22 g de um gás estão contidos em um recipiente de volume igual a 17,5 L, a uma temperatura de 77ºC e

pressão de 623 mmHg. Este gás deve ser: Dados: H = 1, O = 16, N = 14, S = 32, C = 12

a)NO

b)H²S

c)SO²

d)CO²

e)NH³

Option D.

Explanation:

Let's apply the Ideal Gases law to solve the problem.

P . V = n . R . T

First of all we convert the temperature value from °C to K

77°C + 273 = 350K

and the pressure from mmHg to atm

623 mmHg .  1 atm/760 mmHg = 0.82 atm

We replace data: 0.82 atm . 17.5L = n . 0.082 L.atm/mol.K . 350K

(0.82 atm . 17.5L) / (0.082 L.atm/mol.K . 350K) = n

0.50 moles = n

These are the moles that corresponds to 22 g of the gas, so the molar mass will be → 22g / 0.50 mol =  44 g/mol

That molar mass corresponds to CO₂  →  12 g (C) + 16 g (O) . 2 = 44 g/mol

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.