text:
Is dew and water vapour the same?
Is it possible to say water vapour and dew is the same? I know that the water vapour is the gaseous phase of water. But what is the dew? I just wonder how can I categorised it. Is it the liquid  or the gas phase of the water.
Figure below shows dew formed in grass. It seems as a liquid. But is it actually  exhibit in liquid phase or gas phase?

@  santiago : Why the question falls to "off-topic" when I ask for opinions?

@Onthewaytosuccess 'primarily opinion based' is a close reason, the information states Many good questions generate some degree of opinion based on expert experience, but answers to this question will tend to be almost entirely based on opinions, rather than facts, references, or specific expertise. - but, as I mentioned, it is now on topic and has great answers.

Why would you think it's not a liquid?

Let's have a look at the definitions of dew:

dew (the free dictionary)

Water droplets condensed from the air, usually at night, onto cool surfaces.
Something moist, fresh, pure, or renewing: "The timely dew of sleep / ... inclines / Our eye-lids" (John Milton).
Moisture, as in the form of tears or perspiration, that appears in small drops.

Therefore dew is water in the condensed phase, i.e. liquid. Water vapour on the other hand is the gaseous phase. They are definitely not the same.

Then what is the connection between water  vapour and dew?

Water changes its state of aggregation from gaseous (vapour) to liquid (drops, dew), that process is called condensation.

Can I take it as latent heat of vapourization?

@Onthewaytosuccess I don't understand this question.

I mean the heat  produce in the process that you just mention above as   latent heat of vapourization?

Can I take it like that?, I means as  latent heat of vapourization.

@Onthewaytosuccess It might be better if you asked this as a new question, currently, I really do not understand what you are trying to learn. Is it about the terminology? Do you want to know what latent heat is?

Yes ,is it a serious one?

@Onthewaytosuccess I really do not understand what you mean with that either. If you want to talk about any of that, please come to the chat and we might find a way to resolve that...

Let us continue this discussion in chat.

As dew is condensed water vapour [1] and condensation is the change of matter from gas phase to liquid phase [2] dew is clearly liquid water.
So on the one hand they are somehow the same as they both consist of water but on the other hand they are not the same as they are water in different states of matter.


text:
Is the para-position of anisoles more reactive than ortho in electrophilic substitutions?
In my group’s seminar last Friday, we discussed the total synthesis of 6-demethyl-6-deoxytetracycline 1 by Woodward et al.[1] One intermediate of their synthesis is meta-substituted anisole 2 that they subsequently wish to cyclise to create the first two rings of 1 by a Friedel-Crafts acylation. Since the Friedel-Crafts acylation can proceed on both the ortho and para carbons (when viewed from anisole’s oxygen substituent), they protected the para-position using a chlorine atom (3) — introduced by electrophilic aromatic substitution of 2 with chlorine gas. They reported $97\:\%$ yield.

This confused me. I remember listening to a Concepts in Organic Chemistry lecture given at my university where the electrophilic addition to an electron-rich aryl was discussed. Our professor gave us numbers that approximately equalled a $2\::\:1$ ratio of ortho versus para substitution. The reasoning was, that the HOMO has almost equal contributions on both ortho carbons and the para carbon (while having close to no meta contributions), meaning that none of the three be favoured. This is approximately supported by the molecular orbitals of a calculated benzyl anion that the professor has uploaded to his site. (The benzyl anion was used as a model system because it gladly delocates its anionic electron pair into the ring much like oxygens would their unbonded electron pair.)
I do see that 2 has a further substituent in meta position. This substituent (alkyl) displays a weak $+\:\mathrm{I}$ effect meaning that it enhances substitution on the same atoms as the oxygen would. One of these is now ortho to two substituents (position 2) and likely reacts less — I understand that. But I would still have expected a $1\::\:1$ ratio of ortho-oxygen and para-oxygen substituted products rather than $97\:\%$ yield of the para.
How can we rationalise Woodward’s $97\:\%$? Is it wrong to say that the contributions of ortho and para carbons are approximately equal, and if so, why is the para contribution so much stronger that it is almost solely chlorinated?
[1]: J. J. Korst, J. D. Johnston, K. Butler, E. J. Bianco, L. H. Conover, R. B. Woodward, J. Am. Chem. Soc. 1968, 90, 439.

The fascinating part of this is, how they assigned the 97%. They further state that they used the crude product, so how would they even assign the yield? Also following the steps of the procedure, you have 97%, 97% and in the last step just 63% overall yield. So the cyclisation went rouge? Given the reaction time and temperature, one could assume, that the thermodynamically favoured product is formed, which could be the para substituted one...

NMR was also in its infancy and Woodward used to use chemical derivatisation or degradation to determine products (I've not accessed the paper so I can't comment on the specifics here). So only the data presented will highlight if the 97% is plausible. Also, the reaction conditions show no Lewis Acid catalyst but the presence of iodine. Could this explain the regioselectivity, through ICl?

This source suggests some selectivity:  http://www.ncbi.nlm.nih.gov/pmc/articles/PMC411291/?page=2

The 97% yield was of a "crude oil", which when cyclized, gave a crystalline product in 63% overall yield. This chlorination does not exclude other monochlorination products which could have lost in the crystallization.

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sort: votes


text:
Predicting the products of the radical chlorination of propane
Question:

When propane reacts with chlorine, two different monochloropropanes are formed.
a) Name and show the full structural formulae of the two products.
b) What kind of isomerism is shown?

Based on my understanding, the answer for part a is $\ce{2C8H18 + Cl2 -> 2C8H17Cl +H2}$.
Name: chloropropane, formula: $\ce{CH3CH2CH2CH2CH2CH2CH2CH(Cl)}$.
For part b, is it structural isomerism? Are my answers correct?

Let me give you two helpful hints.

Propane

If you use an internet search tool of your choice, you should quickly find out that propane is $\ce{C3H8}$, which has the structure $\ce{CH3CH2CH3}$. The formula and structure you are using is octane. The product you drew, 1-chlorooctane $\ce{CH3CH2CH2CH2CH2CH2CH2CH2Cl}$, is one of the monochlorination products of octane, so you should feel good about being able to correctly apply the reaction type and produce a correct product.

What is the question asking?

A careful read of the intro part of the question When propane reacts with chlorine, two different monochloropropanes are formed indicates that we should be expecting two different organic products with chlorine atoms attached to the carbon skeleton. In other words this reaction should give two products with the formula $\ce{C3H7Cl}$.
However, part a) is not worded well to indicate that you mostly likely are to draw the structures of the two monochloropropane products. You are correct in noting that a radical halogenation reaction will have two products, though the inorganic byproduct is usually considered to be $\ce{HCl}$ and not $\ce{H2}$ or $\ce{H2O}$.
$$\ce{C3H8 + Cl2 -> C3H7Cl + HCl}$$
So let me ask your question in a clearer way:

Propane reacts with chlorine to produce two different monochlorination products with the formula $\ce{C3H7Cl}$. (a) Draw the structures of  both monochlorination products and provide their names. (b) Identify the type of isomeric relationship that exists between the two structures that you drew in (a).

If the answer to (a) was $\ce{C3H8 + Cl2 -> C3H7Cl + HCl}$, I would have tried very hard not to give it to you. Part (a) wants you to draw the structures and provide the names of the two compounds with the formula $\ce{C3H7Cl}$ that can form in this reaction.

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text:
What is CIF (a PCB developer) and what does it do?
I made a PCB (printed circuit board) for the first time, and the process included covering the board with the shape we want, exposing the board to UV light and then putting it in a developer which removes the part of the resin that was exposed to the UV radiation.
My question is, what is this developer? I think it has the commercial name of CIF (http://uk.farnell.com/cif/ar46/developer-ready-for-use-1l/dp/1783557), but what is it really? What does it do to the board and why only to the UV-exposed part?
Thanks!

Developers are usually solutions of $\ce{NaOH}$ or $\ce{KOH}$ (perhaps with some additives). Negative photoresists uses organic developers/solvents. From your description, I assume, that you use a positive photoresist. Positive photoresists are more soluble after exposition to the UV light - polymer bonds are disturbed and therefore more soluble (in areas where exposed). Please see illustration. For more detailed information look for Photoresists and Photolithography.

Yup, NaOH. For future reference, they have to include these things on the MSDS sheet, so you can often look there to see what the major components of these sorts of things are.

Why is it soluble in a base such as NaOH?

The reaction is based on alkaline hydrolysis. But I suggest to ask another question.


text:
Why do soft drinks fizz?
I know it has something to do with carbon dioxide, but what exactly happens and causes the characteristic sound we hear (especially after we shake the bottle)? How do we manage to dissolve a non-polar substance ($\ce{CO2}$) in a polar solvent ($\ce{H2O}$)?

Interesting question! Welcome aboard @AnthonyKelesidis! Just dive in with more questions and answers - take your time to learn the ropes, we are available in chat if you need help, or for a chat (a bit crazy in there sometimes!).

Decrease of pressure causes less-solubility of gas. the gas try to come out from liquid medium.

$\ce{CO2}$ is quite polar.

Because $\ce{CO_2}$ isn't exactly what's being dissolved. When it encounters water, it immediately encounters the following potential reaction:
$$\ce{CO2~(g) + H2O~(l) <=> H2CO3~(aq)}$$
Carbon dioxide is non polar, yes, but carbonic acid readily dissolves (and dissociates, but I didn't care to show that) in water.
The sound we hear is because when we open the bottle, the equilibrium above shifts to the left, creating more carbon dioxide that is being released. Why does the equilibrium shift? It has to do with the pressure in the bottle.
When the bottle is closed, it's under a certain set pressure. When it's closed at the same temperature and pressure for a long time, the reaction above reaches an equilibrium state, where the forward and backward versions of the reaction are happening at the same rate. However, when the bottle is opened, the pressure inside the bottle decreases, as the gas in the bottle suddenly has a lot more space to expand into. A decrease in pressure favors the side of an equilibrium that has more gaseous molecules, which in this case means the carbon dioxide. Ergo, when you open the bottle, the gas is formed, escapes from solution, and rapidly expands outward, making the infamous popping noise.

It may be a little too much hand-holding, but this answer does rely a lot on the meaning of $\leftrightharpoons$ which may not be clear to non-chemists. Explicitly stating immediately after the equation that it’s a two-way street would make the discussion about equilibrium more clear to those unfamiliar with the symbol.

You explain why decreasing the pressure makes it fizz, but now I'm wondering why shaking the drink makes it fizz more.

I think the main effect of shaking isn't to make the gas come out faster, but to make more liquid be carried along with the gas.

States of aggregation should not be subscripted, it is not wrong, but the recommendations (Sec. 2.1.) are different.

There is barely any carbonic acid in carbonated water, the equilibrium constant is just too low
https://en.wikipedia.org/wiki/Carbonic_acid#pH_and_composition_of_carbonic_acid_solutions

@kmm: you didn't read to the part where it says "For a CO2 pressure typical for bottled carbonated drinks ( pCO2 ~ 2.5 atm), we get a relatively acidic medium (pH = 3.7) with a high concentration of dissolved CO2. These features contribute to the sour and sparkling taste of these drinks."

