A sheet of aluminum measures 55.0 cm by 23.3 cm. If the volume is 2.90 cm^3 what is the thickness of the foil? Express your answer with the appropriate units.

0.0023 cm

Explanation:

The following data were obtained from the question:

Length (L) = 55 cm

Breadth (B) = 23.3 cm

Volume (V) = 2.9 cm³

Thickness (T) =?

Thus, we can obtain the thickness of the aluminum foil as illustrated below:

V = L × B × T

2.9 = 55 × 23.3 × T

2.9 = 1281.5 × T

Divide both side by 1281.5

T = 2.9/1281.5

T = 0.0023 cm

Therefore, the thickness of the aluminum foil is 0.0023 cm

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.