text:
Is s-p mixing referring to hybridization or is it the mixing of one atoms s orbital with the other's p orbital?
According to molecular orbital theory s and p orbitals can mix if they are close enough in energy to each other. For period 2 diatomics, this occurs for $\ce{Li}$, $\ce{Be}$, $\ce{B}$, $\ce{C}$ and $\ce{N}$ - in the case of oxygen and fluorine the s and p orbitals are two far apart in energy. The effect can be seen by realizing that s-p mixing causes the $2\sigma_\mathrm{g}$ orbital to be higher in energy than the $1\pi_\mathrm{u}$ orbital.
My question is: Does the mixing of the s and p orbitals occur between the orbitals of one atom (basically hybridization right?) or does the s orbital of one atom mix with the p orbital of the other (and vice versa)?
This has confused me because I realize that concept of hybridization is widely used and functions well so I initially believed that the mixing should occur in one atom - forming hybrid atomic orbitals - and  then bonding. However, after I thought about it some more I began to doubt that this is the case. Firstly, the s and p orbitals on one atom do not have the correct symmetry to mix effectively (the s orbital would simultaneously constructively interference with one lobe of the p orbital and destructively interfere with the other). Also, the molecular orbital diagram of carbon monoxide reveals that s-p mixing must be occurring since the  $3\sigma$ orbital is higher in energy than the $1\pi$ orbital.

This also seems to contradict the idea that the s and p orbitals mix on the same atom because in $\ce{O_2}$ there is no s-p mixing so why would oxygen mix its s and p orbitals when it's bonding with carbon.

Note that in your diagram only the valence electrons are shown. Considering all electrons, you would get two more sigma orbitals lower resulting from the 1s orbitals.

According to molecular orbital theory s and p orbitals can mix if they
  are close enough in energy to each other.

We can make that statement a bit more general so that we can apply it to all orbitals, atomic and molecular, by saying,
According to molecular orbital theory orbitals can mix if they are

close enough in energy to each other, and
have the same symmetry

Does the mixing of the s and p orbitals occur between the orbitals of
  one atom (basically hybridization right?) or does the s orbital of one
  atom mix with the p orbital of the other (and vice versa)?

Good question, you are really thinking about this!  First off, yes, mixing of the s and p orbitals occurring between the orbitals of one atom is called hybridization.
When I think of these problems I

first, hybridize the atomic orbitals (s and p) on an atom to generate hybrid atomic orbitals
second, mix these hybrid atomic orbitals that exist on separate atoms to form a bond, a molecular orbital

But that's not how it really happens.  It is important to realize that hybridization only occurs in response to some type of interaction between two atoms.  In other words, only when atoms begin to approach and interact with one another does hybridization start to occur.  When hybridization starts to occur, bonding between the atoms starts to occur at the same time.  To first hybridize and then bond (like I think of it in my mind) is an artificial construct that allows me to sort things out.  Hybridization on an atom and bonding between two atoms both occur simultaneously.

the s and p orbitals on one atom do not have the correct symmetry to
  mix effectively (the s orbital would simultaneously constructively
  interference with one lobe of the p orbital and destructively
  interfere with the other).

Yes they do.  As this picture shows we can add and subtract an s and a p atomic orbital to form two $\ce{sp}$ hybridized atomic orbitals.

picture source

in $\ce{O2}$ there is no s-p mixing so why would oxygen mix its s and
  p orbitals when it's bonding with Carbon.

I think they are leaving oxygen unhybridized.  So, for example, they are mixing (adding and subtracting) a carbon $\ce{sp}$ orbital with an oxygen s orbital to form the $\ce{1\sigma}$ and $\ce{2\sigma}$ molecular orbitals in your MO diagram.
Edit: Response to OP's comment

Do you know why s-p mixing raises the energy of the $\ce{2σ_{g}}$ orbital?
  (it lowers the $\ce{1σ_{g}}$)

Nice diagrams and a helpful explanation can be found here.  Basically, It has to do with the symmetry of the orbitals, like I mentioned above.  Look at your $\ce{1σ}$ orbital, when you add or subtract it with the $\ce{3σ}$ orbital, the symmetry is such that there is no constructive or destructive overlap. Therefore there is no interaction and there energies will not change.  On the other hand, the symmetries of the $\ce{2σ}$ and $\ce{3σ}$ orbitals are such that they can constructively and destructively overlap.  As a result they interact and their energies change, one is pushed up and one is pushed down.

'According to molecular orbital theory orbitals can mix if they are (1) close enough in energy to each other, and (2) have the same symmetry' That is not correct. (1) The energy of the orbitals does not have to be close. It'll only determine how well (overlap) they mix, not if. (2) They only need to have appropriate symmetry, i.e. a shared symmetry element. Otherwise a sigma bond between a p- and s-orbital would not be possible. As well as hybridisation itself would not be possible.

Hybridisation is nothing that occurs, it is purely a mathematical construct. You can explain any and all compounds without hybridisation.

s-p mixing is with respect to one atom.
It means that in the approximation that a molecular orbital is mathematically a linear combination of atomic orbitals, the best approximation involves a significant amount of the atom's s orbital and a significant amount of that atom's p orbital (plus linear combinations of one or more atomic orbitals from the other atom or atoms).

Firstly, the s and p orbitals on one atom do not have the correct symmetry to mix effectively (the s orbital would simultaneously constructively interference with one lobe of the p orbital and destructively interfere with the other).

This constructive and destructive interference is what actually occurs.  For example, in methane, with respect to the carbon, there is s-p mixing, with the constructive interference being in the direction of a hydrogen atom.

Thanks very much. However why does oxygen mix in carbon monoxide but not in diatomic oxygen? Also is this mixing essentially hybridisation or is there a difference?

@RobChem What's really going on in CO, is for (according to the numbering in your question) $1\sigma$ the linear combination of atomic orbitals is C-2s 13%, C-2p 11%, O-2s 67% and O-2p 8%; for $2\sigma$ C-2s 17%, C-2p 1%, O-2s 25% and O-2p 57%; and for $3\sigma$ C-2s 58%, C-2p 30%, O-2s 1% and O-2p 9%.  J. Phys. Chem. A Vol. 112, p 2437.  So while the diagram in your question shows no s-p mixing for carbon, there really is s-p mixing for carbon and s-p mixing for oxygen.


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How to find the relative abundances of isotopes of an unknown element?
An unknown element $\ce{Q}$ has two unknown isotopes: $\ce{^60Q}$ and $\ce{^63Q}$. If the average atomic mass is $\pu{61.5 u}$, what are the relative percentages of the isotopes?

Related: What is the formula for finding the average atomic mass of an element based on the abundance of its isotopes?

You can reverse engineer the formula used to calculate the average atomic mass of all isotopes.
For example, carbon has two naturally occurring isotopes:
\begin{array}{lrr}
\text{Isotope} &
\text{Isotopic Mass $A$} &
\text{Abundance $p$} \\\hline
\ce{^{12}C}:   & \pu{12.000000 u} & 0.98892 \\
\ce{^{13}C}:   & \pu{13.003354 u} & 0.01108
\end{array}
The formula to get a weighted average is the sum of the product of the abundances and the isotope mass:
$$A = \sum\limits_{i=1}^n p_i A_i$$
For carbon this is:
$$0.989 \times 12.000 + 0.0111 \times 13.003 = 12.011$$
As you can see, we can set the abundance of one isotope to $x$, and the other to $1 - x$.
If $x = 0.989$, then $1 - x = 0.0111$, OR if $x = 0.0111$, then $1 - x = 0.989$.
Therefore, we can simply set up an algebraic equation:
$$A_1(x_1) + A_2(1 - x_1) = A$$
We know $A_1$, $A_2$, and $A$ in your example, so:
\begin{align}
60 x + 63(1 - x) &= 61.5\\
63 - 3 x &= 61.5\\
x &= \frac{-1.5}{-3} = 0.5
\end{align}
Therefore, the element with a mass of $\pu{60 u}$ has a $50\%$ abundance, and the element with the mass of $\pu{63 u}$ has also a $50\%$ abundance.
Proof: $\pu{60 u}(0.5) + \ce{63 u}(0.5) = \pu{61.5 u}$

Solve for $x$:
$$60x + 63(1-x)=61.5$$

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How to use salt spray data in real environments?
I want to convert resultant hours of salt spray corrosion test to estimated values in years for real life environments such as marine. In other words, How could I predict the result of real life tests using salt spray data?

Note that I'm not an expert in this field and wikipedia itself is not a scientifically relevant primary source. The latter might however be different for sources cited there.
In short, you seemingly better shouldn't.
It seems that the German wikipedia article on the salt spray test has collected a lot of critical remarks from different sources. In summary, standardized (ASTM B117 or DIN EN ISO 9227) salt spray tests may be used in quality control, but cannot be transfered to to real world.
According to the wikipedia article, DIN EN ISO 9227 explicitly states:

Nur selten besteht ein direkter Zusammenhang zwischen der Beständigkeit gegen die Einwirkung von Salzsprühnebel und der Beständigkeit gegen Korrosion in anderen Medien. Die verschiedenen Faktoren, welche das Fortschreiten der Korrosion beeinflussen, können sich je nach den herrschenden Bedingungen sehr unterschiedlich auswirken. Dazu gehört z. B. auch die Bildung von Schutzschichten. Die Prüfergebnisse sollten deshalb nicht als direkter Hinweis auf die Korrosionsbeständigkeit der geprüften metallischen Werkstoffe in allen Umgebungsbedingungen betrachtet werden, in denen diese Werkstoffe verwendet werden können.

I'd roughly translate this to: Only rarely, a direct correlation between the resistance towards saline fog and the corrosive influence of other media can be found. Depending on the conditions, different factors promoting corrosion may have a distinctively different impact. This includes, e.g., the buildup of protective layers.
Consequently, testing results should not be taken as a direct hint on the general corrosion stability of metallic materials under all planned conditions.

I have to agree with Klaus.  ASTM B117 is, at best, a QC procedure.  In my experience, most corrosion and coatings scientist who use B117 do so because it is very well established, but it does not correlate to a real-world measure of performance.  ASTM D5894 is a better cabinet test and there is also an SAE procedure developed for automobiles.

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Storing large quantities of KOH
I need to handle large quantities of $\ce{KOH}$; close to 1000 liters of 85 % conc $\ce{KOH}$ for about an hour. Any suggestions on how I go about doing it? Also the temperature during the process might touch 500 degrees Celsius.

why are you doing this?

I'm voting to close this question as off-topic because it asks about safety advice. We are not here for that (and should not IMHO). Please consult a professional company for advice.