@Fizz a pH of 4 would taste a little acidic, but not very. Try diluting one part lemon juice in 100 parts water to verify. The real origin of the sour and sparkling taste is the action of carbonic anhydrase enzymes in our tongue and throat. We know that, because certain medications inhibit those enzymes, and one of the side effects of taking them is carbonated beverages tasting "flat".

Why do soft drinks fizz?

The key word for that is pressure
Manufactures close the bottles/cans by forcing carbon dioxide and water into your soda at high pressures around 40 psi.
When you crack open the can, you release the pressure and allow the gas bubbles to wiggle free from the liquid and rise to the surface. This requires energy because in order for the gas to break free from the liquid it has to overcome the force holding the liquid together.

Source: LIVESCIENCE


text:
Relation between moles, molar mass and mass

You are asked to prepare a 0.8500 M solution of aluminum nitrate. You find that you have only 50.00 g of the solid.
What is the maximum volume of solution that you can prepare?

I'm confusing my molar masses with grams. The molar mass of Aluminum nitrate $\ce{Al(NO3)3}$ is 213.0 g. If I have 50.00 g of the entire solid $\ce{Al(NO3)3}$ does that mean I have 50.00 times the amount of 213.0 g or will I end up Multiplying 50.00 g of Aluminum nitrate by 0.8500 M, then dividing that by the MM of Aluminum Nitrate. But then I don't know how to set up the rest of my dimensional analysis to reach Liters.$%edit$

If you have 50 g of a compound that has a molar mass of 213 $\frac{\rm{g}}{\rm mol}$, then you have $\frac{50~\rm{g}}{213 \frac{\rm{ g}}{\rm mol}}=\frac{50}{213} \rm{mol}=0.234~\rm{mol}$ of the compound.
If you want to prepare a 0.8500 molar solution, that means the concentration you want is $0.85~\rm{M}=0.85\frac{\rm{mol}}{\rm{L}}$.
With those two numbers, the final answer should be apparent.  (Just divide them and make sure the units work out!)

Many thanks! I actually did my dimensional analysis this way before but I was unsure of my answer. It is a confidence booster to see other's punch out the same work and answer for the same problem. :)

You have $x = (50~\mathrm{g} / 213~\mathrm{g})$ moles of aluminum nitrate. You need some volume of a solution where the molarity of aluminum nitrate is $0.85$. If you had $181.05~\mathrm{g}$ of aluminum nitrate in 1 liter of solution, you'd be fine.
But you have $x$ moles. Thus, you need to scale the volume of solution accordingly.
Set $0.85~\mathrm{mol / L} =\frac{ x~\mathrm{mol}}{y~\mathrm{L}}$, using the value you calculate for $x$ as above, and solve for $y$, the solution volume in liters.

Hi! We don't strictly answer homework questions. Instead, we tend to focus on the procedure of the solution. We give hints, not the direct answer, as it's not useful to anyone. Please edit your post in according to our policy.

Can you illustrate how my answer is substantially different from the one immediately above it, from Curt F.? That will help me in the future. I appreciate it.

I agree with Todd that our two answers have substantially similar content (+1!).  I added the HW tag to the question, but since the questioner had actually invested some amount of thought and showed some existing work, I thought an answer similar to mine was merited.  I think the same is true of Todd's answer.

Your answer has less explanations and it doesn't cover anything more than Curt already does. There's not much wrong with it, though I recommend a bit of elaboration. Thank you.

Todd's answer and mine were submitted within 1 minute of each other.


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Determining equilibrium concentrations from initial conditions and equilibrium constant

The equilibrium constant for the following reaction at $\pu{600^{\circ}C}$ is 4.0.  Initially, two moles of $\ce{CO}$ and one mole of $\ce{H2O}$ were mixed in a $\pu{1.0 L}$ container.  Determine the concentration of all species at equilibrium.

Attempt #1:
$$\ce{CO (g) + H2O (g) -> CO2 (g) + H2 (g)}$$
\begin{array}{|r|c|c|c|}
\hline
\text{Initial}~(M)     & 2.0      & 1.0     & 0  & 0 \\ \hline
\text{Change}~(M)      &  -x      & -x      & +x & +x \\ \hline
\text{Equilibrium}~(M) & 2.0 – x  & 1.0 - x \\ \hline
\end{array}

Plug $x$ into the equilibrium expression and solve for $x$. $4.0  = 2[x]$, so $x = 0.85 [1.0 – x][2.0 – x]$.
Determine concentrations: the equilibrium values become 2.0 – 0.85, 1.0 - 0.85, 0.85, and 0.85, giving 1.2, 0.1, 0.85, and 0.85.

Where does the 0.85 come from? Could it be cross multiplied some way or is there another way this is done?
Furthermore, I want to make sure that I've done the 3.00 correctly for $\ce{Fe(SCN)2+}$.

Consider the reaction represented by the equation: $$\ce{Fe^3+ (aq) + SCN- (aq) -> Fe(SCN)2+ (aq)}$$

Attempt #2:
\begin{array}{ccc}
\text{Initial}     & \pu{6.00 M}\ \ce{Fe^{3+} (aq)} & \pu{6.00 M}\ \ce{SCN^{−} (aq)} \\
\text{Equilibrium} & \pu{? M}\ \ce{FeSCN2^+ (aq)}, \ K = 0.33
\end{array}
\begin{align}
\frac{1}{3}x^2 - \frac{124}{25}x + \frac{297}{25} &= 0 &
 x_{(1;2)} &= \frac{\color{red}{\pmb{-}}(-\frac{124}{25})\pm\sqrt{\left(\frac{124}{25}\right)^2 - 4\cdot\frac13\frac{297}{25}}}{\frac{2}{3}} \\
&& x_{(1;2)} &= \frac{124/25\pm\sqrt{\frac{15376}{(625)}-\frac{15601}{1000}}}{\frac{2}{3}} \\
&&           &= \left(124/25\pm{3}\right)\cdot\frac{3}{2} \\
&& x_{(1;2)} &= 12;3
\end{align}

The text

4.0 = [x]2 so x = 0.85 [1.0 – x][2.0 – x]

appears to be an error. The proper equation is:
$$4  = \frac{x^2}{(2-x)(1-x)}$$
The 0.85 is one of two roots of the quadratic equation you'll solve, using
$$x = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a}$$
after you simplify the first expression, above.
As for the rest of your question, please expand: What does

the 3.00

mean?

For that question, you've provided initial concentrations of the species on the left hand side. There is information that is missing in order to complete the problem. If you provide an equilibrium concentration for the right hand side, we can then compute the equilibrium constant (K). Alternatively, you can provide the equilibrium constant and we can then back-out what the concentrations of all the species are at equilibrium. Please provide either K or the concentration of the species on the right hand side at equilibrium.


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Indirect titration with ascorbic acid
We were doing a titration experiment to find out how much ascorbic acid a vitamin pill actually contains. In a flask, there is a solution of water, sulfuric acid, potassium iodide and ascorbic acid. Also starch as an indicator. When adding potassium iodate $(\text{KIO}_3)$, the following reactions will appear:
$$\text{IO}_3^- + 5 \ \text{I}^- + 6 \ \text{H}^+ \longrightarrow 3 \ \text{I}_2 + 3 \ \text{H}_2\text{O}$$
After the iodine is formed, the following reaction will appear:
$$\text{C}_6\text{H}_8\text{O}_6 + \ \text{I}_2 \longrightarrow \text{C}_6\text{H}_6\text{O}_6 + 2\ \text{H}^+ + 2 \ \text{I}^-$$
After using the right amount of potassium iodate, there will be no ascorbic acid anymore, causing to eliminate the second reaction. When there is no ascorbic acid, iodine will not react back to iodine-ions, which causes the starch to give a brown/blue color.
My question is, why isn't this possible without the sulfuric acid?

Look at the first reaction: it requires $\ce{H+}$ ions.

It is possible, sulforic acid is added just to increase oxidation power.


text:
Do we have a reasonable assurance that relative atomic mass is constant throughout the universe?
I know atomic mass can be (over) simplified as the average mass of all isotopes of a given element weighted by the ratio of their natural occurrence. I also know that, as far as we know, refining relative atomic mass is a matter of finding new significant figures, not finding new isotopes. Are we reasonably sure these ratios won't change as we explore the universe?
As an over simplified example, do we have a reasonable certainty that 70% of the Calcium on Mars won't turn out to be Ca-41?

Your question is equivalent to asking "is the isotopic abundance for the isotopes of all elements constant from planet to planet, star to star, galaxy to galaxy, eon to eon, etc"  I think this stack exchange question is identical and has quality answers to boot: http://chemistry.stackexchange.com/questions/33056/is-the-relative-natural-abundance-of-isotopes-of-an-element-the-same-everywhere

BTW Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center.


text:
Cobalt Oxalate / Potassium Permanganate Titration
My class just did a lab today where we mixed cobalt oxalate with $40~\mathrm{mL}$ of $\ce{H2SO4}$ and then titrated it with $\ce{KMnO4}$ (potassium permanganate). I need the mole ratio of cobalt oxalate and $\ce{KMnO4}$ and am not sure how to set up the equation. She provided us with this formula with 3 variables A,B,C -
Co(a)(C2O4)(b) * (c)H2O
but I am not sure what to do with it. I am also confused as cobalt oxalate appears as $\ce{CoC2O4}$ in her equation yet Google says it's $\ce{H2C2CoO6}$.
My table in my lab book is asking me for "Moles Oxalate Present" , "Mass of Oxalate in Sample" , and "Percent Mass of Oxalate in Sample (%)". It is also referring to the "Mass of transition metal oxalate used" which we took during the actual lab. (I guess this lets me know that it solely wants oxalate and to calculate these values aside from cobalt oxalate?)
I am very confused and if any part of this question needs to be cleared up I can certainly do so.
I can provide any/all data that I took during the lab, but explaining conceptually will still do me some good here. I really just don't know where to start with all this data.
My book also gave me this formula for the reaction of oxalate ions with permanganate ions:
$$\ce{5C2O4^{2-} + 2MnO4- + 16H+ -> 10CO2 + 2Mn^{2+} + 8H2O}$$
Will this factor at all into my calculations?

For starters, the Google oxalate must be wrong. It can technically be either $\ce{H4C2CoO6}$ or $\ce{H2C2CoO5}$, but a random search just gave the former. Do you notice an easy way to turn your teacher's formula into the corrected Google formula(e)? The equation you have written at the bottom will most certainly factor in, by the way; and to solve this problem, we need a minimum of three more values (or two values and one bit of information).

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text:
several coefficient differences in UV-vis Spectroscopy
Does anyone know what is scattering coefficient, absorption coefficient and extinction coefficient, and how to separate them experimentally?

