text:
What makes oceanic desalination so expensive?
What specifically makes oceanic desalination so expensive?

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The biggest expense is not necessarily technology related, but rather policy. US Coastal Commissions, legal hearings and environmental impact studies consume a large portion of the budget to get to a completed working facility.

@docscience, what is your evidence to support that claim?

I live in Carlsbad, Ca: home of the first desalination plant in California.  And this project has been delayed for almost 15 years due to approvals. Time is money.

Reason is salt dissolve very easily with water, forming very strong bond. Not only salt there are also other minerals. To break this bond we need technology as well as energy and both of them are very expensive(today).
There are two methods by which we can break this bond:

Thermal distillation:
Thermal distillation involves heat: Boiling water turns it into vapor—leaving the salt behind—that is collected and condensed back into water by cooling it down
Membrane distillation: The most common type of membrane separation is called reverse osmosis. Seawater is forced through a semipermeable membrane that separates salt from water. Because the technology typically requires less energy than thermal distillation, most new plants, now use reverse osmosis.

Today about 10 to 13 billion gallon  water are desalinated worldwide per day. That's only about 0.2% of global water consumption.
Membrane distillation is bit more cheaper than thermal distillation. Even though  large amounts of energy are needed to generate the high pressure that forces the water through the filter. Current methods require about 14 kilowatt-hours of energy to produce 1,000 gallons of desalinated seawater.
So we can conclude that cost of distillation is high because we need notable large amount of electricity to heat water in thermal plant and generate high pressure. Also other technology(like say burner, pumps, etc) are also expensive.
Source: ScientificAmerica and LiveScience

Also note we are trying to decrease the cost of distillation and day by day number of seawater distillation  are increasing.

What I meant by "specifically" was something like "electricity" or "replacement filters," etc. Is that what you're saying? That is, electricity for boiling and cost of filter replacement? Oh, and you misspelled "Scientific."

I have edited question and tried my best to be specific.

Seawater is very corrosive so the materials of construction to resist corrosion are very expensive, this is amplified for thermal / distillation methods. Basically nickel alloys and titanium; and Ti has high fabrication expenses. Polymers have good corrosion resistance but have special fabrication requirements and lack robustness for a unit that will have a life measured in decades.

To answer this question we must consider the huge amount of water consumed. And that the average seawater has $\pu{30g}$ of salt per $\pu{1000g}$ of water; along with the additional $\pu{60g}$ of impurities, sand, mud, dirt, etc.
Reverse Osmosis
This is probably the same technology that you use to filter your water. Unless you use carbon filters that work by  passing water through carbon candles and filtering it. If you however have an RO filter, then you may have realized the water passing back through a tube in to your sink, or the water that you may think "bad". If you switch on the filter faucet to take some water you may have realized how slowly it drips, this is because $90~\%$ of the water is being sent out through the reject stream.
A desalination plant works just like this but with a filter nearly 100 times bigger than yours. This entire processes take about 35 million dollars a day!


text:
Is there any chemical that can destroy PTFE or Teflon?
Polytetrafluoroethylene was discovered by accident. It now is an important material in the industry mainly because of its extremely high bonding energy, which prevents corrosion, halts reaction, and reduces friction (yeah carbon-fluorine bonds!)
And people would have definitely put it to the test, making it contain some of the most vicious and chemically diabolical substances ever created. There is a whole HOST of items it can contain that some chemists have gone so far as to say they were 'evil':

Dioxygen Difluoride
Known as the gas of Lucifer, there is a whole list of people blown up and killed while just trying to work with one of its components, fluorine. It ignites stuff at temperatures that most of the stuff that we breathe in would be in liquid form. No one really knows about its atomic structure (obviously).

Fluoroantimonic Acid
With a staggering pH of -25, it chews through stuff you might not even believe could be corroded; like wax or glass. It can even strip hydrogen off of methane

...There are a lot of other chemical demons it can contain, but this is not the point. Let this suffice: Chemical Resistance Comparison (Spoiler: Fluorine is good at this corrosion thing.)
With this kind of hyper-resistance to about anything chemically destructive, is there anything that can destroy Teflon through only chemical means? A chemical that reacts exothermically to release heat, which melts the PTFE does not count. You get the drift.
Also, I am very curious as to whether there is anything more resilient than Teflon? Polytetrafluoroethylene is made of many carbon-fluorine bonds in series. However, carbon-fluorine is second only to the Si-F bond. Is there an "overclocked" Teflon made of silicon-fluorine bonds that is even stronger?
Now I know that some, but very few, solvents can make a mark on Teflon; but my question has not been answered: Is there any more resistant substances?
(More Teflon bragging; take that aqua regia)

related http://chemistry.stackexchange.com/questions/4624/please-explain-why-dioxygen-difluoride-is-so-dangerous, http://chemistry.stackexchange.com/questions/6183/does-a-sulfuric-acid-nitric-acid-mixture-dissolve-ptfe-ie-teflon?s=7|0.5241

possible duplicate of Are PTFE stir bars resistant to all lab solvents?

FWIW, Wikipedia claims that chlorine trifluoride (your standard go-to stuff for making things, that don't normally burn, burn) reacts with teflon. Frankly, I'm surprised by your claim that FOOF doesn't, since it's generally even nastier. Maybe you need to keep the temperature low enough?

Just some personal experience... a PTFE dish was irrevocably stained (scrape it and it was still brown) by cooking it in a mixture of jet fuel and m-toluidine. So whatever happened, we definitely didn't have a cleanable PTFE dish any more - but it happens (and they were not used again for such a test).
(Of course it could have absorbed the solution like a sponge - but I wouldn't know how they could be separated.)

@Hyperluminal - "With a staggering pH of -25, it chews through stuff you might not even believe could be corroded; like wax or glass. It can even strip methane of hydrogen." ... I thought you said fluoroantimonic acid is an acid. If it's an acid, how can it strip methane of protons?

@Dissenter Methane acts there similarly to basic hydrides.

Try ClF3 most potential oxidiser

Just to add a bit to Ben's excellent answer...

A number of fluorinating agents also react with PTFE, $\ce{XeF2}$ and $\ce{CoF3}$ being examples
Ben mentioned the reaction of magnesium metal.  Typically with metals, they must be in intimate contact with the PTFE surface, so molten metals or metals dissolved in anhydrous solvents will react.

The magnesium reaction is of special interest because it serves as the basis of the thermite flare. A pyrotechnic device commonly used in the countermeasures aircraft use to evade heat-seeking missiles.  The reaction of metals with PTFE is given by the following equation (I think this is the general description for the reaction of metals with PTFE; I'm suspect of the reaction proposed by Ben involving the formation of poly-perfluoroacetylene).
$$\ce{2Mg + -(C2F4){-} → 2MgF2 + 2C}$$
The formation of $\ce{MgF2}$ is extremely exothermic.  The heat given off along with the carbon soot provides a new, much hotter, target for the attacking missile to lock onto.
As to whether there is anything more resistant, I suspect that is unlikely.  The $\ce{C-F}$ bond is shorter (135 pm) than the $\ce{Si-F}$ bond (160 pm) and therefore better serves to encase and protect the carbon backbone.  While there are some other polymers that have better mechanical or thermal properties, I am not aware of any that have better chemical resistance. In Polymers for Electronic & Photonic Application from 2013, the author states, "PTFE is the most chemically resistant polymer known".

I like your reaction better with reducing metals better. It explains why white PTE-coated magnetic stirbars begin to turn black after repeated Grignard or dissolving metal reactions.

Corrosion Resistant Products, Ltd., with the help of Dupont, has established this source of information on what can and cannot eat teflon.
Here's a list:

Sodium and potassium metal - these reduce and defluorinate PTFE, which finds use in etching PTFE
Finely divided metal powders, like aluminum and and magnesium, cause PTFE to combust at high temperatures

These reactions probably reduce PTFE in a manner that starts:
$$\ce{(CF2CF2)_{n} + 2Na -> (CF=CF)_{n} +2NaF}$$

The world's most powerful oxidizers like $\ce{F2}$, $\ce{OF2}$, and $\ce{ClF3}$ can oxidize PTFE at elevated temperatures, probably by:

$$\ce{(CF2CF2)_{n} + 2nF2 -> 2nCF4}$$
Similar things can occur under extreme conditions (temperature and pressure) with:

Boranes
Nitric acid
80% NaOH or KOH
Aluminum chloride
Ammonia, some amines, and some imines

I am aware that this answer does not describe a ‘chemical that can destroy PTFE’. However, since you are also asking ‘is there anything that can destroy Teflon through only chemical means?’, and in order to complete the other answers, I would like to add that PTFE can be easily destroyed by means of radiation chemistry. (Radiation chemistry is the study of the chemical effects of radiation on matter. It is not to be confused with radiochemistry, which is the chemistry of radioactive materials.)
PTFE is exceptionally sensitive to radiation. PTFE experiences significant damage at lower radiation exposure levels than other polymers. In general, PTFE is considered usable without any significant restrictions only for absorbed doses of up to $100\ \mathrm{Gy}$.
The general radiation effects on polymers are formation of gas, cross-linking of polymer chains, and scission of polymer chains. In contrast to most other polymers, all hydrogen atoms in PTFE are substituted by fluorine atoms; therefore, the elimination of hydrogen or hydrogen fluoride and the corresponding formation of double bonds do not occur. The elimination of fluorine due to $\ce{C-F}$ bond break is possible; however, since $\ce{C-C}$ bonds are clearly weaker than $\ce{C-F}$ bonds, $\ce{C-C}$ bond break predominates. Therefore, the primary effect of radiation on PTFE is the scission of the polymer chain (breaking of the large polymer molecule into smaller parts). Due to the absence of π-electrons, the excited states are not particularly stabilized, resulting in high radiation-chemical yields. Because of the absence of unsaturated bonds, the absence of functional groups that can be easily eliminated, and the general chemical inertness, PTFE cannot be cross-linked like an elastomer. Therefore, the scission of the polymer chain is not compensated for by the formation of new bonds. With advancing chain scission, the molar weight of the polymer is dramatically reduced from an initial value of about $6\times10^6\ \mathrm{g\ mol^{-1}}$. It is divided by $4$ after $250\ \mathrm{Gy}$ exposure and by about $20$ for a dose of $1000\ \mathrm{Gy}$.*
The effect of molecular weight reduction is primarily on mechanical properties. Tensile strength is reduced by $25\ \%$ after $500\ \mathrm{Gy}$ exposure and by $50\ \%$ for a dose of about $900\ \mathrm{Gy}$. For doses above $1500\ \mathrm{Gy}$, imminent material failure has to be assumed because of a brittle behaviour with unstable crack propagation.
If possible, PTFE components should be avoided in high radiation environments (nuclear power plants, nuclear fuel cycle facilities, irradiation facilities, or particle accelerators).
* Fayolle, B.; Audouin, L.; Verdu, J.: Radiation induced embrittlement of PTFE. In: Polymer 44 (2003) 2773–2780.

