A balloon of hydrogen is subjected to vacuum. The initial pressure and volume of hydrogen is 0.95 atm and 0.55 L. Calculate the final pressure if the final volume is 1.22 L?

0.43 atm

Explanation:

From the question given above, the following data were obtained:

Initial pressure (P₁) = 0.95 atm

Initial volume (V₁) = 0.55 L

Final volume (V₂) = 1.22 L

Final pressure (P₂) =?

The final pressure of the gas can be obtained by using the Boyle's law equation as follow:

P₁V₁ = P₂V₂

0.95 × 0.55 = P₂ × 1.22

0.5225 = P₂ × 1.22

Divide both side by 1.22

P₂ = 0.5225 / 1.22

P₂ = 0.43 atm

Therefore, the final pressure of the gas is 0.43 atm

answer: - 3.) 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.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)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.

you need to have a picture of the atom that way we can answer it