In most applications, these terms are used interchangeably.
Most absorbance spectrophotometers work by measuring the intensity of the light beam that is transmitted through the sample.  A strong light beam goes in, and a weaker one comes out the other side.  The transmission is the fraction of the light that made it through.
There are two main ways in which the light that goes in gets diminished in intensity before it goes out.  The first is absorption.  Molecules that absorb light of a certain frequency make photons of that frequency disappear.  The energy of those photons excite electrons into higher energy states.  These states may eventually decay, releasing heat or lower-energy photons, but not photons of the same energy that were absorbed.  The propensity of molecules to absorb can be called the absorption coefficient.
Scattering just deflects the incident light to a new angle.  Now, when it exits the sample, it won't come straight out on the side opposite where it went in, and it won't be detected by the detector.  Molecules can cause scattering if they are very large.  The "scattering coefficient" describes a molecule's propensity to scatter light of a certain frequency.
The "extinction coefficient" is -- in most applications -- just equal to the scattering coefficient plus the absorption coefficient, i.e., it's proportional to a molecule's ability to disrupt incident light in some way or other, regardless of the mechanism.
In the usual, transmission-based spectrophotometer, only extinction coefficients can truly be measured.  To resolve these measurements into scattering contributions and absorption contributions would require measuring the intensity of the scattered light, i.e. detectors would need to be present at multiple angles.


text:
What are scattering losses?
In UV spectroscopy, when a beam of light is shone into solution, what is scattering loss during this process?

Scattering is the redirection of the 'beam' of light, as is illustrated in the diagram below:

Source: University of Wisconsin
The description from the document Fundamentals of modern
UV-visible spectroscopy (Figure 46) states that

Scattering causes an apparent absorbance because less light
  reaches the detector

They also detail two main types of scattering in UV spectroscopy (Figure 47):

• Rayleigh scattering: Particles small relative to wavelength
  • Tyndall scattering: Particles large relative to wavelength


text:
I need Iron powder. What's the best way to produce it?
I have Ferric Chloride and Aluminum...
I also have steel filings from grinding steel objects, though some Iron Oxide have be separated (I do have Oxalic Acid and Citric Acid). Though that would give me steel powder, not iron.
Which is the most economical and would yield purest samples?

If you can figure this out, there are some steel makers who would like to talk to you. You can invite them to a conference at your private island.

I suggest buying iron powder. It's rather cheap, isn't it?

I think, that only way is sputtered iron melt. Or sand blasting using iron particles.  Using  wet chemistry, iron oxides will be produced  .

Sigma-Aldrich has a product that is very pure but rather expensive. I suspect you can have high purity or low cost but not both.

You could use a high grade metal file on some clean cast iron or an iron nail and coldest the metal shavings which is similar consistency to a powder. An iron nail would be galvanized though so you'd have to take the coating off first. Impurities may be present from the file though.

Those are unlikely to be near pure iron, which is very soft, more likely  to be high carbon steel.

The iron can be precipitated from the chloride salt by reacting it in solution with naoh and aqueous ammonia to form an iron ammonia complex then react that with glucose which should precipitate the metal .It won't be powder tho.


text:
Aluminium and water safety
I have a 8000 gallon water tank that has always had pristine clear water.   I have a brass float valve at the top input to the tank.  Someone put an aluminum ladder in the tank (15 ft deep tank) and after who knows how long, the ladder was encrusted with a white gooey powder and the water turned milky.  what happened?  Main question is IS the water safe to drink or how do I get rid of whatever is in there.

The white stuff is aluminum hydroxide. The water is fine, filter the solids with a coffee filter if it doesn't settle to the bottom. And remove the ladder.


text:
Boiling dish soap
Is it dangerous to boil dish soap (natural, ex: walgreens brand "ology") while also cooking other foods in the kitchen (ex.pasta)?
I was trying to clean some pots and read online that a good way to clean them is to boil dish soap in them. I was also making dinner at the same time and I am worried that I released toxic chemicals into the air and into my food.
How hot does soap need to be heated to cause it to decompose, and what substances will be formed by the decomposition of the soap ?

You probably don’t have to worry about toxic chemicals in the air/food. Most dish washing liquids/gels/soaps contain chemicals that can be tolerated fairly well.
Nevertheless, boiling dish soap in a pot can potentially generate harmful substances. This may happen when your dish soap is very concentrated (when nearly all water has evaporated) and the temperature in your pot may rise to a point at which the ingredients of your dish soap could decompose or generate noxious fumes. If you make sure that there is enough water in your point at any time, your maximum temperature should not go above the boiling point of water (100 °C / 212 °F) which should be fine for most detergents.

No. This is ill-advised. Changes in heat generate chemical reactions. Raising the temperature of many chemicals can cause them to produce toxic gases or other substances. If you are not familiar with the way each chemical reacts to heat/other chemicals under heated conditions you risk producing many combinations of harmful substances. For instance, many chemicals in cigarettes are harmful and many are not nearly as harmful, but when heat is applied many non-harmful chemicals can produce dangerous carcinogens or even chemically react with the other chemical(because they are under circumstances they were not designed for).

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text:
Benzaldehyde and Cyanide - if you can smell one can you smell the other?
Cyanide and benzaldehyde both have an almond-like smell.
Some people cannot smell cyanide (specifically, HCN). Can they also not smell benzaldehyde?
In other words, do they both trigger the same receptors?

Hm. I think I know where you're getting the link from (they both smell like almonds). Let me see what I can find.

I perceive the smells as decidedly different. But of course when amygdalin is hydrolyzed  both of them usually appear essentially together.Benzaldehyde is easy to get (almond aroma) - so you can check easily whether people can smell that. Cyanide is obviously more difficult. But if you take bitter almonds and hydroloyse the amygdalin in acidic condidtions (e.g. with KHSO4 added), cyanide smell will be available shortly before the benzaldehyde sets in.

Answer is, they are most probably not correlated. Smell requires molecules to bind to receptors, which they do by structure and functional group. CN is a very small linear molecule that does not even share its heteroatoms with benzaldehyde, an aromatic compound several times cyanide's size. Ergo, it's improbable they share the same scent receptors.

Then why do they smell very similar?

Uhm, are there people that cannot smell benzaldehyde? I think not.

@Jori I have a friend who swears he cannot smell ammonia, so I guess anything is possible.

@orthocresol how can some say they cannot detect a stimulus in the first place? How would they know what to detect if they cannot detect it?

@Papul I get where you are coming from, haha. I guess one could imagine what it smells like based on others' descriptions (e.g. amines smell like rotten fish...) but that's also rather subjective.

To me they smell decidedly different (same for my father and brother - my moter smells benzaldehyde but not HCN). I've met several people who cannot smell HCN, but so far noone who cannot smell benzaldehyde (we tested with 50+ people in two seminars).


text:
Atmospheric Nitrogen gas and human body pH buffers
It looks like the human body doesn't have the ability to breakdown atmospheric $\ce{N2}$, as of now $\ldots$ So an enzyme like nitrogenase (found in cyanobacteria) fixes $\ce{N2}$ and produces $\ce{NH3}$. $\ce{NH3}$ is a weak base which qualifies as a buffer component.
If the human body was able to react to $\ce{N2}$ inhaled from the air, would it negatively impact the body's ability to maintain pH levels of vascular systems/cells of the human body given the large amount of $\ce{N2}$ in the atmosphere?

One chemical equation that can be written for nitrogenase is this one:
$\ce{N_2 + 8 \ H^+ + 8 \ e^- + 16 \ ATP \longrightarrow 2 \ NH_3 + H_2 + 16 \ ADP + 16 P_{i}}$
The equation shows that not only does nitrogenase generate the basic molecule $\ce{NH3}$, it also consumes protons (i.e. acid) to form hydrogen as an inescapable byproduct of its mechanism.
So yes, in theory nitrogenase could perturb the human body's ability to buffer pH.
However, the relative rates are always important.  Human requirements for nitrogen are on the order of 80 grams of protein per day.  This translates to about 12 g of (fixed) nitrogen atoms per day, which corresponds to 0.85 mol of N.  If all of that nitrogen were to come from nitrogenase, then 0.85 × 8 ÷ 2 = 3.4 mol of acidity would be lost via proton consumption in addition to the 0.85 mol of base generated as $\ce{NH3}$, or about 5.2 mol of "base" in total.
But human pH homeostasis works even though humans breathe out about 900 g of an acid equivalent ($\ce{CO2}$) per day, or about 20 mol.
So even if nitrogenase completely replaced human nitrogen requirements in some hypothetic bioengineered human, the demands on pH homeostasis would only change by about 25%, maximum.  And considering that pH homeostasis still works even when humans are exercising vigorously, when "instantaneous" $\ce{CO2}$ production rates are far higher, I doubt even a 25% change would be too much to handle.
Thus in conclusion I don't think nitrogenase would wreck human pH homeostasis, unless its activity in the human body was increased to a level far beyond what is necessary to replace dietary N requirements.

It's well pointed out. Sorry I didn't get this part. I doubt even a 25% change would be too much to handle. So you mean, body is able to handle even 25% (hypothetically, looking at these equations)?


text:
What is the concept behind the calculations for the dilution of solutions?
I'm interested in the concept behind this computation and was wondering if anyone could give me some pointers or hints.
I've prepared $0.28~\mathrm{L}$ solution of $0.8500~\mathrm{M}$ of $\ce{Al(NO3)3}$.
I need a $50~\mathrm{mL}$ solution of $0.450~\mathrm{M}$ of $\ce{Al(NO3)3}$ made from my original solution.
For my dimensional analysis what I need should be up front right?
$50~\mathrm{mL}$ of the solution first converted to liters for the bridge conversion to 0.450 moles.
Once I have the molarity of my $50~\mathrm{mL}$ set to $0.450~\mathrm{mol/L}$ I divide my molarity by the molarity of the original solution I am taking this out of in the first place which gives me 0.026 liters of solution
$0.026~\mathrm{L}$ of solution is then what I need out of my $0.28~\mathrm{L}$ $0.8500~\mathrm{M}$ solution in order to come up with my 50 milliliters of $0.450~\mathrm{M}$ of solution? (I feel that something is missing)
My computations (conceptually explained above in my best effort)
$50~\mathrm{mL} \cdot 1~\mathrm{L}/1000~\mathrm{mL} \cdot 0.450~\mathrm{M}/1~\mathrm{L} \cdot 1~\mathrm{L}/0.8500~\mathrm{M} = 0.026~\mathrm{L}$ of $0.8500~\mathrm{M}$ solution needed to prepare $50~\mathrm{mL}$ solution of $0.450~\mathrm{M}$.

I'm not sure the exact question but I suppose you want to find the volume of 1st solution needed to come up with the 2nd solution. I always feel more comfortable with moles for this types of problem. Because mole is conserved (in a way).
So, to make the 2nd solution you need
$$50~\mathrm{ml} \times \frac{1~\mathrm{l}}{1000~\mathrm{ml}} \times 0.45~\mathrm{\frac {mol}{l}}=0.0225~\mathrm{mol}$$
That means you need $0.0225~\mathrm{mol}$ of $\ce{Al(NO3)3}$ from the first solution. You know the molarity and mole number. So the volume is
$$\frac {0.0225~\mathrm{mol}}{0.85~\mathrm{\frac{mol}{l}}}=0.0264~\mathrm{l}=26.47~\mathrm{ml}.$$
But you need $50~\mathrm{ml}$ of the 2nd solution, so you need to add solvent (Water or whatever you used). The volume of extra solvent required is $50-26.47=23.53~\mathrm{ml}$. Now you got the 2nd solution and to check that you have got everything right, you can find the molarity of your new solution.


text:
Are all homogeneous mixtures just solutions?
Or can there be a homogeneous mixture that is not a solution?