The irony here is that one of the earliest uses of PTFE was as valve coatings in uranium enrichment facilities.

A PTFE surface can be chemically altered for bonding to other materials by solvated electrons in liquid ammonia (leaving the bulk PTFE intact).  Classically alkali metals have been used for this, but a more controlled reaction is obtained from an electrolytically generated magnesium solution [Ref. 1].  From the abstract:

Solutions of solvated electrons in the presence of magnesium offer many advantages for the surface treatment of PTFE when compared to the classical solutions of solvated electrons in the presence of alkalis: the polymer remains white instead of black, its surface is not destroyed and presents a controlled hydrophilic character.

Reference
1.
K. Brace, C. Combellas, M. Delamar, A. Fritsch, F. Kanoufi, M. E. R. Shanahan and A. Thiébault, "A new reagent for surface treatment of polytetrafluoroethylene", Chem. Commun., 1996, 403-404.

Similar polyflourines have been decomposed by ball-milling PTFE with dry potassium hydroxide. There s an interesting article by Zhang et al. [1] which is really informative and makes me thing that PTFE would respond similarly. Everything breaks down to the mineral elements with no nasty interims formed. The energy and pressure of the steel balls smashing the KOH into the polyflourines causes the reactions to go with no aqueous stage needed. It could be the answer to reducing the PTFE waste impact on the environment.
References

Zhang, K.; Huang, J.; Yu, G.; Zhang, Q.; Deng, S.; Wang, B. Destruction of Perfluorooctane Sulfonate (PFOS) and Perfluorooctanoic Acid (PFOA) by Ball Milling. Environ. Sci. Technol. 2013, 47 (12), 6471–6477. https://doi.org/10/gf5w4f.

It is a minor point to an excellent question and post, but are you considering pH and Hammett Acidity Function as the same thing for superacids here? The molarity of protons required to have a pH of -25 is pretty mind-boggling. I think given the common confusion over pHs under 0 it's an important distinction to point out. Without doubt H2FSbF6 is going to pH somewhere in the negatives in many, many hydrofluoric diluted concentrations, but the only real way to rate with specificity it is with Hammett. And it could certainly be found to be in the range you ascribe its pH to.
As far as materials that can chemically attack PTFE, in my work in the agrichemical manufacture industry I've yet to come across one, however I find it interesting that when designing piping for concentrated HCl PTFE is consistently rated a 'good' choice for gasketing and EPDM is rated 'better.' Is there a worry with prolonged concentrated HCl exposure? I'm answering a question with a question and my name isn't Socrates, so I apologize, but it is something I've wondered in passing.

Welcome to the chemistry stack exchange . Please take the tour. http://chemistry.stackexchange.com/tour  If you post the question as a second question you are more likely to get an answer.  Your point seemed to be well taken so I see no issue with leaving this as an answer.

HF can attack PTFE slowly, but once it gets going, it can self-propagate. The PTFE decomposes to more HF, which then attacks itself. I find it ironic that PTFE is listed as a container for hydrofluoric acid, but the PTFE is slightly permeable to the acid.

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Are diamonds really forever?
Common saying. Diamond possesses:

ultra hardness, (10 on the Mohs scale; 10000 HV on Vicker's Hard Test (iron merely 30-80))
hyper thermal conductivity, ($2320~\mathrm{W\, m^{-1}\, K^{-1}}$, or over ten times better than the heatsink in your computer!)
extreme pressure resistance, (withstands a crushing 600 gigapascals; or around 2 times the pressure at the center of the earth, enough to snap carbon nano-tubes and graphene or create metallic oxygen or overcome copper's electron degeneracy pressure, making the maximum chamber pressure of a firing pistol seem literally like popping popcorn... I digress)
and excellent luster (what do you expect, it's a diamond) combine to make the gemstone coveted by all.

Diamonds are the stuff of awesome.
But do they really exist forever?
Wikipedia notes that,

Diamond is less stable than graphite, but the conversion rate from diamond
   to graphite is negligible at standard conditions.

Huh. But Wikipedia doesn't mention how long. So how long would it take for this super-material to convert to the stuff I scribble with?
(If you doubt the claims about diamond's seemingly unbelievable properties, check out the link on Wikipedia about diamond and this and this.)
Great point Joe made, that $10^{80}$ is just forever to us puny humans. Being a geek I can't resist the urge to compare the time length $10^{80}$:

Makes the entire lifespan of a red dwarf star seem like the Planck second.
Enough time for you to sift through all the atoms in the entire UNIVERSE at a rate of one atom per second.
Getting $67 worth of US quarters and flipping them, one per second, to get all heads-up.
Chance of macroscopic quantum tunneling!! (I don't know precisely how much, but quite large)

...and this is $10^{80}$ seconds I'm talking about...

More important question what happen when you compress dimond even more. Like 100x more pressure. Does it become a super diamond or black hole dimond or something like that???

@bodacydo I recently read that it was just discovered that they can get stronger under pressure. SciAm online perhaps, but I'm not sure. It might be used for "inner" anvils in a nested pressure anvil situation, it mentioned. I seem to recall that phase changes are involved, with different atomic bonds.

Not directly related to the question, but the "pressure resistance" and "hardness" do not mean that you can't destroy them. Diamonds are brittle - hit them hard enough with a hammer and they shatter. Also, remember that there is a thing called carbon dioxide. It's made of diamond/graphite and oxygen (and there's no lack of that in our atmosphere). Throw a diamond in a fire and it burns away as CO2.

You can destroy diamonds, so they're not really forever - in fact, diamonds burn, releasing just the usual carbon dioxide you'd expect. The ignition temperature is rather high, but the burning is self-sustaining after ignition. Also, while they are incredibly hard, this also leads to them being rather fragile - sure, they've got great pressure resistance, but they don't respond well to shock. You just need  a simple hammer to break a diamond into lots of shards. That said, you'd be hard pressed to find something that scratches a diamond - so for jewelry purposes, they are forever.

De Beers needed a slogan for diamonds that expressed both the theme of romance and legitimacy. An N. W. Ayer copywriter came up with the caption "A Diamond Is Forever," [...] Even though diamonds can in fact be shattered, chipped, discolored, or incinerated to ash, the concept of eternity perfectly captured the magical qualities that the advertising agency wanted to attribute to diamonds. Within a year, "A Diamond Is Forever" became the official motto of De Beers. From The Atlantic 1982

related http://chemistry.stackexchange.com/questions/10226/maximum-pressure-of-covalently-bonded-material

how long would it take for this super-material to convert to the stuff
  I scribble with?

No, despite the fact that James Bond said "Diamonds are Forever", that is not exactly the case. Although Bond's statement is a fair approximation of reality it is not a scientifically accurate description of reality.
As we will soon see, even though diamond is slightly less stable than graphite (by ~ 2.5 kJ/mol), it is kinetically protected by a large activation energy.
Here is a comparative representation of the structures of diamond and graphite.

(image source: Satyanarayana T, Rai R. Nanotechnology: The future. J Interdiscip Dentistry 2011;1:93-100)

(image source)
Note that diamond is composed of cyclohexane rings and each carbon is bonded to 2 more carbons external to the cyclohexane ring. On the other hand, graphite is comprised of benzene rings and each carbon is bonded to only 1 carbon external to the benzene ring. That means we need to break 6 sigma bonds in diamond and make about 2 pi bonds (remember it's an extended array of rings, don't double count) in graphite per 6-membered ring in order to convert diamond to graphite.
A typical aliphatic C–C bond strength is ~340 kJ/mol and a typical pi bond strength is ~260 kJ/mol.  So to break 6 sigma bonds and make 2 pi bonds would require ~((6*340)-(2*260)) ~ 1500 kJ/mol. If the transition state were exactly midway between diamond and carbon (with roughly equal bond breaking and bond making), then we might approximate the activation energy as being half that value or ~750 kJ/mol.  Since graphite is a bit more stable than diamond, we can refine our model and realize that the transition state will occur a bit before the mid-point. So our refined model would suggest an activation energy something less than 750 kJ/mol. Had we attempted to incorporate the effect of aromaticity in graphite our estimate would be even lower.  In any case, this is an extremely large activation energy, so, as we anticipated the reaction would be very slow.
An estimate (see p. 171) of the activation energy puts the reverse reaction (graphite to diamond; but since, as noted above, the energy difference between the two is very small the activation energy for the forward reaction is almost the same) at 367 kJ/mol. So at least our rough approximation was in the right ballpark, off by about a factor of 2. However, it appears that the transition state is even further from the midpoint (closer to starting material) than we might have guessed.
This activation energy tells us that at 25 °C, it would take well over a billion years to convert one cubic centimeter of diamond to graphite.
Note 04/17/20: As mentioned in a comment, the original "estimate" link became  defunct and was replaced today with a new estimate link.  However the original article and estimate can can still be seen on the Wayback Machine and it estimates the activation energy to be 538.45 kJ/mol, reasonably close to our estimate.

I always wondered about that: if mineral diamonds are as much as a billion years old, why are they not mixed state due to some atoms rebinding? If it happens on the surface of the crystal only, then we'd find diamonds encrusted in graphite in the matix, rather than directly in the matrix. It does not seem that diamonds have decayed before being mined.

Taking the calculations shown in the link containing the estimated activation energy and extending them to $\mathrm{298\ K}$ (which is a simple but coarse approximation), then the conversion of diamond to graphite takes place on times of the order of $\mathrm{10^{80}\ yr}$; this is several orders of magnitude higher than the characteristic timescale of solar-mass black hole evaporation ($\approx \mathrm{10^{67}\ yr}$)!!! Compare also with the current age of the Universe, at a measly 13.8 billion years ($\mathrm{1.38×10^{10}\ yr}$).

@JDługosz because they spent most of that time deep in the Earth, where they are indeed stable. It's only recent (geologically speaking) that they were transported to the surface in volcanic eruptions.

Seems to me that $10^{80}$ years is sufficiently long to be synonymous with "forever"...

The link provided under "estimate" is now defunct; is there an alternative source one can check out for more information?

@user39221 Inserted new link for estimate

I add an other answer : it depends what kind of people you are, if you are as crazy as them, of course not !
2012 Christmas Lectures - Burning a Diamond
But ron's answer is better sure ! ;)

Well, it seems that diamond is not forever because upon exposure to sunlight it loses atoms.

This article seems pretty interesting. Could you give a more in-depth summary of it? We tend to avoid link only answers on this site.