@Jori I disagree that safety should be off-topic. Though I can agree that this particular case is asking about a scale where I'm rather uncomfortable having it here. Anyone handling reactions on a thousand liter scale shouldn't ask the internet for help, they should have people that know this stuff already. Anyting else would be dangerous and very likely illegal.

@Jori Safety, covering advices on personal protection during the handling on particular compounds, suggestions for alternative and less toxic reagents and the advice to better not do something at all always had their place here - for good reason. Consequently, I voted to leave the question open.

@MadScientist I should have been more careful with my wording. I agree with you. I was of course talking about the craziness of someone asking about storage of such industrial scale quantities and needing internet for safety advice. Answering this question and leaving it open would IMHO indicate that we think it is OK to ask such things here and rely on our answers. Especially since OP is not asking about theoretical procedures of storage but about actually handling 1000 LITERS of dangerously corrosive $\ce{KOH}$.

@KlausWarzecha Sure I have asked a lot of safety questions  when I first came here and I'm really glad you guys helped me out (and still do!). But there is a difference between asking which type of gloves would be preferable for handling small quantities of sulfuric acid in a home lab for instance, or asking about storing an industrial scale amount of quite hazardous chemicals. Wrongly handling 1000 liters of $\ce{KOH}$, especially at elevated temperature, could very likely get you killed and perhaps also others in the neighborhood of said storage.

Besides that, I think it is very likely, as Mad Scientist already noticed, that such storage is illegal without the proper licenses and "proof of knowledge".

Ok just to clear the air. I am planning to leave my job to open a manufacturing company of my own. And yes i will be hiring professionals to help me out, but i dont think it would hurt to know a few things before so that when they start talking at least i understand a bit of they are saying. to reiterate I am not crazy enough to handle such large amounts on internet advice.
No its not one large 1000 liter container. Actually I will be needing only 500 liters and i plan on dividing it in 5 containers. The facility will be in a industrial area with nothing around within a 700m radius.

Just to add to the above comment. There are 3 of us, one has a masters in chemistry and other a degree in marketing. I happen to be an engineer. And no, I am not going to build anything of such magnitude in my basement :P We have worked at levels of few hundred ml and for the next few months it would be under 10 lt; 500 is a yr or 2 away. Till now we have only performed at a laboratory scale. That too under the guidance of a faculty. Our guide has put us in touch with some people from the industry, while we are evaluating that I at a personal level am trying to gauge what we are getting into.

carefully read everything about 'purity' and carbonate content. While not a problem in lab setting, when it is trivial to add some more $\ce{KOH}$, tossing around such big quantities may cost. Personally, I highly recommend to have a shower nearby and a fresh set of clothes for an emergency, alkaline burns must be dealt with as fast as possible (they are worse then acidic burns).

AFAIK, alkalies dissolve most materials at elevated temperatures. I guess, that in your case the only acceptable materials are steel and silver, most other metals and ceramics will dissolve in molten alkalies.

Thanks Kluas and Steve for your invaluable advice. In spite of all the concerns and noise I feel I am wiser and 'safer.' Closing this thread

Various stainless steels resist boiling saturated aquous $\ce{KOH}$, such as DIN X6CrMo17 (ASI 434), DIN X5CrNi1911 (ASI 305).
Dont't forget about a suitable and sufficiently large reservoir around the tank in case of spills, valve breaks, etc.

ya sure safety first. i will be using only 500 liters but just to be safe i am going for a larger container. would you by any chance know of some reliable suppliers for these tanks?

Unfortunately not. Btw, don't forget about the (local) legal aspects of running such a facility ;)

I know in New Jersey you are required to have a holding tank around an open reaction that is 110% the size of the reacting vessel, or that's at least what my Chem prof said was required when he worked in an industrial lab. Edit: 110% the size of aka 10% larger.

This "stainless" and (expensive=ferritic) steels are a bad choice, and moreover superfluous. Any lye above ph 9 can be stored safely in mild steel. KOH it seems is sold at 45 % maximum, so most of this thread is nonsensy. And first of all, why KOH instead of much cheaper NaOH? Last not least 1000 ltrs is not a big amount, only to an amateur. 50 % NaOH is shipped in 1000 ltr PE containers or in Tank trucks/waggons/barges. If this cryptex has the company of a chem master, why does he ask here?

@Georg there is a difference between Sodium and Potassium. Ever had Potassium Chloride as table salt? Having potassium instead of sodium imparts specific properties to the final product. Hope you are smart enough to understand why not 'Na'OH. It doesn't hurt to ask around, especially when none of us have any solid industrial experience. The one with among the 3 of us who has a degree in chem is on an extended holiday in Norway. So until he comes I am doing some ground work.

The first thing to clear up is – Why do you need to store 1000 L of KOH in liquid form? It is as others have pointed out, very concerning that someone who is planning to store this quantity needs to ask on the internet.
My advice is don't store 1000 L of liquid KOH, store 40 × 25 L of liquid KOH. This will reduce the risk considerably, and help ensure that a single leak isn't a catastrophic event, but is something you can deal with. This is a common philosophy, and is used, for example, in trailers of dangerous gases, which are organized into racks of 20 smaller gas bottles instead of one large tank.
Keep your 25 L containers off the floor in a bunded area for spill containment, and have appropriate means to deal with a spill. I'm not going to tell you the best way to handle a spill because I don't know. – You need to find out. Also, make sure the store is properly monitored to detect any problems.

Now, going back to my initial point: Let's accept that, for whatever reason, you need to store 1000 L of KOH (as you haven't shared with us what you are doing.) Solid KOH is much safer than liquid KOH, so why do you even need to mix up the solution all at once?
One of the causes of the Bhopal disaster (the worst industrial disaster ever, which killed many thousands of people due an uncontrolled venting event) was the unnecessary storage of a toxic and volatile compound (methyl isocyanate.) If the pesticide they were making had been synthesized via a different route, or even if the methyl isocyanate had been made only in the quantities needed, the size of the disaster could have been reduced or even eliminated (despite the extremely poor state of maintenance of the plant.)
I suggest you mix up the KOH in small batches as you need it to reduce the risk

the temperature in the process might touch 500 degrees Celsius

By "process" I assume you mean mixing solid KOH with water, and that you do not actually need to use the KOH at this temperature. If you expect to reach these temperatures, you are doing it wrong. As can be seen from https://koh.olinchloralkali.com/TechnicalInformation/Phase%20Diagram%200-80.pdf KOH solution in all compositions from 0 to 85% is liquid at below 150 °C. Note also that concentrations above 50% are solid at room temperature – so do you really mean 85% conc KOH?
The safe way to do this is to add the dangerous material to the safe material, particularly as water has such a high heat capacity and can lose heat by boiling. You are going to need either a lot of patience, or mechanical screw feeder, to perform the addition at a safe rate. You will also need decent stirring equipment. Also, I would expect most of the heat will be released in mixing the first 50% of the KOH, because after that you are effectively mixing KOH with KOH.
In any case, if you are planning to do this, you should check the thermochemistry yourself and include it in your Risk Assessment, along with any relevant details from industry experience regarding specific hazards of mixing (spattering for example?)
I really hope you make sure what you doing before you do this. Alkalis are a lot worse on human flesh than acids, partly because they turn the oils in your skin to soap, and partly because the spills won't evaporate and therefore persist until something is done about them.
It would be a good idea to dye your KOH so you know where any spills are. I personally have spilt 50L of antifreeze solution in the lab (it was intentionally dyed beforehand in case of spills, and needless to say we used propylene glycol as it is much less toxic than the ethylene glycol used in car radiators.) Whatever you intend, something in your experiment will go wrong.
Please also check the legal aspects in your area. Also, consult with your supplier! The fact that you need to ask on the internet instead of talking to them suggests you are up to something unusual. Would you care to share with us what it is?

EDIT - Now you've clarified you're an engineer, one more very important point. Keep KOH away from aluminium. Alkali will dissolve the oxide layer that protects aluminium from corrosion, and unmask the true (very high) reactivity of the metal. A lot of heat and hydrogen gas will be generated by the reaction $\ce{3 KOH + Al -> K3AlO3 + 2/3 H2}$.
I'm guessing most stainless steels should be fine (but obviously check.) I have used equipment from the catering industry in some experiments, but stay away from stainless steel cookware that has the base thickened with aluminium. Even though it is external, it's not worth the risk.
Also, note that mixtures of acids and bases with water are often most corrosive at a medium concentration, so don't assume that just because a material is OK at 90% concentration, it's going to be OK at 50% concentration.

Hey thanks for suggesting the idea of dyeing. As mentioned in a comment above, it is in an industrial area; the legal aspect is being worked out by another friend. The temperature will be maintained only for an hour when we add the other reactant. Why i am asking is because we have performed the process only at a laboratory level under the guidance of a prof. The prof has set us up with some of his contacts in the industry. But before we scale up, i thought it would be better to seek some advice.

This is the kind of advice i was looking for :). Subtle points; something that is not covered in college education but only people with sharp eye catch. wish we could have you as another guide.

@cryptex If it was not covered during college education, you went to the wrong college. Majority of these risk concerns are 1st year collage training. I am seriously concerned about  people who need this advice: it is a serious red flag that they don't have any relevant knowledge and training.

@Greg I think the problem is they have the wrong degrees. They have a chemist, cryptex is an "engineer" but they don't have a chemical engineer or even an industrial chemist. Chemical engineers study two relevant subjects that are not on the curriculum for either chemists or mechanical engineers: 1. heat transfer and mass transfer (mixing,dissolution,precipitation) 2. analysis of chemical industry disasters. http://en.wikipedia.org/wiki/List_of_industrial_disasters divides them between "energy" and "other." Even then, there's nothing like witnessing a small incident to help you focus in future

@steveverrill Chemists study all this or how to figure out if they don't know, it is basic lab safety. And if you don't know basic lab safety, you shouldn't do experiments, even in 10 000th of this scale.

@Greg I studied chemistry before I studied chemical engineering, and in no way did the chemistry course, even the practicals, prepare me for anything on this scale. The chemical engineering degree didn't really either, because it presented methodologies but there wasn't a lot of practice in applying them. It was, I'm afraid, working on pilot plant (under experienced supervision) that prepared me for working on pilot plant. Despite prior preparation, formal risk assessment and even HAZOP, I've seen a few incidents, all (except perhaps one) of which in hindsight were preventable.

@steveverrill The most issues are not coming from the scale. How to work with alkalies, what are the common types or dangers, what vessels you should use, you have to be careful with spills, with warming it up/boiling, how to dissolve, how to neutralize, don't keep it in aluminium vessel, you cannot boil KOH solution at 500C etc are all basic chemistry knowledge. Without that you are a loose canon in a small scale lab, too.