I think this largely rests on what one means by "homogeneity". It's a concept which changes depending on the scale. The entire Universe is homogeneous on the scales of billions of light years, but below that it is not. A steel plate might seem homogeneous at the metre scale, but view it with an electron microscope and all sorts of aggregates and domains become visible. The limits of homogeneity are somewhat loosely defined.

Homogeneous and heterogeneous classifications are scale-dependent. What we mean when we say a mixture is homogeneous is that there are no visible phase differences on the scale of interest.
For example, milk (when homogenized) and lotions are both examples of colloidal suspensions. There are no phase differences at the human scale, but neither one is a solution. On a more microscopic scale, you could call the suspended micelles phase boundaries, and in that case might not call the mixture homogeneous.
"Solution" always implies mixing on the molecular level (single-phase).
To summarize:

Solutions are always homogeneous mixtures, but homogeneous mixtures are not always solutions.

According to IUPAC gold book:

A solution is a liquid or solid phase containing more than one substance...

So, the concept "solution" is reserved to solid and liquid phases.
In gas phase: Air is an example of homogeneous mixture of nitrogen, oxygen, carbon dioxide and other gases present in air. It's obvious that air is not a solution.

For the liquid phase, I think it is probably true that all homogeneous mixtures are solutions.
But the situation is more complicated for solid phases.  In a diamond crystal, if the $\ce{^{13}C}$ atoms are homogeneously distributed in the crystal, could you really say that the diamond is a "solution" of $\ce{^{13}C}$ in a $\ce{^{12}C}$ "solvent"?  I don't think so.  $n$-doped or $p$-doped silicon is another example.  In contrast, other homogeneous solids, like brass, are correctly regarded as solid solutions.
No one usually regards gas phases as "solutions", so mixtures of gases are probably not solutions.

Your comment suggests another part of the definition of a solution: the components of a solution cannot be covalently bonded together.

@iad22agp - But now you have ruled out things like Si-Ge that forms a continuous solid solution in the diamond cubic structure. It is homogeneous, and is definitely a solution (and a near-ideal solution as well). So I do not believe that the type of bonding has anything to do with the question.

Well, metal bonding is not exactly covalent. I think the key here may be that, when liquefied, a true solution's components able to freely diffuse within the bulk.

Si-Ge is not metallic - it is a covalently bonded semiconductor. And, diffusion in the liquid may be quite different from diffusion in the solid, since the bonding environment may be very different. Again, Si is a metallic liquid, but not a metallic solid.


text:
Which of the five conformers of propyl propanoate is most stable?
I've worked out five different conformations of propyl propanoate:

I think the bisected conformations (E) should be the most stable one.
Are my rationals and suggestions in my scheme correct?

Your basic arguments seem fine to me.  Certainly the s-trans conformations are higher energy than the s-cis conformations (like E), see here for a nice explanation.  As to whether B or E is favored, I think they are very close in energy, but if I had to pick just one, then I'd choose E.  A few minor points, 1) there are 8 expected low-energy conformations, you've only drawn 5, 2) stereochemistry in the ethyl group is not properly displayed in C and D (you show both a methyl and a hydrogen in the plane), but I understand what you mean.


text:
Plastic membrane that lets water in from the outside only?
I'm looking for a type of plastic membrane that would let water through it from the outside, but wouldn't allow water to escape once it's inside the membrane.
Ideally available in a roll.
Does such a plastic membrane exist?

This, if exists, would be such a weird membrane. How would the inner side of membrane act as a hydrophile and hydrophobe?

You start by catching as many Maxwell demons as possible...

Could it be possible to use some sort of fancy surface patterning to make such a membrane? Or perhaps take a page out of water transport mechanisms through cell walls?

A simple passive valve like that of the vains in the human body, but smaller and in the form of a plastic membrane would do this. I dont see how this breaks any laws, but I guess some people are unimaginative. As for its existence, I couldnt say.

The comments above have gently hinted that it would be a violation of thermodynamics for a passive membrane to force a substance in a single direction. Molecules have an equal probability of moving through the membrane in either direction, unless you apply a force to make them move in a specific direction, such as electro-osmosis or active water transport in plants.
Consider that if such a membrane as you posit existed, you could use it to pump water from a lower level to a higher, and then power a water-wheel: perpetual motion. Sadly, Maxwell's demon is getting a bit arthritic to shuttle water molecules, and the dampness only aggravates its condition.

There are plenty of biased pumps- can happen without breaching the laws of thermodynamics- in this case the plastic is a biased sieve so that water molecules on the outside easily enter but can't leave easily the plastic keeps loosing internal energy so it will eventually. fail.

True, there are many substances that adsorb and absorb water, such as silica, calcium chloride, and even cellophane. My understanding of the question was perhaps incorrect: I assumed the membrane was to allow water to pass through in only one direction, but if it's merely to absorb water, there are many hygroscopic membranes, such as those used in HVAC, such as paper-board or even liquid membranes (see http://www.sciencedirect.com/science/article/pii/S037673880000404X).

No passive membrane can achieve this goal but...
A passive structure that permits one-way passage of water vapour can't exist as it would violate the laws of thermodynamics (effectively it would be able to pump water continuously with no input of energy which is obviously impossible).
But practical membranes and active devices can do a useful job of something similar. Gore-Tex, for example, allows the passage of water vapour but not liquid water. So it is widely used in clothing to create breathable but waterproof fabrics. This is a useful goal but not magic: the flow of vapour is not directional but depends on the relative concentration on the two sides of the membrane (but keeping out liquid water is very useful in clothing so even this non-miraculous property is very beneficial compared to impervious membranes that would trap all the sweat and water vapour inside the clothing until it condenses and makes you wet even though the rain has been kept out).
Consider also osmotic membranes. They can often allow selective passage of small molecules like water but prevent the passage of salts or larger molecules. But, operated passively, what happens is that water passes from the side with a higher water concentration to the one with lower concentration (ie from pure water to sea water). They can be used to purify water by "pushing" the impure side driving the reaction in the opposite direction from the thermodynamically favoured one so water travels from the impure side to the pure side.
Some polymers and minerals (eg some zeolytic molecular sieves) can very selectively absorb water. This might look like a potential route to a pump. But they eventually saturate and can't absorb any more (in effect the laws of thermodynamics reassert themselves by equalising the internal "pressure" of water in the pump with the external "pressure" of water outside the pump. There is no way around this that doesn't break the rules of thermodynamics.
The only way to drive one-way passage of water (or anything else) is to provide energy to the system. An amazing example of this is the ionic membrane technology described here which uses a very low current input to continuously "pump" water vapour to dehumidify the inside of electrical junction boxes with no mechanical parts. So, selective pumping is possible, but requires energy.


text:
Identifying organic compounds in Infrared spectrum
What further information would allow you to make a positive identification of an organic compound from an infrared spectrum?

I guess I would use NMR.

@orthocresol the question, however, asked about IR. Do you have any other ideas?

Are you trying to confirm identity or elucidate the structure of a completely unknown compound?

@jerepierre confirm identity

@J.Doe Then what you need is an IR spectrum of an authentic sample.

Well, your question asked for "further" information... I think if you have a clue what sort of compound you're looking at (eg you isolated an intermediate in a synthesis which you know the structure of) then IR is useful as a means of confirmation since peaks will appear/disappear based on the presence/absence of specific functional groups. Although the gold standard, as @jerepierre mentions, would be to compare it to an IR spectrum of the pure compound.

That is like super-general question. Can you be more specific?

If you're analyzing an IR spectrum to find out what compound you have, you'll first want to look past the 1500/cm region to see what functional groups you have. For example, a 1700/cm peak will tell you you have a carbonyl, alcohols will have a peak around 3000/cm, etc. From there, pull up spectra of well known compounds with those functional groups, and compare your compound's 1500/cm to 500/cm region to them. This region is called the "fingerprint region" because it's practically unique for all molecules, so if the spectra match there, they're likely the same compound.

You can compare your IR spactra to a database entry. Usually, the database provides examples of spectra from a high purity sample and that way you can even see how pure your sample is.

This is better written as a comment

Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center. As @santiago (subtly?) points out, we expect our answers to be of a higher quality than this. Please be more elaborate.


text:
Nomenclature of alkoxy and halo groups
CH3-O-CH2CH2Cl
How will this compound be named? Both, methoxy and chloro are given equal preference, since they are prefix functional groups. The answer given in the book is 2-chloro-1-methoxyethane, but why is it not 1-chloro-2-methoxyethane, as the C of chlorine comes before M in the alphabetical order?

This is rather weird. I was expecting your suggestion to be correct.

@M.A.Ramezani Are you sure there's no order of preference for prefix functional groups?

It might be that there is an order of preference after all. According to the following link, ethers have a priority over halo groups: http://www.masterorganicchemistry.com/2011/02/14/table-of-functional-group-priorities-for-nomenclature/

@AnthonyKelesidis Well, I am quite sure there is no order of preference here. Refer to the answer by to this question of mine : http://chemistry.stackexchange.com/questions/33122/nomenclature-of-ether-and-locant-position-enumeration .
It makes complete sense, and also gives me the correct answer.

According to the IUPAC name generator at http://web.chemdoodle.com/demos/iupac-naming, the "correct" name is "1-Chloro-2-methoxyethane".

Your name, 1-chloro-2-methoxyethane, is correct.

Here is a link to the relevant page of the 1993 edition of the IUPAC Blue Book - Nomenclature of Organic Chemistry, which is published online by ACD Labs with permission of IUPAC.
Here are some example ethers and sulfides from that page, and we see a similar case to yours:

There is no difference in preference or priority between halo and alkoxy substituents (or any of the other prefix only substituents). Thus, in a case where there are two such groups and assignment of locants generates the same combination either way, you choose to assign the lower locant number to the alphabetically first.

Please be aware, that there is a new version of the blue book from 2013 and we have already had some questions, where there were significant differences. However, in that case I believe the rule still holds.

The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.5 ALPHANUMERICAL ORDER
Alphanumerical order has been commonly called ‘alphabetical order’. As these ordering principles do involve ordering both letters and numbers, in a strict sense, it is best called ‘alphanumerical order’ in order to convey the message that both letters and numbers are involved
Alphanumerical order is used to establish the order of citation of detachable substituent prefixes (not the detachable saturation prefixes, hydro and dehydro), and the numbering of a chain, ring, or ring system when a choice is possible.
(…)
P-14.5.1 Simple prefixes (i.e., those describing atoms and unsubstituted substituents) are arranged alphabetically; multiplicative prefixes, if necessary, are then inserted and do not alter the alphabetical order already established.