Not upworthy answer, but +1 for the interesting link

Diamonds burn, but the temperature at which they burn depends on whether or not the diamonds are in contact with air. The temperature of diamond ignition in pure oxygen is 690º C to 840º C.  In a stream of oxygen gas, diamonds burn at a low red heat initially. They will gradually rise in temperature and reach a white heat. Then, the diamonds will burn uninterruptedly with a pale blue flame, even after the oxygen heat source is removed. The diamond crystals will gradually decrease in size and finally disappear. The flame at the last moment will flicker brightly and then disappear, leaving not a trace of ash or residue.
For this to take place in an air mixture, the heat must be kept directly applied on the diamonds at all times. If removed, the diamonds will not continue to burn, due to oxygen being diluted with nitrogen that does not support combustion
The most famous gems on Earth have new competition in the form of a planet made largely of diamond, astronomers say.
The alien planet, a so-called "super-Earth," is called 55 Cancri e and was discovered in 2004 around a nearby star in our Milky Way galaxy. After estimating the planet's mass and radius, and studying its host star's composition, scientists now say the rocky world is composed mainly of carbon (in the form of diamond and graphite), as well as iron, silicon carbide, and potentially silicates.
At least a third of the planet's mass is likely pure diamond, info

It seems from the Wikipedia article that the composition of 55 Cancri is hypothetical: "much of which may be in the form of diamond", so "likely" seems a stretch. Still, pretty cool.

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text:
What factors are influencing a woosh rocket?
I asked this over at Physics, but no one seemed to know, so I'll try here:
If I fill a plastic bottle with a small amount of rubbing alcohol, shake it around and hold a match to the bottle neck, I'll get a flame wooshing out of the opening. This can easily be used to propel the bottle upwards, like a rocket. I tried it already, there are lots of youtube videos about this, and NASA has instructions for the experiment as well.
When trying to go for maximum speed or height of the rocket, there are a few factors influencing this experiment. Here are the ones that I can think of:
Some related to the bottle:
volume, shape, aerodynamics, weight, size of opening, possibly shape of the bottle (long and thin or short and fat) When the bottle neck is smaller it creates a stronger thrust.
Others related to the propellant:
How fast does it burn, how dense is it (especially compared to air), how heavily does it vaporize. If the vapor is denser then air it all floats out of the opening of my bottle before I can light it and I don't have as much fuel.
As you might be able to see, I don't know a lot about the propellants.
So here's my question:
What are the factors related to the propellant, that influence the launch height and speed of the woosh rocket.
What's the best propellant to choose for a woosh rocket, and why?

Related: Best Fuel for Potato Gun

Please don't crosspost.

For home projects like this I believe one of the most important things is to ensure you have a good fuel to air (oxygen) ratio. It is surprisingly easy to use too much propellant which can take up air space within the combustion chamber. More fuel doesn't necessarily mean bigger bang.


text:
Correctness of writing H3C as opposed to CH3 in a carbon chain
I have been studying alkanes, alkenes & alkynes for a while, and I see in the examples that all the basic structural formulas start with $\ce{CH3}$, then $\ce{CH2}$, and the last carbon atom is $\ce{CH3}$, but the one that is attached to the single/double/triple bond is $\ce{CH}$. (Please correct me if I'm explaining it with the right terms.)
Anyways, there's a question about drawing the structural formula for compounds.

Draw a structural formula for hex-3-ene.

What my answer is:
$$\ce{CH3CH2CH=CHCH2CH3}$$
I thought this was right, then I looked at the solutions and it was
$$\ce{H3CCH2CH=CHCH2CH3}$$
So now my question is, why does it start with $\ce{H3C}$ not $\ce{CH3}$ like some other examples?

Your answer is the same as the book's. The book wrote the first carbon as $\ce{H3C}$ to stress that the chain continues via a carbon to carbon bond, not bonded a carbon to hydrogen to carbon bond.
As for why it's not like this in other examples, technically it is not necessary to write it this way, but it's strange your book lacks that consistency.

So I'm right too?

You are indeed.

Could you also write 3HC? Just curious. I write from right to left so I put index on left side.

To my knowledge, the number is always written after the atomic symbol when denoting the number of that element.

You can't use 3HC. The number of atoms has to be on the right side of the element's symbol. Otherwise, H2O could mean both 2 Hydrogens + 1 Oxygen and 1 Hydrogen + 2 Oxygen, and this lack of consistency would obviously be very problematic. Also, even though the number of atoms should be subscript, numbers to the left of a formula represent number of moles of the substances when balancing equations.

The "H" part of H3C is to the outside of the center to stress that the carbon atom is oriented towards the other carbon atoms in the center.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the [help]. Somehow I fail to see, how this answer differs from the one that was already given.


text:
How to learn chemistry at home?
I'm simple man who doesn't know much about anything. I just enjoy my life. Lately I like chemistry a lot. I want to learn chemistry at home. How do you recommend I do it?
Do i need to do chemical experiments to become good at chemistry?
I can learn about reactions with books. For example:
O + H2 -> H2O

Also organic chemistry like
C2H5OH + O2 -> CO2 + H2O

What else do I need to learn? What do chemists do all day if all formulas can be calculated on paper?
How does chemist invent new materials?
Can I be theoretical chemist?
Is this a good hobby choice?

Buy a chemistry set! (:

This is relevant; but I'm afraid your question is primarily opinion-based; we are chemists (exclude me, I'm a semi-chemist); not private counselors.


text:
How to calculate the requirements for completing a reaction
What are the most common critical thermodynamic and kinetic variables  required to convert a secondary or tertiary alcohol to an alkane.? Is bond enthalpy data relevant?
What is the process/processes commonly used to then calculate or determine the temperature and duration it would take to complete the reaction from the thermodynamic and kinetic data?

@ron says in the answer to the question about diamonds that from bond activation energy of the c-c bonds it can be calculated that at 25°c it would take diamond a over a billion years to convert to graphite, can someone please explain how this was calculated from the bond enthalpy ?

There isn't enough information with just the bond energies, which are thermodynamic properties. The question you're asking is about kinetics, which requires activation energy parameters.

http://chemistry.stackexchange.com/a/34202/16683 This is the answer referenced. In his answer, Ron cites a geochemistry worksheet from Cornell; this worksheet in turn cites Pearson et al., 1995: "Pearson et al. (1995) estimated that to convert 1 cc of diamond to graphite at 0.1 Mpa and 1000°C would require 1 billion years, but only a million years would be required at 1200°C." If someone could find that paper, then maybe we could answer this question - I can't seem to find it.

Thanks for the comments I'm realy trying to work out how that data was obtained. If it was calculated from the c-c bonds then same style of data could be calculated for the conversion of any material ?

Why is the question marked down?

You can't - at least if the only information you have is just the bond energy.
The problem you're running up against here is the difference between thermodynamics and kinetics. The difference can sometimes be subtle, but (roughly) thermodynamics is normally concerned with state functions, whereas kinetics deals with the process of interconversion of states.
The bond enthalpy is a thermodynamic quantity. It's based off a state function. It's the free energy difference between the bond-broken and bond-formed states. It doesn't matter how you go from the bond-formed to bond-broken states, the free energy difference is the same. In order to calculate it, you only need to know information about the two endpoints, not the details about the states which connect them.
In contrast, reaction rates are kinetic quantities. The rate of reaction is highly dependent on the path (reaction mechanism and conditions) you take to interconvert the states. You can't tell anything about the path if all you have is thermodynamic information about the two end points.
That said, you can theoretically calculate reaction rates if you have additional information about the path you're taking. Most notably you would need to know the activation energy of the reaction you're interested in. This is different from the bond energy, and is related to the free energy of the transition state, the hypothetical high-energy intermediate along the reaction pathway. If you have the activation energy along with some other information for a single-step reaction, you can calculate reaction rates for various temperatures with the Arrhenius equation.
Again, these rates are highly dependent on the exact mechanism of the reaction you're examining. Single step versus multi-step, radical versus electron pair transfer mechanisms, the presence of catalysts, etc. all can change the activation energies of the various steps and the reaction rates. So there is no way to give a general answer about bond breakage rates.

Thanks for a good answer I'll study the things you mentioned ...


text:
What methods could be used to preserve scorpion venom?
As extracting scorpion venom goes, how would one most effectively store and preserve the venom for future research?
My thoughts are lyophilization, however I am not sure that would be the best of all methods in this case. Thoughts?

Bee venom phospholipase A2 is available as lyophilized product and works well. Why do you need scorpion venom?

Medical research specific to the Deathstalker scorpion. No other venom substitutions will suffice.

The enzymes are similarish, so lyophilization stands a good chance at working.

Scorpion venoms are peptides and not really enzymes. So you need not worry about activity. You can lyophilize and store them in -20. Sigma used to have a product - Deathstalker venom  in the form of dry powder. They apparently discontinued it.

I would lyophilize it and store it at minus 80˚C or even liquid nitrogen. Then it should last many years to come without loss of bioactivity.


text:
Determine the rate equation (intermediate reaction)
I was helping some students today, quite successfully, until I came across this problem:
$$\begin{align}
\ce{A &<=>[$k_+$][$k_-$] B + C} \\[5pt]
\ce{C + D &->[$k_1$] E}
\end{align}$$
The first equation is a fast process  with forward and reverse rate constants $k_+$ and $k_-$, respectively.
The second process is slow, with rate constant $k_1$.
The objective is to write down a rate law for the resulting equation:
$$\ce{A + D -> B + E}$$
The technique that had been working was to write down the rate for the slow process...
$$r = k_1[\ce{C}][\ce{D}]$$
Since $\ce{C}$ is a product in the fast reaction, which is an equilibrium equation, we also have:
$$k_+[\ce{A}]=k_-[\ce{B}][\ce{C}]$$
This is where things went awry, because the rate law that we seek is a function of $[\ce{A}]$ and $[\ce{D}]$ only. Considering the stoichiometry, it seemed appropriate to put $[\ce{B}]=[\ce{C}]$. This leads to the substitution $[\ce{C}]=(\frac{k_+}{k_-})^{1/2}[\ce{A}]^{1/2}$. The interface has declared the result to be invalid, so it must be an illegal maneuver. How, then, do I eliminate $[\ce{B}]$ and $[\ce{C}]$ and come up with the desired reaction rate law?

