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High School Chemistry Forum / Re: solvated O^2- anions?
« Last post by Hunter2 on May 27, 2024, 01:43:25 PM »
Na+ is spectator because it is not changed.

The main reaction is
O2- + H+ OH- ( H2O)  => 2 OH-

You get from sodiumoxide sodiumhydroxide.

The solid Na2O will not dissolve as Na2O(aq) it reacts directly to form Na+( aq) and OH- (aq) ions.
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High School Chemistry Forum / solvated O^2- anions?
« Last post by gavindor on May 27, 2024, 01:20:45 PM »
Let's suppose you have MgO or CaO or Na2O

from what I understand, they are all reactive and form hydroxides.. MgO would react at 120C..  CaO would react at eg  25C  . 
 

but let's suppose the temperature was low enough that they don't react.

So presumably they're then insoluble.  (in the insoluble range) e.g. sparingly soluble or practically insoluble.

And you just get crystals?

To the extent that they are soluble , would you get solvated O^2- anions?

And if there's O^2- anions there there'd be, in the case of Na2O,   Na+ cations too?  (albeit at tiny levels we don't factor in).

In an earlier post I wrote

" H2O(l)  + Na2O(s)  --> Na+(aq) + Na+(aq) + OH-(aq) + OH-(aq)

Writing Na2O(s) as aq (and spiltting it up since it's ionic), Na+(aq) + Na+(aq) + O^2-(aq)     (And they would split before reacting so that's fine)

Then certainly there's no phase change, it's clear the Na+ ions  are spectator ions

H2O(l) + O^2-(aq) --> OH-(aq) + OH-(aq)
"

I meant

taking

H2O(l)  + Na2O(s)  --> Na+(aq) + Na+(aq) + OH-(aq) + OH-(aq)

and writing

H2O(l)  + 2Na+(aq) + O2-(aq)  --> Na+(aq) + Na+(aq) + OH-(aq) + OH-(aq)



But The dissociation of Na2O is part of a reaction

Without a reaction, it wouldn't be dissociating.  (or at least not to an extent that a chemist would be interested in).

So, is it wrong to have said that Na+ is a spectator ion there?

And was it wrong to have split up Na2O(aq) into 2Na+(aq) + O2-(aq) ?

'cos e.g. there are no solvated O2- anions. (other than perhaps in the middle of a reaction so then it's not a reactant).

Like the state symbol for insoluble ionic compounds (not yet reacted), (or any insoluble compound) is not (aq). And it's (s) here.  So no solvated ions in the reactants. So no spectator ions?

But then also

Would you then say there's no bronsted lowry reaction? 

'cos if I remove the Na+ then   Na2O + H2O --> 2Na + 2OH-      becomes O2 + H2O --> OH- + OH-

But if I can't cancel out the two Na+  as spectator ions, then I can't get O2 + H2O --> OH- + OH-   So then I can't identify a bronsted lowry reaction


Thanks
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Thank you for your reply Borek. I think it might be a combination of electron distribution (nucleophilic/electrophilic potential of individual C positions), the amount of steric hindrance and the type of reaction applied that determines where the next reaction takes place. Let's see what an organic chemist will say about this.

Z ukłonami, (hope that was correct)
Venoxis
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Undergraduate General Chemistry Forum / Re: cm vs cv (adiabatic equation)
« Last post by mjc123 on May 26, 2024, 02:50:00 PM »
Looks like the problem is the factor of n in your expression of the exponent. You must be consistent in using either molar heat capacities or total heat capacities in both numerator and denominator. You can't use nR/cm.
The exponent is (Cp/Cv - 1), which in total heat capacities is nR/Cv, or in molar heat capacities is R/Cv,m.
I think just using γ-1 as the exponent is simpler and avoids this possible confusion.
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Undergraduate General Chemistry Forum / Re: cm vs cp (adiabatic equation)
« Last post by physikkk2 on May 26, 2024, 11:56:46 AM »
alternative equation: T2=T1(v1/v2)^y-1 , cv is also needed for y
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Undergraduate General Chemistry Forum / cm vs cv (adiabatic equation)
« Last post by physikkk2 on May 26, 2024, 11:40:58 AM »
Hello,

I have to solve the following problem: 2 moles of argon are expanded reversibly and adiabatically. Before expansion, the gas has a temperature of 30 ◦C (303,15 K) with a volume of 2 L. After expansion, the volume is 15 L. The
Molar heat capacity of argon is cm = 12.4 J/mol K. What is the temperature of the gas after
of expansion ? to solve that problem I use the adiabatic equation : T2= T1 * (V1/V2)^(nR/cv)   My question is if cm is the same as cv? When I use cm as cv, I get  20 K for T2 which seems wrong. Thanks in advance
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Analytical Chemistry Forum / Re: Acid chloride TLC
« Last post by wildfyr on May 26, 2024, 09:40:06 AM »
Thionyl chloride would not visualize on tlc and probably degrades immediately.

You should co-spot your starting material to see how it runs. I think this is straightforward enough though, the product on the baseline is the starting material, the one that runs is the ester.

Honestly though, using an excess of thionyl chloride most acid chlorides quantitatively convert to acid chlorides in a couple hours, I personally do not even track such a reaction by TLC. The act of taking some out and forming the ester with it will cause some hydrolysis due to wet solvents so I just trust the chemistry.
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I am not an organiker, so take my answer with a grain of salt.

Regioselectivity in aromatic compounds is in a large part related to deviations in the electron density - which is to some extent related to the presence of resonance structures. In general, the more resonance structures possible, the lower their individual effect (ie the deviations from the equal distribution become smaller). The larger the molecule, the more atoms involved, the more the resonance structures possible - and the less prominent effects.

So if the selectivity of the reaction on a single ring is such that you get a 90:10 mixture of products, you can speak about the reaction being selective. When you get to several rings and you get mixture of 5:6:7:5:5:4:... it is hard to speak about "selectivity" as such.

This is not to say there are no specific examples when the reaction can be selective, it is just that the general trend seems to be working against.
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Is this question really that difficult to answer or is it unclear what exactly is being asked here? I see multiple threads have been answered since my post but unfortunately not mine :-(
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High School Chemistry Forum / Re: Valence electrons and metallic bond strength
« Last post by Borek on May 26, 2024, 03:32:10 AM »
Why do cations with greater oxidation number form stronger metallic bonds with itself?

Do they? Metallic bonds exist in metals between neutral atoms, no cations/oxidized atoms there.
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