An aerosol can contains 800.0 mL of compressed gas at 10.2 atm pressure. When the gas is sprayed into a large plastic bag, the bag inflates to a volume of 4.14 L. What is the pressure of gas (atm) inside the plastic bag if the temperature did not change and all of the gas was transferred completely?

Pressure of gas in the plastic bag is 1.97 atm

Explanation:

It is possible to answer this question by using Boyle's law: Boyle's law describes that, for ideal gases, the pressure of the gas is inversely proportional to the volume it occupies. Is expressed as:

P1V1 = P2V2

Where:

P1 is the initial pressure of the gas (In the problem, 10.2 atm).

V1 is initial volume (800.0mL)

V2 is final volume (4.14L ≡ 4140mL)

And P2 is the final pressure.

Replacing:

10.2 atm×800.0mL = 4140mL×P2

P2 = 1.97 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.

i think it's letter a.