You can't write $[\ce{B}] = [\ce{C}]$ because  $\ce{C}$ is consumed in the second step but $\ce{B}$ isn't. However, you were pretty much on the right track.
Since $\ce{E}$ is produced from $\ce{D}$, you can write $[\ce{E}] = [\ce{D}]_i - [\ce{D}]$ where $[\ce{X}]$ denotes the concentration of $\ce{X}$ at a certain point in time and $[\ce{X}]_i$ denotes the initial concentration of $\ce{X}$. This is assuming that the initial concentration of $\ce{E}$ is $0$, of course.
On top of this, $\ce{C}$ is either produced from $\ce{A}$, or consumed in the production of $\ce{E}$. Therefore you can write $[\ce{C}] = [\ce{A}]_i - [\ce{A}] - [\ce{E}] = [\ce{A}]_i - [\ce{A}] - [\ce{D}]_i + [\ce{D}]$. (There is the reverse step in which $\ce{C}$ produces $\ce{A}$, but the equation still holds, since you can just consider this to be "negative" production of $\ce{C}$ from $\ce{A}$.)
Your rate law is therefore: $r = k_1[\ce{D}]\big([\ce{A}]_i - [\ce{A}] - [\ce{D}]_i + [\ce{D}]\big)$. The initial concentrations are merely constants, so you could view this as being of the form $r = k_a[\ce{D}] + k_b[\ce{A}][\ce{D}] + k_c[\ce{D}]^2$. I can't think of any way to simplify this (the steady-state approximation is not very valid here since $\ce{C}$ is produced quickly and consumed slowly, and in any case it does not yield a nicer expression). I imagine that one could use some form of an integrated rate law to find expressions for $[\ce{A}]$ and/or $[\ce{D}]$, but it would just make things even uglier.

Very nice explanation of how to eliminate C.  But B can also be eliminated by pretty much the same stoichiometric considerations.  You're right that the math is messy though.

The best first step in any kinetics problem is writing down materials balances on all relevant species.
$$\frac{dA}{dt}=-k_+ A  +  k_-B C$$
$$\frac{dB}{dt}=+k_+ A  -  k_-B C$$
$$\frac{dC}{dt}=+k_+ A  -  k_-B C  -  k_1 C D$$
$$\frac{dD}{dt}=-  k_1 C D$$
$$\frac{dE}{dt}=  k_1 C D$$
We want to solve for $\frac{dE}{dt}$ by eliminating $C$.  The assumption that the first reaction is "fast" means we can make the equilibrium assumption on $A$ and $B$ to solve for $BC$
$$BC = \frac{k_+}{k_-}A$$
We can eliminate $C$ by stoichiometric considerations.  The amount of $C$ in the system at any time is just equal to the net forward progress of reaction 1 minus the forward progress of reaction 2.  So $C=A_0-A - (D_0-D)=(A_0-D_0)-(A-D)$.  orthocresol's answer does a great job of explaining why.  But we can also make a similar argument for B: the amount of B is just equal to the net forward progress of reaction 1, or $B=A_0-A$.  So after application of stoichiometry, the equilibrium equation becomes:
$$BC = \frac{k_+}{k_-}A \implies \left[A_0-A \right] \left[(A_0-D_0)-(A-D)\right] = \frac{k_+}{k_-}A$$
That is a quadratic equation in $A$ (I won't show the math) that gives $A$ if you know the constants $A_0$, $D_0$, $k_+$, and $k_-$, and the variable $D$.
Substituting our expression for $C$ into the equation for $D$:
$$\frac{dD}{dt}=  - k_1 C D = -k_1 D [(A_0-D_0)-(A-D)] $$
We could substitute the quadratic solution for $A$ into this equation and get a single differential equation that depends only on $D$ and constants.  Since the ultimate rate we care about, $\frac{dE}{dt}$, is equal to $-\frac{dD}{dt}$, this means we have a rate law. The algebra is messy but that is the path forward.
Some problems are harder than they look!


text:
Bulging Li-ion battery – is it an immediate danger? What’s the likelihood of explosion?
I have a macbook battery that shows as completely dead (doesn’t hold the laptop on even for a second). It now started bulging, and in last week it inflated to about 50 % extra thickness, like this – I can see the inflated battery packs filled with gas.

How big of a danger is to keep a battery like this at home? Will the reaction stop, or will it continue and will the pressure rip through the battery packs?
If that happens, will the battery combust? Internet is full of videos of exploding batteries, but what’s the real likelihood? One in a million? Or one out of two if it swells like this?
I found an advice to keep the battery in a bucket of salty water, does that make sense (I remember that water + lithium = intensive fire)?
How are Li-ion batteries disposed of when I take them into store?
I ask here, because I couldn’t find a qualified answers on the internet, only scare articles about explosions and vague conflicting advices. Thanks!

The battery may have already ruptured due to thermal runaway, meaning that despite it being bulged, pressure from boiling electrolyte has already escaped.  In this case it is also likely that the lithium graphite electrode has already degraded through reaction with air.  If this is true, you are lucky that the previous thermal runaway did not result in a fire.
However, it is impossible to say that the the battery is completely dead without more information, information that would require making measurements that are difficult or impossible for most home chemists.  Thus there may well still be appreciable amounts of flammable electrolyte and air-reactive lithium graphite remaining in the battery. If heated or allowed to react with air, any such material could re-pressurize the system, resulting in a fire or small explosion.  In awareness of this possibility, you should still treat the battery as dangerous.

Also see Apple MacBook battery: Exploded.

@Curt: If you know the answer, the OP had also asked "How are Li-ion batteries disposed of when I take them into store?". I'd like to know too. Stores pay us a small amount when we return used motorcycle batteries to them. Does the same apply for cellphone and laptop batteries?


text:
Stereoisomers of compounds named tribromocyclobutane
Here is the link to the page in which I don't understand the answer to the last question (Question 13). The question states:

Draw all of the constitutional isomers and stereoisomers of compounds named
  tribromocyclobutane (all structures must contain a cyclobutane ring).

The following image exists as a part of the answer.

But why is the following image not part of the answer?

Are the above two images not diastereomers (hence stereoisomers)?

The bottom structure doesn't make sense the way you draw the dibromo carbon (non tetrahedral)

Been reading isomerism for so long that I didn't even care to think about the most basic rules! Thanks for the answer though! :)

The second configuration shown cannot exist simply because of the geometry of the orbitals at the carbon center. The angle between the bromine's (those on the same carbon) is much less than 108° even though there is no forced strain.

Been reading isomerism for so long that I didn't even care to think about the most basic rules! Thanks for the answer though! :)

I think another way to make the point is that for identical substituents at the same carbon is a cyclobutane, there is no reason to even use the wedge notation.  Nothing at all is chiral about the dibromo substituted carbon.  And actually, due to symmetry, I don't think there is anything chiral about the singly bromo-substituted carbon either.


text:
Why is it that the least reactive metals are the best electrical conductors?
Silver, Gold and Platinum are amongst the best conductors of electricity, but also the amongst the most unreactive. Since electrical conductivity depends on the number of delocalized electrons (along with resistivity), if an element gives up its electrons readily (making it highly reactive), then shouldn't that mean it will be a great electrical conductor? Why isn't Cesium the best conductor?

I realize there may be no definite trend with regards to these two properties, but the fact that less reactive materials tend to be better conductors seemed spurious from my reasoning. Am I missing some sort of logic here?

In the trio of metals you quote the most reactive (silver) is also the best conductor. And copper is a better conductor than gold despite being a lot more reactive. Aluminium would be very reactive were it not for the passivating oxide layer on its surface and it is also a very good conductor. are you sure there is a real trend or correlation?

One of the reasons why non reactive metals are good conductors is that they are good at staying as metals. Most metals react with the atmosphere to form oxides. And the majority of oxides are insulators or semiconductors. Of course there are few exceptions to this rule.
Also, just a note: calcium and iron have better conductivities than platinum.

First, why are they "non-reactive"?  They're really not "non-reactive", gold and silver certainly appear in a variety of complexes, but they are highly resistant to corrosion and oxidation, which is what makes them far less reactive with typical reagents than most metals.  They have this property because they have a high electronegativity (EN), so its harder for oxygen or acids/bases to remove their electrons and force them to react.
The requirements for good conductor is partially filled valence bands (alot of electrons) and strong delocalization (electrons can move freely between metal atoms).  Ag, Au, Cu have alot of electrons per atom, and due to the distance from the nucleus and shielding, valence electrons aren't so tightly bound to an individual nucleaus that they can't be shared equally.
The equal sharing is key and makes this bonding is highly energetically favorable because it fills each atoms valence, but leave no Au, Ag, or Cu atom deficient of electrons.  If these metals bonded covalently to a higher EN atom, they would lose some of their electron density, which is unfavorable.  You'll notice however, that gold dissolves perfectly into mercury (slightly lower EN, many free electrons) because of this strong metallic bonding.
The delocalization of electrons is mainly s and p electrons (probably not what you would expect), so heavier atoms that have strong relativistic effects causing the contraction of s orbitals don't act as well as conductors because the sea of s and p electrons isn't as free.  Au, Ag, Cu hit the perfect spot on the periodic table where they have many electrons, don't have a full valence (so want to form covalent bonds), but also have a higher EN than most metals so its preferable for them to bond with themselves.
An atom like Cs, has relatively few electrons to move around its metallic bonding complex, so doesn't fulfill both the requirements for a conductors even though the electrons are highly delocalized.
If you're really interested you should read about the tight binding model and the quantum mechanical approach to metallic bonding.

You should precise with which scale of electronegativity you think because there's a big difference between Pauling and Alfred-Rochow for example.

The simplest answer I can give is that silver (e.g) is a good conductor of electricity because of its sea of delocalised electrons between the layers of metal ions. However it is not highly reactive because you would need a lot of energy to break the metallic bonds within each of those layers.

Is there a more precise explanation you could provide (perhaps I suspect involving periodic properties such as electronegativity/ionization energies or thermodynamic properties such as enthalpies of atomization etc.)? That would be very useful.

I cant give much of an answer at the minute, but to truly understand electrical conductivity of metals, in depth, I would read up on band theory. I'll try to reply when I get chance.

The electrical conductivity of elements does not really depend on its reactivity. Then silver would be one of the best insulators. It actualy depends on the number of free electrons in the atoms. So, diamond has no free electrons thus it is a good insulator whereas silver having a lot of free electrons is the best conductor.


text:
Existence of Deuterium peroxide
Does deuterium peroxide exist? Wikipedia has no reference. Also, the sites for delivery of chemicals like "Pubchem" or "Chemspider" have no reference.
In my book, a preparation of $\ce{D2O2}$ is given,
$$\ce{K2S2O8 + 2D2O -> 2KDSO4 +D2O2}$$
If $\ce{D2O2}$ exist, what are its physical properties (structure and appearance etc.)?

It does exist. A quick SciFinder search brings up around 50 papers, mostly dealing with rotational/vibrational spectra - don't know if larger quantities have been prepared. As for the structure: it's pretty similar to that of hydrogen peroxide, but has a slightly shorter O-D distance. Here is a reference for its structure.

It's simply deuterated H2O2 so of course it exists and is quite similar.

It does exist. At least Google shows one supplier situated in New Jersey.

