What temperature must be maintained to ensure that a 1.00 L flask containing 0.0400 mol of oxygen will show a continuous pressure of 0.981 atm?

The temperature is 298.9K = 25.75 °C

Explanation:

Step 1: Data given

Volume of the flask = 1.00 L

Number of moles oxygen = 0.0400 moles

Pressure = 0.981 atm

Step 2: Calculate the temperature

p*V = n*R*T

⇒with p = the pressure of oxygen gas = 0.981 atm

⇒with V = the volume of the flask = 1.00 L

⇒with n = the number of moles of oxygen = 0.0400 moles

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T = the temeprature = TO BE DETERMINED

T = (p*V) / (n*R)

T = (0.981 atm * 1.00 L) / (0.0400 moles * 0.08206 L*atm/mol*K)

T = 298.9 K

The temperature is 298.9K = 25.75 °C

output?

explanation:

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.