text:
Pre heat coal before burning
So I was wondering if pre-heating coal will improve the efficiency of a steam boiler.
I did find that intake air pre-heater and water pre-heater do increase efficiency, but I could not find anything related to pre-heating coal.
If we were to take the hot exhaust, make a hot iron platform and keep the coal on top of it, it would heat it and maybe reduce the moisture content also?!
Chemistry troll idea or actually feasible?
Note: it is a traditional hand fired boiler
EDIT:
Coal type: Lignite
Application: Industrial/Factory

The update of your question changes everything!
In Germany, around 25% of the base load is generated in lignite-fired power plants.
Lignite (brown coal) is awfully wet (50% water) and it is definitely a good idea to apply drying.
In the power plants, powdered lignite is used. Traditionally, the milled lignite is dried using the hot exhaust gas at temperatures up to 1000 °C, but modern plant like the one in Niederaußem operated the fluid bed drying at temperatures as low as 110 °C.
In summary, pre-drying of lignite is a good idea, if not essential, but it open to question how efficient the drying process of stacked lignite is.

its not for a house but for a factory.

and for the heated surface, place it near the boiler and divert some of the fume exhaust to heat the surface.

what about increasing the temperature of the coal so it absorbs less heat from the already burning coal and less heat is wasted ?


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Equivalence ratio explained
To let you know where I'm coming from: I haven't studied chemistry since high school. I am now studying something else, but I have a dataset coming from your field and I would like some help in understanding just the very very basics of what it is.
What I have is $\ce{NOx}$, nitric oxide and nitrogen dioxide in an engine exhaust and the equivalence ratio. It's from the dataset ethanol in the package SemiPar in R, and the original reference is Brinkman (1981). You'll find a figure at the bottom.
What I want: I want to be able to, extremely briefly, explain what this is describing.
My understanding of the equivalence ratio is that if it is below 1, then there is too little air (compared to the stoichometric mix). If it is above 1, that means that there is too much fuel, and not all of it can be used leading to what is called incomplete combustion.
Is this correct? Would you be able to provide a better explanation? Imagine you're trying to explain it to a five-year-old, that's about how much of this I know!
If you've read this far, please also help me with appropriate tags for the question. I'm not sure what to choose.

Reference:
Brinkman, N.D. (1981). Ethanol fuel – a single-cylinder engine study of efficiency and exhaust emissions. SAE transactions Vol. 90, No 810345, 1410–1424.

You are correct.  There is a theoretical ratio of air-to-fuel that provides exactly enough air to completely burn the fuel, converting it to $\ce{CO2}$ and $\ce{H2O}$.  This theoretical ratio can be calculated knowing (a) the exact composition of the fuel and (b) how much oxygen is in the air (generally ~21%), and (c) the formula weight(s) of the molecules that are in the fuel.  For fuel that are pure octane, the combustion reaction is:
$$\ce{\frac{25}{2}O2 + C8H18 -> 8CO2 + 9H2O}$$
Which means 12.5 moles of oxygen are required for every mole of octane.  For 12.5 moles of oxygen you need about 60 moles of air because air is only 21% oxygen.  That works out to 1740 grams of air (~29 grams per mole).  One mole of octane is 114 grams.  So the theoretical, or stoichiometric ratio of air to fuel is 1740/114 or about 15.
Most motor fuels are not pure octane.  The additives complicate the math a bit, but the stoichiometric ratio is around 14 for many fuels.
The equivalence ratio is just the actual air-to-fuel ratio divided by the theoretical.

My understanding of the equivalence ratio is that if it is below 1, then there is too little air (compared to the stoichometric mix). If it is above 1, that means that there is too much fuel, and not all of it can be used leading to what is called incomplete combustion.

You said the second part backwards.  If the ratio is below one, there is too little air (i.e. too much fuel), and if it above one, there is too much air (not enough fuel).  So incomplete combustion should happen more at lower equivalence ratios.
Also, the exhaust temperature will be maximized for equivalence ratios close to 1.  Otherwise, there are extra reactants, either fuel or air, which remain after combustion and absorb some of the combustion energy, resulting in a lower temperature increase.  Since temperature is important for NOx formation, I figured it was worth mentioning.

Thanks for your answer! I think I'll be able to give just a brief description of the meaning of the data I'm using now. Regarding the incomplete combustion, I more or less just took that from here: http://www.engr.colostate.edu/~allan/thermo/page9/page9.html where it says "If it is < 1, the combustion is lean with excess air, and if it is >1, the combustion is rich with incomplete combustion." But not mentioning incomplete combustion and simply saying that <1 means too little air and >1 too much air is probably enough for me.

That page seems to be speaking of "fuel-air ratios" instead of "air-fuel ratios".  People sometimes use one and sometimes the other.  It just flips things around.  So their "equivalence ratio" is the opposite of ours.  If you check the wikipedia page, their (fuel-to-air based ratio) is phi or $\phi$, and the one I used in my answer (air-to-fuel based ratio) is lambda or $\lambda$.

There are roughly 2 definitions of equivalence ratio (ER). φ=AFR(stoich)/AFR(actual) and λ=1/φ
I personally don't understand why they are both referred to as ER but they are so be careful. In the graph that you've shown, ER>1 refers to a rich mixture (not enough air to completely burn). I can say that due to the fact that NOx formation is reduced since there is no oxygen available for oxidation of nitrogen at high temp and pressures. ER<1 is a lean mixture meaning there is more air than needed. This means N2 gets oxidised.
AFR(stoich) is a property of your fuel but AFR(actual) is the condition under which you do the combustion. These are usually for internal combustion engines and by the looks of your graph, it's a compression ignition engine.
feel free to drop me an email if you don't understand what I just wrote! :D
It's quite a lot of stuff to get your head around. alireza.parpaei@gmail.com

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Electron Configuration of an element with no. of electrons > 18
Let's say that we have an element, say $\ce{Ca}$.
According to law, the last shell should contain only upto 8 e-.
Hence, electronic configuration of $\ce{Ca}$ would be 2,8,8,2.
But, according to the 2n2 rule, the M-shell can hold upto 18 e-.
So, doesn't this mean that the e- configuration of $\ce{Ca}$ can also be written as 2,8,9,1 or say, 2,8,7,3 or something?
But it's not so... Why?

But, according to the 2n2 rule, the M-shell can hold upto 18 e-. So, doesn't this mean that the e- configuration of Ca can also be written as 2,8,9,1 or say, 2,8,7,3 or something?

Experimentally, the ground state electron configuration is 2, 8, 8, 2, so there is no other correct way to write it.  2,8,9,1 and 2,8,7,3 would be incorrect.
Really the only way to know is through experiment or possibly very high level calculations that include relativity and quantum electrodynamics.  As atomic number increases, the problem becomes more and more complex.  It isn't necessarily true that one configuration accurately describes the ground state.
For example for Lawrencium (element 103), for which no experimental determination has been made, you will see some periodic tables that say it has a 7p electron (8 electrons in the P shell and 3 electrons in the Q shell) and some that say it has a 6d electron (9 electrons in the P shell and 2 electrons in the Q shell).
See The low-lying level structure of atomic lawrencium (Z=103): energies and absorption rates to appreciate how difficult it is to really calculate such information.

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Why is a hydroxyl group more activating than a methoxy group in electrophilic aromatic substitution?
Why is phenol more reactive towards electrophilic substitution than methoxybenzene? Isn't the lone pair on the methoxy group oxygen more available for participating in resonance into the ring than the lone pair on hydroxyl oxygen due to the +I effect of the methyl group? Or is it because of the steric hindrance of the methoxy group that will favour lesser substitution at the ortho- position of the benzene ring?
Does the hydrogen bonding of phenol have anything to do with this?

First of all we must have to note that the difference in reactivity between anisole (methoxybenzene) and phenol is small. Small enough for the undergraduate texts I own to not mention anisole separately from phenol, just referring to it as having "phenol-like reactivity" (i.e. strong ortho-para directing). We can more or less measure the difference in ortho-para directing strength by looking at the proton NMR spectra of anisole and phenol.

Source.

Source.
As you can see they are almost identical to the point where you can ask yourself if there is any practical difference in reactivity at all. The A protons nicely reflect the ortho-para direction since they haven't shifted from benzene's 7.26 ppm signal and the B and C protons are all upfield as expected.
Also I don't think steric hindrance is a big factor either since aromatic electrophilic substitution goes first trough a $\pi$-intermediate and a single methyl-group is not that bulky.

Both of your images have some text in chinese. Kindly translate it for the average viewer to understand. Thank you.

If you got the spectra from SDBS, they should be acknowledged.

Mechanism of electrophilic aromatic substitution to both phenol as well as anisole involves attack of the electrophile to the electron rich $\pi$ nucleus of the aromatic system.
In anisole the resonace stabilisation from the lone pair of electrons on the oxygen atom is somewhat disturbed through a hyperconjugation (electron donation from the $\pi$ lone pair on oxygen to the antibonding molecular orbital of the C-H bond of the methyl group of the anisole takes place).
This effect is however absent in phenol and hence the aromatic ring  is more electron rich thereby phenol is more reactive than anisole towards electrophilic aromatic substitution.(The same reason makes methanol more acidic than water.)

This is because phenolic oh group is having a negative charge ( phenol is acidic) extra which is absent in methoxy oxygen. So oxygen in phenol can get into resonance easily than oxygen in methoxy benzene..

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Why did the solution turn green?
I mixed L Arginine with l Citrulline in Hibiscus and green tea water which was red in color. Once the red liquid hit the mix of powdered Aminoacids the solution changed green. Can someone tell me why? Thanks so much for any info!

Both citrulline and arginine are nitrogen-rich, basic amino acids. A basic acid sounds confusing, but that's the way it is.
The colour change indicates a change in the pH of the solution (the tea mix).
The pH indicator that changes the colour is in the hibiscus flower and belongs to the class of the anthocyanes.
Under acidic conditions, the dye exists as a deeply-red flavylium cation. At a slightly higher pH, the dye is reversibly converted to a colourless form. Here, the colour of the mixture is determined by the other ingredients. When more base (read: more amino acid) comes to action, further colour changes are conceivable and the double deprotonated form of some anthocyanes are known to be bluish-green.

Dear Klaus,thank you so much for answering my question. So does this mean that the mixture created is now more acidic? Also since I am drinking this do you see any problems or should I just drink the Amino acids with water? Thank you so very much,Jeff

Jeff, actually it should be just the opposite: the amino acids make the tea is more alkaline. Yes, alkaline (basic) amino acids are confusing, I'm a bit reluctant to answer the medical part oft the question, that's not really my field of work. but since the amino acids are a food supplement, I'd try to drink the with water and check if the taste of the tea is better then. Btw, if you think that the answer was helpful, you might have to accept it. Please have a look at this for the background.


