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.
Learn more:
Answer:
C) 0.457
Explanation:
The ratio between O2 and H2O is 1:2 according to the balanced equation. You can find how many moles is O2 by : 5.12/22.4 = 0.22857 ( 1 mole = 22.4 litters)
Moles of H2O will be 0.22857 * 2 = 0.457142.
Therefore answer C)
Answer:
(slow)xy2+z→xy2z (fast) c step1:step2:xy2+z2→xy2z2
Explanation:
Step1: xy2+z2→xy2z2 (slow)
Step2: xy2z2→xy2z+z (fast)
2XY 2 + Z 2 → 2XY 2 Z
Rate= k[xy2][z2]
When the two elementary steps are summed up, the result is equivalent to the stoichiometric equation. Hence, this mechanism is acceptable. The order of both elementary steps is 2, which is ‘≤3’; this also makes this mechanism acceptable. Furthermore, the rate equation aligns with the experimentally determined rate equation, and this also makes this mechanism acceptable. Therefore, since all the three rules have been observed, this mechanism is possible.
I believe its B. <span>The chemical formula for water H2O represents 2 hydrogen and 1 oxygen in the compound. </span>