Therefore, the correct alphanumerical order of citation in your example corresponds to the name ‘x-chloro-y-methoxyethane’.
Furthermore,

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

Note that Rule f takes precedence over Rule g.
According to Rule f, the name of the given compound could be ‘1-chloro-2-methoxyethane’ as well as ‘2-chloro-1-methoxyethane’ since both names correspond to the locant set ‘1,2’ (all prefixes considered together in a series of increasing numerical order).
According to Rule g, however, the compound is named as ‘1-chloro-2-methoxyethane’ rather than ‘2-chloro-1-methoxyethane’ since ‘chloro’ is cited first as a prefix in the name.

So, you mean the book is wrong?

@ag_dhruv The name “2-chloro-1-methoxyethane” given in your book is wrong in so far as it is not the preferred IUPAC name; it is not even a valid name for general (IUPAC) nomenclature. Nevertheless, this name is unambiguous, and interpretation of this name would yield the correct compound. Therefore, it is not completely wrong.


text:
How to calculate the mass of iron with a temperature of 325 °C needed to melt ice and raise the temperature of water from 0 to 22 °C?

A large well-insulated container holds a mixture of $75\ \mathrm g$ of ice and $100\ \mathrm g$ of water at $0\ \mathrm{^\circ C}$. Using the data given below, calculate the mass in grams of solid iron, $\ce{Fe}$, at $325\ \mathrm{^\circ C}$ that you would have to add to this mixture in order to melt all of the ice and raise the temperature of the resulting $175\ \mathrm g$ of water to $22\ \mathrm{^\circ C}$.
Specific heat capacity of water is $c_p(\ce{H2O})=4.184\ \mathrm{J\ g^{-1}\ ^\circ C^{-1}}$
Specific heat capacity of iron, $\ce{Fe}$, is $c_p(\ce{Fe})=0.45\ \mathrm{J\ g^{-1}\ ^\circ C^{-1}}$
Molar enthalpy of fusion (melting) of water is $\Delta_\text{fus}H_\mathrm m(\ce{H2O})=6.01\ \mathrm{kJ\ mol^{-1}}$
$M(\ce{H2O})=18\ \mathrm{g\ mol^{-1}}$

My answer was that you needed $25041.67\ \mathrm J$ to melt the ice and $16108.4\ \mathrm J$ to raise the temperature to 22 degrees Celsius.
Therefore $mc \Delta T(\ce{Fe}) = -(25041.67\ \mathrm J + 16108.4\ \mathrm J)$ and dividing this by the mass of iron and the change in temperature results in a mass of $301.8\ \mathrm g$.
Is this correct?

I can not see anything wrong with your calculations at all.  For posterity, the following is a rough method to confirm:
Melting ice
Enthalpy of fusion $\Delta H_\text{ice} = 6.01\ \mathrm{kJ\ mol^{-1}}$
Mass of ice $m = 75\ \mathrm g$
Molar mass of $\ce{H2O}$ $M = 18\ \mathrm{g\ mol^{-1}}$
Hint: usually, when you are given mass and molar mass, then the amount of the substance is required.
Note that the Enthalpy of fusion ($\Delta H_\text{ice}$) is measurement in $\mathrm{kJ\ mol^{-1}}$, and as the ice is measured in grams, and you are provided with the molar mass, you'll need to use the formula:
$$n = \frac{m}{M}$$
hence:
$$n_\text{ice} = \frac{75\ \mathrm g}{18\ \mathrm{kJ\ mol^{-1}}}$$
$$n_\text{ice} = 4.1666\ \mathrm{mol}$$
Therefore, the enthalpy of fusion:
$$\Delta H_\text{fus} = 6.01\ \mathrm{kJ\ mol^{-1}} \times 4.1666\ \mathrm{mol} = 25.04167\ \mathrm{kJ} = 25041.67\ \mathrm J$$
Heating water
Specific heat capacity of water $c_p = 4.184\ \mathrm{J\ g^{-1}\ ^\circ C}$
Mass of water $m = 175\ \mathrm g$
Temperature change $\Delta T = 22\ \mathrm{^\circ C}$
So:
$$Q = mc_p \Delta T$$
$$Q = 175\ \mathrm g \times 4.184\ \mathrm{J\ g^{-1}\ ^\circ C} \times 22\ \mathrm{^\circ C} = 16108.4\ \mathrm J$$
Total energy used by the water/ice mixture
$$Q_\text{total} = 25041.67\ \mathrm J + 16108.40\ \mathrm J = 41150.07\ \mathrm J$$
Iron
Specific heat capacity of iron $c_p = 0.45\ \mathrm{J\ g^{-1}\ ^\circ C}$
Mass of water $m = X$ (unknown)
Temperature change $\Delta T = 325\ \mathrm{^\circ C} - 22\ \mathrm{^\circ C} = 303\ \mathrm{^\circ C}$ decrease
As, according to conservation laws:
$$Q_{\ce{H2O}} = Q_{\ce{Fe}}$$
Then:
$$41150.07\ \mathrm J = X \times 0.45\ \mathrm{J\ g^{-1}\ ^\circ C} \times 303\ \mathrm{^\circ C}$$
Or:
$$41150.07\ \mathrm J = 136.35\ \mathrm{J\ g^{-1}}\times X$$
Then:
$$X = \frac{41150.07\ \mathrm J}{136.35\ \mathrm{J\ g^{-1}}}$$
$$X = 301.8\ \mathrm g$$


text:
Palladium Nanocube
What is the relative permittivity and relative permeability of a palladium nanocube?
If I change the size of this $\ce{Pd}$ nanocube, will these values be changed or remain the same?

For part (1), you should be able to find values on the internet. For (2), relative permittivity and relative permeability are intensive properties and do not change with the size of the system. However, both quantities are functions of the applied electric and/or magnetic fields.

I'm not so sure - increasing the size of the nanocube may allow higher order plasmon wavefunctions to exist, thus changing the complex dielectric constant. An easier example - it is well known that the color of nanoparticles is dependent on their size.


text:
How many coordinates should be used for the potential energy surface of a non-rotating linear molecule? (3N-6) or (3N-5)?
Take, for example, $\ce{CO2}$.  It is pretty clear that in the absence of rotation, the energy of this system depends only on the two bond lengths and the bond angle.
So if I wanted to build a potential energy surface for this molecule that was valid near and far from equilibrium, it should depend on these coordinates.  But how do I rectify this with the fact that $\ce{CO2}$ has four vibrational modes? As we learn in undergraduate chemistry, it has two stretches and two degenerate bends.
If I build a Hamiltonian in the three valence coordinates, could it possibly describe the ground state energy correctly?  That is, would it contain the zero-point energy for all four modes?  Would it be able to find the right low-lying excited states of the bending mode(s)?
EDIT:
Wildcat has brought up the point of symmetry and point groups below, and perhaps that can help get to the heart of my question.  When we say that $\ce{CO2}$ is a linear molecule, that means that the lowest energy configuration has a bond angle of $\pi$.  But in no way do I want to restrict the symmetry of the molecule.  I want to work with a potential surface that is valid for symmetric and asymmetric geometries.
But here is where I get confused.  Let's say I did restrict it to be linear.  I would then have a two-dimensional potential, a function of the two bond lengths.  Of the seven normal modes, I only keep two: the symmetric stretch and anti-symmetric stretch.  The ground state energy would be the sum of the zero-point energy of these two modes.   So therefore the number of modes is equal to the number of coordinates I used to write my Hamiltonian.
If I further constrict the symmetry, and impose the $D_{\infty h}$ point group, then the potential is now a function of one parameter.  I will now only have one vibrational mode, the symmetric stretch, and the ground state energy is the zero-point energy of this mode.  So, again, the number of modes is equal to the number of coordinates I use to write my Hamiltonian.
But what if the symmetry were not restricted in any way?  I can write the potential as a function three internal coordinates, be they bond lengths and angle, othogonormal rectilinear coordinates, jacobi coordinates, etc.  This potential, and the Hamiltonian I construct with it, should find the same modes described above, as well as the bending modes.  But will it find both bending modes?  I don't see why it should.  Based on the two reduced-dimensional examples above, I would expect the number of modes to equal the number of coordinates.  I would expect therefore, not to find the right ground-state energy, etc.
Specifically, I am concerned about the accuracy of dynamical simulations performed using this three-dimensional potential.
Likewise, what about a tetratomic liner molecule like $\ce{C2H2}$? I would use three bond lengths, two bond angles, and one dihedral angle, for a total of 6 coordinates - but it has 7 vibrational modes.  Would I model the dynamics properly?

So, yes, you'll get the "correct" vibrational eigenstates. In quotes because the very separation of nuclear motion into vibrational and rotational is approximate, and thus, if you include only vibrational contribution (ZPE) into the total energy for both linear and non-linear configurations, that would be "not quite right" since you have different number of vibrational degrees of freedom in these cases. Better include corrections from all three types of nuclear motion: translational, rotational, and vibrational.

No, from you last edit, I see that you do not understand the matter, so I repeat one more time: the number of vibrational degrees of freedom is always either $3N-5$ for a linear configuration or $3N-6$ for a non-linear one, while the number of structural parameters is always $3N-6$. So, for $\ce{CO2}$ molecule you always have $33-6=3$ structural parameters and either $33-5=4$ or $3*3-6=3$ vibrational modes (for a linear configuration and a non-linear configuration respectively).

The Cartesian Hessian ($\mathbf{H}_C$) can be obtained by from the Hessian computed in internal coordinates ($\mathbf{H}_i$) through a non-linear transformation
$$\mathbf{H}_C = \mathbf{B}^T\mathbf{H}_i\mathbf{B}$$
where $B_{ij}=\frac{\partial R_i}{x_j}$ and $R_i$ and $x_i$ is an internal and Cartesian coordinate, respectively.
In the case of the degenerate bending mode you can obtain the components of $\mathbf{H}_C$ that you need from $\frac{\partial^2 E}{\partial \theta^2}$, where $\theta$ is the bond angle, through suitable choices of $B_{ij}$'s.
So the Hessian generated in the three internal coordinates has all the information you need.

It is pretty clear that in the absence of rotation, the energy of
  this system depends only on the two bond lengths and the bond angle.

Yes, that is true, if you do not impose any restriction on geometry (such as, for instance, the linearity), the $\ce{CO2}$ molecule will indeed have $3N-6=3*3-6=3$ internal degrees of freedom and the same number of independent structural parameters (as which you can naturally choose two bond lengths and the angle between them). However, the point is that when you start to impose some restrictions on the geometry (symmetry), the number of internal degrees of freedom can increase by one (if you require your molecule to be linear), while the number of independent structural parameters will stay the same (though, few of them will be fixed and not varying anymore).
As a result the following three quantities which you mixed up are usually different:

the number of internal degrees of freedom: $3N-6$ (for a non-linear molecule) or $3N-5$ (for a linear molecule);
the number of independent structural parameters: always $3N-6$;
the number of varying (i.e. not-fixed) independent structural parameters: depends on the point group.