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Applying Raoult's Law to a condensed gas mixture
So, here's the question which I'm having trouble solving :
Suppose that we condense a gas mixture. At 25 celsius, what will be the pressure of benzene and of toluene over the obtained liquid.
We know that vapor pressures of pure benzene and pure toluene at 25 celsius are 12,7 kPa and 3,79 kPa respectively.  We also previously prepared a solution that has equal molar fraction of benzene and toluene. We had to determine the vapor pression of benzene and toluene. (this is an ideal solution)
This gave us:
 $$p_{benz}=x_{benz} * p^°_{benz}= 0.500 * 12.7 \text{ kPa} = 6,35 \text{ kPa}$$
$$P_{tol}=x_{tol} * p^°_{tol}= 0.500 * 3.79 = 1.90 \text{ kPa}$$
Total pressure: $P = 6.35 + 1.90 = 8.25 \text{kPa}$
Finally, there was another question which asked us what was the composition, in molar fraction, of the vapor in balance with the solution of benzene and toluene at 25 celsius.
This gave us
$$x_{benz} =p_{benz} /P_{total} = 6.35 / 8.25= 0.770$$
$$x_{tol} =p_{tol} /P_{total} = 1.90 / 8.25 = 0.230$$
By the way, here's the answer to my question :
$$p_{tol} = x_{tol} * p^°_{tol} = 0,770 × 3,79  = 2,91 \text{ kPa}$$
$$p_{benz} = x_{benz} * p^°_{tol} = 0,230 × 12,7  = 2,92 \text{ kPa}$$
So, where does the Xtol= 0.770 and Xbenz=0.230 come from ? Thank you

You have confused the vapor-phase mole fraction with the liquid-phase mole fraction.  The liquid-phase is mole fraction in usually denoted as $x_i$ for compound $i$.  The vapor-phase mole fraction of $i$ is $y_i$.
Raoult's law is that the partial pressure over an ideal solution is equal to the pure-component vapor pressure times the liquid mole fraction, $p_i=x_i*p^°_i$.
Separate from Raoult's law we have the ideal gas law. It says that that says that partial pressures are proportional to mole fraction in the vapor phase, i.e. that $p_i=y_iP$.  So when you took the ratios of $p_{benz}/P$ and $p_{tol}/P$, you were solving for the vapor-phase mole fractions, i.e. for $y_{benz}$ and $y_{tol}$.
Your mistake comes in taking those $y_i$ values and treating them as $x_i$ values.  That's a mis-formulation of Raoult's law.
You can read more about Raoult's law and the ideal gas law for binary ideal mixtures in my answer to a previous chem.se question.


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Could PNP (Silicon) transistor be toxic when exposed to air\light?
The story goes as follow:
While I was in the electronic room I got bored and crushed a PNP transistor (Model CDIL 2N2907) using a rotating pipe jaws just like in the picture bellow.
Now, it took me some time to realize that I hear something like a sizzle or a tiny fire sound.
I looked at the crushed transistor and I saw a hole in the bent metal.
I am almost clueless in Chemistry, but I guessed that there was some kind of reaction
with the substance when exposed to air (Normal air in Temperature room about 20C)
I looked at the datasheet of the exact part: CDIL 2N2907
and I couldn't find any Chemistry info apart for the presence of silicon.
I do know that a N-doped semiconductor \ P-doped semiconductor are being used and in this case the semiconductor is silicon.
My question: Does (doped) silicon become toxic when exposed to air? Maybe there are other well known substances (insulator) being used with silicon when creating PNP that could react to air?
Also, the room had light (Fluorescent white light). Could it had any effects?
Thank in advance for any information on the matter.
Edit : The sound of sizziling came after I already crushed the transistor. Of course there was sound while crushing - of metal and more.
Rotrating pipe jaws:

Breaking transistors out of boredom is stupid! You will definitely die - some day and due to old age. Case closed.

The sound was likely (case, die, substrate, wirebonds) breaking. As for toxicity, I suppose if you pulverized the silicon finely enough, made sure it was oxidized, and inhaled deeply you could get silicosis.

Maybe, although silicosis is generally the result of long-term exposure to silica dust in environments where it's difficult to avoid breathing it (construction, mining).

The semiconductor itself is harmless, but, since you heard some sizzling, it is possible that a getter, such as an alkali metal, was in the metal-cased component. There is a patent for getters to be used with semiconductors, http://patentimages.storage.googleapis.com/pdfs/US3259490.pdf, though I've never seen one in use.
More likely, you heard the glass header breaking.

The 2N2906A in the TO-18 package is mentioned here: http://www.semiconductormuseum.com/Transistors/Motorola/Corrigan/Corrigan_Page7.htm The "Transistor Museum" is a fun site if you like this kind of thing.


text:
Concentration Inquiry
I just found a random example so I could highlight my inquiry:

A chemist wants to make .5L of a 20M solutionusing $\ce{NaOH}$. The solution will be made from solid $\ce{NaOH}$ and water. How many grams of
  $\ce{NaOH}$ are needed to make this solution?"

In this case, I am finding the moles of $\ce{NaOH}$ that would equate to the number of moles in 0.5L of a 20M solution. But, is it saying that (x grams of $\ce{NaOH}$ + x volume of $\ce{H2O}$) make up the total 0.5L ? or am I adding x grams of $\ce{NaOH}$ to .5L of $\ce{H2O}$ and not considering the changes in volume?

If the final volume of the solution is .5L then what is important here is how many moles/grams of the substance would be used to make the solution. In general when making a solution, the solution is raised to the final volume after the mixture. So whether .5L of $\ce{H2O}$ is used, or .4998L of $\ce{H2O}$ is used doesn't matter.

is it saying that (x grams of $\ce{NaOH}$ + x volume of $\ce{H2O}$) make up the total 0.5L

That would be the exact way!
Molarity is defined as $\mathrm{\frac{moles_{solute}}{\mathbf{total}\ volume_{solution}}}$


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Aluminum deoxidation gas
I am new to chemistry. I am looking for a gas that would de-oxidise aluminium.
Which gases would work for that and at what temperature?
Details: I need to sinter aluminium powder. But because the top layer of aluminium is oxidized when exposed to the atmosphere, I need to remove the oxide film before the sintering would work.
Thanks so much,
Sergey

Since Al2O3 is very stable, powerful acids such as HF and chromic acid are used to remove the oxide surface. Since you're working with fine particles and wish to use a gas, rather than liquid, you'll need to consider what will become of the removed oxides. Gaseous HF would leave fluorides behind. Perhaps atomic hydrogen might work, but you'd have to remove the resultant water before the Al reoxidizes.
It is easier to prepare powdered aluminum in an inert atmosphere from bulk aluminum. There are a number of sites showing how to grind aluminum using anything from a file to a coffee grinder; you'd need to adapt this to run in argon, perhaps with a glove-box.
Sintering might be difficult, since Al is so soft that pore spaces would close.

Thanks so much for the answer.  At what temperature H would take O from Al?  Would other gases, like N or CO work?  If yes, at what temperature?  Should I use aluminum alloy instead of pure aluminum? Thanks a lot!

It would be easier to produce your own unoxidized Al powder than to remove the surface. The mention of atomic hydrogen was hypothetical rather than practical... you'd need to generate a hydrogen plasma, perhaps with RF, at low pressure and temperature so as not to melt the aluminum powder. As for sintering, an alloy would be better than pure Al, as it could be harder and resist compaction (I assume you want to sinter to a frit, rather than merely solidify).

I do want it to merely solidify, and compaction is desirable for my application.  Hydrogen plasma sounds too complicated, so I'll look into production of my own unoxidized Al powder. Thanks a lot!

Sorry for misunderstanding the intent... you might be interested in this process of Al sintering, similar to injection molding of plastics: http://www.aluminiumindustry.org/en/powder-injection-moulding-aluminium.html .

Yes, it's somewhat similar, but they use a binder. Unfortunately they don't tell how they do the deoxidation.

It looks like Hydrogen will not displace aluminum even in plasma state because it's below aluminum in the activity series: http://antoine.frostburg.edu/chem/senese/101/redox/faq/activity-series.shtml


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How to calculate pH of a solution with two dominant equilibria?
Say you have acid X for which $K_a(X)$ = 1E-7 and acid Y for which $K_a(Y)$ = 0.8E-7. In separate solutions, it would be easy to calculate $[H^+]$ because there would be only one dominant equilibrium. But in solution together, both contribute significantly to $[H^+]$, and as a result, both equilibria must be driven to the left by Le Chatelier's Principle. How do you calculate the pH of such a solution? Since both contribute to $[H^+]$, the resulting $[H^+]$ is not additive, but rather a bit less than that because of suppression.
Assume X and Y are 0.1M and monoprotic.

The first step is to write down the variables of interest, and the equations that relate them.  If there exactly as many (non-redundant) equations as variables, then the system might be solvable.
In this case, the variables are:

$\ce{[H+]}$
$\ce{[X^{-}]}$
$\ce{[Y^{-}]}$
$\ce{[HX]}$
$\ce{[HY]}$
$\ce{[OH^{-}]}$

The equations are:

water dissociation equilibrium: $K_w=\ce{[H+][OH^{-}]}$
HX dissociation equilibrium $K_{a,x}=\frac{\ce{[X^{-}][H+]}}{\ce{[HX]}}$
HY dissociation equilbrium $K_{a,y}=\frac{\ce{[Y^{-}][H+]}}{\ce{[HY]}}$
charge balance: $\ce{[H+]}=\ce{[OH^{-}] + [X^{-}] + [Y^{-}]}$
X mass balance: $\ce{[X^{-}] + [HX]}=\text{constant}=0.1\text{ molar}$
Y mass balance: $\ce{[Y^{-}] + [HY]}=\text{constant}=0.1\text{ molar}$

So, "all" you have to do is solve those equations for the six unknown variables.  That isn't easy to do with pen and paper, but it can be done.  Or you could use a computer to numerically solve the equations since you know all the numbers $K_w$, $K_{a,x}$, $K_{a,y}$, and the two different 0.1 molar total concentrations.
Once you solve the equations for $\ce{[H+]}$, the pH is just calculated from $pH = -\log_{10}\ce{[H+]}$.

Curt's answer is correct, but as a practical matter considering the particular concentrations and ionizaiton constants in the question, we can reduce the problem to 3 equations and 3 unknowns as:
$K_{ax}=\frac{[X-][H+]}{0.1M}$
$K_{ay}=\frac{[Y-][H+]}{0.1M}$
$[H+]=[X-] + [Y-]$
And substituting:
$[H+]= \frac{0.1M~K_{ax}}{[H+]} + \frac{0.1M~K_{ay}}{[H+]}$
$[H+]^2 = 0.1M~K_{ax} + 0.1M~K_{ay}$
$[H+]^2 = 1.8 \times 10^{-8}M^2$
$[H+] = 1.34 \times 10^{-4}M$
$pH = 3.9$
So how accurate are the approximations?
Instead of zero, [OH-] is $7.46 \times 10^{-11}M$
So really $[H+]=[X-] + [Y-] + 7.46 \times 10^{-11}M$
About a 1 part in 1,000,000 error in [H+] from neglecting the autoionization of water.  If the concentration of acids was very small instead of 0.1M, and pH closer to 7, this error could be much larger.
The error in approximating [HX] and [HY] as 0.1M is about 1 part in 1,000

I like this answer, but could you spell out the assumptions you are making? E.g., compared to Curt's answer you ignore the ionization of water, and you fill in 0.1 M for the concentration of the acids when we know they are ionizing as well. I think those are what gives people problems when tackling these types of problems.

@jerepierre I'll try to add more explanation, but the main sources of error would be approximating pH as -log[H+], instead of -log (H+ activity), approximating that there exists a pKa that is a product of concentrations only (rather than activities or concentrations and activity coefficients) and is independent of pH; and to a lesser extent neglecting the concentration of water in the equilibrium.  The after considering all those approximations that would apply to both what Curt and I have done, there are my approximations.


text:
Finding Kp at another temp with /\Hrxn
Took a pic of the paper I tried it on it on, has the exact problem description written at top, and what I tried. Figured that would be easier then typing all that. The answer is supposed to be 2.0.

