How many ml of c16h34 are needed to react with 0.050 mol o2, assuming complete combustion occurs? the density of c16h34 is 0.80 g/ml.

0.56 mL

Explanation:

Let's consider the combustion of C₁₆H₃₄.

C₁₆H₃₄ + 24.5 O₂ → 16 CO₂ + 17 H₂O

The molar ratio of C₁₆H₃₄ to O₂ is 1:24.5. The moles of C₁₆H₃₄ that react with 0.050 mol of O₂ are:

0.050 mol O₂ × (1 mol C₁₆H₃₄ / 24.5 mol O₂) = 0.0020 mol C₁₆H₃₄

The molar mass of C₁₆H₃₄ is 226.44 g/mol. The mass of C₁₆H₃₄ is:

0.0020 mol × (226.44 g/mol) = 0.45 g

The density of C₁₆H₃₄ is 0.80 g/mL. The volume of C₁₆H₃₄ is:

0.45 g × (1 mL / 0.80 g) = 0.56 mL

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.