@Greg Sure, but the issues are much bigger on pilot scale, and there are additional issues to consider. For example, neutralizing a small spill of KOH in the lab, the heat evolved is not a huge issue, whereas on pilot scale it could well be. You need to plan & document what to do specifically for a spill of KOH - is it even practical to neutralize immediately, or do you contain? Another example from my experience: heating a non-toxic aqueous polymer solution at 10L was fine, but at 100L it unexpectedly caused a health hazard due to volatile decomposition products from localized overheating.

@steveverrill This answer (I am commenting) contains only marginal information about industrial size processes. The bigger problem is that OP seems to be not able to handle lab scale experiments, neither.

@Greg pretty sure i went to the right college (top 5). steve might be right; may not have the right degrees. Though I am elec. engineer I work at an I-bank. The project was something we undertook last year in college and have only now decided to scale up. So yes we certainly lack a lot here.

I am not being naive, I do have first hand experience from my early days in college what conc KOH can do even when you are near a bath. Like i have made it abundantly clear,we are in touch with industry experts. I am one of those who try not to get influenced by others and believe in looking for wise opinion. To an extent i had an economical motive; my firm doesn't employees to have a second job. So I have to calculate if my savings would see us through plus i wanted to be more involved . You are free to downvote or simply ignore this thread; never asked for any of you guys to build our plant.


text:
Diels-Alder rate of reaction
Why is cyclopentadiene more reactive than 9-anthracenemethanol when acting as a diene?
Both molecules are locked in an s-cis conformation, which is favourable. I'm thinking that it's due to the extensive conjugation in 9-anthracenemethanol?

Yes, you are exactly right.  Anthracene is better described by this structure (where there are no fixed single and double bonds, but rather a delocalized aromatic structure)

than a structure like this (where the single and double bonds are fixed and distinct from one another).

So in anthracene, the aromatic character must be disrupted to some degree in order to undergo the Diels-Alder reaction.  In cyclopentadiene, there is no aromaticity to disrupt, the double bonds are basically fixed like in buta-1,3-diene (and, as you point out, locked in the required s-cis conformation).
A final note, benzene won't undergo a Diels-Alder under normal conditions because its aromaticity is completely disrupted and lost during the process.  In anthracene the aromaticity remains in two of the rings, a much smaller price to pay.

Thank you ron. I'm still co fused about the need for 3 molar equivalents of n-methylmaleimide versus 1 molar equivalent of 9-anthracenemethanol?

In organic solvents a 1:1 ratio of diene to dienophile is typically used.  Are you running your reaction in water?  The reactants are less soluble in water and therefore tend to cluster around themselves, effectively increasing their concentration and accelerating the reaction.  Because the maleimide does have some solubility in water an excess is used so that there is 1 equivalent of maleimide that is not dissolved in the water and available to react with the 9-anthracenemethanol outside of the water.


text:
How do things heat one another?
For example, how does a gas stove heat a metal pot?
A combustion reaction between the gas and the air takes place, and this releases energy due to new chemical bonds being formed.
Do the atoms in the metal pot absorb these photons, which then excites the atoms' electrons and creates kinetic energy within the atoms (thus, heat)?
Or, are the atoms in the combustion reaction just moving around so fast that they're bumping into all the atoms of the metal pot, generating kinetic energy?

You already have the gist of it in your latter suggestion. Every time the microscopic phenomena behind combustion and heat transfer are brought up, I can't help but link this pair of amazing clips by Richard Feynman discussing how it works in simple terms. Understanding flames at that level makes them even more beautiful!

In short, conduction, convection and radiation.


text:
Basicity of Group 15 hydrides
Why does the basicity of Group 15 hydrides ($\ce{NH_3,~ PH_3,~ etc}$) decreases down the group ?
I know that's because of its Lewis Base nature due to the lone pair but wouldn't the more EN element be less reluctant to give the lone pair ?
 And nitrogen is most EN in that group .

It can be explained by the very helpful Bent's rule. As nitrogen is more electronegative than phosphorus, there is a bigger s character of the sp3 hybrid orbitals that are used to form the bond to hydrogen than the ones of phosphorus. As a result, the sp3 hybrid orbital that holds the electron pair, has a bigger p character for nitrogen than for phosphorus, meaning that the pair is more distant from the nucleus and more susceptible to donation.

If s character is more for N , p character should be more for P right ?

In the sp3 orbitals used to form the bond with hydrogen the s character will increase for both nitrogen and phosphorus (they are both more electronegative than hydrogen), but much more for nitrogen. The s character for a standard sp3 orbital is 25%, for phosphorus it should be little higher than that and for nitrogen much higher. On the other hand, the sp3 orbital containing the electron pair will have a much higher p character for nitrogen than for phosphorus.


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Permanent Balloon?
Is it possible to create a container for helium that it can't permeate through, like a permanent balloon? Or, if not, what kind of material would have the lowest permeability for that application?
I don't just mean stuff that we could build today either, for instance, could it permeate through diamond? Like if we had molecular manufacturing or something, in the future, and built a box out of diamond, of the appropriate dimensions (like total mass vs volume inside) so that it would float, and constructed it so that there were no openings...well, say we build it almost all the way and then stop, and there's a hole in the top, and we drop some solid, frozen helium in, the amount we would need it to be filled with, and then built the rest of the box, sealing it in (all in a vacuum chamber of course, since that's how those hypothetical fabricators are typically depicted).  Would that be permeable? Atomically precise diamond lattice or whatever, would that be permeable?
Hypothetically speaking...

Yes, this is feasible in practice for a fair definition of "permanent". Monolayers (sheets a single atom thick) of graphene are impermeable to any gas, including helium. To be more precise, the bulk permeability rate for helium through graphene is very low, of the order of $10^{-15}\ mm\  g\ cm^{-2}\ s^{-1}\ bar^{-1}$. For comparison, it means a stack of graphene layers $100\ \mu m$ thick (about the thickness of paper or a human hair) will leak less helium than a $10\ cm$ thick slab of glass
In fact, the original article which investigated gas retention of graphene monolayers specifically mentioned making microscopic balloons, including helium balloons. I don't see why a macroscopic balloon wouldn't work, other than the difficulty involved in preparing a large flawless sheet of the material. There is currently a ton of research behind large scale production of graphene, so something of the sort may actually happen eventually.
Even more interestingly, a closely related yet quite distinct material called graphene oxide has been shown to also create films which are essentially impermeable to helium, just like graphene, but water passes through the graphene oxide film unimpeded, as if there were no barrier at all!


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What is the reason for including excited states in the wave function in post-Hartree Fock methods?
I'm new to computational chemistry, so perhaps this question has an obvious answer. I'm wondering what the intuitive reason for constructing Slater determinants that involve excited states is, when one tries to describe correlations, starting from the Hartree-Fock picture. I understand that the single determinant in the Hartree-Fock model only describes exchange (antisymmetry of fermionic wave functions).
I am trying to understand the logic behind why constructing Slater determinants involving excited states can give us a picture of correlation. I guess I don't understand the relation between excited states and correlations (especially thinking about it from the point of view of DFT, which is interested only in the ground state, but correlations are still important).

related: http://chemistry.stackexchange.com/questions/8722/why-are-excited-slater-determinants-used-to-describe-electron-correlation?rq=1, http://chemistry.stackexchange.com/questions/7944/excited-states-and-electron-addition-removal-energies-within-hartree-fock-ci-e?rq=1

I think that here it is not accurate to say "starting from the Hartree-Fock picture", instead of it, I find better to say "starting from the Hartree-Fock method". When this method is carried out we get a lot of orbitals and that allow us to construct a lot of Slater determinants, even if, we can be only interested in one of them.
Given a complete set of functions $\{f_i(x)\}$, any function $g(x)$ can be thought as a linear combination
$$g(x) = \sum c_i f_i(x).$$
For a two variable function $g(x_1, x_2)$, fixing $x_2$ we have
$$g(x_1, x_2) = \sum c_i(x_2)f_i(x_1).$$
But as $\{f\}$ is a complete set:
\begin{align}
c_i(x_2) &= \sum d_{ij} f_j(x_2) &\text{and}&&
g(x_1,x_2) &= \sum d_{ij}f_i(x_1)f_j(x_2).
\end{align}
This reasoning is trivially extended to a function of $n$ variables.
For an $n$-fermions system, the wave function depends of $n$ spin-orbitals ($n$ variables). To get a mathematic valid form, we only need to antisymmetrize the generalization of the expressions above.
Take in mind that none of these determinants are the true state (they are not eigenfunctions of the true hamiltonian). The same happens with the orbitals (that are eigenvectors of the Fock operator).
So, the use of excited determinants arises naturally from a math view point. Just because we already have functions for the determinantal expansion, and we know that they are pretty good in some sense, because the are the best we can get at least for one determinant (the Hartree Fock solution).
It is not needed to think configuration interaction as a mixture of stationary quantum states (ground and excited states), as said above they truly aren't.
Direct to your question: We can think the problem just as a decomposition of the wave function, somewhat like Fourier series, where the functions involved are the Slater determinants, just because it is convenient. But, this doesn't imply a physical meaning.

Response to comment:

Also this discussion is quite old, I would like to add that the term "picture change" is frequently used in relativistic quantum chemistry, when (effective) Hamiltonians are compared to each other. E.g., when approximating the four-component (4c) Dirac equation by an effective two-component (2c) formalism, one usually introduces further approximations. In this respect, "starting from the Hartree-Fock picture" is a correct terminology, as it implies certain approximations on the Hamiltonian. However, the better way is usually to end up with a certain picture, compared to starting from it... (TheFox)

I agree: By choosing an effective Hamiltonian one is working on a class of approximations that build up a picture for a set of analogous systems. When considering correlation we have in mind a "picture" of the system. Being that H-F is a mean field method, it belong to some picture, but I feel that when dealing with Coupled Cluster the H-F method is just a part of a bigger procedure consisting in finding some functions in order to obtain a solution for another picture. In such sense, while the H-F method itself can be understood as the method for obtaining a solution of the H-F picture, in this context is just part of method for solving a problem in another picture. I added the word here for strictness.

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text:
Calculating charge of amino acids (am I doing it right?)
Please consider this picture of a peptide chain:

I've done some examining on the different amino acids and concluded that the chain is as follows:
Lys-Leu-Gly-Ser-Citrullin (variant of arginine).
I would now like to calculate the charge of this peptide at neutral pH.
I'm using this table to orient myself: http://www.sigmaaldrich.com/content/dam/sigma-aldrich/life-science/biochemicals/migrationbiochemicals1/common-amino-acids-table.jpg
So here's my train of thought:
As we're at neutral pH the +1 and -1 of the mandatory amino and acid groups will cancel out for all amino acids.
Lysine has a pKx of barely 11 which is some higher than 7 so it is basic in our solution. It means that it will accept a proton so Lysine has a charge of +1 in our respective solution.
Leucine, Glycine and Serine have no charge on their x group so their charge is 0 and does not contribute anything to the total charge.
Citrullin, I read, is uncharged at neutral pH, opposed to Arginine which is, so it does not contribute either.
So my total answer is +1. Is this correct? I am in no way 100% sure of my method but it would be great to get a confirmation that I am doing it correctly or to be corrected on my possibly faulty method.
Thank you!