I don't really get what you are asking. Do you want us to check your calculation? It might be easier for you to just take a picture and post it here, but it is much harder for everybody else. Also this way, this is not helping anyone how looks for a solution of a similar problem. Please consider the users helping out here are doing this in their free time (just for fun and giggles).

You used the wrong version of $R$.  Try using $R = 8.314 J mol^{-1} \ K^{-1}$.


text:
What is the molecular geometry of NHF₂?
I had to draw the molecular shape for $\ce{NHF2}$ as a Trigonal Planar like this

But when I check to make sure it was right, I came across a source saying

NHF2-difluoroammonia--tetrahedral, almost trigonal
  (Source 1)

After looking some more I found a source that now said it was

Trigonal pyramidal, Bond angle of 107 degrees Bond angle of F-N-F > F-N-H
  (Source 2)

Can someone clarify which one it is so I can understand.

Consider the presence of an electron pair on nitrogen; that should guide you to predict the shape of such as molecule.

You can look up the molecule on chemspider, where you have a little applet for the 3D structure. Or you can download a coordinate file from NIST and view it in a molecular viewer, like Avogadro. Or keep on reading for some deeper insight.
As Klaus already pointed out, if VSEPR is a valid concept, one would arrive at the conclusion, that the molecule is trigonal pyramidal, like ammonia. I performed a DF-BP86/def2-SVP calculation on this to analyse the electronic structure a bit more.

One reason why VSEPR does fail (a lot) is that lone pairs are more stable in orbitals with high s character. Another point that is often observed is that orbitals with higher p character are directed to more electronegative elements (Bent's Rule).[1,2] As a result of that, the bond angle between electronegative elements gets smaller. This can be observed here, too. From the ideal sp³ arrangement we would expect an 109.5° angle, but here it is about 5° smaller. As a consequence also the hydrogen-nitrogen-fluorine angle has to decrease. In this case the bond angle becomes much smaller because of an intramolecular hydrogen bond.[3]
The NBO analysis consequently builds the same picture:
$$\begin{array}{llll}\hline
\text{type}&\text{main contribution}&\text{other contributions}\\\hline
\sigma\text{-Bd}~\ce{N-F}& 64\%~\ce{F}~(14\% s + 86\% p) & 36\%~\ce{N}~(15\% s + 85\% p) \\
\sigma\text{-Bd}~\ce{N-H}& 68\%~\ce{N}~(25\% s + 75\% p) & 32\%~\ce{H}~(100\% s) \\
\sigma\text{-LP}~\ce{N} & 100\%~\ce{N}~(52\% s + 48\% p) &  \\
\sigma\text{-LP}~\ce{F} & 100\%~\ce{F}~(75\% s + 25\% p) &  \\
\sigma\text{-LP}~\ce{F} & 100\%~\ce{F}~(10\% s + 90\% p) &  \\
\sigma\text{-LP}~\ce{F} & 97\%~\ce{F}~(100\% p) &  2\%~\ce{N}~(7\% s + 93\% p) & 1\%~\ce{F2}~(15\% s + 85\% p)\\\hline
\end{array}$$
The planar (trigonal) structure on this level of theory is about 33 kcal/mol higher in energy. It is the transition state for the nitrogen inversion, which is therefore very slow.

Notes:

Bent's rule on wikipedia
Bent's rule on our network: Utility of Bent's Rule - What can Bent's rule explain that other qualitative considerations cannot?
The bond length is much smaller than the sum of the corresponding van der Waals radii, suggesting a hydrogen bond. Compare A. Bondi J. Phys. Chem., 1964, 68 (3), pp 441–451.
$r_{\mathrm{vdW}}(\ce{HF})=r_{\mathrm{vdW}}(\ce{H})+r_{\mathrm{vdW}}(\ce{F})= (106 + 140)~\mathrm{pm}= 246~\mathrm{pm}$

what programs did you use in the calculation/analysis for this post?

@permeakra Gaussian 09, rev. D.01 and the built-in NBO 3.1; pictures are chemcraft.

I favorited the question because of this answer!

+1 Martin's work would have been a nice publication a few years back: http://pubs.acs.org/doi/abs/10.1021/ja00427a013

@DavePhD Haha, that's true, but a 'few' years back you would not have done it in an hour ;) Thanks for the link, I was looking for a publication, but not quite able to find one in the allotted time.

@DavePhD Any motivation to nail this analysis with CCSD(T) and a large basis set? Martin and I could knock that sucker out in a few months (the computations I mean). :)

@LordStryker Knock it out of the park, mate. Large basis set? Go BSL (or is it CBS)! Why not include quadruples?! Seriously, if I had the capacity, I would totally run this...

Assuming for a moment that VSEPR is a valid concept,
difluoramine ($\ce{NF2H}$) is represented by $\bf{AX_3E_1}$, and so is ammonia ($\ce{NH3}$). Try to walk your way down from there.


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How to report pH values?
I have measured the pH in the water-soluble fraction of a powder. I weighed in a certain amount of powder and a certain amount of water, and extracted the water-soluble fraction. I then measured the pH in this extract.
How do I report this? The pH depends on the amount of powder and the amount of water. Should I give the weighed-in amounts along with the pH-value? Or is there a way to "normalize" the pH-values to make them comparable for all? Just giving the pH values make no sense, as no one can compare anything to them.

Imagine your report to be the experimental part of a publication. How much detail is needed to reproduce your results without mental acrobatics and mind reading. Sometimes, more is more.

Will it become more clear for a reader if you use an an abbreviated, "coded" form to express what you have actually done?
If not, why waste your time to encode it and waste the time of the reader to decode it (and maybe still leave space for ambiguity)?
I'd be fine with a report that precisely describes the experimental procedure and the observations.

Yes, always make everything as clear as possible. You sound like my professor in academic writing. Thanks!

@AndersMB Thanks! I take that as a compliment :)

I would say "pH of x% solution" if you didn't separate the extract from the insoluble fraction.
I would say "pH of y:1 water extract" you performed a separation.

I am slightly confused. Lets say I added 7g powder to 45g water, and separated the extract from the insoluble powder. How would you report that?

I would say "pH of the 6.4 to 1 water extract" or "pH of the 6.4:1 water:powder extract" or "pH of the 45:7 w/w water:powder extract"


text:
Identify "decoupling sites" in glucose oxydase reaction

Given the following coupled reactions:
  $$\ce{glucose + O2 + H2O + glucose~oxydase -> gluconic~acid + H2O2}$$
  $$\ce{H2O2 + TMB(star) -> TMB}$$
Identify:

A potential reagent or product that could act as a decoupling site leading to a falsely high glucose concentration.
A potential reagent or product that could act as a decoupling site leading to a falsely low glucose concentration.

Note: The second reaction is just hydrogen peroxide reacting with $\ce{3,3',5,5'-Tetramethylbenzidine}$, to form the more absorbent version (I can't remember if it's reduced or oxidized).
The experiment involves running the above reaction on a sample with an unknown amount of glucose, then running the sample through a spectrometer to determine the absorbance and thus the concentration of glucose.
The only thing I can think of is that if the sample warmed up too much you might lose some of the oxygen, resulting in a lower than expected glucose measurement. Though you'd have to lose enough oxygen for it to become the limiting reactant, which I don't see as being too plausible.
I can't find anything anywhere else and I'm stumped.


text:
Equation for salt formation for ammonium oxalate?
I'm having trouble finding out the salt formation equation for $\ce{(NH4)2C2O4}$
I thought it was $\ce{NH3 + H2C2O4}$ but that doesn't seem to balance out.
$\ce{HC2O4-}$ seems to work, but are ions allowed in salt formation/neutralization formulas?

It's 2 $\ce{NH3 + H2C2O4}$ - the reaction won't balance itself - you have to do it by putting stoichiometric coefficient. On the other hand ammonium hydrogen oxalate - $\ce{(NH4)HC2O4}$ also exists, so reaction with different stoichiometry is another option.

Where am I wrong? First of all, note that the ammonia is dissolved in water. So there is an ion that's involved in the reaction, not the ammonia itself. And second, the charge of the ion involved requires a different coefficient for ammonia. Notice that the $\ce{H+}$ cations in Hydrogen oxalate are spectator ions.
The formation of Ammonium Oxalate will be a precipitation reaction. (Or some may put it an acid-base one) Answer these in order to reach a final net ionic reaction:

What are the ions that are present in the final product (ammonium oxalate)?

 The ions involved are ammonium ($\ce{NH4+}$) and oxalate ($\ce{C2O4^{2-}}$).

Notice the charges of those ions:

 Ammonium has a 1+ charge (that's written correctly as +, but of course, I wanted to emphasize) and oxalate has a 2-.

The other ions are spectator ions. These two ions actually form the precipitate. But the final formula (where your misunderstanding rises from) is

 $\ce{(NH4)2C2O4}$

You can now write the net ionic equation and get it balanced.

How come in the case of $\ce{(NH4)2SO4}$ the reaction goes like $\ce{2NH3 + H2SO4-}$ --> $\ce{(NH4)2SO4}$?

@quidproquo Sulfuric acid doesn't have a negative charge. And, ammonia is aqueous in water, thus ammonium is what that's participating in the reaction.

$\Huge{\mathrm{(NH_4)_{\color{red}{2}}C_2O_4}}$

It's really hard to tell what OP misses. Probably the whole concept of ionic compounds. I gave it another shot. Hopefully it'll work for them.

I finally remembered that ammonium salts come from ammonia and does not produce water. I had trouble balancing because I thought water was part of the products. $$\ce{2NH3 + H2C2O4 -> (NH4)2C2O4}$$

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


text:
What is the meaning of the "dot" notation in chemical formulas?
For example for 4 given elements A,B,C,D, what's the meaning of $\text{AB} \cdot x \text{CD}$?

By writing $\ce{AB.xCD}$ chemists mean that there are CDs are found in the crystalline framework of AB. The most common example of this is water trapped inside the crystal structure of ionic compounds. (See water of crystallization in wikipedia)
An example that's often taught is $\ce{CuSO4.5H2O}$. See that 5 that's a representative of $x$? It means that water can be bound only with intermolecular interactions, but it exists in the structure in a stoichiometric ratio.

Source
Also, according to this article

In summary, when a dot is used to break a formula into subunits, it may signify ignorance of how the subunits are structurally related, as in our inorganic example; or it may correspond to actual significant structural subunits, as in our organic example; or it may represent the combining ratios of the binary starting materials required for the synthesis of the compound, as in our phase diagram example.

As a result, for example in direct addition reactions, where there is ignorance about the structural relation of the compounds, we use a dot.

http://en.wikipedia.org/wiki/Adduct I don't think use of the dot is limited to crystals.

I'll improvise.

What is the phase of water- solid, liquid or gas? If it is in liquid form, can we see the liquid water?

@Apurva copper(II) sulfate pentahydrate is a crystalline solid. Water does not form a separate phase, and its molecules are trapped inside the gaps in the crystal. If you heat it, the lost water would then be water vapor (with a white $\ce{CuSO4}$ powder), but nothing of the sort can be said before it happens.

According to the linked wikipedia, it is mentioned ""water of crystallization" refers to water that is found in the crystalline framework of a metal complex or a salt, which is not directly bonded to the metal cation." Here, 4 molecules of water are directly bonded to the metal cation. So, what is the water of crystallization? 1 or 5?

@M.A.R. When we heat blue vitriol, the water is removed in the steps of: 2 molecules, 2 molecules and then finally 1 molecule to give the anhydrous powder. Can we explain this from its structure?

