ - why not go one step more and have "Oxygate" $\ce{OO4^{-2}}$?
I would guess that such a thing, if it could be created at all, would break up into an molecular oxygen and ozone $\ce{O2 + O3}$ with the electrons doing whatever they please.  But sulfate and the heavier VI central atoms don't break up.  I've never encountered a good clear explanation about this.

Would like to tag this with 'oxygen' and 'anion' but alas none such exist yet.

Why doesn't HH4 not exist? This is elementary textbook level.

@DarenW: Neither tag makes much sense as a tag. Also see http://meta.chemistry.stackexchange.com/questions/6/should-we-tag-questions-with-elements-they-involve

It's not necessary that an element shares some chemical property with all the others in its group. If it did share all chemical properties with the rest of the group, then we could just as well call the elements (something akin to) isotopes of each other. You'd have a pretty boring periodic table then.
What's happening here is that the most striking difference between elements in a group is at play. That difference being the number of electrons.
Our electron shells are broken into "orbitals". Every orbital can hold 2 electrons. $s$ orbitals come in sets of one, $p$ in sets of 3, $d$ in sets of 5, etc. Not every shell is made up of all orbitals--the first shell has only one $s$ orbital, the second shell has $s$ and $p$, and so on. Electrons fill into these orbitals by a triad of rules (which I won't go into here)-- the Aufbau principle, the Pauli exclusion principle (which is implicit in my statement that each orbital holds two electrons), and Hund's rule.
Now, an interesting thing is that bonding can only occur with orbitals with close energies. Thus, if the $4s$ orbital is involved in bonding, the $1s$ orbital will not be.
OK. Now, let's first look at Sulfur's outer electron configuration. There are 2 electrons in a $3s$ orbital, 4 in the $3p$ orbitals (so there is enough space for 2 more electrons here), and you have some "vacant space" in the $3d$ orbitals.
In the sulfate ion:

sulfur is "accepting" 6 electrons in this bond. There's some nifty mixing of orbitals that goes on, but basically, those electrons need some place in sulfur's outer shell to hole up--two electrons per orbital as usual.
Since we have 6 extra electrons, we need 3 orbitals. We have one "left over" $p$ orbital, and 5 $d$ orbitals, so we have enough space to accommodate those electrons and form a bond.
OK. Now, let's try the same thing for Oxygen. It has, in its outer shell, a full $2s$ orbital, 2 full $2p$ orbitals, leaving an "extra" $2p$ orbital empty. As mentioned before, the second shell only has $s$ and $p$ orbitals, so there obviously aren't any $d$ orbitals. You've probably noticed the issue here-- we have six electrons and only one extra orbital to stuff them in. This isn't possible, so oxygen never forms $\ce{OO4-}$. The other chalcogens($\ce{Se, Te}$, etc) all have $d$ orbitals in their outer shells, so they have no trouble forming such ions.
Note that I've grossly oversimplified the bonding process, deliberately leaving out a few crucial aspects (which would otherwise take extremely long to explain). If you want to know about the mechanism behind bonding, you may want to read up on orbitals first, then VSEPR theory, and then MO theory.

sulfur is "accepting" 6 electrons in this bond.  || It was repeatedly proven, that it does not.


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Is ash fireproof?
At high temperatures (almost) everything burns to ash. Is ash 'fireproof'?
Is further combustion of ash possible for example by combining with hydrogen, oxygen, etc?

Short answer no, ash can be burnt. It may contain unburned particles that did not receive enough air (oxygen) to burn completely.
Although one may consider ash to be product of complete burning. In which case it won't burn easily. But given the right conditions it can burn.
In fact even water can burn, there is quite a bit of research in burning aluminum with water (especially for under water applications like submarines), this produces aluminum oxide and hydrogen.