Cs2Se + Sc Reaction: Resultant Chemical

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The reaction between Cs2Se and Sc is uncertain, as the conditions, such as the presence of air or a solvent, can significantly influence the outcome. Scandium, being a metal, is likely a solid, while the state of Cs2Se is not specified, complicating the analysis. The discussion suggests consulting electrode potential tables to explore possible redox reactions. There is a call for input from a knowledgeable inorganic chemist to provide a definitive answer. Overall, the resultant chemical from this reaction remains unclear without further details on the reaction conditions.
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I guess straight forward enough question, what would be the resultant chemical if Cs2Se + Sc = ?
 
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I thought someone who knows more about chemsitry might reply but, I can't seen that reacting. Scandium isn't going to pull anything from Ceasium.
 
PH7SICS said:
I guess straight forward enough question, what would be the resultant chemical if Cs2Se + Sc = ?

You did not say if this reaction should be in air, or any specified solvent. The conditions could make a difference. Scandium is a metal, probably a solid (not sure); Cs2Se, what state of matter? Is it dissolved in something? My unrefined guess would be to check in a table of electrode potentials to look for a reasonable redox reaction. Maybe a true inorganic chemist would read your question and give an expert answer.
 
Thread 'Confusion regarding a chemical kinetics problem'

Thread 'Confusion regarding a chemical kinetics problem'

TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...
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