Looks good to me.

I second Lighthart's "Looks good to me"

Let's break out the Henderson-Hasselbalch equation to be sure.

The Henderson-Hasselbalch equation is usually used to determine the pH of a buffer, but we can work backwards to get the ratio of an acid to its conjugate base at any pH.
$$\ce{pH}=\ce{p}K_a +\log{\left(\dfrac{[\ce{A-}]}{[\ce{HA}]}\right)}$$
$$\log{\left(\dfrac{[\ce{A-}]}{[\ce{HA}]}\right)}=\ce{pH}-\ce{p}K_a$$
$$\dfrac{[\ce{A-}]}{[\ce{HA}]}=e^{\ce{pH}-\ce{p}K_a}$$
Let's take the three possible cases:

$\ce{pH}>\ce{p}K_a$

If the pH is greater than the pKa, then
$$\ce{pH}-\ce{p}K_a>0$$
$$e^{\ce{pH}-\ce{p}K_a}>1$$
$$\dfrac{[\ce{A-}]}{[\ce{HA}]}>1$$
$$[\ce{A-}]>[\ce{HA}]$$
For OH and SH functional groups in proteins, when $\ce{pH}>\ce{p}K_a$, then the group is deprotonated and has a negative charge.
For amino groups, $\ce{HA}$ is $\ce{RNH3+}$, so for amino groups at $\ce{pH}>\ce{p}K_a$, they are not protonated and neutral.

$\ce{pH}<\ce{p}K_a$

If the pH is less than the pKa, then
$$\ce{pH}-\ce{p}K_a<0$$
$$e^{\ce{pH}-\ce{p}K_a}<1$$
$$\dfrac{[\ce{A-}]}{[\ce{HA}]}<1$$
$$[\ce{A-}]<[\ce{HA}]$$
For OH and SH functional groups in proteins, when $\ce{pH}<\ce{p}K_a$, then the group is protonated and neutral.
For amino groups at $\ce{pH}<\ce{p}K_a$, they are protonated and positively charged.

$\ce{pH}=\ce{p}K_a$

In this case you will find that:
$$[\ce{A-}]=[\ce{HA}]$$
Half of the groups are protonated and the other half are not.


text:
Oxidation of acetophenone to benzoic acid
According to the solutions manual, the oxidation of acetophenone to benzoic acid can be achieved through addition of chromic acid and application of heat.
The solutions manual refers me to a section in the book which explains that when a benzylic hydrogen is present, a strong oxidizing agent combined with heat can perform oxidative cleavage.
Problem is that I don't see any benzylic hydrogens on acetophenone. Neither does its enol tautomer have any benzylic hydrogens ...
So, how is the oxidation of acetphenone to benzoic acid possible given what the book and solutions manual has told me? Self-consistency aside is this even a possible reaction?

The mechanism of benzylic oxidation is not well understood. For acetophenone, consider what would happen if the enol tautomer were to undergo oxidative cleavage of its $\ce{C=C}$ bond, say using $\ce{KMnO4}$?

@BenNorris is that possible? It wasn't introduced in the text so I'm confused about a) why the text would introduce a reaction it doesn't even cover itself, unless I'm missing something and b) what exactly do we know strong oxidizers such as KMno4 and H2CrO4 to do?

@ron what's the "Good Book"?

What book is this?

@jerepierre - Organic chemistry by Brown.

@jerepierre frankly I think it just boils down to the fact that the people who prepared the solutions manual weren't all the same people who wrote the text so it's not surprising that there is this discrepancy. The task of writing the solutions manual likely largely fell on the shoulders of some poor grad students.

From a SciFinder search, there are ~10 examples of oxidations of aryl methyl ketones resulting in a carboxylic acid using a chromium based reagent (usually dichromate but also chromium trioxide). This suggests the reaction is at least possible, although I think including it in a textbook is unwarranted.
One reaction slipped through my search that is informative (below). Notice that in the presence of a benzylic position that does contain a hydrogen, the methyl ketone is unchanged, while the other group is oxidized.

Taken together, it seems that chromium oxidation of an acetophenone to a carboxylic acid is slower than alkyl benzylic oxidation.

I'd like to clear up one issue that has not been addressed in comments, but provides a weak answer.
There are NO benzylic hydrogens in acetophenone.  So it is not possible based on what the textbook tells you.

you've nailed one of the points I was trying to make ... why does my text fail to be self-consistent? Well, I guess I already know the answer ...

This may be  Bayer-Villager variant, but I've been away from teaching for awhile, and never used this reaction in my lab work, so it is not familiar to me.  My intuition tells me chromic acid is not strong enough, but I'd never bet my intuition against a good literature search.

That Cr[6] compounds oxidize benzylic hydrogens does not mean that they do not oxidize ketones. The presence of a H seems sufficient but not necessary. It depends on the time and conditions. I did my Senior research on chromic acid oxidation of methyl-2-propylphenylcarbinols and the alkenes and alkanes of the same carbon arrangement. The conditions were glacial acetic acid, 25 degrees Celsius, a relatively mild condition. The products were good yields of acetophenone and iso-butyrophenone. [This is from memory I do not have the thesis any more.] Compounds with no benzylic H especially the carbinol were rapidly oxidized to the ketones. Our conclusion was that oxidation of a side chain yielded ketones with carbinols possible intermediates, carbon-carbon bonds were easily broken, and that ketones were more difficult to oxidize.
There are many references about acidic dichromate oxidizing side chains to acids. It is the reagent of choice for chemical oxidation demand [COD] measurements because it seems to be a more thorough oxidizer than permanganate. However, that could be also because it agrees more closely with biological oxygen demand [BOD]. The general opinion is that quaternary benzyl carbons are not oxidized altho a quick search gave no specific data. Tertiary alcohols are oxidized and ketones seem to be less reactive but still oxidizable. This seems to be a subject that could use some lab research; it is a subject for many specific catalysis studies.


text:
Where did all the elements come from?
I heard somewhere (I think is was a show called "How The Universe Works") that the big bang was the result of a hydrogen atom and a helium atom. Is this true? If so, where did all the other elements come from?

Are you sure it said that the big bang resulted from hydrogen and helium, rather than that the hydrogen and helium resulted from the big bang? The first isn't remotely true; the second is merely omitting a huge amount of detail.

This is a very broad question, so below I mention just the very key points.
Big Bang
According to the prevailing cosmological model for the early development of the Universe, known as the Big Bang theory, the Universe begun at about 14 billion years from a singularity that can be thought of a state of an infinite density and temperature, and since its creation the Universe began to expand and cool.
Right after the Big Bang the Universe was immensely hot, the energy density was incredibly high and no particles existed, but the Universe was extremely rapidly expanding and cooling, so that soon enough energy was allowed to be converted into various elementary particles: quarks, leptons, bosons.
Baryogenesis
It is generally assumed that initially there was an equal numbers of particles and corresponding antiparticles, but at some point an unknown physical process referred to as baryogenesis lead to a very small excess of quarks and leptons over antiquarks and antileptons.
As the Universe continued to decrease in density and fall in temperature quarks and gluons combined to form baryons such as protons and neutrons, and the small excess of quarks over antiquarks led to a small excess of baryons over antibaryons, which in turn after mass annihilation resulted in the predominance of matter over antimatter in the Universe.
Big Bang nucleosynthesis
After annihilation it was just the matter of time for the Universe to cool down to a point where remaining protons, neutrons and electrons could combine to form atoms.
Already a few minutes after the Big Bang neutrons combined with protons to form the lightest nuclei (H-1, H-2 or D, H-3, He-3, He-4, Li-6, Li-7, Be-7, and Be-8) in a process called the Big Bang nucleosynthesis, and roughly 378,000 years after the Big Bang the Universe cooled to the point when the formation of light neutral atoms (mostly H and He with trace amounts of Li) became energetically favored.
Eventually, most of the protons in the Universe were bound up in these neutral lightest atoms and matter was uniformly distributed in the Universe.
Stellar nucleosynthesis
But the gravitational attraction of the nearby matter by slightly denser regions and the subsequent growth of such regions resulted in formation of gas clouds, stars, galaxies, and other large structures.
The formation of stars meant that the extreme conditions (temperatures and pressures) for the nuclear fusion were recreated and light elements (H, He and Li) from which stars were formed started to turn into heavier one by a process referred to as the stellar nucleosynthesis. Stellar nucleosynthesis has its endpoint, namely, when nuclear fusion reaches nickel-56 which then undergoes radioactive decay into iron-56 the process stops. The reason is that iron-56 has one of the highest binding energies of all of the isotopes, and thus, no more energy can be extracted from an iron-56 nucleus, either by fusion or by fission. But it is not the end of the story.
Supernova nucleosynthesis
When the core of a star converts to iron-56, the nuclear reactions stop. And that is a big problem, since so far the star balanced the forces of gravity pulling its materials inward by nuclear reactions pushing them outward, but now all the reactions pushing outward stop. What happens then depends on how massive the star is. Supergiants, that have masses from 8 to 12 times the Sun, gravitationally collapse in supernova explosions accompanied by a large burst of neutrons. These neutrons are captured by heavy nuclei, such as iron-56, and the resulting neutron-saturated isotopes then undergo beta decays producing stable isotopes of elements heavier than iron.

So are you saying hydrogen and helium were results of the Big Bang? I was told that the two atoms themselves colliding caused the Big Bang. If that was true however, wouldn't that mean that there was something before something as opposed to  nothing before something?

@DavidJohnson: I don't know where you heard that but it is nonsense. How do you think the entire Universe could be spawned by a collision between two atoms?

@DavidJohnson, this site is about chemistry, not cosmology, so your last question is clearly off-topic here. You could ask about "something before something" on Physics.SE, but before doing so I strongly recommend to find a credible source that "two atoms colliding caused the Big Bang", since phrases like "I've heard that", "I was told that" irritates scientists. :D

@Wildcat: Phrases like "two atoms colliding caused the Big Bang" irritate scientists! No, jokes aside, he won't be able to find a credible source for the claim, but that's what he was trying to generate with this question to satisfy a curiosity, which seems reasonable to me. At least, it would have been, on the correct website.

@DavidJohnson, and besides, questions like "What was before the Big Bang if there was one?" at best lie on the science-phylosophy boundary if not totally beyond science.