Thus, for instance, if you require the $\ce{CO2}$ molecule to be linear the number of internal degrees of freedom will be $3N-5=3*3-5=4$. It does not mean though that you need 4 independent structural parameters to describe it, 3 is still enough. And besides, by requiring the $\ce{CO2}$ to be linear you make it to belong to $\mathrm{C_{\infty v}}$ point group which fixes one of the parameters, the angle between bonds. So, while you have 4 internal degrees of freedom, you still need just 3 structural parameters to describe the molecule one of which is fixed in addition to that. So, if you want to scan PES, you have to vary just two of them: two bond lengths.
You can decrease the number of varying structural parameters even further: to just 1, namely, just the bond length, by making an additional (but quite reasonable) assumption that two $\ce{C-O}$ bonds in the $\ce{CO2}$ molecule have the same length, i.e. that the $\ce{CO2}$ molecule belongs to $\mathrm{D_{\infty h}}$ point group. Note that the number of internal degrees of freedom and the number of independent structural parameters will still be 4 and 3 respectively.

Thanks for pointing that out.  I've edited the question to clarify and remove symmetry from the discussion.  How would you answer for carbonyl sulfide?

@largeboulder, if you want to scan PES for the carbonyl sulfide molecule and you want the molecule to be linear, you will need to vary just 2 geometrical parameters (two bond lengths). But you'll still have $3N−5=4$ internal degrees of freedom and 3 geometrical parameters exactly as in the case when you require $\ce{CO2}$ molecule to be just linear.

it isn't that I am requiring the molecule to be linear - it is that the minimum of the potential is at a linear geometry.  But what I want to do is construct a potential that is valid near and far from the equilibrium, thus disregarding all symmetry.  So if I construct a potential as a function of 3 internal coordinates (bond lengths and angles, rectilinear, jacobi, etc.) and then I use this potential to find the eigenstates (both low-energy and high-energy), is that sufficient?

@largeboulder, in principle, you can vary all three parameters for $\ce{CO2}$ molecule, but I'm pretty sure the global minimum on the resulting PES will be the same as on PES obtained by constraining $\ce{CO2}$ molecule to $\mathrm{D_{∞h}}$ point group and varying just the bond length. The two PESes will, of course, be different, but the global minimum should be the same.

@largeboulder, but I still could not figure out your last question. Do you want to know will there be difference in ZPE at this equilibrium geometry with and without taking the symmetry of molecule into account? No, there should not be any difference: same number of vibrational modes contributing the same amount of energy. I'm pretty sure about that.

@largeboulder, if you would like to do the calculation of vibrational modes for a non-linear geometry, you'll indeed have just 3 vibrational modes rather than 4. Conceptually, there is nothing wrong with that: you'll have one fewer vibrational degree of freedom but one more rotational degree of freedom. Yes, this rotational degree of freedom would not contribute to the (vibrational) ZPE, but it will do so into the total energy.

@largeboulder, do you understand, by the way, that the very separation of nuclear motion into vibrational and rotational one is just an approximation?

You need to incorporate all $3N-5$ modes (both bends and both stretches) in order to accurately calculate the ZPE even if, per Jan's answer, you can neglect one of the bending modes in constructing an accurate model of the PES. Even though the bends appear identical in mind's eye, both of their (degenerate) ZPE's must be included in order for the ground state calculated by the model to be accurate.
The way I think about it is, imagine a $\ce{CO2}$ molecule is oriented along the $z$-axis, with the carbon at the origin. If you only allow one bending mode, say with both oxygens translating toward $+x$ and the carbon toward $-x$, you have to hold the orientation of that motion fixed and your system will be unable to accommodate vibrational movement of any atoms in the $y$-direction. Accurate evaluation of the ZPE requires a model that allows bending along both axes.

The maddening thing about vibrating linear molecules in general, though, is that most linear combinations of a certain pair of these degenerate bending motions actually contain a rotational component.  Using $\ce{CO2}$ as an example and ignoring anharmonicity, if the bending vibrations are $90^\circ$ out of phase with one another, the oxygen atoms gyrate around the $\ce{O=C=O}$ axis just like they would in a rotating molecule with a bent equilibrium geometry (imagine the orange dot is $\ce{C}$ and the green dot is a projection of the positions of the two $\ce{O}$'s):

Source
Of course, the radius of the circle of projection of the motion of the oxygens will be quite small, I believe less than a degree, but it still is a rotation about the molecular axis.
Phase angles other than $90^\circ$ result in some combination of a bending motion and a rotation, where only in the $\theta=0^\circ$ case does the rotational aspect vanish:

For any linear molecule with three or more atoms, the degenerate bends involving the three atoms at each terminus of the molecule are the pair that can generate rotation.  For $\ce{CO2}$ that means the entire molecule; for acetylene each vibration involves the two carbons and one of the hydrogens; etc. EDIT: The more I think about it, the more confident I am that every pair of degenerate bends in a linear molecule actually has this capability to generate rotation. It's just that any bends not centered at the central atom (if it exists) will cause the molecule to precess, as well as rotate, due to the asymmetry of the two 'arms' of the bending motion.
So, to some extent, I would actually argue your question is ill-posed: Linear molecules are always rotating! They're just rotating in a manner inconsistent with the standard rigid-rotor assumption, and (I think?) at energies much higher than those one would expect to see in conventional rotational spectroscopy.


text:
Electronic density distribution in hydrogen bond
I'm currently studying about the atomic configuration of electrostatic interactions. Concept is there that in case of interactions like in N-H...N-H (amides), the stability depends on the position of resonating electron between H...N nuclei (THE NATURE OF THE CHEMICAL BOND. II. THE ONE-ELECTRON BOND AND THE THREE-ELECTRON BOND - Linus. Pauling).
Given that hydrogen bonds formed between N-H...N-H or N-H...O-C are partial covalent in nature so the issue of directionality comes in effect.
I'm just trying to understand, from the view-point of the electron density shared by H...N to form the bond, that how (if possible) atomic configurations/geometry affects the electron cloud distribution ?
I'm basically trying to understand the hydrogen bond forming amides (currently interested in) from the electronic density distribution's point of view; to calculate the single-point-energy using dft.
Thank you.

I have a little trouble understanding your question. Do you want to know more about the background of hydrogen bonds, or are you more interested in dipole-dipole interactions? I cannot grasp what you refer with 'resonating electron', do you refer to a description of the electron density in this structure. What kind of stability are you referring to in the first place? Dimerisation?

@Martin-マーチン Yes, trying to ask about amide hydrogen bond. By the term resonating electron I meant the electron shared between two nuclei to form the electrostatic interaction. I was trying to say about the stability induced by the sharing of electron.

(1) Are you sure you want to know more about hydrogen bonds in amides, where I think the NH...OC interaction would be dominant. Are you possibly talking about amines? (2) I still do not understand what you want to know. It appears you want to know more about why bonds are formed, i.e. electrons are shared between nuclei. That dives deep into molecular orbital theory. Or do you want to know the prerequisites for an hydrogen bond? (3) I do not understand in which aspect your question is linked to the geometry. If you can clear some things up, please [edit] your question.

By DFT you can get value of electron density between hydrogen bonded atoms, but it's not connected with classical meaning of resonance - or electrostatic interaction.

@Martin-マーチン Sorry for my poor representation. I'm basically trying to understand the hydrogen bond forming amides (currently interested in) from the electronic density distribution's point of view.

@Mithoron I'm just trying to calculate the single-point-energy using dft here.

There is no covalent bond formation when the hydrogen bond is formed, no "resonating electrons" (if I understand correctly what you want to ask). You can understand most of it from electrostatics, esp. if all you do is some basic DFT calculations.

@Greg H-bond generally occurs due to the sharing of a single electron's density, whereas covalent bond is 2 or more electron sharing phenomena.

@diffracteD Single electron density? Please, clarify what you want to say, because it makes no much sense in this form.

@Greg Try the literature link in the question. The density of electron from electronegative acceptor which is shared by electropositive H atom.

@diffracteD Try something newer and relevant. It is a 90  years old paper. Also, it is an irrelevant paper: H-bond doesn't do anything with single electron or three electron bonds.

@Greg I agree, but I'm trying to explain the electronic density's behavior of electronegative atom(acceptor) in presence of a near partial-electropositive H atom(forming a perfect geometry for H bond). I'm trying to figure out too the relation.

There are a few descriptors in 'conceptual'  DFT just to characterize hydrogen bonds. E.g.: non-covalent interaction (NCI): note that hydrogen bonds are often consider as NON-covalent bonds, Density Overlap Regions Indicator (DORI).
One can also look at the  deformation density contributions originating from natural orbitals for chemical valence within ETS-NOCV theory.

I was just pointed out that from an energy decomposition analysis point of view, hydrogen bonds do have a covalent character, see e.g. this chemistry open paper http://onlinelibrary.wiley.com/doi/10.1002/open.201402132/abstract


text:
POCP: How many ozone molecules can be created by one ethene molecule?
One ethane molecule can create two molecules of ozone, one ethanol molecule can create one molecule of ozone:

In the following table the POCP is twice as high for ethene than for propane:

From this fact I expect four molecules of ozone from ethene, but in my own formulated reaction mechanism I get only three ozone molecules:

How many ozone molecules can be created by one ethene molecule and if my proposed mechanism is wrong what is the correct one?

I think this question cannot be answered so easily - your proposed reaction mechanism, however, looks good. A detailed explanation and "main reaction mechanism" can be found here, page 192 (just search for "ethene" in the book). As you already indicate ethene reacts with a hydroxyl radical, followed by the addition of oxygen. The reaction with NO generates the HOCH$_2$CH$_2$O radical which can either dissociate or react with another oxygen molecule.
Further reactions of the oxidized products occuring during this pathway can also be found in the above reference, see page 372 (unfortunately only some parts of the book are available for free).

Thank you for your research you've done. So it makes sense to me that the values are some average values. (If maximally three mols ozone molecules can be created by one mol ethylene I expect to create in average 0.75 mol ozone by one mole ethanole.)

You're welcome. They do mention that there are different models/mechanisms to estimate/simulate the amount of ozone that will be created on average, but I have never dealt with those (probably an atmospheric chemist needed for that).


text:
Which temprature is taken in Gibbs free energy equation, of the system or the surrounding?
Which temprature is taken, the temprature of the system or the temprature of the surrounding? I think it should be the temprature of the surrounding. This is because if we take the temprature of the system then change in Gibbs free energy would always be zero as then, change in enthalpy of the system would always be equal to change in entropy of the system multiplied by temprature of the system.

"Which temperature is taken" ... for what?   At equilibrium, $dG=0 \implies dH = dS/T$ and the temperature of the system is equal to that of the surrounding.

The edit you made is a massive revision to the question!  I feel like my answer no longer applies.  What is your actual question?  The old version is a duplicate of a previous question, but this version doesn't seem to ask an actual question.

I took the liberty of rolling back the edits.  Ravi, if you like your revised version better I recommend you submit as a new question.  (Asking more questions is also a way to get more points here on chem.se!)

