What is the pressure in a 20.0 l container filled with 0.896 moles of nitrogen gas at a temperature of 42 oc?

Hello!

We have the following data:

P (Pressure) =? (atm)

V (volume) = 20.0 L

n (mol number) = 0.896 mol

R (gas constant) = 0.082 atm.L / mol.K

T (temperature) = 42 °C (in SI the temperature is given in Kelvin, needs to turn into Kelvin)

TK = TºC + 273.15

TK = 42 + 273.15

TK = 315.15

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Applying the data to the Clapeyron equation, we have:

I Hope this helps, greetings ... DexteR!

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.

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