According to what i have heard and read, the after the BIG BANG, the first element was
$\ce{H}$ which fused to form $\ce{He}$ and after that the first compound was $\ce{H_3^+}$ and $\ce{HeH^+}$. Then these elements underwent fission and heavier elements were formed.
Then stars were formed inside which we have super HOT cores where still heavier elements were formed. The highest being $\ce{Fe}$ . After wards after Supernovae these elements got kind of thrown of and also planets were formed from Stars and hence we today have all these elements.
Some other elements like the Actinoids were man made by nuclear reactions.
Hope this helps a little!


text:
What is the partial molar volume of epsomite (magnesium sulfate heptahydrate)?
I am trying to calculate the expected volume of an aqueous solution of epsomite and NaCl (308 deg K but willing to go with 298 K as close enough, and 1 atmosphere pressure).  I realize the two may interact, but as a first approximation I'm looking for the partial molar volumes of both. I found NaCl's in a paper (Archer (1992) J Phys Chem Ref Data 21(4): 793-829), but I can't find the values for epsomite.
I have a known quantity of salts and am trying to figure out how much solution I'll end up with after adding enough water to get a particular overall specific gravity. For the curious, I've got 600 pounds of epsomite, 50 pounds of NaCl, and I'm shooting for a specific gravity of 1.25.  Knowing the amount of water to add is straightforward and, unfortunately, not sufficient for this particular situation. I really need to calculate the total final volume after the salts have dissolved.
Thank you.

As you know, partial molar volume of given substance is the change in solution volume after one mole of this substance is added. When you add $\ce{MgSO4.7H2O}$ to water it dissolves to magnesium cation and sulfate anion but the hydration water simply mixes with the rest of water. So, I suppose that partial molar volume of the magnesium sulfate heptahydrate is the sum of partial molar volume of magnesium sulfate and seven times molar volume of water. If $\widehat{V}_i$ denotes partial molar volume of substance $i$, and $V_i$ denotes molar volume, then: $$\widehat{V}_{\ce{MgSO4.7H2O}}=\widehat{V}_{\ce{MgSO4}}+7V_{\ce{H2O}}.$$
Nevertheless, partial molar volume is function of composition and commonly quoted values are for very low concentration (adding substance to large quantity of water). If you want to obtain solution with higher concentration, finding its exact volume requires knowledge about $\widehat{V}_i$ as a function of composition and such function has to be integrated.

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text:
Whose concentration is more 1 molar or 1 molal?
I was reading about concentration terms like molarity, molality, etc. and then I came across a question: which concentration is more:

1M(molar) of solute x dissolved in water vs.
1m(molal) of solute x dissolved in water

When I Google it, I did not find exact answers but did find some related information.
My approach is:

Convert 1M into molality and then compare if it is less or more than 1m. (?)
The density of water is 1kg per L.
When solute is added to it, density would definitely increase(?)
And therefore 1L of that solution will weigh more than 1 kg.
now if we calculate molality of 1M solution it would be less than 1m . (?)
am I correct ??
Density would always(?) Increase after adding solute
**Hence 1m will always be more than 1M **

density of water solutions can be smaller or bigger than dens. of pure water

user14530, I think this is a good question so I took the liberty of editing the wording with the hope of making it clearer to our on-site experts here.  Let me know if I've altered your meaning in any way.

No it is more beautiful now .

One thing to be careful of is that molarity involves the volume of the whole solution, while molality involves the mass of just the solvent.

To convert from molal to molar, one generally needs to know the density of solution as well as the formula weight of solute.  Suppose there is a single solute $x$ in water, and that $x$ has a formula weight of $f_x$ g/mol.  Suppose also that the solution of $x$ has a density of of $\rho$.
Let the molarity of $x$ in the solution be $C_x$ mol/L.  Then:

the mass concentration of $x$ in g/L is $C_x f_x$.
the weight fraction of $x$ in grams of $x$ per total gram of solution is thus $w_x = \frac{C_x f_x}{\rho}$.  The weight fraction $w_x$ must be a number between 0 and 1.
The weight ratio of $x$ in grams of $x$ per gram of water is $\frac{\frac{C_x f_x}{\rho}}{1-\frac{C_x f_x}{\rho}} = \frac{w_x}{1-w_x}$.
The molality is simply the weight ratio divided by the formula weight, $\frac{1000}{f_x}\frac{\frac{C_x f_x}{\rho}}{1-\frac{C_x f_x}{\rho}}$.  The factor of 1000 is because molality is moles per kg of solvent, not moles per g.
That last expression can be written as  $\frac{1000}{f_x}\frac{\frac{C_x f_x}{\rho}}{1-\frac{C_x f_x}{\rho}} = \frac{1000}{f_x} \frac{w_x}{1-w_x}$

This last formula shows that molality values will always be higher than molarity values.  Because $w_x$ has to be between zero and 1, then the molality formula will divide by a smaller number than then molarity formula.  One molar water "dissolved" in water (say we are dissolving distilled water in tap water, or dissolving isotopically labeled water into regular water) has a molality of around 1.02.

My approach is:

Convert 1M into molality and then compare if it is less or more than 1m. (?)

Good approach.

The density of water is 1kg per L.

True, but as noted in the comments, the density of solutions in water can be either higher or lower than the density of pure water.

When solute is added to it, density would definitely increase(?)

Not necessarily true.

And therefore 1L of that solution will weigh more than 1 kg.

Not true, and this is the point where I think your logic mislead you.  To make one liter of a 1 molar solution, less than one liter of pure water is usually required.  This is because part of the liter of solution is taken up by solute.  So not all of the liter needs to come from pure water.  And since water has a density of 1 kg/L, if we need less than one liter of pure water, we need less than 1 kg of pure water.  This then means that our one mole of solute is being dissolved by less than kg of solvent: thus the molality of a one-molar solution is greater than one.
The neat thing is that this is true regardless of the density or formula weight of the species being considered.

Molality is defined as the number of moles of solute per kilogram of solvent.i.e we have to weight 1kg of water (solvent) before adding solute. so 1 molal will be less concentrated.


text:
Can water electrolysis detect the water impurities?
A salesman came to my house to demonstrate his water purifier machine which is a RO Purifier. To show the difference between his purifier machine and the machine I own, he took some water from my machine in a glass. Then he pour water from a bottle he bought in another glass claiming that this was a purified water from his machine (little suspicious). Then he put a small electrolysis machine (that one we used for experiments in school) on both glass and started it. After a minute, there was lot of uhh... particles looking like mud floating in my water, while in his glass, there were very few in dissolved state.
So he said that those particles were salts and bad impurities which my purifier can not filter out, and his purifier can, thus having few particles in his glass.
So my question is, can this method detect the impurities in water, or is it just some dirty sales tactic? Because as I remember, i was taught in school that electrolysis does not has to do anything with water, it just transfer cathode to anode or something like that (sorry for my bad chemistry knowledge :P)
Please help me out!

Which kind of "impurities" do You filter out  with the "purifier" You have already? What is wrong with the water coming into Your house?

[…] can this method detect the impurities in water […]

In principle, yes. The electrolysis can, as an example oxidize iron ions ($\ce{Fe^2+ -> Fe^3+}$) The latter forms a brownish, voluminous oxide hydroxide, $\ce{FeO(OH)·n H2O}$.
The WHO suggests an upper limit of 0,2 mg/l iron in drinking water.
Am I right that your drinking water comes from a well? Did you perform any analyses on the the water quality or consult local authorities concerning possible health risks and their prevention?

[…] or is it just some dirty sales tactic […]

I can't comment on the dirty part, but salesmen need to sell. We don't know whether the reference sample was actually produced by reverse osmosis or just a sample of distilled water (which you would not want to drink over a longer time).
Even if reverse osmosis can reduce the mineral content in your drinking water, you might want to check whether the additional costs for operation are really worth it.
Again, try to obtain some information from local authorities.

Depending on the electrodes used and the voltage applied, the cloudy material may well be caused by degradation of the electrodes themselves. Here's an interesting related video. Notice that after becoming more conductive, the water allows more current through, which speeds up corrosion of the copper wire. So naturally, after a certain amount of time at a given voltage, regular tap water with salts inside will become cloudier than distilled/deionized/RO water.

@Klaus Warzecha Thanks for the suggestions! I live in a city, and the main water source is a dam. Government purifies it before supplying, I also use water purifier, but still there was impurity in my water, hence I was very surprised.

It is entirely normal and indeed desirable for your drinking water to contain some dissolved substances ("impurities"). Unless you are in very poor living conditions, a combination of public water treatment and a home filter will be more than enough to ensure a supply of safe drinking water from the tap. Techniques such as reverse osmosis are only necessary in exceptional cases, like making seawater or stagnant surface water drinkable.


text:
What is the actual reason behind PCl₅ existing as an ionic solid?
Why does $\ce{PCl_5}$ exists in the solid state as $\ce{[PCl_6]^{-}[PCl_4]^{+}}$? What is the problem with it that it doesn't exist in its original form in solid state? Like why isn't it just $\ce{PCl_5}$ in solid state as well?

In the gas phase ions cannot be stabilized through solvation or lattice forces so ions are destabilized relative to non-ionic entities in the gas phase.  $\ce{NaCl}$ is a case in point, in the gas phase it exists as an equilibrium of monomers and dimers.  Yet in the solid phase it exists as a highly ionic crystal.  This example demonstrates that going from a gas to a solid can dramatically affect the relative stabilities of ionic and covalent structures.
The same is true in the case of $\ce{PCl_5}$.  In the gas phase it exists as a covalent compound with a trigonal bipyramid structure.  However, in the solid phase, it turns out that enthalpic affects favor the ionic tetrahedral/octahedral structure $\ce{[PCl4]^{+}[PCl6]^{-}}$.  Simply said, the ionic structure turns out to be thermodynamically more stable in the solid phase where positive and negative charges can be arranged to stabilize the entire lattice; whereas in the gas phase the positive and negative ions cannot be stabilized so the covalent form is favored since it is now lower in energy than the ionic form.
Another reason why the ionic form is preferred in the solid phase is that packing regular geometric shapes such as tetrahedra and octahedra together allows for a more stable arrangement since they can be packed closer together and in a more ordered arrangement than non-regular (kind of an odd shape) trigonal bipyramids can be arranged.
Edit: Response to OP's comment on ion stabilization
If we generate an ion in the gas phase, it will have a certain energy associated with it. If we put that ion in a solution its energy will be much less.  We say the ion has been stabilized by placing it in solution.  Some solvents will stabilize the ion more than other solvents.  This relates to the dielectric constant of the solvent which is a measure of the solvent's polarity.  The higher the dielectric constant for a solvent, the more polar the solvent and the better its ability to stabilize and lower the energy of an ion dissolved in it.
Here is a picture showing a sodium ion dissolved in water.  Note how the negative ends of the dipoles in the water (the negatively polarized oxygen end of the water molecule) molecules congregate and orient themselves around the positive ion.  This orientation of the negative ends of multiple water molecules around a positive sodium ion is what stabilizes the ion in solution.  There is no such stabilization in the gas phase.  In the gas phase it is just a "naked" ion without any external stabilization, so in the gas phase an ion will have a higher energy.

image source see the "Solvation od Ions" section
Here is another picture showing showing (on the right) sodium chloride in a crystal lattice.  The sodium and chloride ions pack around each other to perform the same function as a solvent.  The opposite charges arrange themselves around each other to minimize electrostatic repulsion between like ions and maximize the stabilizing electrostatic attraction between oppositely charged ions.  Through these types of arrangements, the "lattice" can stabilize ions much like a solvent can stabilize ions.  Consequently, both solvents and crystal lattices stabilize ions and lower their energy compared to the ion's energy in the gas phase where no such stabilization exists.

image source

Could you please elaborate on the thermodynamic stability attained by making it exist as $[PCl_4]^+[PCl_6]^-$? And moreover why is closed packing of tetrahedral and octahedral so favorable? I do get that charge separation may make the bond stronger but won't the chlorines repel with each other more?