The use of dots in inorganic chemistry
Let's take the example of copper sulfate penta-hydrate: $\ce{CuSO4. 5H2O}$.  The dot here is used essentially as an expression of
ignorance to indicate that, though the parts of the
molecule separated by the dot are bonded to one another in some fashion, the exact structural details of that interaction are not fully expressed in the resulting
formula.
Using Alfred Werner’s coordination
theory that indicates
that four of the five water molecules are actually
bonded directly to the $\ce{Cu}$ in the form of a complex
aquo ion  $\ce{[Cu(H2O)4]^2+}$, then the formula is better
expressed
as  $\ce{[Cu(H2O)4][SO4].H2O}$
So, the dot here signifies ignorance of how the
subunits of a formula are structurally related.
The use of dots in phase diagrams
Let's take the example of the ternary phase diagram of $\ce{Na2O}$,$\ce{Al2O3}$,$\ce{SiO2}$. The formulas of
the various resulting complex sodium aluminosilicates
are all expressed in the form $\ce{a Na2O.b Al2O3.c SiO2}$
in which it is implicitly understood that these initial starting components do not exist as such within the resulting compounds. They represent the combining ratios of the binary/ternary starting materials required for the synthesis of the compound.
For more details, please see http://www.che.uc.edu/Jensen/W.%20B.%20Jensen/Reprints/134.%20Dots.pdf

Let's mention a formal definition for this notation from the IUPAC Nomenclature of Inorganic chemistry Recommendations (Red Book 2005) (external links added):

IR-2 GRAMMAR
IR-2.5 DOTS, COLONS, COMMAS AND SEMICOLONS
(…)
(c) Centre dots in formulae of (formal) addition compounds, including hydrates, adducts,
  clathrates, double salts and double oxides, separate the individual constituents. The dot is
  written in the centre of the line to distinguish it from a full stop (period).
Examples:

$\ce{BF3.NH3}$
$\ce{ZrCl2O.8H2O}$
$\ce{CuCl2.3Cu(OH)2}$
$\ce{Ta2O5.4WO3}$

(Please note that according to the original text typography, there should be no whitespace around the dot and around the following numeral – stoichiometric coefficient; here it comes (if it does) from the chemical formatting script, which is not 100% perfect)


text:
Will trenbolone acetate in the presence of methanol yield trenbolone acetate if distilled water is subsequently added thereafter?
I have 2 grams of trenbolone acetate. I was told to add it to 20 ml of methanol. Then wait until it is fully dissolved. Then distilled water is to be added yeilding a precipitate that is supposed to still be trenbolone. Can anyone shed light on this?
Here is the link.

The method will not convert trenbolone acetate to trenbolone, the ester is not cleaved under these conditions.

It is however possible to separate the steroid from a mixture with other, water-soluble compounds this way.
Unless you are a veterinarian and use trenbolone or its acetate as a growth promoter for cattle, you might want to consider the legal situation in your country. As far as Germany is concerned, the possession of as much as 2 g might lead to real trouble. Here, the non-punishable, non-significant amount is in the 150 mg range (for its use in sports).


text:
Molar mass of a mixture of two solids
I wonder, if I wanted to know the molar mass of a some white crystalline solid that is a mix of $\ce{C6H12O6}$ and $\ce{C12H22O11}$, would I just do this :
$$170.1559 \; \ce{g/mol} + 343.2965 \;\ce{g/mol} = 522.4524 \; \ce{g} / 2 \ce{mol}$$
Thus : $261,2262 \; \ce{g/mol}$.
Would this be a correct operation ? Or it wouldn't make any physical sense ? Thank you!

Does the crystalline contain fifty percent of each?

Glucose + sucrose? The idea of an averaged molecular mass seems pointless to me.

Sounds like a chemical mixture of sucrose and glucose.  In which case, you only talk about the molar mass of the two sugars separately.  The mass or number of moles of the mixture can be determined by knowing the ratio of sucrose and glucose in the mixture.  FYI... the molar mass of glucose is 180 g/mol (not 170).


text:
Why are water droplets shaped like that?
With nothing to do, I stared at the droplets that get condensed on the glass panel of my window.
Upon examination, these droplets appear to be in some sorta pattern. There are big droplets as well as small ones that fill the space between the big ones.

The ones in the middle, between rows of small droplets are usually big one.
All these intrigue me to think about what are that factors that contribute to this pattern.
My guess: I think vapor gets condensed on dust particles thus forms those droplets in the picture. But what accounts for the different sizes of the droplets?

The reason water takes up a spherical form is due the water's cohesive forces. Basically the water 'droplets' will take the shape of a sphere as an effort to reduce surface area.

From the Wikipedia article on surface tension:

Surface tension is responsible for the shape of liquid droplets. Although easily deformed, droplets of water tend to be pulled into a spherical shape by the cohesive forces of the surface layer. In the absence of other forces, including gravity, drops of virtually all liquids would be approximately spherical. The spherical shape minimizes the necessary "wall tension" of the surface layer according to Laplace's law.

In short, the more surface tension is, the rounder shapes of water you get. And the opposite goes for gravitational potential energy: The lesser gravitational acceleration results in more spherical droplets of water.
The symbol for surface tension is $\gamma$.

$\gamma (\ce{H2O}) = 72.8~\mathrm{dyn~cm^{-1}}$ (at $20~\mathrm{^\circ C}$)
   $\gamma (\text{mercury}) = 465~\mathrm{dyn~cm^{-1}}$ (at $20~\mathrm{^\circ C}$)[1]

That's the reason you hardly ever see mercury drops out of their spherical shape.
In short

The spherical shape minimizes then necessary "wall tension" of the surface layer according to Laplace's law.[2]

Oh and I almost forgot: This great article - Antonin Marchand, Joost H. Weijs, Jacco H. Snoeijer, and Bruno Andreott, Why is surface tension a force parallel to the interface?. Physique et Mecanique des Milieux Heterogenes, 2012 - is very nice in case you wanted to do additional study in the case.

References:

R Nave. Surface Tension. URL (accessed April 25th, 2018)
R Nave. Surface Tension and Droplets. URL (accessed April 25th, 2018)

I think that Doeser knows well what surface tension is and his question is much more specific

But I don't think that way @Mithoron. The OP says: "I think vapor gets condensed on dust particles thus forms those droplets in the picture." We have to correct that misconception and then go on. But, I can expand if they aren't satisfied.

Is surface tension the cause of the effect, or the name of the effect? From that wikipedia link: "surface tension results from the greater attraction of water molecules to each other (due to cohesion) than to the molecules in the air (due to adhesion)."

@JohnSnow It's better to put it this way: There is a war between the surface tension tribe and gravity tribe (among other factors). :) You see, gravitational potential energy always tends to make water droplets get more, say, flat and surface tension is getting them rounder. And not only is surface tension one of the causes (and the main cause) of the shape water droplets take, it also affects the size of the water droplets.

I would like to add that the sizes in this picture are also a result of droplets getting big enough to touch each other, at which point they merge. Then the shape changes toward making a sphere again, but with the water from both droplets. The ones at the bottom are likely bigger because gravity has pulled more smaller droplets together. At the very bottom, it seems they get large enough for gravity to "win," and they just run down the window.

I think @MysteriousWhisper actually hit the nail on the head here. Anyway I also want to add something: it is the spherical shape that is adopted because for a given volume of water, its surface area is the smallest. The effect of surface tension is to reduce the surface area; this is shown by the equation $đw = \gamma,\mathrm{d}\sigma$ where $\sigma$ is the surface area (for derivation, see Wikipedia) - essentially, you need to put in work to increase the surface area, which leads to an increase in the Helmholtz free energy of the system at constant volume.

Falling drop of a liquid is always spherical in shape due to surface tension. The inward forces on the surface molecules of the liquid droplet tend to cause the surface to volume ratio as small as possible. Since surface to volume ratio is minimum for the spherical shape that’s why falling drop of a liquid is spherical.

for a given volume sphere is having minimum surface area as compare to circle it means sphere have less surface tension as compare to circle (surface tension have tendency to the liquid to keep it free surface area minimum)
 which in turn means minimum surface area


text:
What is "mer", the first ingredient in a type nail polish?
The first ingredient in a type of nail polish is "mer". What is this? Is it some kind of chemical? How is it used? How does it work?

In the organic lab, sometimes we would run reactions that produced a tar-like substance in the reaction flask as an undesired by-product.  It was just a gooey polymer, and it happened often enough that this substance was just referred to as "mer", short for polymer.  "Mer" was therefore a fairly common term used in many (most?) labs.
Since "mer" is listed as the first ingredient in nail polish, that indicates that it is the major component in the formulation.  If you google "nail polish", you'll find a Wikipedia article. Its opening paragraph states:

Nail polish consists of an organic polymer with various additives

I suspect they use the term "mer" to refer to whatever polymer is the main proprietary ingredient in their formulation.  The Wikipedia article also notes that

Nail polish consists of a film-forming polymer dissolved in a volatile organic solvent. Nitrocellulose that is dissolved in butyl acetate or ethyl acetate is common.

Once the volatile solvent evaporates, you are left with a hard polymer coating on your fingernail.


text:
HOMO and LUMOs of Carbon based Compounds
Why are the HOMO and LUMOs often $\pi$ orbitals in carbon based compounds that contain multiple double bonds or conjugation such as ethene and butadiene?

You can visit this page to find out how to make your future posts better.

Why are the HOMO and LUMOs often π  orbitals in carbon based compounds

The heat of hydrogenation of ethylene is -32.5 kcal/m

image source
In other words, ethylene is 32.5 kcal/m less stable (higher in energy) than ethane and this amount of energy (heat) is given off when we hydrogenate ethylene and convert it to ethane.
This tells us that pi bonds are, generally, less stable (higher in energy) than sigma bonds as shown in the following diagram (just forget about the delta bond orbitals, they don't play into this discussion).

image source
This diagram is a pictorial answer to your question.  Since pi bonds are generally higher in energy than sigma bonds, they will become the HOMO in most molecules.
Since bonding and antibonding orbitals are arranged symmetrically about the zero energy point, whatever type of orbital is the HOMO, then the antibonding counterpart will be the LUMO.  If a $\ce{\pi}$ orbital is the HOMO, than a $\pi^{*}%$ (pi star) orbital will be the LUMO.


text:
Why isn't this a compound?
I don't know what to call it I'll just provide a link. From what I can tell it is impossible, but why?
http://www.chemicalize.org/structure/#!mol=structureId%3A1609384473364&source=calculate

Wow, trying to put cycloheptatrienyl inside cycloheptane - terribly strained bonds.

Apart from anything else the central ring won't be aromatic which is implied by the drawing. Try drawing with single/double bonds to see the problem.

@matt_black It could be as a tropylium type cation.

Chemicalize says it cannot calculate. I have no idea what your structure is.

This structure is impossible as either the molecule is planar and then the C-C bonds on the outer ring are extremely long (and therefore weak), or you distort from the sp2 geometry (120 degree bond angles) of the central ring carbons so that the outer ring is beneath the inner one, which would also require a very high energy state. The only stable circumscribed ring structures you'll find are macromolecular.

So if this were made it would decompose or react instantly?

