Answer : The molecule of ethane present in 64.28 L of ethane gas at STP is, molecule.
Solution :
At STP,
22.4 L volume of ethane present in 1 mole of ethane gas
64.28 L volume of ethane present in of ethane gas
And, as we know that
1 mole of ethane molecule contains molecules of ethane
2.869 moles of ethane molecule contains molecules of ethane
Therefore, the molecule of ethane present in 64.28 L of ethane gas at STP is, molecule.
Answer:The correct answer is ;
The oxidation state of nitrogen in NO changes from +2 to 0, and the oxidation state of carbon in CO changes from +2 to +4 as the reaction proceeds.
Explanation:
In an oxidation recation addition of oxygen atom takes place or loss of electrons takes place.
In an reduction reaction removal of oxygen atom takes place or gain of electrons takes place.
In the given reaction , the nitrogen atom is present in +2 oxidation state in NO molecule and present in 0 oxidation state in molecule. Hence, nitrogen is getting reduced that is reduction reaction. NO is oxidizing agent
In the given reaction , the carbon atom is present in +2 oxidation state in CO molecule and present in +4 oxidation state in molecule. Hence ,carbon is getting oxidized that is oxidation reaction. CO is a reducing agent.