Answer:
Initially
of nitrogen dioxide were in the container .
Explanation:
Volume of the container at low pressure and at room temperature =
Number of moles in the container = 
After more addition of nitrogen gas at the same pressure and temperature.
Volume of the container after addition = 
Number of moles in the container after addition=
Applying Avogadro's law:
(at constant pressure and temperature)



Initially
of nitrogen dioxide were in the container .
The question is incomplete, complete question is:
Study this chemical reaction:

Then, write balanced half-reactions describing the oxidation and reduction that happen in this reaction.
Oxidation:
Reduction:
Answer:
Oxidation taking place in given reaction :

Reduction taking place in given reaction;
Explanation:
Redox reaction is defined as the reaction in which oxidation and reduction reaction occur side by side.
Oxidation reaction is defined as the chemical reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.
Reduction reaction is defined as the chemical reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.


In the given reaction, iron(II) ions are getting reduced and zinc metal is getting oxidized to zinc(II) ions.
Oxidation :

Reduction ;
Answer:
D
Explanation:
due yo high energy from the ocean to the earth
I would choose C bc if u look at it closely u can notice it goes up steadily and then BANG it decreases a lot
Answer:
Se =[Ar] 3d¹⁰ 4s² 4p⁴
Explanation:
The noble gas notation is used for the shortest electronic configuration of other periodic table elements.
For example:
The atomic number of Argon is 18, and its electronic configuration is,
Ar₁₈ = 1s² 2s² 2p⁶ 3s² 3p⁶
The atomic number of selenium is 34, its electronic configuration is,
Se₃₄ = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁴
By using the noble gas notation, electronic configuration of selenium can be written is shortest form.
Se =[Ar] 3d¹⁰ 4s² 4p⁴
This electronic configuration is also called abbreviated electronic configuration.