@tox123 Yeah, most likely.


text:
Conversion of standard InChI to InChI with fixed-hydrogen-layer
I am trying to convert a standard InChI string to a non-standard InChI string with fixed-hydrogen-layer:
Standard InChI for Uracil: InChI=1S/C4H4N2O2/c7-3-1-2-5-4(8)6-3/h1-2H,(H2,5,6,7,8)
Desired outcome: InChI=1/C4H4N2O2/c7-3-1-2-5-4(8)6-3/h1-2H,(H2,5,6,7,8)/f/h5-6H
My question is if this is at all theoretically possible without further information and if not, are there other ways to achieve this goal, e.g. using an existing database?
The ultimate goal is to convert the InChI string with hydrogen layer into a mol file to be able to map chromatogram peaks against it, so I installed OpenBabel which can do the trick, but it can not solve the aforementioned conversion between the InChIs.
I already found the cactus website which allows to convert from InChIKey (which I can generate from the standard InChI that I have) to the correct non-standard InChI (with hydrogen layer) via this URL:
http://cactus.nci.nih.gov/chemical/structure/{INCHIKEY_HERE}/inchi.
The problem here is that it returns multiple correct non-standard InChI strings (with fixed hydrogen layer). One of them is the one I am looking for but I don't see a way to figure out how to pick the correct one. As you can imagine I want to run this in an automated fashion for hundreds of conversions.
Thank you very much for your input.

First off, I'm not sure why you didn't ask your Open Babel question on the mailing lists. It certainly should be able to do what you want.

But I think you're trying to pick a particular tautomer. That should be your focus. There are multiple tautomers possible, so why are you singling out one?

Geoff, for the example above, I am looking to get the specific tautomer that is Uracil. I am starting my search with a keyword like Uracil and by this I get the standard InChI string for it, which I then want to use to convert to the one with hydrogen layer but specific to the initial search term. That is why I single one out.

I don't think you understand. There are multiple tautomeric forms of uracil. You would have to have some sort of "standardization" procedure, generate all tautomers, etc.

Geoff, is there a way to calculate the most common tautomer? This could be the solution. Thanks

No. You could calculate the most stable using force fields or QM, but "common" is likely dependent on solvent, pH, etc.

Thanks for your help, can you just tell me quickly what QM stands for and if you have a link to information on how to use either method to do the actual calculation, I would be very grateful.


text:
Converting mm Hg to mm H2O
So here's my problem:
What is the osmotic pressure, noted in mm of $\ce{H2O}$, of a solution with 125 micrograms of vitamin B12 ($\ce{C63H88CoN14O14P}$) in 2.50 mL of $\ce{H2O}$ at 25 degrees Celsius.
The "hint" that comes with the problem says: Weak osmotic pressures are often measured and given in mm of $\ce{H2O}$. Use here the density of water (1,00 g/mL) and of $\ce{Hg}$ (13,6 g/mL). In this case, the constant $R$ has a value of $\pu{62,36 mm Hg * L / mol * K}$.
Here:
$$
2.50 \,\text{microgram} = 0.000125 \,\text{grams}
$$
$$
pi = cRT
$$
$$
\frac{(\pu{0.000125 g * 1 mol / 1355.3652 g})}{(\pu{2.50 mL / 1000})} * \pu{62.36 mm \ce{Hg} * L / mol * K * 298.15 K} = \pu{0.685890194 mm \ce{Hg}}
$$
So, I blocked here and I saw that the solution was:
$$
\pu{0.685890194 mm \ce{Hg} * 13.6 g/ml / 1.00 g/ml = 9.32 mm \ce{H2O}}
$$
So, here's two questions:

Why did I have to use R = 62.36 and not the usual R = 8.31?
Also, what's happening in the conversion exactly? How was I supposed to figure out what they did to convert from mm $\ce{Hg}$ to mm $\ce{H2O}$?

I wanted to make a joke but I won't do it. We need to be serious :-)

You CAN use R=8.31 but it has the (standard) unit of (m3 Pa K−1 mol−1), simply convert your volume from liter (aka dm3) to meter cube (m3) and the pressure will be in Pascal (Pa)
If you use R=62.36, which has a unit of (L mmHg K−1   mol−1), as long as your   volume is in liter, the pressure will be in mmHg
But the question ask for mmH2O.
Pressure is simply force on an area (F/A), weight on an area (mg/A), density * volume on a area (rho*V /A) = density * height (rho*h)
Thus P = density (Hg) * h (Hg) = density (water) * h (water)
h (water) =  (13.6 / 1.00) * h(Hg)


text:
Finding equilibrium concentrations, trouble simplifying equation
For $\ce{CO (g) + H2 (g) <=> CH2O (g)}$
with Kc = 0.068 @ 273K
initial concentrations being $\ce{[CO]}$ = 1.25M, $\ce{[H2]}$ = 2.00M, $\ce{[CH2O]}$ = 1.00M
I did $\frac{1.00}{((1.25)(2.00)} = 0.4$, which is more then Kc, so it's leaning to the left, will produce more reactants, and products will lower. So the I did:
$\frac{1.00 - x}{(1.25 + x)(2.00 + x)} = 0.068$
Multiplied the bottom and got $X^2 + 3.25x + 2.5$
so
$\frac{1.00 - x}{X^2 + 3.25x + 2.5} = 0.068$
I then tried multiplying the denominator by 0.068 and rearranging to solve for $x$ from there, but that's where I got stuck. $X$ is supposed to be 0.65.

Everything you do is correct and it is only the final algebraic transformation missing:
\begin{align}
0.068 &= \frac{1-x}{(1.25+x)(2.00+x)}\\
0.068 &= \frac{1-x}{x^2 + 3.25x + 2.5} &&|\cdot (x^2 + 3.25x + 2.5)\\
1-x   &= 0.068(x^2 + 3.25x + 2.5) &&|-1 +x\\
0.0   &= 0.068x^2 + 1.221x -0.83\\
\end{align}
Now you can solve for $x$ according to
\begin{align}
0&=ax^2+bx+c & x_{1,2}&=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\
\end{align}
And that gives me the values $x_1\approx0.656$ and $x_2\approx-18.6$. The latter one of course makes no sense.


text:
Effect of tear gas and laughing gas together
What will happen if someone uses tear gas, such as CS $(\ce{C10H5ClN2})$ and laughing gas, $\ce{N2O}$, in auditorium? Means will people start laughing and crying together? Or there will be other effect?

I think you should do some research on both classes of chemicals because their colloquial names don't tell the full story.  Reformulating this in terms of their pharmacology would be interesting, but right now it's based on a false premise.

This question is an interesting thought question.

Admittedly, I had to smile when I first read your question.
Unfortunately, the effect for people exposed to both agents is not comparable to hearing a joke and a sad story at the same time. There is no cuteness in that.
Flooding an enclosed space with $\ce{N2O}$ is not a good idea! $\ce{N2O}$ is an interesting drug, used for sedation in dentistry. Here, a mixture of $\ce{N2O}$
and oxygen is fed to the patient through a nose mask, while the patient can freely exhale. Note that $\ce{N2O}$ apparently isn't metabolized, but just exhaled.
Feeding a heavier-than-air gas to a room without proper fresh air supply and circulation will lead to accumulation of this gas. Whether this could lead to (cerebral) hypoxia and permanent damages is open to calculation.
The use of CS in enclosed spaces is even worse. It is actually considered a chemical weapon, banned from war zones and only used in civil riot control. Oh, the irony! Note that it actually isn't a gas but a solid, dispersed either in solution or from granades. It is a strong irritant to all mucuous membranes and the skin.
Now imagine a densely packed auditorium full of people with burning sensation in their eyes and on their skin, crying, shouting, coughing, vomiting and running around in panic for an exit or relief and you know that this isn't the place to be!

+1. I never understood why CS is legal to use (and own in some countries) while being banned by the UN in war as a chemical weapon. Admittedly, I actually never understood war regulations at all...

@Jori, without CS in the hands of the militias who think of themselves as policemen, the mafias who construe the elaborate systems we live in would have no means to keep the few thousand people in each country who see through it all from putting them all in jail, and bringing the systems back to serving honest citizens.


text:
calculate the final concentration of this equation
A student mixed 260ml of 1.2 M lead(II) nitrate with 300ml of 1.90M potassium iodide. What is the final concentration of $\ce{NO3^{-}}$ ?
the answer is 1.11M , but he used the moles of lead(II) nitrate to find out the moles of $\ce{NO3^{-}}$ ,, and I think he should had used the moles of potassium iodide since it is the limiting reactant.
could anyone explain ?

Is the concept of the limiting reagent reagent useful here? Actually, you are looking at a spectator ion, that does not participate in the precipitation but remains in solution!

Related: What does Pb(NO₃)₂ + KI form?

[…] he should had used the moles of potassium iodide since it is the limiting reactant.

Did any reaction take place upon mixing solutions of $\ce{Pb(NO3)2}$ and $\ce{KI}$?
If so, did the reaction change the number of moles of $\ce{NO3-}$ in solution by precipitation, oxidation, reduction, etc.?
Would the result of the calculation be different if just 300 mL of water were added to the $\ce{Pb(NO3)2}$ solution?

Update I
Now for the rocket science (solubilities at 20 °C):
\begin{array}{cr}
\mathrm{compound} & \mathrm{solubility\ [g\cdot L^{-1}]} \\ \hline
\ce{Pb(NO3)2} & 522\\
\ce{KI} & 1430\\
\ce{KNO3} & 316\\
\ce{PbI2} & 0.76\\
\end{array}
The rather low solubility of $\ce{PbI2}$ suggests that on combination of the two solutions, the following may happen:
$$\ce{Pb^{2+} + 2 NO3- + 2 K+ + 2 I- -> 2 K+ + 2 NO3- + PbI2 v}$$
So again:

Does this in any way change the absolute amount of  $\ce{NO3-}$ in solution?
What does change for the concentration of $\ce{NO3-}$ when 300 mL of another solution is added to 260ml of 1.2 M $\ce{Pb(NO3)2}$ solution?

Update II

Initial $\ce{Pb(NO3)2}$ solution

 260 mL of a 1.2M solution of $\ce{Pb(NO3)2}$ contain $1.2\ \mathrm{\frac{mol}{L}}\ \cdot\ 2\cdot\ 0.26\ \mathrm{L} = 0.624\ \mathrm{mol}\ \ce{NO3-}$

After mixing

 Addition of 300 mL of the $\ce{KI}$ solution results in the precipitation of $\ce{PbI2}$, but does not change the absolute amount of  $\ce{NO3-}$ in solution. However, it changes the total volume and hence the concentration, which calculates to $\frac{0.624\ \mathrm{mol}}{(0.26 + 0.3) \mathrm{L}} = 1.11 \ M$

authors: Çhãø Pûrbähñå Phükøñ

authors: Χριστινα Μουστα'ι'ρα


text:
What is the meaning of parentheses in a condensed molecular formula?
In the formula $\ce{(CH3)3-C-CH(CH3)-CH2-CH3}$, what do the parenthesized portions represent? What is the molecular structure, and how do I name the compound?

The parentheses just have a grouping function: $\ce{(CH3)3-C -}$ represents a carbon atom to which 3 methyl groups are attached, $\ce{-CH(CH3)}-$ is a carbon atom in the chain to which one hydrogen atom and a methyl group are attached.
Draw the molecule, find the longest chain for the base name, determine the position of the substituents and you're done.

I think that the bad bad nasty (CH₃)₃−C− notation shouldn't be encouraged. I could live with (H₃C−)₃≡C−. (But the correct one in Loong's answer.)

The Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008), include recommendations for contracted labels (labels representing more than one non-hydrogen atom) and entire formulas (formulas may be considered as labels not connected to any bonds).
Such labels and formulas rely on the fact that many elements have consistent and well-understood bonding patterns. They can be interpreted with a simple application of valence rules; i.e. they are interpreted to fill as many valences as possible. For example, when reading the contracted label for a carboxylic acid group ($\ce{-COOH}$) from left to right, the carbon atom has three remaining valences. The next atom is oxygen, which has two available valences. Both valences are used to form a double bond with the carbon atom, which leaves the carbon atom with one remaining valence. The third atom is another oxygen, but only one of its valences can be used to create a single bond with the remaining valence on the carbon atom. That fills all of the valences for the carbon atom, but leaves one remaining valence on the oxygen atom. Finally, the fourth atom is a hydrogen atom. It fills the remaining valence on the oxygen atom.
Explicit single bonds are generally not necessary within contracted labels and should be avoided. Therefore, the formula $\ce{(CH3)3-C-CH(CH3)-CH2-CH3}$ that is given in the question should be written as $\ce{(CH3)3CCH(CH3)CH2CH3}$. Note that the first explicit single bond in $\ce{(CH3)3-C-CH(CH3)-CH2-CH3}$ is even wrong since a $\ce{(CH3)3-C-{}}$ group is not possible.
Parentheses are used to indicate branching. The corresponding IUPAC recommendations read as follows:

Simple branching patterns may be implied by the basic valence rules described above, and do not require special notation. More complex branching may be clarified by placing parentheses around all elements within a branch. One valence for the first element within the parentheses is used for connecting the previous atom outside the parentheses; subsequent atoms within the parenthesized section are then bound to the first or subsequent atoms, even if an atom outside the parentheses has remaining open valences.

For example, the $\ce{{}-CH(CH3)-{}}$ segment of the formula that is given in the question includes parentheses that are placed around the atoms of a methyl branch. One valence of the carbon atom within the parentheses is used to form a bond with the previous carbon atom outside the parentheses. This leaves one remaining valence on the carbon atom outside the parentheses when reading from left to right. The three hydrogen atoms within the parentheses fill the remaining valences on the carbon atom within the parentheses.
Therefore, the complete formula $\ce{(CH3)3CCH(CH3)CH2CH3}$ corresponds to the structure of 2,2,3-trimethylpentane.


text:
Ideal gasses reaction under const temp and pressure
Conceptual problem:

Let $A + 2B \rightarrow C + 2D$ be a reaction under constant pressure ($P$) and temperature ($T$), that goes all the way towards the products. $\ce{A,B,C,D}$ are ideal gases. Let $\Delta H$ be the molar enthalpy change in the reaction.
What is the change in the internal energy?

The total amount of moles in the reaction is constant throughout it, can we say that $\Delta U=\alpha nR\Delta T$ is the term for the internal energy? (if so that will be 0).
Another idea is that $PV=nRT \rightarrow V=\textrm{const}$, and then the internal energy equals the sun of chemical work and heat which we know.
All help appreciated.

Got it!
as $H=U+PV$ and $PV = nRT=const$ we ger $\Delta(PV)=0 => \Delta U= \Delta H$


text:
Should double or triple bonds have preference for low locants in IUPAC nomenclature?

Both double and triple bonds are present in the hydrocarbon shown above, which has the name hexa-1,3-dien-5-yne. Why is the numbering chosen such that the alkene has lower locants than the alkyne? Is there a rule which stipulates that the alkene is a more 'senior' functional group?

The appropriate passage in the 2013 Blue Book is as follows:

P-31.1.1.1 The presence of one or more double or triple bonds in an otherwise saturated parent hydride [...] is denoted by changing the ending ‘ane’ of the name of a saturated parent hydride to ‘ene’ or ‘yne’. Locants as low as possible are given to multiple bonds as a set, even though this may at times give ‘yne’ endings lower locants than ‘ene’ endings. If a choice remains, preference for low locants is given to the double bonds. In names, the ending ‘ene’ always precedes ‘yne’, with elision of the final letter ‘e’ in ‘ene’. Only the lower locant for a multiple bond is cited, except when the numerical difference between the two locants is greater than one, in which case the higher locant is enclosed in parentheses.

Therefore, the preferred IUPAC name for the above compound is hexa-1,3-dien-5-yne, as the double bonds receive the lower locants (1,3) than in the alternative name hexa-3,5-dien-1-yne.
Note that this preference for double bonds over triple bonds is only relevant when there is no difference between the locant set for the multiple bonds. In the following example, the locant set (1,4) is lower than (2,5) and hence hex-4-en-1-yne is the preferred name, even though it has a higher locant for the double bond:


text:
Which is more: dissociated ions in water or undissociated?
While studying equilibrium we get to know that there are $1*10^{-7} \ce{~~OH-}$ and $\ce{H+}$ ions in pure water. However, I think there are more undissociated molecules.

related http://chemistry.stackexchange.com/questions/19277/dissociation-of-water-into-h-and-oh-does-the-law-of-mass-action-hold-at-nanos

Wait.... What exactly are you asking?

[…] there are $1\cdot10^{-7}\ \ce{OH-}$ […] ions

Are you sure that this is an absolute number or rather a concentration in $\mathrm{\frac{mol}{L}}$? IN the latter case, what would be the concentration of undiscociated $\ce{H2O}$ molecules in 1 liter of water?


text:
How to apply the Henderson–Hasselbalch equation when adding KOH to an acidic acid buffer?
Will the Henderson–Hasselbalch equation be able to calculate the pH when you have an acidic solution and you add a basic solution? Will that be possible or can you only use it for adding an acid to an acidic buffer to calculate the pH, or only a base to a basic buffer? If it can, how is the pH calculated, if you add 0.025 mol of solid KOH to 1 L of an acidic acid buffer solution of pH 3.97.

In case of very dilute buffer solutions, you shold try the Brönsted–Charlot equation: https://en.wikipedia.org/wiki/Charlot_equation

Asker123, when you originally asked this question (before people edited it), did you mean that we start with pure pH 3.97 acetic acid solution and add base in an attempt to create a buffer, or did you mean that we start with pH 3.97 buffer?

If the initial solution is just acetic acid and water, and the pH is 3.97, this means that it is a very dilute solution, less than 0.001M acetic acid. See http://depts.washington.edu/chem/facilserv/lecturedemo/pHofAceticAcid-UWDept.ofChemistry.html
and Does the number of H+ ions in solution go up on dilution of a weak acid?
Therefore, of the 0.025 moles of OH-, almost all (more than .024 moles) will remain unreacted.
pH = 14 - pOH
pH = 14 + log(0.025) = 12.4
In a different situation, where you added less strong base than weak acid, the Henderson-Hasselbach equation would be helpful, but it is not helpful in this example.

You are adding OH- to the solution

yes, because acetic acid concentration will be less than 0.001M, almost all the OH- will stay OH-

Alright then but can you find the pH of a weak acid buffer when you add OH- with an equation?\

Yes, use the Henderson Hasselbach and don't add more OH- than weak acid

The question makes no comment about the concentration of the buffer. It should however be noted, that pure acetic acid is no buffer, and your answer only accounts for this case.

@Martin-マーチン on this SE, you always have to dig back through the overenthusiastic editors' efforts to find what the real question was.  There was no "buffer" in the original title. The original body did not say "acidic buffer".  OP wants to know "the pH when you have an acidic solution and you add a basic solution" and said "Has pH of 3.97 and is 1L. Acetic acid to be precise." So, I interpret pH 3.97 "acetic acid to be precise" as pure acetic acid solution.

@DavePhD it said buffer in the question, that's why I edited it this way, and I was not overenthusiastic while editing. I would have closed/deleted the question on sight and it only came to my attention because of a flag. I'm sorry this caused inconveniences.

@Martin-マーチン ok, please delete my answer.  Moderator approval needed to delete accepted answer.

@Martin-マーチン I'm not sure why this answer would require deletion, but the question would not.

@Martin-マーチン I'll ask OP to tell us if he meant pure pH 3.97 acetic acid solution, or something else.  I thought his intent was: start with pure pH 3.97 acetic acid solution and add base, in an attempt to a create buffer.  If OP indicates pH 3.97 buffer of unspecified composition/concentration was intended, then my answer should be deleted.  Otherwise the question should be re-edited.

I actually like your answer, and find it helpful. The question is a bit vague on details, but with your answer I would not want to delete it. My initial point was simply that I thought the question was unclear. Given your very first statement there is no misinterpretation possible. In doubt, I'd just leave it as is.

I agree. Even though the question is highly non-ideal (confused/poorly grounded in more than one way, at minimum), I think this answer was really well done in response to it. Firmly in favor of keeping it here.


text:
Regioselectivity in electrophilic substitution of pyrrole
In the electrophilic aromatic substitution (EAS) of pyrrole, why is the alpha position preferred over the beta position so much?

The following figure shows the resonance structures we can draw to describe the intermediates produced by electrophilic at the 2- ($\alpha$) or 3- ($\beta$) positions in pyrrole.

image source
You can see that the intermediate produced by attack at the $\alpha$-position can be described by 3 resonance structures.  Whereas attack at the $\beta$-position produces an intermediate that can be described by only 2 resonance structures.
Therefore, the intermediate formed by $\alpha$ attack is more stable and the activation energy leading to it will be reduced.  Consequently, electrophilic attack at the $\alpha$ position in pyrrole is preferred.


text:
When does a solute form ions?
If I have $\ce{HCl}$ and I put it in water, it will form ions. But, if I take $\ce{HCl}$ and put in benzene, it won't. How do you know that? If I had other substances, how would I know if it forms ions or not?

In water, HCl reacts to form $\ce{H3O+}$ and Cl-.
Then, water molecules surround the respective ions in a stabilizing orientation.
The partially negatively charged O atoms of water interact with the $\ce{H3O+}$.
The partially positively charged H atoms of water interact with the Cl-.
None of the above can occur in benzene because benzene is very reluctant to accept a proton and because benzene is non-polar.
Generally speaking, polar solvents have an ability to stabilize ions that nonpolar solvents do not.


text:
Why is snow white?
I know that this is a rather ambiguous question; but my question is, whenever we take water and freeze it in the freezer, it still tends to stay clear. Since snow is just frozen water, why is it white? Is it due to contents of the air - i.e. dust - that make snow this color?
Or is my freezer just weird?

Because the "color" is a description of frequency. Snow reflects a combination of frequencies of light, and their sum is turning out to be white.

Ok well then what makes this reflect this "combination" and normal ice cubes don't tend to do that?

Because ice is translucent.

Translucent but not completely white right? So why even an individual snowflake appear white?

This might be of interest.

The difference between snow and ordinary ice cubes is mainly about the size of the particles. Snow is made from small, irregular crystals with many edges at a very small scale. Light is refracted or scattered by the edges (or the interface between air and the edges). Snow is white because the scattering effect of those edges dominates what happens to light shining on the snow. In a large block of ice like an ice cube there is very little refraction or scattering as, for any ray of light, there are only two edges (one when light enters the cube, one when it leaves). So transmission dominates (and there is little colour as ice only weakly absorbs visible light).
This is common in many other compounds. Titanium dioxide is a transparent mineral but is used as the primary ingredient in many "white" products like paint. The secret is to use TiO2 particles that are just the right size to maximise the scattering at the particle edges thereby creating "whiteness" as all light is equally scattered.

