Electrolysis of water<span> is the </span><span>decomposition reaction, because from one molecule (water) two molecules (hydrogen and oxygen) are produced. Water is separeted into two molecules:
</span>Reaction of reduction at cathode: 2H⁺(aq) + 2e⁻<span> → H</span>₂(g<span>).
</span><span><span>Reaction of oxidation at anode: 2H</span></span>₂<span><span>O(l) → O</span></span>₂<span><span>(g) + 4H</span></span>⁺(<span><span>aq) + 4e</span></span>⁻.<span><span>
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A.) half - life of a radioactive substance is defined as the time during which concentration of the substance becomes half the initial value.
the answer to your question is
volume
Explanation:
-Filter help — delete some big unreacted, undesirable species (norit is probably from what you are sorting through, its only carbon which cleans up things)
— extract with DCM because you are probably in an aqueous phase, and some butanoate is in it
- Anhydrous sodium absorbs excess of water (dries the material)
-evaporation in the hood to clear the DCM and crystallize the product.
Answer : The rate constant at 785.0 K is, 
Explanation :
According to the Arrhenius equation,
or,
![\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7BK_1%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= rate constant at 
= 
= rate constant at 
= ?
= activation energy for the reaction = 262 kJ/mole = 262000 J/mole
R = gas constant = 8.314 J/mole.K
= initial temperature =
= final temperature =
Now put all the given values in this formula, we get:
![\log (\frac{K_2}{6.1\times 10^{-8}s^{-1}})=\frac{262000J/mole}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{785.0K}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7B6.1%5Ctimes%2010%5E%7B-8%7Ds%5E%7B-1%7D%7D%29%3D%5Cfrac%7B262000J%2Fmole%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B600.0K%7D-%5Cfrac%7B1%7D%7B785.0K%7D%5D)
Therefore, the rate constant at 785.0 K is,