The Gibbs free energy is a property of a system, so you should use that system's temperature when calculating it.  The Gibbs free energy is the amount of work that could be extracted from the system in a isothermal, isobaric process: i.e. the temperature and pressure of the system won't change.  Thus Gibbs energy has to do only with the temperature of the system, not of the surroundings.
There is a related but distinct variable called exergy or sometimes "availability".  This variable is given the symbol $B$, and is also a thermodynamic state function.  It is defined as $\Delta B = \Delta H - T_0 \Delta S$, where $T_0$ is the temperature of the surroundings.
Exergy represents the maximum amount of work that could be extracted from a system when that system is brought into thermal equilibrium with the surroundings, for example using Carnot engines.  Thus exergy indicates work obtainable through a combined (a) isothermal change in Gibbs free energy at temperature $T \neq T_0$ and (b) a temperature change occurs when Carnot engines are used to transfer heat from the system at $T$ to the surroundings at $T_0$ until thermal equilibrium is obtained.


text:
How is dissolving a physical change and not a chemical change?
When salt dissolves in water it dissociates and it is no longer in a lattice structure. The electrostatic forces are no longer there and are replaced by water-ion interactions. To me that sounds like old bonds breaking and new bonds forming. Shouldn't the reaction look something like this? $$\ce{NaCl + H2O -> Na+ + Cl- + H2O}$$

A process where bonds are broken and made is classified chemical, a process where no chemical bonds are involved is classified as physical.  What about processes like the one above where there is only breaking or making?  Maybe the issue is in the classification of hydration.

This is indeed a bit of a grey area. Consider the following: chemical or physical?
$$\ce{Na+ Cl- (aq)  + Ag+ NO3- (aq)   ->  Ag+Cl- (s) + Na+ NO3- (aq)}$$
There is a tendency in more advanced chemistry to view the formation / breaking of ionic lattices as a physical change, because there are no differences in the electron shells of the atoms. Ions are considered as independent species. As such we could consider the sodium and nitrate irrelevant and reduce the above to simply
$$\ce{Ag+ (aq) + Cl- (aq) -> Ag+ Cl- (s)}$$
Note that I have shown the charges all the way through, to indicate that there are no covalent, dative, or metallic bonds made or broken.
Reactions such as these are definitely "chemical":
$$\ce{Na + \frac{1}{2}Cl2 -> Na+ Cl-}$$
(metallic bonds in $\ce{Na}$ and covalent bonds in $\ce{Cl2}$ broken, electron transferred from $\ce{Na}$ to $\ce{Cl}$.)
$$\ce{C2H4 + H2 -> C2H6}$$
(double bond of $\ce{C2H4}$ opened up, bond of $\ce{H2}$ broken, new covalent bonds formed.)
$$\ce{SOCl2 + 2H2O -> H2SO3 + 2HCl}$$ (covalent bonds broken, new bonds made.)
This last reaction sets up another point. It is usually done with excess water, so the products (which are themselves acids) will then dissolve in water, thereby protonating it:
$$\ce{H2SO3 + H2O -> H3O+ + SO3H-}$$
$$\ce{HCl + H2O -> H3O+ + Cl-}$$
These particular dissolutions are more likely to be seen as a chemical change than the dissolution of $\ce{Na+Cl-}$, because there is a rearrangement of covalent bonds (even though it's only a $\ce{H+}$ that moves from one molecule to the other.)
Finally, $\ce{H2SO3}$ has another equilibrium, again involving covalent bonds being broken and made, and therefore a chemical change:
$$\ce{H2SO3 -> H2O + SO2 (g)}$$
Conclusion
Making or breaking of metallic, covalent or dative bonds, or the formation/destruction of new ions or molecules, is always considered a chemical change.
The simple dissolution of an ionic salt, where none of the above mentioned bonds are made or broken, and no new ions or molecules are formed/destroyed, is frequently considered a physical change.
Note the care taken to always write $\ce{Na+Cl-}$ (and never $\ce{NaCl}$) in this answer.

I have improved the formatting of your post using $\LaTeX$. For more information on how to do this yourself please see here and here.

When solvating $\ce{Na+}$, you create dative bonds from water’s oxygen to sodium. That kind of counteracts your conclusion. Otherwise I like your answer.

@Jan Another grey area. Whereas H+ (which has extremely high charge density) definitely exists in aqueous solution strongly associated with one water molecule, forming the molecular ion H3O+, Na+  exists in aqueous solution surrounded by several water molecules, forming a weak association with each of them. A better example of what you are referring to would be a small multiply charged ion such as Zn2+ or Al3+ with higher charge density, which form closer to dative bonds with water, and undeniably form dative with bonds with OH- ions when an acidic solution is basified.

@LevelRiverSt What about $\ce{CO2 (aq) \rightleftharpoons CO2(g)}$? Is this a reaction (it has also an equilibrium constant)?

The reaction you have written is correct. The physical vs. chemical change issue has to do with the presence/absence of substances with new compositions not previously present appearing after the change, and I would argue that the dissolution reaction you present represents both a physical and chemical change: Physical due to the change in properties/state of NaCl, and chemical due to the appearance of aqueous species not present initially.

The dissolution of NaCl in water is a physical change. Because, it's a reversible process. If water evaporates, you get back the salt.
The dielectric constant $\epsilon_r$ of water is 80 compared with 1 for air. So, the attractive electrostatic forces between ions are weakened by a factor of 80 when salt is added to water. We say the salt NaCl is dissociated. The molecules of water are rearranged around the dissociated ions and we get the hydrated ions, $\ce{Na^+_{aq}}$, $\ce{Cl^{-}_{aq}}$

What if we have a solution that can't be separated? How would the dissolution be considered a physical change in that case?

Let's consider the dissolution of sugar in water. Here, the molecules of sugar do not dissociate. But, they are stabilized in the solvent by intermolecular forces (Hydrogen bonds and Van der Waals forces). The process is always reversible. It's a physical process.

So is your reason ‘reversable = physical process’?

yes. Physical change: a reversible process+formation of  new substances.

Some chemical changes are reversible. For example any acid base reaction is reversible(although the forward is preferred over the reverse). The formation of an alkyl halide is another reversible chemical change

You are right. But when you have an acid base reaction, it's either quantitative or equilibrium. What I mean by irreversible: you can not undo the process (ex. iron rusting: chemical change). While a reversible process: you can undo the process (ex. melting ice: physical change).


text:
Is the hybridization only related to atomic orbitals but not to the molecular orbitals?
What is Hybridization?

(Spelling: Hybridisation (British English) / Hybridization (American English))
  The hybridization is a concept that describes atomic orbitals. In other words: Hybridization is an approach to model hybrid orbitals mathematically. The geometry of a molecule determines the hybridization and not vice versa.

IUPAC Gold Book's definition of hybridization:

Linear combination of atomic orbitals on an atom. Hybrid orbitals are
  often used in organic chemistry to describe the bonding molecules
  containing tetrahedral ($sp^3$), trigonal ($sp^2$) and digonal ($sp$) atoms.

For a certain molecule structure we can allocate the atomic orbitals of one atom to hybrid orbitals which are also localized on this atom and which cannot be redistributed over the molecule to over atoms, e.g. a hydrogen can only have a s orbital but no sp hybrid orbitals because it has no p atomic orbitals.
The allocation can be done using the following relations:

$\ce{1s + 3p ->4sp^3}$
$\ce{1s + 2p + p ->3sp^2 + p}$
$\ce{1s + 1p + 2p -> 2sp + 2p}$

We say: "The atom X is best described by a $\ce{s^\text{x}p^yd^z...}$ hybridization."
Combining two orbitals / two hybrid orbitals / one hybrid orbital (e.g. $\ce{sp^2}$) with a orbital (e.g. $\ce{s}$) we obtain bonds. $\sigma$-bonds with $\ce{sp}$ orbitals (ethyne) of the atomic centers have lower energy than $\sigma$-bonds with $\ce{sp^2}$ orbitals (ethylene), lower in energy means: they are better acceptors.
What is my question now?
I'm thinking if we can also describe a bond (a molecular orbital) with a hybridization, e.g. if we can combine a $\ce{sp^2}$ hybridized carbon with a $s$ atomic orbital of hydrogen to make two $\ce{sp}$ molecular orbitals (a bonding $\sigma$ orbital with $sp$ character and an antibonding $\sigma^\star$ orbital with $\ce{sp}$ character; just be saying for a $\ce{C-H}$ bond: $\ce{s} + \ce{sp^2} = 2 \ce{sp}$, what's probably wrong).
I guess this question has to be answered with No, because we cannot say a bond has hybridization.
The question might arise why I'm thinking of the hybridization of molecular orbitals.
Over several years I've heard statements like this:
The $\ce{C#C}$ bond of ethyne ($\ce{HCCH}$) has higher s character (50%) than the $\ce{C=C}$ bond of ethylene ($\ce{H2CCH2}$) (33%) that's why ethyne has a lower $\mathrm{p}K_\mathrm{a}$ than ethylene (25 versus 44 at standard conditions in water as solvent) because a bond with higher s character can better stabilize a negative charge, because the s orbitals are penetrating deeper into the shell as the p orbitals do.
Again, I think hybridization is only a good model for atomic orbitals (=localized on atom) not for linear combinations of atomic orbitals (=bonds).
Can we really say the $\sigma$ bond in ethylene has $sp^2$ character?

Hybridisation is a concept devised by man to ease explanation of chemical observations. It is not a 1:1 equivalent of anything seen in molecules but rather a way of rationalising what one has seen — one that is as mathematically correct as molecular orbitals (that span the entire molecule) are.
There are two fundamentally different approaches on how to describe ‘bonds’. The first one is the molecular orbital approach. It considers electrons that can freely move around the entire molecule, in a so-called potential energy space: the potential energy derives from the (negatively charged) electron being close to (a series of positively charged) nuclei which attract each other. As you may remember from your very first chemistry classes ever, the mass of even a proton is $2000$ times that of an electron, so a carbon nucleus is some $24000$ times heavier than an electron. According to Newton, the nuclei have a lot more inertia than the electrons do, so it is reasonable to assume that whichever fixed positions the nuclei take, all electrons can then assume favourable movement patterns.
Quantum mechanics, and further quantum chemistry, give us methods to calculate what movement patterns these electrons might assume — although technically the equation is not solveable, because it also must contain electron-electron interactions, there are good approximations out there. The wave functions that one can derive from all this information correspond to molecular orbitals you might have seen before: They abide by the molecule’s symmetry (will always fit into one irreduceable representation of the molecule’s symmetry point group) but extend over the entire molecule (even the orbitals representing atomic core electrons can be thought of being delocalised somewhat).
Using the quantum mechanical method of calculation (operators) you get an energy for each molecular orbital. Thereby you have ordered your molecular orbitals by energy (approximately equalling the number of nodal planes) — this gives you HOMO and LUMO that you might have heard of before.
To reduce this back into ‘bonds’, we need to use mathematical tricks. Simply spoken: All wave functions can be treated as vectors, and vectors can be linear combinated to give other vectors — meaning we take $0.5$ times orbital $\Psi_1$, $0.01$ times orbital $\Psi_2$ and so on for each molecular orbital, add them up, divide them by a normalisation factor, and we get a new orbital that has a different shape (and a new energy). If we do this in a sensible way, we can manage to reduce molecular orbitals to those that only ‘span’ across one bond — because we have cancelled out all wave functional contributions from other parts of the molecule. This would correspond to your typical $\sigma$ and $\pi$ type bonding, nonbonding and anti-bonding orbitals.