ID 1130   Deuterium Peroxide (30% in D2O) D2O2    98  5   5.0g / 10.0g

Though for some reason I could not find any reputable supplier like Cambridge Isotopes or Sigma-Aldrich (strange!). As far as physical properties are concerned, as pointed out by @snurden there exists a significant amount of literature. Here is one article from the Canadian Journal of Chemistry that includes study of many properties (behind a pay wall though).
Abstract:

And to answer final part of your question:


text:
Physical significance of double-well potential in quantum bonding
I'm currently studying about the hydrogen bond's Potential Energy Surface (PES) pattern.
As per literature, single point energy obtained from Density Functional Theory (DFT) using MP2 approximation performed with Gaussian software (www.gaussian.com) showed the pattern of Lennard Jones Potential (LJ-potential). A Lennard Jones is a one-well potential http://www.cmbi.ru.nl/redock/:

However, in certain literature it is also shown to obey the double welled potential model (PES_vs._atomic-distance response).
Having said that, I'm basically wondering about the physical significance of a double-well potential model with respect to interacting-atoms, in case of hydrogen bonds. Any insight or reference towards deciphering the concerned fact would be highly appreciated.

thank you @jerepierre for the necessary edit. But I'm using Nwchem package not Gaussian package.


text:
Similarities and Differences between Resonance and MCSCF treatments
Recently, there has been a question by Voldemort concerning different resonance structures of $\ce{NCO-}$, requesting an explanation why one resonance structure would be more preferred than another.
There has been a little discussion on that between ron and Martin - マーチン that took my interest:

ron: "[...] structure #1 ($\ce{N#C-O-}$) is more stable (preferred) than structure #2 ($\ce{^{-}N=C=O}$)."
Martin - マーチン: When talking about resonance structures, I think stable is the wrong word. I agree with your analysis in terms of contribution. Therefore #1 should have the highest contribution of all.

Now while I agree with ron that structure #1 is the most preferred resonance structure (referring to electronegativities), I regret with him concerning his attitude towards the "stability" as Martin does.
But I mainly do this, because I used NBO6 with the Natural Resonance Theory (NRT) keyword on the Natural Bond Orbitals (NBO) produced of an MP2/6-311+G(2d) optimized wavefunction to find the contributions of all relevant resonance structures - $66~\%~(\ce{N#C-O-})$, $30~\%~(\ce{^{-}N=C=O})$ and $3~\%~(\ce{^{2-}N-C#O+})$ - and not because I have a superior understanding on all underlying theories.
For me, the resonance stuff lookes a lot like MCSCF treatment of some VB wavefunction, which means that the complete molecule is described best from it's contributing resonance structures.
Now on a more complete comparison, what are the main similarities and differences between both theories and as they don't exist but are rather part of an explanation, can resonance structures indeed be described as "stable"?

Related, but still without answer: http://chemistry.stackexchange.com/q/33879/4945

Great question - it engages with the difficult task concerning conceptualization of molecules, and how our models and jargon we use can often cause confusion while also helping to explain significant phenomena.
To begin, it is important to emphasize the fact that all these theories are models only (as you said, they are "part of an explanation") - we attempt to describe complex systems of quantum phenomena in a way that helps us explain chemical interactions in experiment (and/or what we would then expect from computational studies). Instead of focusing on all the minutiae of both resonance theory and MCSCF, I'm going to aim for using the differences between their orbital theory choice (VB or MO) to help with this distinction, and then approach the "stable" vs. "contribution" remarks. For further details on VB and MO theory, Martin - マーチン 's comment towards another question has a fantastically detailed answer, that I would recommend taking a look at here by hBy2Py.
Resonance theory relies on VB as a means to explain what is occurring in a molecular system, illustrated by the drawings seen in organic chemistry and later on with increasingly complex systems relying only on this idea of atomic orbitals (simple constructs localized around atoms). Resonance as a concept is applied to help us visualize the idea of electron delocalization in a way that also lets us select "better" structures to draw (based on electronegativities, geometric preference, etc...), which match up to chemical expectation/intuition. However, when drawn or used as the basis for a calculation, the orbitals are then limited to those around the atom, (with some hybridization) and how they overlap with neighbors, without much freedom to mold the orbital shapes (leading to computational difficulties due to non-orthogonality).
MCSCF, or other configuration interaction theories, build on the MO idea, which greatly expands the types of orbitals utilized when describing the bonding interactions between atoms, and captures the idea of delocalization much more effectively in doing so, a feature that applying the idea of "resonance" is supposed to recover in VB, but does not do quite as well as MO. Not perfect, but one of the best models so far, and continuing to be improved upon even today (or, built on, such as with new basis sets).
In essence, yes, those "resonance structures" do not exist by themselves, and thus there is no true "stability" to them, nor in that case much of a "contribution" as it is still not the whole picture. However, use of "contribution" helps to avoid implying that the resonance structures, drawn only to illustrate how VB theory handles delocalization, are localized/exact organizations of the electrons and atoms at a particular moment. In practicality, you will likely hear these interchanged, but it is best to keep in mind the idea trying to be captured: electron location is highly complex, with a range of local/delocal orbital characteristics that could be "best" depending on the system.
As further reading just in case, and as references for the topics discussed above:
Resonance Theory: if you go down to the section on misconceptions, it will show the common errors made when describing resonance, which you brought up nicely here in terms of computation as well. This is where use of "stable" may imply one of these misconceptions, and "contribution" may help a bit (but could also lead to the second misconception)


text:
Defining enantiomer of morphine by R/S designation
How to know if the following chemical structure of morphine is an R or S? What I basically need to know is: which chirality center do we use to make the decision?
$\hskip2in$

The structure that you are showing is neither R nor S.
As it has five chirality centers you need to look at all of them.
$\hskip3in$
You can see in the above figure that there are three R and two S centers in the configuration of morphine that you posted.


text:
Covalent Bonding in odd electron sharing molecules
Why in odd electron sharing molecules (like NO ) the bond containing 3 electrons is considired to be half covalent bond?

I will answer to your question, but first let me show you that NO has a $2.5$ bonds, and I will use what I made for NO to answer your question. Here we go ! :)

Proof that NO has a $2.5$ bonds :
First the electron configurations of these two elements are :
$\ce{{}_7 N :1s^2 2s^2 2p^3}$ which have $5$ valence electrons in $4$ orbitals (one $\ce{s}$ and three $\ce{p}$)
$\ce{{}_8 O :1s^2 2s^2 2p^4}$ which have $6$ valence electrons in $4$ orbitals (one $\ce{s}$ and three $\ce{p}$)
Then we can build the orbitals-diagram of $\ce{NO}$ and I will suppose that its not a correlated diagram, and because of electronegativity of these elements verifies $\ce{\chi(N)<\chi(O)}$ then we have:

Then the electon configuration of $\ce{NO}$ is $\ce{(1\sigma)^2 (1 \sigma^{\star})^2 (2\sigma)^2 (1 \pi_{x})^2 (1 \pi_{y})^2 (1 \pi_{x}^{\star})}$
The bond order $i$ is : $$i=\frac{\text{number of electrons in bonding MOs}-\text{number of electrons in anti-bonding MOs}}{2}$$
Here we have $$i=\frac{2\times 4-3}{2}=\frac{5}{2}=2.5$$
Quod Erat Demonstrandum.

So now I did it for $\ce{NO}$ I have this configuration : $\ce{(1\sigma)^2 (1 \sigma^{\star})^2 (2\sigma)^2 (1 \pi_{x})^2 (1 \pi_{y})^2 (1 \pi_{x}^{\star})}$.
Then you said $3$ electrons bond considered has $0.5$ bond. As I understand your question, its the case if the configuration is $\ce{(1\sigma)^2 (1 \sigma^{\star})^2 (2\sigma)}$ (for example but it could be an other) then you have $3$ electrons in bonding bonds and $2$ in anti-bonding bond.
Then the bond order of your molecule is : $$i=\frac{3-2}{2}=0.5$$
Which conclude my answer.

NB:I'm seeking common examples of this kind.

I would not consider the 1s electrons of nitrogen and oxygen as valence electrons and for your analysis this is also not really necessary.

@Martin-マーチン Oups I made a mistake, I use my phone, thank you ! :)

No problem, I assumed it was just a typo ;)


text:
Difference between catenation and allotropy
Can anyone tell me what the difference between catenation and allotropy is? I would like the basic explanation.

I gave a short description of the two terms, but imo you really could just Google this and it might help you more. The Internet would have much better and in-depth explanations for this kind of question which I basically interpreted as "define this" since catenation and allotropy are just different concepts.

Catenation refers to the ability of an element to form long chains. The most notable example is carbon (catenation of carbon is what gives rise to the entire field of organic chemistry).
Allotropy is an entirely different concept; it refers to the natural existence of an element in different forms, i.e. bonded differently. For example, solid carbon exists in the forms of diamond and graphite (and more).
In a sense, the two are related because the ability of an element to catenate often leads to the formation of many different allotropes, since the chains can be linked in different ways. Apart from carbon, another example of an element that can catenate well and forms multiple allotropes would be phosphorus. On the other hand, nitrogen and oxygen prefer to form triple and double bonds respectively, instead of single bonds; this means that catenation is unlikely and that is why these elements have few allotropes (nitrogen exists as diatomic $\ce{N2}$, oxygen as diatomic $\ce{O2}$ and ozone).


text:
Calculating enthalpy per gram mole of oxygen bond in haemoglobin

A solution of $\pu{5 g}$ of haemoglobin in $\pu{100 cm^3}$ of solution shows a temperature raise of $\pu{0.031 ^\circ C}$ for complete oxygenation. Each mole of haemoglobin binds $\pu{4 mol}$ of oxygen. If the heat capacity of the solution is $\pu{4.18 J K^{-1} cm^{-3}}$, calculate $\Delta H$ per g of oxygen bond.

Attempt:

for $\pu{100 cc}$, heat capacity is $\pu{418 J K^{-1}}$
for $\pu{\frac{5}{64000}mol}$ of haemoglobin, $C\Delta T=\pu{12.952 J}$ is released
hence for $\pu{1 mol}$, $\pu{165862.4 J}$ energy is released
This energy is released when four bonds are formed with oxygen. Hence for 1 bond, $\pu{41.47 kJ}$ energy is released.

The answer given in the book is $\pu{-41.47 kJ}$.

