Find the number of atoms in a copper rod with a length of 9.75 cm and a radius of 1.15 cm. the density of copper is 8.96 g/cm3. express the number of atoms to three significant figures.

There are 3.44 × 10²⁴ atoms in that copper rod.

What is the volume of the rod?

Cross-sections of this rod are circles with radius 1.15 cm. Area = π · Radius² = π · 1.15² = 4.15476 cm², which is the same along the entire rod.

Volume of the rod = Area of cross-section × Length = 4.15476 × 9.75 = 40.5089 cm³.

What's the mass of the rod?

Mass = Volume × Density = 40.5089 × 8.96 = 362.960 g.

How many moles of atoms in the rod?

The rod is made up of copper atoms. According to a modern periodic table, the atomic mass of copper is 63.55. In other words, the mass of one mole of copper atoms will be 63.55 grams. Note the unit.

Number of moles of copper atoms

= Mass of copper atoms / Mass of copper atoms in each mole

= 362.960 / 63.55

= 5.71141 moles.

How many copper atoms in that many moles of copper atoms?

By definition, there are about 6.023 × 10²³ particles in each mole of that particle. 6.023 × 10²³ is approximately the value of the Avogadro's Number, . In other words,

Number of atoms

= Number of moles of atoms × Avogadro's Number.

= 5.71141 × 6.023 × 10²³

= 3.44 × 10²⁴ atoms. Rounded to three significant figures.

h let me think

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.