The second method is the hybridisation approach. This time, you take the general shape of the molecule and consider each atom by itself. For each atom, you can assume a hybridisation based on its geometry: If there are three other atoms in a triangular shape around it, all sharing the same plane with the atom in question, we can assume it to be sp² hybridised.
Once we have done this, we need to take a look at neighbouring atoms to create ‘bonds’ again. Usually, there will be two hybrid orbitals ‘pointing’ at each other, ready to create a bond. Using the same technique we have to create the hybrid orbitals, we can mix these two hybrid orbitals to create another orbital. This orbital is not considered to be of a new sp-hybrid type, but rather gets a new type of name: It can be $\sigma$, $\pi$ or $\delta$ (to name the most common ones). These Greek letters closely resemble s, p and d, and the bond orbitals in question share a common symmetric trait with the corresponding atomic orbitals: $\sigma$ orbitals can be rotated freely around the bond axis, $\pi$ orbitals have one and $\delta$ orbitals two nodal planes along the bond axis (the bond axis is completely part of the nodal plane).
If we did this correctly, we will arrive at orbitals of similar shape and energy as with the above method, only we are much, much faster, which is why this method is often taught in schools.
Onto the question of how do the $\ce{C-C}$ (and $\ce{C-H}$) bonds differ in $\ce{C2H4}$ and $\ce{C2H2}$:
You just saw how we decided what orbitals we want to combine to form a bond. In $\ce{C2H4}$, we are combining a carbon’s sp² orbital with a hydrogen’s s-orbital to create a $\sigma$ bond. For simplicity, we can label this bond $\sigma_{\ce{sp^2-s}}$. Similarly for $\ce{C2H2}$, only that this one has to be labelled $\sigma_{\ce{sp-s}}$. You see that you created bonds with different hybrid orbitals and therefore the bonds will have different ‘shapes’ and energies. This is what is meant with

The $\ce{C#C}$ bond of ethyne ($\ce{HCCH}$) has higher s character ($50\,\%$) than the $\ce{C=C}$ bond of ethylene ($\ce{H2CCH2}$) ($33\,\%$)

Technically, this is short for:

The hybrid atomic orbitals of the carbons of ethyne have a higher s character ($50\,\%$) than those of ethene ($33\,\%$), so the $\sigma_{\ce{C-H}}$ bonds will retain that property somewhat.

Long story short: No, we cannot say that a $\sigma$ bond has sp² or sp³ character; we can (and should) only say that the atomic orbital of one side of that bond has this character.
And: It is wrong to assume $\ce{sp^2 + s} = \ce{2 sp}$, because that would recentre a p-orbital from the carbon to the hydrogen.

Let me be the nitpicker this time: The two approaches you describe are not exclusive and lead to an identical bonding picture, if treated with the proper quantum chemical approach, where the hybridisation approach is covered by valence bond theory. In short: You can extract the VB picture from the MO ansatz and vice versa.

@Martin-マーチン would it be correct to say that MOT is a more fundamental approach and can make predictions about bond strength, geometry etc.... while hybridization of VBT is a reverse-engineering approach and can only rationalize them?

@OVERWOOTCH No. Fundamentally, at their respective limits: MOT = VBT. That is also why you can approximate one with the other. What is in the between is too complex to deal with in a comment or maybe even an answer.


text:
New finding in cobalt compound that absorbs oxygen - is it really a breakthrough?
Very recently Chemical Sciences published results of the discovery of a new cobalt compound that can absorb oxygen at presumably a higher capacity than what has been achievable before. The paper is here. I don't have access to the full publication, and since I'm not a chemist I don't really follow the abstract too well. But from the popular news there are claims that 10 liters of the compound can absorb "a whole room of oxygen". Not very quantitative.
So my main question - is this really a breakthrough in terms of absorption capacity? And to what degree?
Also - how can the absorption/release of oxygen be controlled?
Edit: I found a link to the full article here.

They compare it to mioglobine and say it has good reversibility. Seems it's better than complexes synthetised earlier.

@Mithoron to reverse the absorption, do you have to heat the compound?

Yes, they were heating and cooling.

@Mithoron  There is allot of hype in the news that this will be a technology breakthrough - perhaps competing with SCUBA compressed gas cylinders. But if you have to supply heat to recover the O2 then you need a battery or fuel - which adds additional volume and weight. Not to mention diving depth would be limited to less than 33 ft of seawater - to prevent oxygen toxicity.


text:
Virial equation of state - Taylor expansion
I don't get how Taylor expansion over $ \frac{1}{\overline{V}} $ of Virial equation of state is caried out: $Z = \frac{P\overline{V}}{RT} $ which yields
$$ Z = 1 + \frac{B_{2V}(T)}{\overline{V}} + \frac{B_{3V}(T)}{\overline{V^2}} + ...
$$ I thought I understood calc2 Taylor expansion of form $$f(x) = \sum\limits_{k=0}^\infty \frac{f^{(k)}(0)}{k!} x^k $$ so maybe I was in illusion. What's is $Z$ function of? At what expansion point do we expa

I have the same question and I still don't get it.
Z = PV/RT (N=1)
And I've read that the expansion has to be when P→0, or when V→∞, so (1/V)→0.
Ok, fine, but I can't expand in a series of Taylor in terms of P or V because in this form they don't have a second (and third, fourth and so on) derivative:
Z'(P) = V/RT
Z''(P) = 0
Z'(V) = P/RT
Z''(V) = 0
From what I know, its a expansion (Maclaurin) of:
1/(1-x) which gives (1 + x + x² + x³ +...)
And I don't see a way to have PV/RT to be written in this form.
Tomorrow I'll ask my professor and I'll come back if I get a clear answer.

Here, Z is explicitly a function of temperature.

At what expansion point do we expand?

Can you clarify what you mean? I can state that it's common to truncate this particular expansion at the second term, so we'd have:
$$\ce Z = 1 + {B_{2V}(T)\over \overline{V}}$$
The virial coefficient in this case accounts for deviation from the ideal case by accounting for attractive and repulsive forces between the particles in the system.

I'm sorry I have badly formulated my question, but thanks for the answer. I mean if we expand around some particular point (like 0 or any other fixed value). I have reformulated my question. So Z is function of T, how do we carry out expansion over $ \frac{1}{\overline{V}}? $

Are you asking about the mechanics of the Taylor series expansion specifically?

Exactly, this is where my confusion is.

The unity term corresponds to the $k = 0$ case, and the next term (where $k = 1$) is the first derivative term (the negative sign from differentiation is incorporated in $B_{2V}$).

The expansion in a Taylor series of is not made starting from an ideal gas equation, it's actually made from the equation for the pressure obtained from statistical mechanics, $pV = kT\ln Z$, in the grand canonical ensemble.
The derivation of the virial expansion is shown in McQuarrie (ref 1). I had misunderstood (probably you too) based on what I read in some site or videos. The ideal gas equation is obtained from the virial expansion and not the opposite.
References

McQuarrie, Donald A.. The Virial Equation of State from the Grand Partition Function, Section 12-1, p. 224, in Statistical Mechanics, ISBN 06-044366-9.


text:
Free software for calculating the net charge of an amino acid as a function of pH?
Does a free software for calculating the net charge of amino acids as a function of pH exist?

I like MarvinSketch. You can run it in your browser or on your computer. Under Insert > Template, you can find all the amino acids. Once you've inserted one, select Calculations > Protonation > Isoelectric point and click OK. You'll get a window that looks like this showing the net charge versus pH (I picked a pretty boring one, but other amino acids produce more complex curves):

Needs Java. Thats a bummer. :(

I came accidentally across this old not fully answered question. Recently I wrote a web application mainly for the calculation of physico-chemical property of proteins (Prot pi). But you can also use this tool to calculate/estimate the titration curve of single amino acids. You can enter the single letter code for your amino acid (e.g. k for Lysine):

And there you get the net charge in function of the pH:


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to convert ppm gas (N2O) concentrations in microgram nitrogen per gram dry soil basis?
What calculations are needed to convert  $\ce{N2O}$ gas concentrations in ppm to micrograms of nitrogen per gram dry soil basis? I measured the concentration of $\ce{N2O}$ in soil samples. I extracted a small gas sample from a container (the soil is kept in this container; the container has an head space of 80 cc). Then I injected the small gas sample in the GC and I had a measurement in ppm. I want to know how many grams of N are present in my ppm measurement, taking into account that the gas comes from 5 grams of soil.

I assume the ppm value you are reporting is ppm by volume (or equivalently for ideal gases, ppm by mole).  If that is true, then:

The total moles of all gas in the container can be calculated from the ideal gas law:

$$ N_{tot} = \frac{PV}{RT}=\frac{(101325 \rm{~Pa})(\frac{80}{1000^2} \rm{~m^3})}{{(8.314 \rm{~\frac{J}{mol\cdot K}})}{(298 \rm{~K}})}$$
Here I have assumed that the temperature was 298 K, or about 25 °C.  Note that the units work out because pascals ($\rm{Pa}$) are equivalent to joules per cubic meter, $\rm{\frac{J}{m^3}}$.  Also, I assume your experiment was done at approximately atmospheric pressure, which is 101325 Pa.

The moles of $\ce{NO2}$ in this 80 cc of gas can be calculated from the ppm measurement:

$$N_{\ce{NO2}}=\frac{\rm{ppm~value}}{1,000,000}\times N_{tot}$$

The mass of the nitrogen, in grams, in the $\ce{NO2}$ gas can be calculated from the formula weight of a nitrogen atom, $\rm{FW}_{\ce{N}}$:

$$m_{\ce{NO2}}={\rm{FW}_{\ce{N}}}\times N_{\ce{NO2}}=14\frac{\rm{g}}{\rm mol} \times N_{\ce{NO2}}$$

The mass of soil used was 5 grams, so the mass ratio of $\ce{NO2}$ to soil, $x_{\ce{NO2}}$  can be calculated by simply dividing the masses:

$$x_{\ce{NO2}} = \frac{m_{\ce{NO2}}}{5 \rm{~grams}}=\frac{m_{\ce{NO2}}\times \frac{1,000,000 \rm{~\mu g}}{\rm{g}}}{5 \rm{~grams}}$$
The last part of the equation above means that if you want the mass ratio in units of micrograms (of $\ce{NO2}$) per gram of soil, you have to multiply by 1,000,000.
Also, the question says you want a "per gram dry soil basis", but you didn't mention if the 5 g of soil you used is "dry".  If it isn't, then you will first have to figure out how many grams of dry soil you added to your reaction, if it wasn't 5.


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Getting the pressure using the equation of states - will the answer be the same?
When I get the pressure using the different equations of state (e.g. van der Waals, Redlich-Kwong, etc.), will the answer still be the same?

In general no, the pressure will not be the same, although there will be intersection points where (by coincidence) they will be the same, and there will be regions of the phase diagram for some equations of state (EOS) and for some substances that might be close enough that it doesn't matter.
For example, the van der Waals EOS gives pressure values at high temperature and low pressure for noble gases that are close to what you would obtain with the ideal gas law.
Since it is typically easy to calculate pressure from an EOS given temperature and volume, you can try this yourself and see how much of an impact the choice of EOS makes for your particular system at a particular point in phase-space.