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:
Volume of a substance can be determined by dividing mass of the substance by its density.
That can be mathematical shown as:
Density=Mass/Volume
So, Volume=Mass/Density
Here mass of the substance given as 24.60 g
Whereas density of the substance is 2.70 g/mL
So,
Volume=Mass/Density
=24.6/2.7
=9.1 mL
So volume of the substance is 9.1 mL.
I just took a test with this question and got the answer wrong for saying ethane. The correct answer is propane.
So the molarity equation is moles of solute/liters of solution. so i’m pretty sure the answer should be 0.63/0.70= .9
Answer:
2 moles of KCl will be produced
Explanation:
Given parameters:
Number of moles of K = 2 moles
Unknown:
Number of moles of KCl produced = ?
Solution:
To solve this problem;
Obtain a balanced chemical equation:
2K + Cl₂ → 2KCl ;
Since K is the limiting reactant, its amount will determine the extent of this reaction.
From the balanced equation;
2 moles of K will produce 2 moles of KCl
Given that 2 moles of K reacted, 2 moles of KCl will be produced
