Answer:
31.31× 10²³ number of Cl⁻ are present in 2.6 moles of CaCl₂ .
Explanation:
Given data:
Number of moles of CaCl₂ = 2.6 mol
Number of Cl₂ ions = ?
Solution:
CaCl₂ → Ca²⁺ + 2Cl⁻
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.
In one mole of CaCl₂ there are two moles of chloride ions present.
In 2.6 mol:
2.6×2 = 5.2 moles
1 mole Cl⁻ = 6.022 × 10²³ number of Cl⁻ ions
5.2 mol × 6.022 × 10²³ number of Cl⁻ / 1mol
31.31× 10²³ number of Cl⁻
Pure water is called distilled water or deionized water.
Answer is: D. Cl (chlorine).
The ionization energy (Ei) is the minimum amount of energy required to remove the valence electron, when element lose electrons, oxidation number of element grows (oxidation process).
Barium, potassium and arsenic are metals (easily lost valence electrons), chlorine is nonmetal (easily gain electrons).
Alkaline metals (in this example, potassium) have lowest ionizations energy and easy remove valence electrons (one electron), earth alkaline metals (in this example, barium) have higher ionization energy than alkaline metals, because they have two valence electrons.
Nonmetals (in this example chlorine) are far right in the main group and they have highest ionization energy, because they have many valence electrons.
