How many atoms are present in 3588.3 amu of Al?

800× 10²³ atoms

Explanation:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

18 g of water = 1 mole = 6.022 × 10²³ molecules of water

1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen

In a similar way,

27 amu of Al = 1 mole = 6.022 × 10²³ atoms of Al

For 3588.3 amu,

3588.3  / 27 = 132.9 mol

132.9 mol ×  6.022 × 10²³ atoms / 1 mol

800× 10²³ atoms

false

because it does not have the potential to be entirely used up

idk what the question is about