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DJF90
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I've worked through a thermodynamics question and cannot finish. The reaction is 1/2 N2 + 3/2 H2 <=> NH3. All I know is that 1 mole of ammonia
was allowed to come into equilibrium at a given temperature and 1 atm pressure, and I have calculated the mole fractions for each component. So how do
I go about finding the moles of each component?
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Nicodem
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You also need the equilibrium constant at the given temperature (http://en.wikipedia.org/wiki/Equilibrium_Constant). The equation for the equilibrium constant is derived from the reaction stoichiometry. Without
this information we can not help you. Alternatively you could calculate the equilibrium constant from the free energy (deltaG=-RT*lnK), which can be
calculated from the reaction enthalpy (deltaH), reaction entropy change (deltaS) and temperature (http://en.wikipedia.org/wiki/Gibbs_energy): deltaG=deltaH-TdeltaS
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brew
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It seems abit odd that it is a thermodynamic question and no temperature is given, no standard conditons etc. Perhaps it is just an equilibrium
problem, and if at equilibrium Ke would have to equal 1.
Perhaps change the equation back to its origin N2 + 3H2 > 2NH3 and from there write an eqilibrium constant expression where K is one and NH3 is one
mole, hence if able to, solve for the other reagants algebraically. It is late in the day, and perhaps I am wrong anyway.
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DJF90
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Sorry I've worked through the problem, I'm just at the end of it and stuck with mole fractions that I am unable to convert into moles. Here is the
full question:
Find the amounts of (i) N2, (ii) H2 and (iii) NH3 present when 1 mole of NH3 is allowed to come to equilibrium at the temperature where Kp=1 and 1 atm
pressure.
Basically what I did was:
Kp = pi (x) [p(x)/p(standard)] raised to v(x), the stoichoimetric coefficent of x.
The partial pressures used were (1-X)p (where (1-X) is the mole fraction) for NH3, 0.25X for N2, and 3/4X for H2. after some algebraic manipulation,
it was found that X = 0.7930..., which gives mole fractions of 0.2069, 0.1982, and 0.5947 for NH3, N2 and H2 respectively.
Now my problem is how to change these mole fractions into moles?
I know that:
n(NH3) / [n(N2) + n(H2) + n(NH3)] = 0.2069
n(N2) / [n(N2) + n(H2) + n(NH3)] = 0.1982
n(H2) / [n(N2) + n(H2) + n(NH3)] = 0.5947
But as I dont know what the sum of moles is at equilibrium (the denominator in the above fractions) I cannot work out the moles of the individual
components. There is probably some logical thing I can do to work it out that is staring me in the face but I just can't see what to do. Sorry for the
confusement earlier.
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solo
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I need to find the CAS number for Sodium Naphthalene, `i've found Sodium 1 naphthalenesulfonate and ` Sodium 2 naphthalenesulfonate in Aldrich I
wonder if it's the same or even if these are good substitutes and the same sodium naphthalene save the different location of a sulfate ......mmmmm, is
this what's called the complex of sodium naphthalene..........solo
It's better to die on your feet, than live on your knees....Emiliano Zapata.
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sparkgap
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| Quote: | Originally posted by solo
I need to find the CAS number for Sodium Naphthalene, `i've found Sodium 1 naphthalenesulfonate and ` Sodium 2 naphthalenesulfonate in Aldrich I
wonder if it's the same or even if these are good substitutes and the same sodium naphthalene save the different location of a sulfate ......mmmmm, is
this what's called the complex of sodium naphthalene..........solo |
The CAS number for sodium naphthalide (or naphthalenide, depending on who you ask) is 3481-12-7 . 
sparky (~_~)
"What's UTFSE? I keep hearing about it, but I can't be arsed to search for the answer..."
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kclo4
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What is the most common way of determining alkaloid concentrations in plant material?
are the methods very different when using the wet material vs the dried material?
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Nicodem
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Most common method would be GC for volatile enough alkaloids (a FID detector is fine, but GC coupled with MS is better) and HPLC for all other UV
active alkaloids. You need a pure sample for standard to determine the alkaloid both qualitatively and quantitatively. In some cases the alkaloid can
be determined qualitatively already by making the TLC of the alkaloid fraction of the plant matter by comparison with a standard, but this is nearly
not a 100% confirmation of its presence, neither does it tell anything about its concentration (but at least it is easily done by the average
amateur).
