At a certain concentration of n2 and o3, the initial rate of reaction is 0.770 m / s. what would the initial rate of the reaction be if the concentration of n2 were halved? be sure your answer has the correct number of significant digits.

The question is incomplete, here is the complete question:

The rate of certain reaction is given by the following rate law:

At a certain concentration of and , the initial rate of reaction is 0.770 M/s. What would the initial rate of the reaction be if the concentration of were halved? Be sure your answer has the correct number of significant digits.

Answer : The initial rate of the reaction will be 0.192 M/s

Explanation :

Rate law expression for the reaction:

As we are given that:

Initial rate = 0.770 M/s

Expression for rate law for first observation:

....(1)

Expression for rate law for second observation:

....(2)

Dividing 1 by 2, we get:

Therefore, the initial rate of the reaction will be 0.192 M/s

answer

the ions in solute supperate from eachouthere

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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