This problem is providing two reduction-oxidation (redox) reactions in which the oxidized and reduced species can be identified by firstly setting the oxidation number of each element:
Reaction 1: 2K⁺I⁻ + H₂⁺O₂⁻ ⇒2K⁺O⁻²H⁺ + I₂⁰
Reaction 2: Cl₂⁰ + H₂⁰ ⇒ 2H⁺CI⁻
Next, we can see that iodine is being oxidized and oxygen reduced in reaction #1 and chlorine is being reduced and hydrogen oxidized in reaction #2 because the oxidized species increase the oxidation number whereas the reduced ones decrease it.
In such a way, the correct choice is C.
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Answer:
The correct appropriate will be Option 1 (Acid anhydrides are less stable than esters so the equilibrium favors the ester product.)
Explanation:
- Acid anhydride, instead of just a carboxyl group, is typically favored for esterification. The predominant theory would be that Anhydride acid is somewhat more volatile than acid. This is favored equilibrium changes more toward the right of the whole ester structure.
- Extremely responsive than carboxylic acid become acid anhydride as well as acyl chloride. Thus, for esterification, individuals were most favored.
The other options offered are not relevant to something like the scenario presented. So, the solution here is just the right one.
Answer:
A) The catalyzed reaction passes through C.
Explanation:
<span>P*V/T=constant
so P*V= constant*T
if T doesn't change then
P*V= constant
so 150kPa*0.8L=75kPa*xL
xL=150kPa*0.8L/75kPa=1.6L
hope it help</span>
Answer:
19.K, potassium
Explanation:
it has all properties of metals
