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 on: Yesterday at 10:44:53 AM 
Started by blokeybloke - Last post by Enthalpy
Writing just Na for the solid is usual, but it's confusing. In that case, it means "one atom taken from solid sodium", which is a molecule.

Na (s) for the solid would be better practice. Feel free to write it that way more often than most people do. Usually, it serves to differentiate from other states, like Na (g) gaseous.

If you noticed that and keep aware of it, you have a chance to avoid a common mistake, in thermochemistry. The reference state of Na (and many more) is the solid. Thermo quantities, for instance the enthalpy of formation, are given by comparison with that state. Though, an atom of Na in the solid has already bonds, which must be broken before forming other molecules, and this takes energy. The writing "Na" hides this.

 on: Yesterday at 08:56:50 AM 
Started by blokeybloke - Last post by chenbeier
The formulsa are used to show the molecule of a compound. Na ist for sodium generally, but in combination with chlorine you get NaCl the easierst compound for sodiumchloride .

 on: Yesterday at 07:13:14 AM 
Started by saippo - Last post by rolnor
In the E1cB case you have a carbanion thats kicks out the OH. The carbanion is more basic then OH- and this is a driving force I think.

 on: Yesterday at 07:09:01 AM 
Started by Shannon Dove - Last post by pgk
Not exactly!
Under conditions like in the Miller’s experiment, in neonatal planets and space objects, formation of organic molecules could be considered as “random made”.
But their existence also depends on their stability to heat and UV irradiation, as well as on their chemical and physical properties.
As an indicative example, imagine the new-born planet Mars. Due to the intense volcanic activity, there was an initial atmosphere rich or “random made” methane and oxygen. Within time, oxygen combusts methane to CO2 and water. But Mars’ gravity is low and cannot hold the light gases, such as CH4 and H2O with molecular weight 16 and 18, respectively and which easily escaped with the help of solar wind. The remaining atmosphere of CO2 and O2 is not quite dense and cannot adequately filter UV irradiation, which catalyzes the reaction between oxygen and chloride salts. As a consequence, only gaseous CO2 and perchlorate salts are detected on Mars today, together with traces of oxygen and indications of traces of methane.   
PS: The above is a hypothetical scenario and not a detailed and "official" history of the Mars’ atmosphere. 

 on: Yesterday at 06:46:02 AM 
Started by saippo - Last post by saippo
Thanks so much for the replies! They helped me clear up what I was confused about.

 I think I'm not sure why in an E1cB elimination, the LG can be poor (ex. OH)?

 on: Yesterday at 06:06:43 AM 
Started by AllylicAzideUMN - Last post by pgk
Another technique is recrystallization, followed by slow cooling at room temperature (just, allow to stand at room temperature). Larger crystals tend to form that way.
This doesn’t always works but in many cases, it does.

 on: Yesterday at 05:52:36 AM 
Started by Richard Jeong - Last post by pgk
It doesn’t seem a bad idea because both Ni and Co halides are used as dimerization catalysts.
Besides both Ni and Co belong in the same period of the periodic table as Ti.
Furthermore, Ni-Raney is a dehalogenation catalyst.
It might work.

 on: Yesterday at 04:12:43 AM 
Started by Shannon Dove - Last post by Shannon Dove
I appreciate everyone's response, thank you for taking the time to consider what many think to be a stupid subject.
I have another question for you all.... wouldn't organic molecules identified in space be considered "random made"?

 on: Yesterday at 03:00:09 AM 
Started by xshadow - Last post by xshadow
I don't understand the formula that brings from specific conductivity "κ"  to equivalent(or also molar)conductivity Λe=

My  textbook says:

Λe= 1000*κ/(z*C)

Where(the textbook says...)
Z= psitive ion charge
C=concentration charge

Now if I have NaCl 0.01M
C (Na+)= 0.01M
Z(Na+)= (+)1

BUT if I have a weak electrolytes such as CH3COOH 0.01M...

C is its analitical concentration , 0.01.lM

OR ,again,the "real" ion concentration ,so  C*αch3coo- = 0.01M * α

I need help,thanks !!

 on: Yesterday at 02:40:46 AM 
Started by ostudent - Last post by rolnor
I think A and B have to little conjugation to be fluorescent?

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