I've edited my answer to address your comment. But basically ions aren't very stable in the gas phase and can be very stable in the solid phase.  As to the packing question, regular shapes with high symmetry can be arranged in a more ordered fashion than an irregular shape like a trigonal bipyramid.  The more ordered an ionic  solid, the lower its energy because this allows multiple positive charges to be arranged around a negative charge and vice versa.

"But basically ions aren't very stable in gas phase and can be very stable in solid phase". I didn't get this. The entropy increases when we convert a solid to a gas. It is more thermodynamically favorable for stability. And the axial bonds aren't so irregular. The molecules could be stacked a little sideways to accommodate them. As I said before, I do get that charge separation is favorable, since negative will attract positive and vice versa. I'm asking about the fact that how do the new shapes make packing more efficient.

Are you saying you don't understand why ions are destabilized in the gas phase and stabilized in the solid phase?

Also look at the figure at the top of page 620 here.  Notice how the octahedron tucks in snugly between the two tetrahedrons, I don't think you could have any similarly efficient arrangement with only trigonal bipyramid structures.

I understood your packing point. Thanks for the link. And yes I didn't get why ions are destabilized in gas phase but stabilized in solid phase.

I've edited my question to try and better address the stabilization question

Solvents make the liquid state stable because they try to bear the strain present in the bond if any. There is a lot of difference between ions in a solvent and in solid state. An ion in solid state "only has to face itself" and there is no external factor to make it more stable. In a liquid, presence of other ions makes the liquid state stable. I fail to understand your analogy between the "lattice" stabilizing the ions and the solvent stabilizing the ions.

"An ion in solid state "only has to face itself" and there is no external factor to make it more stable." No, it also faces other ions or other molecular dipoles and the electrostatic attraction stabilizes the ion.   "In a liquid, presence of other ions makes the liquid state stable."  No other ions don't make the liquid stable, other ions make the other ions stable and the liquid stabilizes all ions through various (dipole H-bonding, etc.) interactions.  Read the articles linked to my image sources.

The links go towards Diels-Alder reaction and solubility of liquids. They nowhere touch this topic. And how can a solid face other ions/molecular dipoles ? I am not talking about the solid in a solvent. I'm saying that a solid, just in air, not in any solvent or anything. There is no external factor to make it more stable. And in my above comment the "in presence of other ions" meant the ions of the solvent. The solvent makes the solute stable. Just wanted to clarify that out because I didn't get your last statement.

As I wrote next to the link, you need to look further down the page at the "Solvation of Ions" section.  A solid faces other ions internally, within its structure.

Would chatting perhaps be helpful?

Let us continue this discussion in chat.

OK I'll see you there.

Sorry @ron I had to leave the chat. Please see my messages in the chatroom.

see my response in chat


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Thermochromic pigment that changes at around 50 °C / 120 °F?
I wish to find a pigment that can indicate when a battery contact is getting too hot, and all the thermochromic pigments from hobby stores change color below body temperature, at 31 °C is the highest I found.
What kinds of thermochromic pigment can be used for electronic components, that would change color at 40–50 °C / 120–140 °F?
The best thing would be easily obtainable and would give an rough indication of temperature transitions from 30 to 55 °C

The easiest way to obtain such a pigment is to mix a blue and red thermochromic pigment together, one that changes at 32 and another that changes at 50 dregrees. They aren't very easy to find, for the moment i have found the 32 degrees pigments on ebay, and a 45 degree pigment on aliexpress:


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How to Run GAMESS and Avogadro on Command Line?
I always use GAMESS and Avogadro on my own laptop. Recently I installed them on our university supercomputer and started using them by remote logging-in.
On the laptop everything was super easy but unfortunately I can not find any documentation for using them from Terminal Command Line ( I mean sending a job and receiving the result all from Terminal Command Line ).
It would be great if someone knows a good documentation or tutorial for this subject.

Your using Linux terminal?

Avogadro isn't a command line program. The best option is to run it on your laptop and transfer files as needed to the facility, by SSH / SCP for example.

@Mithoron    I use Mac terminal on my laptop endpoint but the other side ( the supercomputer ) is using Linux ( intel ) with these modules : slurm, intel. Ifort. IMIP

Contact the HPC helpdesk at your university or search for some instructions compiled by them.
We cannot know whether your local cluster uses PBS (TORQUE/MAUI), SLURM or anything else to reserve cores, nodes, memory and cpu time to run jobs.

Yes you were right.  I called them and received the needed information .  Thanks !


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How does NMO reform Osmium tetroxide after reaction with an alkene?
After Osmium tetroxide adds to a double bond and is then hydrolysed the following species is formed:
Because Osmium tetroxide is toxic, NMO is used to transform the above back to Osmium tetroxide. NMO is:
What is the mechanism for this?

N-methyl-morpholine-N-oxide (NMO) is used to reoxidize and regenerate osmium tetroxide ($\ce{OsO4}$) in what is known as the "Upjohn Dihydroxylation" procedure.
Mechanistically, the redox occurs when a lone pair of electrons on the reduced $\ce{Os(VI)O3}$ are used to form a bond to the morpholine N-oxygen; the electrons from the former $\ce{N-O}$ bond are, in turn, transferred to the morpholine nitrogen.  The $\ce{OsO3}$ gains an oxygen and is oxidized back to osmium (VIII) tetroxide and the NMO is reduced to N-methylmorpholine

image source
Osmium tetroxide is both toxic and expensive. Using the Upjohn procedure allows the reaction to be run with a catalytic amount of osmium teroxide and 1 equivalent of NMO.


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How fast do electrons move around the nucleus?
Electrons, as we all know, are incredibly small. Smaller things do tend to move faster, right? So exactly how fast considering how small they are? Also, does the electonegativity difference between two atoms alter the speed of the electron?

What do you mean by electonegativity between two atoms?

Electronegativity is the tendecy to attract shared elecrtons toward itself. I was wondering if an electron is being pulled between 2 atoms, would that alter its speed?

So you mean electronegativity difference - you should edit this. Electrons are very fast, but rater because of their low mass than size.

about (1/137)c for hydrogen atom ground state.  I'd write an answer but there is already a good one here: http://physics.stackexchange.com/questions/20187/how-fast-do-electrons-travel-in-an-atomic-orbital

A mind boggling 7.8 million kilometers per hour.

1 mm per second in Uranium atom.

The ratio of the speed of an electron traveling in the first Bohr orbit to the speed of light is given by the handy equation
$$V_\mathrm{rel}=\frac{[Z]}{[137]}$$
where $Z$ is the atomic number of the element under consideration and 137 is the speed of light in atomic units, also known as the fine structure constant.  Consequently a 1s electron in the hydrogen atom will travel at approximately 0.7 % the speed of light. In silver ($Z=47$) the 1s electron will travel around 34 % the speed of light, while the 1s electron in gold ($Z=79$) will travel at about 58 % the speed of light.
Once we get up to around silver, the electrons are traveling at relativistic speeds and this can dramatically impact properties of the atom. For example, the relativistic mass of an electron is given by
$$m_\mathrm{rel}=\frac{m_\mathrm e}{\sqrt{1-(V_\mathrm{rel}/c)^2}}$$
where $m_\mathrm{e}$, $V_\mathrm{rel}$ and $c$ are the electron rest mass, the velocity of the electron, and the speed of light respectively. The following figure provides a graphical representation of how the electron mass increases as the electron velocity increases.

The following equation relates the ratio of the relativistic radius of the first Bohr orbit $R_\mathrm{rel}$ to the normal radius $R_0$, to the relativistic velocity of the electron
$$\frac{[R_\mathrm{rel}]}{[R_0]}=\sqrt{1-(V_\mathrm{rel}/c)^2}$$
As the relativistic velocity of the electron increases, the orbital radius contracts (the above ratio becomes smaller).  For silver, the first Bohr radius contracts ~6 %, while for gold the contraction is ~18 %.
Look at these earlier Chem SE answers to see the interesting physical effects that atoms can exhibit when their electrons travel at relativistic speeds.

francium
gold
mercury

Just to add the importance of relativistic effects when you move down to heavier elements, DFT calculations on heavier atoms need to consider this relativistic effects when computing the electronic properties of molecules (materials), and can lead to interesting effects when included. I would highly recommend reading about Topological Insulators if you need a full physics description of why is it so.

Well, if you consider the ground state of hydrogen atom (Bohr's model), you can calculate the speed by using
$$\frac{m_ev^2}{a_0}=\frac{1}{4\pi \epsilon}\frac{e^2}{{a_0}^2}$$
You get
$$v=e\sqrt{\frac{1}{4\pi \epsilon m_ea_0}}$$
Plugging these values, you get speed to be approximately 2187691.264 m/s, or in other words, 7.8 million kilometers per hour.
It's pretty fast, especially for something thats stuck within a volume of $6.21×10^{-31} m^3$. In fact, at this speed, the electron could, in fact, circumnavigate the globe in 18.4 seconds! Pretty mind-boggling I guess.

If they actually moved in tight orbits, electrons would continuously radiate energy until they fell into the nucleus. Niels Bohr postulated that there were somehow stable orbitals and "ignored" the motion, the beginning of quantum theory (along with Einstein's work on the photoelectric effect). See Bohr model.
When an electron is accelerated (or decelerated), as opposed to staying in one orbital, it emits bremsstrahlung (see Bremsstrahlung).