The heat flows from each Hb molecule during the reaction at constant p and T, so $q_{p, rxn} = \Delta H_{rxn}\cdot\Delta X$.
The energy of the reaction is $\Delta H_{rxn}$ as per mole of reaction.
The reaction itself is Hb + 4O$_{2}\rightarrow$ Hb$\cdot$(O$_{2}$)$_{4}$.
The reaction is exothermic.
The number of moles of Hb present is $$n_{\rm Hb} = {5\over 64,000} = 0.078\;{\rm mmol},$$
all of which reacts: $\Delta X = -\Delta\,n_{\rm Hb} = n_{\rm Hb}$.
That much heat raises the temperature of the water (assuming all the heat is contained in the water):
$$0 = \Delta H_{\rm water} + q_{p, rxn} = C_{p}\Delta T + q_{p, rxn}$$
from which it follows that
$$q_{p, rxn} = -C_{p}\Delta T = -100\times 4.18\times 0.031 = -13\;{\rm J}$$
and
$$\Delta H_{rxn} = {q_{p}\over\Delta X} = {-13\;{\rm J}\over 0.078\;{\rm mmol}}= -167\;{{\rm kJ}\over{\rm mol-rxn}}$$
There are 4 moles of O$_{2}$ used so
$$\nu_{\rm O_{2}} = 4\;{{\rm O_{2}}\over{\rm rxn}}$$
and on an oxygen basis
$$\Delta H_{rxn} = -42\;{\rm kJ}$$


text:
Could a vacuum airship be possible?
According to Wikipedia, a vacuum airship is a hypothetical airship that is evacuated instead of using a lighter gas. Really, once I first saw this, I thought, "Why did I never think about this before?!" Indeed, whenever I pictured an evacuated container, I thought it would just sit there.  But no. Instead here on Earth, a theoretical vacuum airship could lift 1.28 grams per liter of evacuated space, a 14% boost over helium, owing to the density of air.
Of course, the air is also the destroyer of this powerful capability; it exerts 14 psi, or ~0.1 MPa!! According to Wikipedia's calculations, a hemispherical shell would have to withstand $$450,000 kg^{-1} m^5 s^{-2}$$ which is over 3 times the buckling pressure of diamond.
Give up? Not yet. There is a couple of interesting ideas I thought about:

Why hemispherical? Wouldn't other shapes do well also? For example, a sphere, or a cylinder (maybe a cylinder with radius=height). From what I know (not much..) about spherical items, I suppose it would distribute the pressure evenly. Wouldn't that mean that diamond could easily sustain 100 kPa (withstands up to 600 GPa)?
Is there no other way around this obstacle? Could we make use of any exotic forces or structural methods? (One interesting suggestion: Gradually evacuate the helium-filled airship, slowly going up, so that when we reach a total vacuum the airship is basically in space? In that case there wouldn't be any powerful pressure from air...)

I you have answers, you may ask this question here http://physics.stackexchange.com/ ;)

http://physics.stackexchange.com/questions/195122/could-a-vacuum-airship-be-possible?noredirect=1#comment416256_195122 @Shadock

I'm closing this question since it is cross-posted to physics.se where it has an upvoted answer.

@Martin-マーチン Ok!


text:
Does the Bohr model violate the uncertainty principle?
In a book, it is stated that one reason for the failure of Bohr's theory was the fact that it violates the uncertainty principle. Is this fact true? How so?

There is an issue of semantics here.  The Bohr model predates the Uncertainty principle, so one can't really speak to the impropriety 'violation' implies.  However, Bohr's model is not compatible with the Uncertainty principle, and the two really can not be rigorously combined to model the quantum mechanics of atoms.

The Heisenberg uncertainty principle states that
$$\Delta x \Delta p \geq \frac{\hbar}{2}$$
When you think about it, the whole of classical mechanics violates this principle. Think of a particle with mass $m$, a defined position $x_0$ at time $0$, a defined velocity $v_0$ at time $0$, and a constant acceleration $a$. The laws of kinematics state that at any time $t$, the position $x_t = x_0 + v_0t + \frac{1}{2}at^2$, and the momentum $p_t = m(v_0 + at)$, with zero uncertainty in both quantities, i.e. $\Delta x \Delta p = 0$. That works in most real-life cases because on a macroscopic level, quantum effects are negligible.
The Bohr model was derived via the classical assumption that the electron orbits the nucleus, much like how the Earth orbits the Sun, except that the origin of the centripetal force was electromagnetic instead of gravitational. It then went on to derive the energy levels of the electron in a hydrogen atom using the laws of uniform circular motion and electrostatics. This means that it is theoretically possible to know the position and momentum of the electron simultaneously with no uncertainty - a violation of the uncertainty principle.

In the Bohr model of the atom the energy (thus its momentum) of the electron and the radius of its orbit (thus its position) are precisely defined quantities. This is a direct violation of the Uncertainty Principle. Since the energy levels of the hydrogen atom had been shown to be experimentally well defined by Bohr’s model ($\Delta$p ~ 0), we are left with the conclusion that we know very little about the exact location of the electrons in space ($\Delta$x ~ infinity).

Source


text:
Why does zinc sulfide glow?
Zinc sulfide is used in road signs because when light from cars hit the road signs, it glows. What actually is the reason for the glowing of road signs (actually glowing of $\ce{ZnS}$)?

In the Wikipedia article Phosphorescent paint,

Phosphorescent paint is commonly called "glow-in-the-dark" paint. It is made from phosphors such as silver-activated zinc sulfide or doped strontium aluminate, and typically glows a pale green to greenish-blue color. The mechanism for producing light is similar to that of fluorescent paint

The process is summarised in the article Fluorescence vs. Phosphorescence,
The explanation from the article:

Phosphorescent materials produce light in a similar way as does fluorescence materials. A visible difference between these two types of luminescence, the ability of phosphorescence materials to glow after the excitation energy source is removed.

The website has a diagram exhibiting the process (in a bit of a simplistic, but understandable way - just replace the 'UV' with 'car headlights etc'):

Essentially, photons are emitted from every step back from the excited state to the ground state, unlike fluorescence, where the photon is emitted when the electron goes straight to the ground state.

Nice answer for the care put in writing it but it seems a bit generic, at least for Chen SE.


text:
Why some cloths become rough after drying?
I line dry my cloths and every now, they become rough after drying. This mostly happens with socks. They are regular cotton socks and once I start to wear them they become soft again. Any idea why this is happening?

I'm not sure it's appropriate site for this question.

I believe it is down to water relaxing the fibres in the material. Upon drying in one position the fibres become temporarily fixed in that position. I have also read that the detergents used in washing can cause clothes to dry stiffer. A little experiment such as soaking your clothes in hot water should kill bacteria and then see if they dry stiff. This would give an idea as to whether or not its down to the detergent or just the water.
Interestingly, a recent study suggests that fabric softeners reduce the wettability of materials. So if you use fabric softener on your towels, over time they will become less effective at drying (being towels).


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Why does chlorine trifluoride ignites concrete and sand?
Concrete and sand are highly unreactive to most of the chemical attacks. But chlorine triflouride is so dangerous that it will ignite concrete, sand and other fire retarding substances. Which property does ClF3 possess to cause this ignition and why?

The related posts (1, 2) by Derek Lowe on his blog In the Pipeline are practically required reading for anyone interested in chemistry! His prose is interesting and very enjoyable.

@NicolauSakerNeto From that first post: "It is also hypergolic with such things as cloth, wood, and test engineers...." Not often one laughs and shudders at the same time...!

Concrete, sand, and asbestos are all, in large part, silicon dioxide (SiO₂). It's energetically favorable for available fluorine to take the place of oxygen, producing SiF₄. This occurs even in weak solutions of hydrogen fluoride, and it's why HF solutions are used for etching glass.
ClF₃ has a much greater amount of potential bond energy than HF or even F₂. This means that, when it decomposes, such as upon contact with silicon dioxide, it gives off a lot of energy, which is released as fire.

Chlorine trifluoride is very unstable so it's energetically favourable to react with almost anything to produce more stable fluorides; releasing oxygen, chlorine and compounds of the two.

You really should elaborate.


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Can our chemical elements be different in other universes?
Let's pretend that in other universe we have an element with atomic number 6, like our carbon, but in this other universe the element is completely different from our carbon, like a new element with other properties.
Is this a possible scenario?

Speculating about the physical and/or chemical properties of some other universe does not seem to be a very useful chemistry question.

This is a very tough question to answer precisely, as you have to dig deep into nuclear physics and quantum field theory. As Jon mentions, it might be more of a physics question. I can try to make a small answer with an interesting reference, though.

One could try to imagine a universe with different underlying physical rules, and work out the implications of this on chemical properties. But, so little detail is known even about how quantum mechanics, etc. leads to the chemical properties of the elements in our actual universe, that confidence in the accuracy of such an analysis must necessarily be ... rather low.

I agree with what @JonCuster said. The body should actually start from "let's pretend that there are other universes" passage, since it is a matter of warm debate are there in fact some other universes (whatever they are) or not. Before cosmologist will settle the things down the question has little sense (at best), therefore, I vote to close it.

I admit this question is far out of my depth, but it's interesting enough that I feel I should give it a shot.
TL;DR - The question is so hard as to be almost unanswerable in detail. Things can be very different, though not arbitrarily different, and the exact line between the two is very difficult to define. There have been some mild investigations on the chemistry of toy universes, however, one of which I describe below.
The problem with thinking about other universes is that there are potentially a lot of different universes which can arise. Famously, string theory currently posits that there are something like $10^{500}$ "valid" universes, each with different details regarding physical laws and physical constants. Furthermore, universes are very complex objects in the sense that there is a ton of interrelation between everything, so even minor changes to details can create wildly different results. Not just that, but there apparently are many parameters to fiddle with, which would create wildly different periodic tables, often apparently breaking any possibility of life. This all leads to fine-tuning problems.
For example, if the attraction between two protons were about 2% larger, then every proton in the Universe would be bound in the form of a diproton nucleus, i.e. the nucleus of helium-2 ($\ce{^2He}$, which is just barely unstable in our Universe), meaning the entire periodic table would collapse to a single element. Apparently there's no very interesting chemistry in that universe! Now, what exactly would have to be tweaked in order to get a 2% stronger attraction between protons while not breaking everything else, I don't know.
So it seems thinking about chemistry in different universes is far too complicated to approach in almost any situation, but still there are some investigations. One particular case I have come across is what would happen in a universe without weak interactions.
Most of Chemistry ultimately has to do with studying different aspects of electromagnetic interactions. However, the stability of atomic nuclei depends immensely on strong interactions (aka strong nuclear force), and to a lesser extent on weak interactions (aka weak nuclear force). The authors in the article linked above analyzed the effects of taking a universe similar to ours, but removing the weak interaction. There are several differences with respect to the evolution of the universe (big bang nucleosynthesis, stellar nucleosynthesis, etc) and the abundance of the elements, but thinking just about their stability, it seems that in such a universe, a large portion of the periodic table would still exist.
Additionally, given that the weak interaction mediates the decay of neutrons in our Universe, its absence means neutrons are completely stable subatomic particles (in our Universe free neutrons decay with a half-life of around 8 minutes). This also means that neutron-rich nuclei are more stable to decay in the hypothetical weakless universe. In our Universe, deuterium ($\ce{^2_1H}$)is the most neutron-rich stable hydrogen isotope, while tritium ($\ce{^3_1H}$) is stable enough, with a half-life of about 11 years, to be chemically important (heavier hydrogen isotopes can essentially be neglected). However, in the weakless universe, it seems that even hydrogen nuclei as rich as $\ce{^6_1H}$ are indefinitely stable, and there are possibly even more neutron-rich isotopes. See the following figure:

(Source)
For comparison purposes, the most neutron-rich stable isotopes for each element in our Universe are $\ce{^2_1H}$, $\ce{^4_2He}$, $\ce{^7_3Li}$, $\ce{^9_4Be}$, $\ce{^11_5B}$, $\ce{^13_6C}$, $\ce{^15_7N}$ and $\ce{^18_8O}$.
Due to this added nuclear stability, the periodic table in the hypothetical weakless universe could actually have more stable elements in it. In our Universe, elements beyond about the 110th are quite unstable, being produced naturally in small amounts, if any, and they will not likely have great significance in Chemistry. However, highly neutron-rich nuclei could make ultraheavy elements more accessible, even in natural conditions such as supernovae.

