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 same genes or slightly different versions of the same gene can be found on each chromosome in a pair. They form a line and split off bits of themselves, which they barter with one another. In sexual reproduction, crossing over is the first method that genes are shuffled to develop genetic variation.
Your answer is 2 goes to the right and twice to the left
Zn(s) + 2HCl(aq) = ZnCl₂(aq) + H₂(g)
zinc + hydrochloric acid = zinc chloride + hydrogen
Quantitative observations include numerical data. Ex: 32 degrees, 10 inches, etc.