As matt_black said, it's about the size.
If you take a tall, thick glass tube, it'll be transparent, right? It'll transmit the light falling on it. But make it thinner and thinner(think human hair) then it'll appear white, as it'll refract the light passing through it.
As the light from the sun(which is a combination of all colors) is white, snow appears white.
Many people think that old people's hair are white, but they're not. They get their color(black, brown, yellow, red, whatever) from a pigment called melanin, and in white hair this pigment is absent, leaving only a transparent rod of hair. It's so small it appears white.
About the ice cube part, if you've observed, sometimes there's a ice cube that's transparent on the outside, but in the middle there's a white core. That happens because it has shattered in the middle, probably because of anomalous expansion of water.


text:
Why is ionic strength (mu) calculated differently for these 2 examples?
In the following examples; I don't understand why $\ce{Na_3PO_4}$ is a much more complicated process when calculating $\mu$ compared to $\ce{NaH_2PO_4}$ and $\ce{Na_2HPO_4}$. Specifically, I don't understand how to determine which species to include in the $\mu$ equation. All of the examples I've come across just behave just using water and the original species at most. What is more confusing also, is why doesn't the $\ce{NaH_2PO_4}$ and $\ce{Na_2HPO_4}$ include the concentrations and charges of their conjugates as the last example with $\ce{Na_3PO_4}$ does.Thanks for any help!!! Here is the work:
$0.1000M$ of each reagent to start.
$\ce{NaH_2PO_4}$
$$\text{Ignoring} ~H^+ \text{and}~ OH^-. [C] << 0.1M$$
$$\mu = \frac{1}{2}\sum([Na^+](+1)^2 + [H_2PO_4^-](-1)^2)$$
$$\mu = \frac{1}{2}\sum([0.1M](+1)^2 + [0.1M](-1)^2)$$
$$\mu = 0.1000M$$
$\ce{NaH_2PO_4}$
$$\text{Ignoring} ~H^+ \text{and}~ OH^-. [C] << 0.1M$$
$$\mu = \frac{1}{2}\sum([Na^+](+1)^2 + [H_2PO_4^-](-2)^2)$$
$$\mu = \frac{1}{2}\sum([0.2M](+1)^2 + [0.1M](-1)^2)$$
$$\mu = 0.3000M$$
$\ce{Na_3PO_4}$
Without using Quadratic (which is just to keep things simple):
$$[OH^-]=0.044M$$
$$[PO_4^{3-}]=0.056M$$
$$[HPO_4^{2-}]=0.044M$$
$$[H^+]=\frac{K_w}{0.044M}=2.27*10^{-13}M$$
$$\mu = \frac{1}{2}\sum([Na^+](+1)^2 + [PO_4^{3-}](-3)^2+[HPO_4^{2-}](-2)^2+[H^+](+1^2))$$
$$\mu = \frac{1}{2}\sum([0.3M](+1)^2 + [0.056M](-3)^2+[0.044M](-2)^2+[2.27*10^{-13}M](+1^2))$$
$$\mu=0.5120M$$

The third pKa of phosphoric acid (going from disodium to trisodium phosphate) is 12.32.
When trisodium phosphate is added to water it is a relatively strong base, beginning to titrate H20 to OH-, to a significant degree.  In other words, to be above the pKa of 12.32, the concentration of OH- needs to be at least 0.02M.
In the other examples, since the other two pKas of phosphoric acid are 2.15 and 7.2, the pH in those examples will be about midway between 2.15 and 7.2 for the monosodium salt and about midway between 7.2 and 12.32 for the disodium salt, so the concentrations of H+ and OH- are insignificant compare to the other ions.

So if the concentrations of H+/OH- are insignificant. You're saying it's safe to assume that the conjugate is just as insignificant?

Also, if I were to calculate Mu for H3PO4; then I would include the conjugate since the pKa is ~2.15 or close to 0?

@PiZzL3 "You're saying it's safe to assume that the conjugate is just as insignificant?" Yes, because one OH- ion is created for each conjugate acid created, when you state with purely the base.  "Also, if I were to calculate Mu for H3PO4; then I would include the conjugate since the pKa is ~2.15 or close to 0?" Yes.


text:
condensation of tellurium on inside of ampoule?
After heating an ampoule of tellurium around its melting point in a furnace for about a day, I've noticed drops of the element on the upper inside surface of the ampoule, as if it condensed there. I've also noticed similar behavior with other intermetallic compounds. Is it common for something like tellurium to condense and resolidify in drops when an ampoule of it is heated (like water as shown here)?

This is a general phenomenon. All substances at any temperature above $0\ K$ will shed a few atoms from the solid or liquid, forming a gas however dilute. The tendency for this to happen varies with temperature, and it can be measured by determining the equilibrium pressure of a substance inside an evacuated chamber. The key concept behind this is vapour pressure. There are several previous Chem.SE questions regarding vapour pressure, so you may find something useful, such as perhaps this.
As a rule of thumb, the higher the boiling/sublimation point of a substance, the lower its vapour pressure at any given temperature (in fact, the definition of the boiling/sublimation point is the temperature where the vapour pressure equals the pressure exerted by the surroundings onto the solid/liquid substance, which is in many cases equal to $1\ bar$). Tellurium is a relatively low boiling-point solid, with a vapour pressure of $1\ bar$ at $988\ °C$. According to this article, the vapour pressure of tellurium at its melting point ($450\ °C = 723\ K$) is approximately $70\ Pa$. This means there is a fair amount of tellurium vapour present inside the ampoule at this temperature. It is entirely conceivable that some of this vapour could deposit in small droplets on the upper surface of the glass ampoule, outside the main body of the liquid. Note that theoretically the deposition of tellurium on the upper surface should also happen at room temperature, though the equilibrium vapour pressure of tellurium at $25 °C$ is only about $5 \times 10^{-4} Pa$, which would make the transfer of a macroscopically visible amount of tellurium an extremely slow process.

I think this best answers the question, but I am still confused about whether the top surface of the ampoule would have to be cooler than the rest of the ampoule. I heated things slowly, and the furnace was big and the ampoule reasonably small, so I'm guessing that temperatures would be roughly the same throughout the ampoule. I wonder if the light from the heating elements heated the tellurium before the glass, causing the condensation, and if shielding the  ampoule from the light would stop the condensation from occurring.

@user1704042 No, even if the entire ampoule were in perfect thermal equilibrium, you could still expect to see matter transfer upwards. Once you have the dilute tellurium vapour inside the ampoule, it is in dynamical equilibrium with the solid. The solid will continue to sublime slightly, and the vapour will continue to deposit. Even though it is kinetically easier for the vapour to deposit on the surface of the solid tellurium (which works as a good nucleation surface), there's no reason some of it can't deposit on the ampoule glass instead and begin nucleating tellurium crystals there.

@user1704042 If you feel like testing it out, you can experiment with pure naphthalene. The solid has a larger vapour pressure at a much more accessible temperature (melts around 80°C). If you pack some pure naphthalene into the bottom of an ampoule or a tightly closed vial, and heat it homogeneously in your furnace to 60-70°C, you should be able to see small crystals form above the previous level of the solid. Give it enough time and it should coat the entire interior surface of the container.

The melting point of tellurium is approximately 450 °C; hence, I guess, the hot part of the melt could have a temperature of roughly 500 °C.
At this temperature, the vapour pressure of tellurium is roughly 100 Pa (0.1 % of the normal atmospheric pressure). By way of comparison, the vapour pressure of water at room temperature is 2–3 kPa.
Thus, a small but noticeable amount of tellurium could evaporate during the period of one day if the tellurium vapour is continuously removed by condensation on colder parts of the ampoule.
The actual transport rate depends on the temperature gradient, the pressure of other gases in the ampoule, and the geometry of the ampoule.


text:
Washing a solvent?
Why would you wash a solvent with $\ce{NaOH}$ solution?
Does $\ce{NaOH}$ solution help water and solvent separate like $\ce{NaCl}$ solution does (salting out)?
Would you wash with hot or cold $\ce{NaOH}$ solution/water?
Anybody know anything about washing a solvent please explain basics to me? Thank you

In my experience, washing is quite literal in meaning. I use solvents to remove an impurity that is soluble in the solvent while the product is not. If you're putting a strong base into solvent, I would think it's to remove an acid and allow separation by forming an aqueous layer. I'm no professional, so I'll just leave this comment here incase it helps.

In general, the term washing means to remove something. For example, you have a water insoluble solvent containing some ionic contaminants, and you wash this with water to remove those water soluble ionic compounds.
As the first comment suggests, washing with a strong base like $\ce{NaOH}$ suggests that you are trying to remove an acid. A strong base would remove any acid, strong or weak. The base would be in an aqueous layer, and so there would also be the removal of water-soluble compounds.
The key is that your solvent must be non-miscible in water! In order to wash, your solvent and water cannot mix, so that the water pulls out the water soluble compounds and purifies the non-water soluble solvent.


text:
Signs of Exchange and Correlation Potentials
The exchange and correlation potentials refer to those defined in density functional theory. (See also http://en.wikipedia.org/wiki/Local-density_approximation)
Define the exchange potential as $V_{x}(n) = \partial E_x(n)/ \partial n$, when $E_x(n)$ is the exchange energy density of the whole system(not per particle). Define a similar quantity $V_c(n)$ for the correlation contribution. My question is, are the two potentials (generally) of the same sign?
Another question: It seems that the prediction of the ionization potential using the -ve energy of HOMO is not good (See the section Exchange-correlation potential on http://en.wikipedia.org/wiki/Local-density_approximation). Using a simple slater exchange potential and a reasonable correlation potential (for a uniform electron gas), the result (?; not sure about its correctness; feel free to point it out if you think it's wrong) is about 0.7*13.6 eV, which is not good. Of course one need not use DFT for hydrogen; it however serves as a test case. Is there any reason to believe the results will improve as the atomic number increases?
Thanks a lot.

This would really be two questions. Can you separate your second (about ionization potentials in DFT) into a distinct question? Then this is about exchange vs. correlation potentials and there are separate pages and discussions for each.

What is the ''-ve energy''?

"Of course one need not use DFT for hydrogen; it however serves as a test case." No, it doesn't. Systems with one electron are those where any standard DFT must inevitably fail. $\ce{H}$ might be the smallest testcase, but that's about it.


text:
C-H bond energy in substituted methanes
What happens to the $\ce{C-H}$ bond strength in methane as the degree of halogenation is increased (e.g. from $\ce{CH4}$ to $\ce{CH3X}$ to $\ce{CH2X2}$...)
Also, what is the effect of varying the halogens from fluorine to iodine?

According to David R. Lide, ed., CRC Handbook of Chemistry and
Physics, 90th Edition (CD-ROM Version 2010), CRC Press/Taylor and Francis, Boca Raton, FL, the C-H bond dissociation energies for increasing chlorination of methane are as follows:
$D\left(\ce{H-CH3}\right) = (439.3 \pm 0.4) \; \text{kJ/mol}$
$D\left(\ce{H-CH2Cl}\right) = (419.0 \pm 2.3) \; \text{kJ/mol}$
$D\left(\ce{H-CHCl2}\right) = (400.6 \pm 2.0) \; \text{kJ/mol}$
$D\left(\ce{H-CCl3}\right) = (392.5 \pm 2.5) \; \text{kJ/mol}$
The effect of different halogens in H-CX3 is shown below:
$D\left(\ce{H-CF3}\right) = (445.2 \pm 2.9) \; \text{kJ/mol}$
$D\left(\ce{H-CCl3}\right) = (392.5 \pm 2.5) \; \text{kJ/mol}$
$D\left(\ce{H-CBr3}\right) = (399.2 \pm 8.4) \; \text{kJ/mol}$
$D\left(\ce{H-CI3}\right) = (423 \pm 29) \; \text{kJ/mol}$

Is this purely experimental or Is there a reason for the lowering of BE after each chlorine is substituted ?