Bohr didn't ignore motion - in his model orbits were circular, and haven't introduced orbitals.

The point is that a circular - or any - orbit would continually radiate energy until the electron fell into the nucleus. Bohr was forced to sidestep that issue.

The electrons do not move around the nucleus. Furthermore, they don't have a velocity. But they have a kinetic energy, which is not $1/2 mv^2$. This apparent paradox is explained by quantum mechanics. This theory will not be developed here, because it's too long. It would require a lot of pages !


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What allows all 3 states of H₂O to be present at the same time?
How is it possible for there to be a glacier floating on water but under clouds? Depending on the temperature, shouldn't there only be one state of water?

just to clarify. clouds are liquid water in very small droplets (or sometimes very small ice crystals). otherwise you wouldn't be able to see them.

At the triple-point of a substance, all three phases can be present. Since it takes additional energy to go from ice at 0 C to liquid water at 0 C, there can be a considerable overlap for the two substances, e.g. 1 g ice with 99 g water or 99 g ice with 1 g water, but for thermodynamic equilibrium, the triple point is very narrow.
BTW, the first well-know American chemist, Josiah Willard Gibbs, pioneered in defining the relationship of temperature, pressure and state. See  Gibbs' phase rule for more information.

Unless you're at water's triple point (273.16 K, and a vapor pressure of 611.73 Pa), you must have fewer than three phases if all phases are in equilibrium.
However, it's easy to have ice, water, and water vapor present simultaneously under other conditions if the phases are not all in equilibrium with each other.
Note: Water has other triple points, but they involve multiple forms of ice.


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Eco Friendly Cleaning Solution
I would like to find a solution that can kill roof algae, fungus, mold, mildew effectively without using environmentally unfriendly products but don't know enough about chemistry to make a solution myself.

How about peroxide?

This question is a bit broad because "environmentally unfriendly" is not well specified.  For example, algae, fungus, and mold are all naturally found in the environment, and so from a certain perspective, anything that kills them will be environmentally unfriendly.

I suggest to stay with commercially available, read-made solutions. These products are thoroughly tested and certified.
They typically contain around 5% of various benzalkonium chlorides, such as N-Myristyl-N-benzyl-N,N-dimethylammonium chloride, as biocides.
UPDATE As far as the mildew is concerned: The orange ladybird (16 white spots) and the 22-spot ladybird apparently feed on mildew. If these two are naturally found in your region, get some. This is as eco-friendly as it can get to keep mildew in the garden close to a minumum.


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Are all strong bases alkalis?
In my text book, it says that strong bases are alkalis because they fully dissociate in aqueous solution. But I'm not too sure of this. Can you not have a weak alkali? And can you not have a strong base which is not an alkali?
Thank you in advance :)

Are all strong bases alkalis?

No, quaternary amine hydroxides such as tetramethylammonium hydroxide can also be strong bases.

Can you not have a weak alkali?

According to "Acids and Bases" by Kristi Lew, beryllium hydroxide is the one and only exception to the "alkali bases are strong" rule.
I would add that many of the pKb values of the alkali hydroxides are very close to zero.  So you should only think of the them as strong in the sense of strong means pKb near or less than zero.  If you think of strong as full dissociated even in a concentrated solution then they (NaOH for example) would not be strong by that definition.
Also, for the alkaline earth metal hydroxides (which have two hydroxides per metal ion), it is only the first pKb that is strong.


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Except pure alloys, are there any compounds with more metal elements proportion of atoms than nonmetal elements in proportion of atoms?
Except pure alloys, are there any compounds with more metal elements in proportion of atoms than nonmetal elements in proportion of atoms?
For example, Aluminium oxide has 2 metal elements & 3 nonmetal elements. Do we have a compound with X metal elements & Y nonmetal elements, where Y is more then X.

Pure alloys have no nonmetals, so we do not count them.

It's proportion of atoms of elements... Your asking about Na2O, Li3N and other compounds like that? Please ask precisely - it seems that this answer could have been closed as imprecise.

In addition to the answers by Nicolau and DrMoishe.
Most prominent examples are the binary oxides of the alkali metals $\ce{A2O}$: lithium oxide $\ce{A~=~Li}$, sodium oxide $\ce{A~=~Na}$, potassium oxide $\ce{A~=~K}$, rubidium oxide $\ce{A~=~Rb}$ and caesium oxide $\ce{A~=~Cs}$.
Of course there are also higher homologs of these compounds like lithium sulfide $\ce{Li2S}$.
Especially for rubidium and caesium there are also a number of suboxides, i.e. $\ce{A9O2}$ and others. This is a whole class of compounds whith a rich variety.
There are also the nitrides of the alkali and earth alkali metals, e.g. lithium nitride $\ce{Li3N}$ and magnesium nitride $\ce{Mg3N2}$.
There are a couple of phosphides known, that fit your description, e.g. $\ce{K3P}$, $\ce{K4P3}$, $\ce{K5P4}$. These compound are also known from transition metals, like $\ce{Cu3P}$ or $\ce{Ni5P2}$.
Like the above, there are also some very common carbides, like aluminium carbide $\ce{Al4C3}$.
And I am very certain, there are many more.

There are some rare but very interesting cases in which two metals react when mixed instead of forming an alloy, creating a non-metallic compound made only with metal atoms. Examples are caesium auride ($\ce{CsAu}$), caesium platinide ($\ce{Cs2Pt}$) and several barium platinides ($\ce{BaPt}$, $\ce{Ba2Pt}$, $\ce{Ba3Pt2}$, and possibly others), all of which best described as ionic salts rather than alloys. These are a remarkable consequence of relativistic effects in heavy atoms.

They are not rare, they are rarely discussed in common chemistry.

@permeakra Really? Those are just about the only ones I know. Could you provide some more examples? I'd be curious to look them up.

Let me put it this way: there is a significant amount of known solid phases with most atoms having high coordination numbers and having non-trivial stochiometry. Some of them exhibit metallic conductivity, some do not. Most of them are united under 'umbrella term' 'intermetallics', but a lot of borides, carbides, silicides and nitrides follows similar rules. Since the family includes exclusively solid state compounds they are hard to study and classify, so there is very little easily googleable info on them.

As typical examples: $\ce{Fe3C,  Fe3Si, FeSi, FeSi2, TiN_{1-x}}$ as having obvious practical use. Less known, but also significant $\ce{Ni3Al, Nb3Sn}$. A lot of obscure phases guarded by packing.

Again, the compounds alone seem to be pretty common, a lot of metal pairs and triples raise several solid phases. However, most of them have little practical applications and so the field is not widely known, especially since they are mostly 'synthesized' by alloying and detected by X-ray phase analysis.

Still, some phases can be dissolved in liquid ammonia (AFAIK, some phases with $\ce{Pb, Bi}$-containing polyhedral 'aniones').

Thank you for the information, I like having my worldview expanded! I had thought the examples I cited were uniquely ionic among intermetallics, but indeed the field seems so vast as to contain many cases like you cite. I also hadn't read about Zintl ions in a long while, so I forgot about those; even if monoatomic metallic anions are rare and presumably difficult to stabilize in a simple binary/ternary solid, anionic metallic clusters are not so limited.

Sure, e.g. sodium aluminate, Na2Al2O4 (anhydrous formula), is a common chemical (it's the "gray stuff" on aluminum dishware cleaned in an automatic dishwasher with alkaline detergents).

Petzite, $\ce{Ag3AuTe2}$, is an uncommon telluride containing mineral that seems to fit the bill.
Sodium oxide $\ce{Na2O}$ is well known, as is potassium oxide, rubidium oxide, etc.
Some one should write a parser and see if this list from Wikipedia has any more examples.  I didn't see any that weren't already mentioned here on a quick skim.


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Is delta H a state function?
My teacher says $H$ is a state function but $\Delta H$ is not. But I think both of them should be state function. Is he correct?

$\Delta H$ is a function of two states, the initial state and the final state.
For a given final state, there can be infinite $\Delta H$ values depending upon what the inital state was.
For a given inital state, there can be infinite $\Delta H$ values depending upon what the final state is.
Therefore, $\Delta H$ is not a state function.

Then is it a path function?

It only depends on the inital state and the final state, regardless of the path between the two states

I generally agree with DavePhD's answer but I also want to give the opposite view (!).  Everyone agrees $H$ is a state function, but energy (and thus enthalpy) has no built in zero-scale.  So really even in defining $H$, there is some built-in standard state that you are comparing the state of interest to.  For example, the standard state might be all electrons and all nuclei infinitely separated from each other (this standard is commonly used in quantum chemistry).
But we could also specify other standard states. We could say the standard state is one where all electrons and all protons and neutrons are infinitely separated...but that's possibly only useful for nuclear chemistry.  We could also a standard state where all the atoms in a sample are separated into pure chemical elements.   So to compute "the" enthalpy $H$ of say $\ce{CH4}$ using this standard state, we first separate its atoms to $\ce{C}$ and $\ce{2H2}$, our standard state, and then see how much the enthalpy changes.
Regardless of how we definite the standard state, enthalpy is a state function.  But "the" enthalpy $H$ that we compute from the last standard state is the same as the enthalphy change of formation, $\Delta H_{form}$!  So if those two things are equal, and one of them is a state function, then the other one must be a state function too!  So $\Delta H$ is also a state function.

That is just a way to say that dX is actually conceptually same with X if we insist that dX always should be used only with a single standard state. While technically you can call it correct, I think it is not helpful: 1) the OP's question is referring to a misunderstanding that you do not correct or explain, only bury; 2)  dH is more often than not used for comparing specific, non-standard states

Greg, I'm not sure what misunderstanding I have buried -- I do think whether $\Delta H$ is a state function is intimately related to standard states, and also that if $X$ is a state function then its differential $dX$ and by implication $\Delta X$ should also be a state function.  If you have more to add (and by no means do I think my answer is the best or final word!), you should add your own.  I would probably upvote it, just like I did DavePhD's.

As I said, dH is generally not referring a standard state. Also whether something a state-function or not is nothing to do whether it has a "built in zero". This distinction in thermodynamics (Q, W, dH are not state-function etc) actually has a meaning you fully ignore.

When I ignore something, I always try to ignore it "fully".

The "built in" scale for energy is $E=mc^2$. If we have a block of U238 at a certain T and P it has an energy and an enthalpy.  We can heat it, cool it or take it down other paths, and each path will correspond to a delta H, that is not solely a function of the initial state.

A state function is a property of a single state of the system.
A change in a state function is associated with a process that involves two states of a system.
You may be thinking that the change in enthalpy will be a state function because you can take any path you like between the initial and final states and still get the same enthalpy change. But that's beside the point; the enthalpy change isn't a property of a single state of the system, so it isn't a state function.

