Consider the following reaction. I– 2 H2O2 (l) 2 H2O (l) + O2 (g) A solution contains 15 mL 0.1 M KI, 15 mL of DI water and 5 mL of 3% H2O2. After the decomposition of H2O2 is complete, you titrate the solution with 0.1 M AgNO3. If the catalyst, I–, is not consumed in the reaction and is completely recovered, what volume of the 0.1 M AgNO3 is required to reach the end point?

Explanation:

The given chemical reaction is:

From above equation   serves as catalyst which is not consumed by the reaction and also it is completely recovered; as a result to that , the full volume of KI will definitely react with AgNO₃.

Given that :

the volume of potassium iodide

the molarity of potassium

the volume of distilled water

The volume of 3%

Molarity of AgNO₃

Let take an integral look with the reaction between KI and AgNO₃; we have

At the end point; the moles of KI will definitely be equal to the moles of AgNO₃

So;

Thus; the volume of 0.1 M AgNO₃  needed to reach the end point is 15 mL

answer:

the answer is 6 valence electrons.

explanation:

valence is the # of electrons on the outer most shell of an atom.

oxygen has 6.

although, the valency of oxygen is 2, because it only needs to make 2 covalent bonds, or gain 2 electrons, to achieve the stable octet configuration of a noble gas.

that's it!

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