Sir, are these values experimental?

@MarckTagapan The values are taken from the CRC Handbook, which is not the original source but a huge data collection. The reference given in the CRC Handbook is the Comprehensive Handbook of Chemical Bond Energies, by Yu-Ran Luo, CRC Press, 2007, which again appears to be a data collection. Thus, I do not know the original experimental data and derived calculations.

A remark in the CRC Handbook reads: ‘The D values in polyatomic molecules are notoriously difficult to measure accurately since the mechanism of the kinetic systems involved in many of the measurements are seldom straightforward. Thus, much lively controversy has taken place in the literature and is likely to continue for some time to come. We will continue updating and presenting our assessment of the most reliable BDE data every year.’


text:
What will be the major product on chlorination of m-nitrophenol?
I came across a question asking for the major product on chlorination of m-nitrophenol. In my opinion the deactivating nature of nitro group should control the regioselectivity, but I found out the answer to be 4-chloro-3-nitrophenol. But why is the activating nature of the hydroxyl group dominating over the deactivating nature of the nitro group?

But why is the activating nature of hydroxyl group dominating over the deactivating nature of Nitro group?

This is the precise reason why Picric acid is formed. ( Check here )
The activating nature of the $\ce{-OH}$ group is strong enough to counter the deactivating nature of the nitro groups.
Look at Ron's comment on the question over here.
Even methyl group sustains deactivating nature of nitro group in the formation of TNT. So no wonder phenolic group, a better activator, dominates.
The phenolic group increases electron density on ortho para (making it o-p director) while nitro groups decrease e-density on the o-p positions (making it a meta director). When both these effects act simultaneously, the question is who wins and is the density on o-p carbons less than corresponding meta ones? (as that would decide where the electrophile will attack)
Resonace structures of the groups gives the solution.
In nitro group:

There are three charged atoms bonded together in the $\ce{-NO_2}$ molecule. This is less table than the positive charge bearing oxygen atom in phenolic group's resonance structure:

Hence phenolic group wins!!
Coming back to the given starting reactant:

So now we consider the dominating ortho-para effect of phenolic group over the meta-directing effect of nitro groups to get 4-chloro-3-nitrophenol:

Well, I had already understood that OH is dominating more an activating group over the deactivating nature of nitro but I still didnt get my answer "why". Anyways, thanks for coming up to help :-) My question was "why is it a better activator"

@DhrubaBanerjee I edited it to include the reason.

@DhrubaBanerjee This reasoning is a theoretical approach but experimental values confirm it .


text:
Pathway from Sucrose to Cholesterol
I was reading Linus Pauling's book How to Live Longer and Feel Better, and in various places in the book he discusses a correlation between increased sucrose intake and increased blood plasma cholesterol levels, as reported in a study by Milton Winitz.
Pauling writes:

... the metabolism of sucrose yields in it's first step equal amounts of glucose and fructose.  ... Fructose metabolism goes in part by a different route such that it produces acetate, which is a precursor of the cholesterol that we synthesize in our liver cells.

I was wondering specifically what the metabolic pathway from sucrose to cholesterol is and given say 100 grams of sucrose, how much will be turned by the body into cholesterol.

Welcome to chemistry.SE! If you had any questions about the policies of our community, you can visit the help center or take a tour of the website.

As you have known that metabolism of sucrose yields glucose and fructose.So formed glucose undergoes glycolysis producing pyruvate or lactate(under anaerobic condition). In case of fructose, it is converted into fructose-1-phosphate and which on action of aldolase enzyme produces dihydroxyacetone phosphate and glyceraldehyde-3-phosphate which are the intermediate products of glycolysis and finally converted to pyruvate.

The end product of glycolysis i.e. pyruvate is converted into acetyl coa in presence of coenzyme A , which is the precursor for the synthesis of cholesterol.

For the synthesis of cholesterol , acetyl coa undergoes different enzymatic reactions finally forming cholesterol at the end.

Any idea what percentage of the sucrose ends up as cholesterol?

Well it depends on the requirement of glucose in our body. If we have to expense large energy then less cholesterol is formed and viceversa. According to research article http://www.jbc.org/content/263/19/9366.full.pdf ,about 20-30% of sucrose is converted to cholesterol in fibroblast cells.

By the way, what program generated that graph?  Or did you scan it?

Well I found the article online so I am not aware of the inside procedure and instruments used in the research. But they might have really high class and significant instruments for calibrating the data and programs for analysis..

I mean the image of the process you used in your answer, how was it created?

They are well established and accepted pathways of biochemistry. As for their development, I dont have exact idea but they are outcomes of profound and hard study in biochemistry.

Our body uses the so-called "mevalonate" pathway to synthesize cholesterol (and many more related terpenes) from acetate.  Sucrose is 12 carbon atoms, acetate is 2.  Even if we allow both glucose and fructose to be converted to acetate, which is possible but not commonly attained because of the way metabolism is regulated) we will get only 4 acetate molecules, containing 8 carbon atoms.  The other 4 carbons are lost as $\ce{CO2}$.
Cholesterol has 27 carbon atoms.  But the pathway for its synthesis goes through lanosterol, which has 30.  So each mole of sucrose (which gives rise to 16 carbon-moles of acetate) can make 8/30 = 26.7% mole of cholesterol.
That calculation assumes that only carbon is limiting, and that the emitted $\ce{CO2}$ cannot be recovered.  If you relax those assumptions, you can calculate the theoretical yield in more detail.  Here is one paper that goes through very similar calculations for production of terpenes.
BTW, some bacteria have a different pathway for terpene biosynthesis that has a higher theoretical yield, because it doesn't start with acetate, it starts with glycolytic precursors before pyruvate dehydrogenase.  This completely different pathway is called the DXP pathway (or the MEP pathway or the non-mevalonate) pathway.

So given your number of a theoretical maximum of 26.7% of the sucrose being converted to cholesterol is CCR's comment that a study showed between 20 to 30% being converted sound reasonable to you?

Since 26.7 is between 20 and 30, yeah, sounds reasonable to me!


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Microorganism preventative solution safe for using on a roof and will last in the sun
Information is scarce in the roof cleaning industry and the big guys are making a killing selling chemicals that are harmful to the person using it and everything around the home. I would like to find something safe that I can use to prevent microorganisms from ever popping back up on my customers roofs.

What chemicals are in the materials sold by the "big guys"?  How unsafe are they?  What degree of safety is important to you?  I ask because it will be impossible to find something 100% safe, i.e., complete without risk.  (Even water can kill you if you use it wrong.)  What kinds of microorganisms do you want to control?  Surely not all of them...just fungi?

I want it to be safe enough for people and also safe enough for the come household plant. If it has the possibility to kill or damage any plant it will be bad for business.


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Why is ethyne and not ethene used for welding?
Ethyne (Acetylene) is used with $\ce{O2}$ for welding. My question is that why is ethene not used? What properties of ethyne make it suitable for welding?

An acetylene molecule is composed of two carbon atoms and two hydrogen atoms. The two carbon atoms are held together by what is known as a triple carbon bond. This bond is useful in that it stores substantial energy that can be released as heat during combustion.The triple bond which makes the oxy-acetylene flame the hottest of all gas flames is also responsible for two rather exceptional properties of acetylene gas.
The first is: If free gaseous acetylene is subjected to severe shock, or a source of ignition, some of the triple bonds may break, releasing enough energy to cause all the other molecules in the enclosed volume to decompose into carbon and hydrogen with explosive force. The force of such an explosion is not so great as that released by the explosion of most mixtures of acetylene and oxygen, or acetylene and air, but it is substantial, and can be withstood only by extra-heavy-wall steel tubing.
The other property of acetylene is that the flammability range of mixtures of air and acetylene is broader than that of any other fuel gas/air mixture. Acetylene/air mixtures can be ignited when they contain anywhere from 2.5 percent acetylene to 80 percent acetylene. Mixtures of methane (the principal component of natural gas) and air are flammable when they contain as little as 5 percent methane and not more than 15 percent methane.
To be suitable for welding operations, a fuel gas, when burned with oxygen, must have the following:

High flame temperature.
High rate of flame propagation.
Adequate heat content.
Minimum chemical reaction of the flame with base and filler metals.

Among the commercially available Oxy Fuel welding fuel gases, acetylene (combination of hydrogen and carbon) most closely meets all these requirements. Other gases, fuel such as MAPP gas, propylene, propane, natural gas, and proprietary gases based on these, have sufficiently high flame temperatures but exhibit low flame propagation rates.

One additional factor is that the burning of acetylene actually occurs in two steps, and this creates the inner and outer flame envelopes. It's useful for welding to have a hot center and a cooler outer area, because preheating, post-heating, and gas shielding can be products of the outer flame while welding occurs mostly in the inner flame.

It sounds from the first paragraph that it's because the C-C triple bond is so strong that you get a lot of energy from burning acetlyne.

Could you provide some sources?

dicyanoacetylene/oxygen flame is hottest.

@Mithoron you took it another way. Dicyanoacetylene/oxygen flame is the hottest among natural chemicals but oxyacetylene flame is the hottest among fuel gases(as i have written in the answer) and is the third hottest natural chemical flame after dicyanoacetylene and cynogen.... http://en.wikipedia.org/wiki/Acetylene(please view info in application ) and http://www.linde-gas.com/en/products_and_supply/gases_fuel/acetylene.html. Thank you!

@CCR technicality ... Bond breaking is not exothermic. If disturbing a container of acetylene gas broke some bonds and that was it, then there would be no explosion. I think that you mean is that the result of breaking the bonds - the formation of new bonds - is what causes the explosion

Another interesting fact about acetylene. The molecule is so unstable, storing the solitary gas under pressure presents a high risk of spontaneous explosion. Therefore commercial welding tanks dissolve the acetylene gas in acetone and furthermore use a spongy stone within the tank to limit energy dissipation from sloshing. Welding regulators typically limit the pressure to under 15 psig to keep the acetylene somewhat stable outside the tank.

[The triple bond in acetylene] 'is useful in that it stores substantial energy that can be released as heat during combustion." This is incorrect.  Bonds don't store energy -- you don't add energy to form bonds; rather, it's the the opposite: energy is released in forming bonds.  I.e., atoms that are bonded are in lower-energy states than free atoms.


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Amine addition to nitro group
What is the mechanisms of this reaction? How feasible is this reaction? Can it happen under relatively mild conditions if it is intermolecular?

It's actually a hydrazine that adds to the nitro group. Usually one wouldn't bother to isolate the nitrosubstituted phenyl hydrazine and instead react 2-nitrochlorobenze with aqueous hydrazine under alkaline conditions.

That would just make a nucleophilic aromatic substitution.

Exactly - for the first step.  But would it stop there or  rather continue to furnish HOBt?

Okay, thanks for the info, but I am far more interested in the mechanism of the attached reaction.

A reasonable mechanism for the transformation is shown below. Since this is under basic conditions, I've written a mechanism that avoids cationic intermediates. Deprotonating the hydrazine gives a very nucleophilic nitrogen that can attack the nitro group. Proton transfer gives a nitrogen anion, which can kick out hydroxide. Two additional proton transfer steps lead to the product, HOBt.

A SciFinder search didn't reveal any examples of the intermolecular variant of this reaction, although I didn't find this exact intramolecular reaction either.