@NicolauSakerNeto has nicely covered the possible effects that a change in nuclear physics would have on the periodic table.  I wanted to point out two other possibilities:  a different spin for the electron, and a massive photon.
Electron spin
In particle physics, particles come in two main types:  "bosons" and "fermions".  The electron in our Universe is a fermion, with "spin 1/2";  however, you can in principle also have fermions with spin 3/2, 5/2, 7/2, ...  Bosons, meanwhile, have spin 0, 1, 2, ...  In terms of familiar quantum numbers, a particle of spin $s$ can have a quantum number $m_s$ anywhere between $-s$ and $+s$, in integer increments.  So the electron in our Universe can have two possible values for $m_s$ (-1/2 or +1/2);  a spin-one particle can be in one of three "spin states" (-1, 0, or +1);  and so forth.
Now, bosons (unlike fermions) do not obey the Pauli Exclusion Principle;  in fact, they are more likely than not to be found in the same state.  (This is a consequence of something called the spin-statistics theorem.)  This means that a Universe in which the electron was a boson would be a very boring universe indeed;  instead of filling orbitals in a sequential way, every "electron" would be in the ground state of the atom.  (This neglects interactions between the electrons themselves, but this would be a good first-order approximation.)  You therefore wouldn't get unpaired electrons in orbitals like you do in our Universe; bonding between atoms, if it exists in this Universe, would be very different from our own.  My guess would be that you'd effectively have a Universe full of noble gas atoms, with nothing around that wasn't monatomic.
A Universe with a spin-3/2 electron would be more interesting.  These do obey the Pauli Exclusion Principle;  and so every electron in an atom would need to have a different set of quantum numbers.  However, you could now have four electrons in every orbital ($m_s = -3/2, -1/2, +1/2$, or $+3/2$);  and so the periodic table would look very different indeed.  I wouldn't want to hazard a guess as to what it would look like in the end;  the filling order of the orbitals would be very different in this Universe, and it would be a hard question to determine (for example) how many electrons would go into the $2s$ orbital before the $2p$ orbitals started to fill.
Massive photon
In our Universe, the photon is a massless particle;  through the magic of quantum electrodynamics, this leads to the familiar Coulomb potential for an electron orbiting a proton:
$$
U(r) = - \frac{Ze^2}{r}.
$$
However, if the photon had a mass, then this potential would be replaced by something called the Yukawa potential:
$$
U(r) = - \frac{Ze^2}{r} \exp{\left(-\frac{mc}{\hbar}r\right)}.
$$
In particular, the Yukawa potential (unlike the Coulomb potential) does not allow for bound states with arbitrarily large angular momentum; there is a maximum possible angular momentum above which no bound states exist.  This would again change the structure of the periodic table drastically.  The filling order of the orbitals would certainly be changed.  It's not even clear to me that a heavy nucleus (with atomic number $Z$) would necessarily always have $Z$ orbitals for electrons to fill, and so you could end up with the odd effect of heavy atoms tending to exist as charged ions rather than being neutral.


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Why is MP2 failing to calculate proper IR frequencies for [Co(CO)4]- and [Fe(CO)4]2-?
I was doing some IR calculations on different 3d transition metal carbonyl complexes during some practical courses where we were introduced to effective core potentials. While doing this we found that MP2 has some problem with the calculation of at least the asymmetrical $\nu_\text{CO}$ band (that was the only one we were interested in).
We made the calculations for $\ce{[Ti(CO)6]^2-}$, $\ce{[V(CO)6]-}$, $\ce{[Cr(CO)6]}$, $\ce{[Mn(CO)6]+}$ in octahedral symmetry and $\ce{[Fe(CO)4]^2-}$, $\ce{[Co(CO)4]-}$ and $\ce{[Ni(CO)4]}$ in tetrahedral symmetry. The method was MP2, as stated in the title, and the basis set was 6-31G(d). The program we used was Gaussian$~$09 Rev. A.02 but that was not the problem, as I did calculations with other programs with similar results. All optimized structures are considered to be true minima as there either is no imaginary frequency or it is calculated to be $\ge -50~\mathrm{cm^{-1}}$ with was stated to be a problem of the insufficient DFT grid but not further questioned.
The following graphic shows that most of the calculated values are acceptable and near their experimental values but two of them, namely $\ce{[Co(CO)4]-}$ and $\ce{[Fe(CO)4]^2-}$ differ significantly from the experimental values by nearly $300\ldots1000~\mathrm{cm^{-1}}$.
$\hskip1.4in$
After this I varied some parameters of the calculation that could influence this deviation as was:

additional diffuse functions for the better description of ions
tight optimization criteria (T) as this might reduce the imaginary frequency
no frozen core (F) just to try it
solvents: PCM(DCM), PCM(Heptane) as experimental values aren't taken in gas phase
I also optfreq'ed with B3LYP/6-31G(d)/MDF10 (B) to see if it's the method itself

The results can be seen in the following table:
$$\begin{array}{cccccccc}
\hline
\text{Basis~Set} & \mathrm{Solv.} & \mathrm{d(MC)} & \mathrm{d(CO)} & \nu_\text{CO}(calc.) & \mathrm{Int.} & \nu_\text{CO}(exp.) & \Delta\nu_\text{CO}\\
 &  & \mathrm{Ang.} & \mathrm{Ang.} & \mathrm{cm^{-1}} & \mathrm{km\, mol^{-1}} & \mathrm{cm^{-1}} & \mathrm{cm^{-1}}\\
\hline
\mathrm{E}/6{-}31\mathrm{G}^\ast & & 1.802 & 1.184 & 2830 & 2.0\cdot10^5 & 1890 & 940\\
\mathrm{EFT}/6{-}31\mathrm{+G}^\ast & & 1.799& 1.187& 2966& 2.7\cdot10^5 & 1890 & 1076\\
\mathrm{E}/6{-}31\mathrm{+G}^\ast & & 1.800 & 1.187 & 2996 & 2.9\cdot10^5 & 1890 & 1106\\
\mathrm{E}/6{-}31\mathrm{+G}^\ast & \mathrm{DCM} & 1.795 & 1.186 & 4671 & 0.0 & 1890 & 2781\\
\mathrm{E}/6{-}31\mathrm{G}^\ast & \mathrm{Hept.} & 1.799 & 1.184 & 3203 & 4.3\cdot10^5 & 1890 & 1223\\
\hline
\mathrm{BE}/6{-}31\mathrm{G}^\ast & & 1.767 & 1.169 & 1978 & 1322 & 1890 & 88\\
\hline
\end{array}$$
What can be seen:

Within MP2 there is no significant effect on $\ce{M-C}$- or $\ce{C=O}$-bond lengths.
The diffuse functions shift the frequency to higher wave numbers and
the impact of the solvent is big to huge. As without the solvents the
deviation was calculated to be somewhere around
$+1000~\mathrm{cm^{-1}}$ it is now at $+2800~\mathrm{cm^{-1}}$!
Also the intensities are horribly wrong ... they are somewhere around $2$
to $4\cdot10^5~\mathrm{km\, mol^{-1}}$ instead of somewhere around
1000 or in the special case of DCM they are completely zero.
B3LYP predicts everything within common error ranges

What are your ideas why MP2 fails with this special task?

"All optimized structures are considered to be ground states". You probably meant "true minima" out there rather then "ground states".

Well, to start from: how big are MP2 energy corrections for problematic as well as for non-problematic cases?

sure ... the true minima ... changed that

@Wildcat the E2 energy corrections for all calculations alle within $(-65.5 \pm 3.2)~\mathrm{eV}$ and the values for the problematic ones (Fe, Co) are very close to the mean with a difference of $+0.5\ldots+1.7~\mathrm{eV}$

I recommend to make calculations for models shifted in both directions on the vibration coordinate and perform  NBO charges and occupation analysis for M-C-O  for several metals both withing DFT and MP2 framework (and maybe vanilla HF as well). This should at least give some info for speculations. Comparative analysis of molecular orbitals also may help

As a hypothesis, MP2 overestimates electron transfer from M to O atom (supported by increase of error in solvent), though I'm not familiar enough with methods to say why.

@PH13, well, just a couple of Hartrees, which should be fine. But could you also check the values of T2 amplitudes? These are basically the coefficients of doubly-excited determinants. And they should be below 0.1, otherwise perturbation theory is not reliable.

@PH13, the failure of MP2 might be due the multi-reference character of a molecular specie. The quick way to check that in Gaussian would be to run single-point CCSD calc requesting the T1 diagnostic (T1Diag). If T1 is bigger than 0.02, you might have a multi-reference situation for which neither MP2 nor CC are reliable. DFT is also unreliable in this case, but it might work sort of by accident. :D

1st row TM have always been problematic for a HF reference wf, mainly because of strong static correlation effects as mentioned. MP2 is not expected to work well. Typically, hybrid-functional are also expected to have larger errors than GGAs for 1st row TM, but at least they can deal better with static correlation. With such small basis sets, all  kind of effects can occur.

Transition metal chemistry is a graveyard for MP2 :D ||| I think the better question is almost, why MP2 is not failing for the other complexes ;)

Could you please check by a simple CAS(2,2), 2 electrons in 2 orbitals, the occupation numbers? This would be a indication for the reliability or not of the HF determinant used for MP2.

Some months later but here we go...
I agree with most of the comments that multiconfigurational effects should be present for this system, and maybe they are very strong. But I tend to think that this is not the cause of the huge difference with experimental values. The fact that B3LYP returns reasonable values should be taking into account. Surely it "could be" just a coincidence but let's face it, I is very unprovable that all the results are reasonable due to coincidence (it is very simple to check it out, just try some others well proved functionals).
My guess is that you found the correct nuclear coordinates for the minimum of the PES, but the SCF part of the calculation have serious problems and it converged to some unreasonable electronic "state". This would imply that the MP2 results are ridiculous.
As HF and B3LYP are very different in the SCF part, B3LYP converged correctly.
You could try different initial guesses (Hückel, PA, a previous converged calculation) and different methods (like RHF, OHF, UHF etc.). Other basis sets can also help (a little larger if possible). I would try to avoid effective core potentials! (I understand that this can be a problem for your teaching purposes). Convergence method is of course of main importance (consider quadratic or very damped convergence) . I found almost impossible that calculating inner electrons correlation helps in this case.
Of course that the refinements that you tried are useful, but I think that just to improve quantitatively the results.
I hope that the ideas above help you. I would like to know if you effectively solved your problem.
Regards

Thank you for your answer. My current plan is to further investigate on that during this summer. Besides all comments to my question, your answer gives me other good points to look at.