Whether the material is wet or dry it does not make any difference for the determination itself, just the homogenization of the material and its
extraction to give the alkaloid fraction is different.
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hector2000
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does phenylacetone dissolve in acetone?what about ether?
Chemistry=Chem+ is+ Try
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bquirky
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Hello,
I have a short question for the short question thread about salt solubility in acids,
It should be simple but its been puzzling me.
If you have say a %50 solution of sulfuric acid and then dissolve as much metal salt (say zinc sulfate) into it as possible. so you end up with a
saturated zinc salt solution with a very low PH.
What is "supposed" to happen when a piece of zinc metal is then added to the solution ?
the sulfuric acid should want to react with the zinc but zinc sulfate is not as soluble in water as sulfuric acid.
Will the extra zinc sulfate precipitate onto the surface of the zinc and pasivate it ? or will the perciptate form somewhere else in the solution
because of a temperature gradient ?
any ideas as to how these kind of processes might work ?
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kclo4
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Thank you for your reply Nicodem. I guess if I wanted to determine a rough approximation of alkaloid content I'd probably just have to do an
extraction and measure the end alkaloid mass relative to the begging plant material.
bquirky, I don't know if this would really help you out on answering your questions, but you might want to read about the common-ion effect.
http://en.wikipedia.org/wiki/Common-ion_effect
having the acid-water solution saturated with zinc sulfate would lower the levels of H+ in solution and that might play a roll in its reaction with
zinc.
[Edited on 16-1-2009 by kclo4]
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Sedit
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Ok its about time i start beating down the short questions threed because i have a bunch and reading just isnt cutting it any more 
For chlorination efforts would a fluoresent lightbulb with the Phosphor blown off to shine the UV/Blue light thru suffice for dissociation of
chlorine?
It is generated by mercury vapor so im assuming that the output should be satisfactory for this application.
Is this correct?
BTW: i plan on clearing the Phosphor using a small tesla coil to clean the glass without breaking it for anyone woundering
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not_important
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Do you know that the Tesla discharge would do the job and still leave the lamp functional?
Light in the blue to UV range works, for moderate to large diameter reaction vessels the shorter wavelengths are strongly absorbed enough that they
aren't as effective.
You want uncoated bulbs like these
http://www.hardwareandtools.com/invt/u252122?ref=gbase
http://www.prolighting.com/h33cd400.html
http://www.peclamp.com/buymvclear.html
[Edited on 17-1-2009 by not_important]
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Sedit
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"Do you know that the Tesla discharge would do the job and still leave the lamp functional?"
 Yes sir iv been a mad scientist well before found this forum. Been building
tesla coils for at lest 10 years now.
Would have to say i prefer the smaller ones that you can play with over the ones that one has to be a little cautious with.
Give me a little 2-3 incher diameter and i can clean a floresent light bulb in about 15 seconds leaving a pale blue glow that floresent objects shine
in.
[Edited on 17-1-2009 by Sedit]
[Edited on 17-1-2009 by Sedit]
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not_important
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The high pressure in out and out mercury vapour lamps increases the output in the near UV and blue end of the spectrum, as compared to the low
pressure of fluorescent lamps which radiate about half their light at 254 nm - too short for efficient driving of the chlorination reactions. Rough
spectra and other information can be seen here
http://www.crystec.com/senlampe.htm
http://www.lamptech.co.uk/Documents/M3%20Spectra.htm
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Sauron
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This may or may not turn out to be a short question.
I want to prepare t-butyllithium. The lit. says I need a 2% alloy of sodium in lithium, preferably as dispersion in mineral oil.
I have not been able to find a supplier for this (98%Li 2% Na).)
I have hundreds of grams of Li wire in mineral oil. And a loy of Na.
Should I made my own?
Li mp 180 C. Na as I recall less than that. A melt of the two under a high boiling mineral oil stirred with a PTFE paddle ought to produce a fine sand
dispersion of the alloy (2 g Na for every 98 g Li.)
Is this a plan? Or am I missing something?
Aldrich has a "high sodium" Li but it is 0.5% Na, while the lit. is quite specific about 2% Na.
[Edited on 17-1-2009 by Sauron]
Sic gorgeamus a los subjectatus nunc.
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appetsbud
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can someone point me in the direction of a Paraformaldehyde synthesis? preferably from an aqueous formaldehyde solution
is it just as simple as heating with a little conc. acid and then removing the water via distillation?
thanks.