Fred, can you say a few words about the (interesting) new picture in your avatar?

@ron. Ha. It's just a recent picture of me--this is the avatar I use on my Moodle sites. Funny, after I switched to this one, I got 37% fewer instant messages from students--I wonder why?


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Why is enthalpy of atomization of Mn exceptionally low?
Transition elements usually have high values, especially so for greater number of unpaired electrons. Why the exception here?

According to "Atomic Structure and Periodicity" by Jack Barrett, the reason is a combination of the $d^5$ electron configuration (half filled subshell) in the gas state and, relative to chromium, "the extra electron in the metal occupying an anti-bonding region of a band, s or d".

I didn't understand that? The d5 configuration should help, isn't it?

It should help stabilize the atomized gas phase

Oh so what about the other d5 config elements then?

In each row of transition metal elements, there is a drop in atomization enthalpy when you get to d5


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How do I know that an acid and conjugate base are present in equimolar amounts?
Acetylsalicylic acid is a molecule in several popular drugs such as Magnyl and aspirin. It has a $\mathrm{p}K_\mathrm{a}$ of 3.5.
I'm about to calculate the $\mathrm{pH}$ of a $\pu{0.010 M}$ solution of acetylsalicylic acid.
I've read up on how to do this and in one video the author talks about "equimolar amounts".
He says that now that we have equimolar amounts, we can write: $$\ce{\frac{[H+][A-]}{[HA]}} = \ce{\frac{[H+]^2}{[HA]}}$$
Which makes sense as there are equally many $\ce{H}$ and $\ce{A}$ ions. However when can I assume that an acid disassociated equally? Also, I thought that when it did this then it was called a strong acid? Why would I be using this formula at all for a strong acid?
If I can assume that acetylsalicylic acid disassociates into equal acid and base ions, then I can calculate my problem, but I'm skeptical as to assuming this is true.

Think about the dissociation reaction. How could there be differing amounts of proton and conjugate base? With respect to strong vs. weak acid, a strong acid is one that dissociates completely. Acetylsalicylic acid is a weak acid, but that doesn't change the reaction stoichiometry, just where the equilibrium lies.

Do you mean that we can always say that there are equally many protons and conjugate bases? Why would the formula exist as [H][A]/[HA] at all and not just as [H]^2/[HA]?

Because you could add [A] from a different source, i.e. a salt. If you're considering that [H] and [A] can only come from the acid, then they must be equal.

Oh, that makes sense. You should write this as an answer so I can choose best answer.

I'm really confused here!

No, you are not! You only (temporarily) mixed up two things, both of with you have already understood separately.
$$\ce{HA + H2O <=> H3O+ + A-}$$
1. Position of the equilibrium
This is described by the equilibrium constant $K$, which is the quotient of the rate constants for the forward and the back reaction: $K = \frac{k_{forward}}{k_{back}}$. For strong acids, of which you know that they completely dissociate, the equilibrium is on the product side.
2. Equimolarity
There's nothing out of nothing! If one molecule of $\ce{HA}$ dissociates is water, there exactly one molecule $\ce{H3O+}$ and exactly one molecule of $\ce{A-}$ generated. Not more, not less.
Q: How could another molecule of $\ce{A-}$ appear out of nothing?
A: It can't and it doesn't happen - not even in the LEGO universe.

It is just a proton transfer reaction from $\ce{HA}$ to $\ce{H2O}$; the overall number of atoms does not change!
If is however possible that some $\ce{A-}$ may come from a different source, just by adding a salt $\ce{MA}$ of the conjugate base (think sodium acetylsalicylate) to the solution of your acid. jerepierre has mentioned this in his comment! But for the moment, this is another story.


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How do transmembrane protein alpha helixes bind to the membrane?
I would post this in biology but as the question is biochemical in nature I decided this would be a more appropriate platform.
What I'm asking is how exactly does a transmembrane protein bind to the cell membrane? I realize that the transmembrane alpha helix is about 20 amino acids long with mostly hydrophobic amino acids as the inside of the membrane is hydrophobic.
What I don't realize is how exactly this alpha helix binds to the membrane and sticks to it? I've tried searching on the subject but I can't seem to find the answer. I found out that the alpha helix satisfies the hydrogen bonds internally leaving no polar R groups for the membrane as most groups are hydrophobic, but that only explains how it doesn't bind to the membrane.

The helix and the inside of the membrane are both hydrophobic. Look up the hydrophobic effect and London dispersion forces.

Well, it binds via hydrophobic interacions - non-polar R groups are attracted to hydrophobic interior of the membrane via dispersion forces, and also don't disrupt the structure of water as if they would outside of membrane. It's the same thing which holds molecules of the membrane together.


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Safe yet strong exothermic reaction?
I'm studying chemical engineering at uni (Belgium), and this year (my 2nd) we have to develop and build a stirling engine. It has to be powered by a relatively safe exothermic reaction (not dangerous for bystanders, so no formation of toxic fumes,...) and the whole thing is electrically controlled but that's not important here,
I've come up with this so far:

Sodium acetate crystallization
Hydratation of anhydrous copper sulphate
calcium chloride and water
sodiumhypochlorite oxidizing sodiumsulfite

An 'open flame' reaction is not allowed sadly enough..
I'm hoping to at least reach 70-80 degrees Celsius (158-176° Fahrenheit)
Chances are quite big that there are better reactions but I really can't think of anything else..
Thanks A LOT in advance everyone!

Here are some more options:

Pd-catalyzed oxidation of dilute hydrogen gas in air: $\ce{2H2 + O2 -> 2H2O}$  If you use a dilute hydrogen gas mix (say 95% nitrogen and 5% hydrogen), no flaming combustion can occur in air.  But oxidation still happens, at least when catalyzed by palladium particles. This is how microbiological anaerobic chambers maintain a zero-oxygen environment.  You can probably buy some catalyst fairly cheaply from a microbiology anaerobic chamber company like Coy labs.
Phosphorus pentoxide hydration.  Wikipedia says this reaction, $\ce{P4O10 + 6 H2O -> 4 H3PO4}$, has $\Delta H_{reac}=-177\text{ kJ/mol}$.  There is no flame and no fumes, but the reaction does create phosphoric acid, which is corrosive and can be dangerous.
A modification of your sulfite / bleach idea: use sodium dithionite instead of sodium sulfite.  Sodium dithionite ($\ce{Na2S2O4}$) is air-unstable but if you can work out a way to handle without air exposure it will be more exothermic than sulfite.  Instead of dithionite itself you could use the much easier to handle Rongalite and hypochlorite.  Possibly, using calcium hypochlorite suspensions in water (mostly insoluble) will allow you to use higher concentrations of hypochlorite and Rongalite without the reaction rate and temperature skyrocketing (since the rate will be controlled by $\ce{Ca(OCl)2}$ dissolution).

Your first suggestion reminded me of hand-warmers, an interesting bit of old technology.

Nicolau Saker Neto, I think you meant the first suggestion in the original question, not in my answer.  Hand-warmers use sodium acetate crystallization.

I'd avoid working with phosphorus pentoxide when you have so many other options. It's extremely corrosive so you would have to be careful handling it, and phosphoric acid fumes are an irritant (although if you're just using a few grams probably not so much of an issue) Source: http://www.cdc.gov/niosh/docs/81-123/pdfs/0506.pdf (phosphoric acid fumes tolerated below 5 mg/m^3


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How can inconsistencies in these sodium chloride solutions be explained?
I'm entirely chemically-naive and am confounded when trying to convert $\ce{NaCl}$ brine concentrations to moles per litre.
The situation is that I'm performing linear multiple regressions using published data form multiple studies. Because the studies report their brine concentrations using different units I am trying to convert all of the concentrations into molarity (moles per litre). To do that I'm using the conversion tool at EndMemo - here's a screen shot:

But in more than one place in the literature people report different M values ... for example, below is an extract from one of the studies. Their M values don't match what I'm getting using the online converter (for example, 4 g/L $\ce{NaCl}$ is coming out as 0.068 M/L...hopefully I'm missing something very basic here?)

(Broseta D., Tonnet N., and Shah V., Geofluids, 2012, 12 (4), 280-294.)
As an aside, I'm not finding consistency with the unit symbols for Molarity and litre (sometimes they are capitalised, sometimes they are not...are they referring to different things or is this another example of inconsistency)?

Welcome to chemistry.se! If you have questions about how to beautify your posts, have a look at the [help]. Do you want to know more about this site, please take the [tour].

As for the last part, the notation for litre is $\mathrm{L}$, but $\mathrm{l}$ is also allowed. The amount concentration (molarity) may be abbreviated with $\mathrm{M}$, but since this is not official or standard notation, everybody does what she or he wants, so it's just inconsistency. (I personally do very, very much dislike the use of an abbreviation for molarity.)

I do also believe, they messed up in the paper you excerpted, your calculation is correct and their use of concentration questionable. As $\mathrm{M}$ refers to $\mathrm{mol\cdot L^{-1}}$, there should not be an extra $\mathrm{L^{-1}}$. Could you add the full reference to the paper, so we can look up the context.

The extract is from page 283 of: Broseta D, Tonnet N, Shah V. 2012. Are Rocks Still Water-Wet in the Presence of Dense CO2 of H2S? Geofluids 12, 280-294

The original unit symbol for the litre (a non-SI unit that is recognized by the CIPM as having to be retained for use together with the SI) is a lower case l because it is not derived from a proper name of a person. Only this original symbol is used by ISO, IEC and corresponding national standards. However, by way of an exception, the CGPM has approved the two symbols l and L due to the risk of confusion between l and 1 in some fonts. According to IUPAC, also by way of an exception, the symbol may be either L or l. …

… The term molarity is deprecated. The correct quantity name is amount-of-substance concentration. The word concentration is normally used alone where there is no risk of confusion. The quantity symbol is a lower case c and the unit name is mole per cubic metre or mole per litre (or any decimal multiple or submultiple), thus the unit symbol is mol/m³ or mol/l.

example, 4 g/L $\ce{NaCl}$ is coming out as 0.68 M/L...hopefully I'm missing something very basic here

Yes, you are off by a factor of 10.
4 g/L $\ce{NaCl}$ is 0.068M (0.068 moles per liter of solution).
As far as the journal article, I would used the 4 grams per liter value (and other grams per liter values) and convert to moles per liter properly and not rely on the 0.08 (small capital "M" l-1) value.
If the value is critical to your research, email the corresponding author of the paper and and ask about it in a polite way.

Thank you for the clarification and the advice, I'll do as you suggest. I had written the wrong value in the original post- I did have 0.068 for 4 g/L (and 0.68 for 40 g/L) - and have updated the original post accordingly.