@pH13, If it is not a problem for you, it would be interesting that you share the output file to play with it, I'm curious about what happened. If a nuclei coordinate file (XYZ file or something like that).


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Can you push a liquid and a gas into a container at different pressures?
I have a gas ($\ce{CO2}$ for example) and I want to infuse it into a liquid ($\ce{H2O}$ for example). I need a higher pressure of $\ce{CO2}$ to do this. So I push in water into a sealed container (at 7 psi) and add $\ce{CO2}$ (at 14 psi).
Is this possible? Will it back up? What if I use a check valve? Can I use a non-electrical actuator of some sort to let liquid in, and then add gas 'automatically'? What about something that functions like a carburetor?
I need to let the new solution out after it is mixed and also allow a consistent flow if I leave the exit open. If I have a liquid at 7 psi and a gas at 14 psi and I am trying to push them both into a sealed container, how is this done? Letting it out? Doing it on the fly?
I know you do NOT carbonate water at these pressures. Just an example with real gas/liquid.

Pressure is due to the random impact of molecules on the surface, regardless of whether water or carbon dioxide or anything else. Once equilibrium is reached, all components are at the same pressure.
That said, there's no problem using a moving stream of $\ce{CO2}$ to propel and to mix with water at much lower pressure, using Bernoulli's principle. The automotive carburetor and spray atomizer operate on this basis.
The alternative, of course, is to add the water to the container through a check-valve, then add $\ce{CO2}$ at a higher pressure.
The choice is a continuous process vs. a batch process.

The continuous process is the objective. I have been looking into injectors https://en.wikipedia.org/wiki/Injector
and Aspirators
https://en.wikipedia.org/wiki/Aspirator_%28pump%29
but I'm not sure if they would work like I need. Thoughts? For the record, the automotive carburetor was my initial thought, but I couldn't find anything small for my application.

A "perfume" atomizer is smaller than the carburetor, and you could make a really tiny one from glass tubing with minimal flamework, e.g. pull a point, bend it a bit, and mount it in a T-tube. The moving CO2 in the straight top of the T would pull up the water in the vertical. You would need a check valve in the water line.

I was think of just 3D printing one... But designing it like an aspirator.
The purfume atomizer was suggested elsewhere as well though. It merits a closer look!

We flow gases and liquids into reactors at 80 bar quite easily.  This is not into a sealed (batch) container, but into packed tubes with a back pressure regulator(ultimately)vat the end. The gas pressure is higher than the liquid pressure, which does not lead to gas in-flow due to check valves on the liquid feed.
In a sealed container, it would be expected that the headspace pressure would eventually match the gas pressure. If the liquid pressure was lower and the liquid feed couldn't hold the higher pressure, something would fail.
A mass flow meter can control your gas delivery accurately. A tube packed with inert material is a cheap way of getting good mixing.

Interesting. So it sounds like you're suggesting letting the gas pressure be higher, but regulating the flow of gas so that it doesn't overwhelm the liquid?
If I need to achieve continuous flow, this might be an issue. In the end I need more volumes of liquid than gas, but a higher pressure of gas than liquid to force it into the liquid.

Imagine a glass 2/3 full of liquid, then fill that last 1/3 with N2 and to 1 bar higher than the pressure the liquid was pushed in at.
Think Henry's Law. I need to do this, continuously.
In my other comment I mentioned aspirators and injectors. What do you think?

You may want to consider delivering your desired gas in a condensed form, if you want to dissolve nitrogen (using liquid nitrogen) or carbon dioxide (using dry ice).
You would be able to reach high pressures in a sealed container easily by just adding the dry ice or liquid nitrogen to your water. The nice thing is that you won't need any pumps, regulators, valves, etc. A safety valve would keep from accidentally shattering the container.
While this method doesn't lend itself to a continuous flow, and you'll have to carefully regulate temperature to avoid freezing, you may save a lot of time/money by making a series of batches quickly.

I like it! However, this will be something the customer uses. They cannot prepare batches and I can't effectively keep it under pressure after it leaves my hands to go to them. I need the continuous production no matter what  :/


text:
When Compounds are Considered Compounds and Their Types
I'm a bit new in chemistry, so I need help with a couple of things. Although I don't mind complicated answers, please try to keep it around high school level, and not rocket science. Thanks!

Can someone describe exactly what makes a compound a compound? Is it just the simple definition of "multiple elements combined" or can someone get a more specific?
Sodium chloride is an ionic compound, but it's not made out of "molecules". So does that mean Sodium Chloride isn't an actual molecule (like $Na^+$ and $Cl^-$ are combined as separate ions but they can't be combined in a 1 to 1 fashion)
Can someone explain to me exactly what the law of gaseous reaction is, (it doesn't show up on internet specifically, but it's a law by Gay-Lussac).

Thank you in advance!

This should probably be three different questions.

I didn't want to spam the forums, so I wanted to condense it, sorry if it's bad.

Small quibble - this is not a forum

Don't worry; if you post them as three different questions they can be dealt with better and extensively. Also, it would really be more appreciable if you provide your thoughts and your confusion more elaborately.

Agreeing with the earlier posts. Stack Exchange is not a forum. It is a question and answer site. You ask questions, we give answers. The community votes. Three questions means three times as many opportunities to earn rep (which is good!). For more information about how our site works, consider taking our [tour] or visiting our [help].

A molecule is formed when two or more atoms join together chemically e.g. Na and Cl.
A compound is a molecule that contains at least two different elements e.g. Na and Cl.
All compounds are molecules but not all molecules are compounds e.g. H2 and O2 are molecules but are not compounds because bond is between two of the same elements.
Im not sure what you are asking in question 3. Gay-Lussacs Law is generally associated with the combining volumes principle. e.g. 1 volume of H2(g) + 1 volume of O2(g) = 2 volumes of H2O(g). However, you may want to consider the combined gas law, not just Gay-Lussac law. This combines Gat-Lussacs law with Boyles' and Charles' Law.

I hope this helped, but I assume this is easily accessible via Google (etc.).  In future, I would recommend trying to find the answers yourself first and then ask what you are confused on.


text:
How to determine whether a solvent is polar or unpolar?
I'm going to get different solvents with different bonding types (polar and non polar) and put different solutes with an "unknown" bond type to test whether or not the solute is polar or non polar The question is, can I improve this in anyway or take it further than I've already described. It sounds too easy to me and I want to take it a step (if not a few steps) further.

Is this project for school students? if students have prior knowledge then you can explain things like eutectic mixtures and azeotropes

One thing you should definitely do is a control experiment where you check some known with certainty solutes as to their polarity. This will establish a baseline hypothesis experimentally; i.e. you've been taught that "like dissolves like", but you should demonstrate this to yourself experimentally. Once you do this, then your testing with unknown solutes is more meaningful because you then have similar experimental observations.
And you may find some surprises when you do this! For example, soaps are interesting compounds because they are largely nonpolar molecules with polar "ends". This is why they can dissolve oil (nonpolar) in water (polar), and thus is why soap "cleanses".


text:
Electrode Potential Half Reactions
Does flipping a reaction as written change the sign of the value of E or is it a completely different value?
For example if I have the value for the half reaction $\ce{Cl2 + 2e- -> 2Cl-}$ but need the value for the opposite reaction ($\ce{2Cl- -> Cl2 + 2e-}$), do I just put a negative sign and treat it like $\Delta H$ values?

Yes, you are correct - just flip the sign. As quoted from the venerable UC Davis ChemWiki:

To determine oxidation electrodes, the reduction equation can simply be flipped and its potential changed from positive to negative (and vice versa).

They give the value for your (forward) reaction as +1.358 V, so for the reaction in question, you'll use -1.358 V.


text:
Reduced-pressure (vacuum) evaporation/drying of 15 liters of aqueous solution
I have 15 liters of aqueous solution inside an aluminum stock pot which rests upon a hot plate. I want to reduce the boiling point from 150°C to 100°C using reduced pressure.
What is the most economical way efficiently to dry the solute? Economical means does not require specialty equipment (i.e., laboratory vacuum pumps, expensive cold traps, etc.)?
I own a water aspirator pump (attaches to sink, 27.4 in Hg maximum vacuum, 6 liters/minute) and a 2-stage electric vacuum pump (15-micron, 3CFM) that I use for degassing epoxies. I don't know whether either of these will be of any use here.

The aspirator pump would be unlikely to be damaged by the distillate (and can be cheaply repaired or replaced), but the electric vacuum pump likely will require a cold-trap to prevent the vapors from destroying it.
The aspirator pump would seem the way to go, but you must factor in the cost of water in your location. This Nalgene aspirator uses ~6.5 liter/minute. A recirculating aspirator pump might be the best compromise for a production environment, where efficiency is important, and it might be feasible to save and recycle the distillate.


text:
Can a reverse osmosis system remove radioactive elements in contaminated water?
I am curious to know if a reverse osmosis system can clean water from radioactive elements like cesium-134 and cesium-137. If it can, how efficient can it be?

I don't know about cesium in particular, but according to the EPA's web site (see the section called "How will radionuclides be removed from my drinking water?"), both isotopes should be covered under the beta particle section.
I cannot quantify the efficiency part of your question, but presume it's acceptable to the EPA (which may or may not instill confidence to you in terms of whatever methodology they're using).

The following treatment method(s) have proven to be effective in removing radionuclides at levels below their MCLs:
Beta particle and Photon Radiation: ion exchange and reverse osmosis;
(Gross) Alpha Emitters: reverse osmosis;
Radium 226 and Radium 228 (Combined): ion exchange, reverse osmosis, lime softening;
Uranium: Ion exchange, reverse osmosis, lime softening,  coagulation/filtration.

Radioactivity does not apply so much to an element as it does to a specific isotope of that element.  Many elements have both radioactive and non-radioactive isotopes.  In general, if a reverse osmosis membrane removes a particular element or compound it will remove all isotopes of that element to a similar degree.

"The best solution may be the one used routinely to treat water at the Savannah River Site. The process combines activated carbon, reverse osmosis, and ion exchange. If one doesn't get the iodine-131, two others have a chance to capture the radiation through other means."
https://www.forbes.com/sites/jeffmcmahon/2011/04/07/how-to-remove-radioactive-iodine-131-from-drinking-water/#18325ad52a66