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Sauron
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There is a very recent US patent 7005083 on process for preparing alkyllithiums, that encompasses and describes preparing lithium-sodium alloy
dispersions using an autoclave at a temperature above the mp of Li (180 C). The patent also states that such Li/Na alloy dispersions are commercially
available from Postin Products Inc., in Faith, N.C., USA.
So I guess I can prepare it myself as I have a stirred heated autoclave.
[Edited on 17-1-2009 by Sauron]
Attachment: US7005083[1].pdf (360kB)
This file has been downloaded 693 times
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hissingnoise
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Paraformaldehyde is, I think, produced by simple evaporation of formalin without recourse to acid catalysts.
Distillation can be used to minimise losses.
Formalin normally contains small quantities of methanol to inhibit this polymerisation.
I have unused formalin which contains a fairly large precipitate of the polymer.
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Sauron
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There is an ACS monograph on Formaldehyde in the forum library for free download and it covers paraformaldehyde in detail.
Sic gorgeamus a los subjectatus nunc.
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panziandi
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Sauron I was reading your post on t-BuLi and thought oooh "high sodium lithium" from Aldrich but when I got to the end of your post I realized it was
no good! A quick google for "lithium sodium alloy" turned up a few springerlink journals but my proxy server sign in is playing up bastard thing! I
think certainly making your own in an autoclave is the way forward.
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Sauron
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Making t-butyl chloride is easy. t-BuOH and conc HCl.
Per the patent, preparing the alloy dispersion is quite straightforward.
From there on calls for good technique with argon atmosphere and anhydrous conditions. But that is not a surprise.
Commercial solutions of t-Butyllithium are rather pricey, typically 800 mls of a 1.7 M solution in pentane works out to about $300 US by the time I
get it. Too damned much for something I can prepare myself.
The US company in NC has no website and no published email address. Amazing in this day and age.
Here is the JOC paper from 1960 that is referenced from Paquette's book, this was the lit. I was referring to above. The prep of t-BuLi is in
experimental section.
[Edited on 17-1-2009 by Sauron]
Attachment: jo01080a035.pdf (82kB)
This file has been downloaded 952 times
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Sedit
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Thank you not_important
So judging from the links you sent me A floresent light prepared that way would be effective since a small fraction is produced in the bandwidth area
I would need but very far from efficient in terms of time and electrical input
How about those small hand held Black lights these seem to have there output in the early UV range. Just curiouse because I found on laying around
last night.
Just found the frequency of the type of blacklight I have.
320-400 nm
I just dont know the strenght of it but that should be effeciant. Experimentation will tell
[Edited on 17-1-2009 by Sedit]
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not_important
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I suggest reading this http://www.faizkaskar.8k.com/light1.html particularly near the end when chlorination of toluene is being discussed. Then consider that a clear
(no phosphor) high pressure mercury lamp, such as used for outdoor lighting, has peaks around UV (365nm), violet (406nm), blue (435nm), green (546nm),
and yellow (578nm) while the low pressure fluorescent ones most of their energy in the 253 nm line. relying on the phosphor coating to convert that
to the desired output.
By this http://www.standardpro.com/product/categories/fluorescentgui... blacklight bulbs have most of their output betweem 350 and 400 nm with the peak
around 370 nm.
With fluorescent lamps the actinic bulbs used with aquariums or the bulbs designed for curing plastics and coatings are likely more effective than a
blacklight bulb.
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gnitseretni
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I wanted to make some picric acid to make DDNP, so I got me some extra strength aspirin(500mg) and dissolved it in acetone. Unfortunately I ran out of
acetone so I added some MEK. I'd say I had about 1 part acetone and 2 parts MEK.
Anyways, after boiling for a while and about one third of the original volume left, the solution started taking on a brown color. I took a syringe and
sucked up about a ml or two and while swinging over to squirt it into a separate container, the stuff in the syringe solidified on me before I could
do so.
What could this stuff be? It looks just like caramel.
I got another syringe and sucked up some more and quickly squirt it out onto the concrete. I thought it was completely solid, but found after hitting
it with a hammer that it was a thick shell with fine brown powder inside. The powder was bone dry.
Anyways, I did this a week ago and don't have the stuff anymore so I can't do any tests to see what it is. I just thought I'd post it to see if anyone
knows what I may have made.
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