A gas has a volume of 1000.0 mL at a temperature of 20.OK and a pressure of 1.0 atm. What will be the new volume when the temperature is changed
to 40.0K and the pressure is changed to
0.5 atm?

4000mL

Explanation:

Using the combined gas law equation;

P1V1/T1 = P2V2/T2

Where;

P1 = initial pressure (atm)

P2 = final pressure (atm)

V1 = initial volume (mL)

V2 = final volume (mL)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the information given in this question:

V1 = 1000mL

T1 = 20K

P1 = 1.0atm

V2 = ?

P2 = 0.5atm

T2 = 40K

Using P1V1/T1 = P2V2/T2

1 × 1000/20 = 0.5 × V2/40

1000/20 = 0.5V2/40

50 = 0.5V2/40

50 × 40 = 0.5V2

2000 = 0.5V2

V2 = 2000/0.5

V2 = 4000mL

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.