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Answer is: 5.22·10²² atoms of Iodine.

m(CaI₂) = 12.75 g; mass of calcium iodide.

M(CaI₂) = 293.9 g/mol; molar mass of calcium iodide.

n(CaI₂) = m(CaI₂) ÷ M(CaI₂).

n(CaI₂) = 12.75 g ÷ 293.9 g/mol.

n(CaI₂) = 0.043 mol; amount of calcium iodide.

In one molecule of calcium iodide, there are two iodine atoms

n(I) = 2 · n(CaI₂).

n(I) = 0.086 mol; amount of iodine atoms.

Na = 6.022·10²³ 1/mol; Avogadro number.

N(I) = n(I) · Na.

N(I) = 0.086 mol · 6.022·10²³ 1/mol.

N(I) = 5.22·10²²; number of iodine atoms.

Answer: [tex]5.2\times 10^{22} atoms[/tex] of iodine are present in 12.75 grams of [tex]CaI_2[/tex].

Explanation:

[tex]moles=\frac{\text{given mass of the compound}}{\text{molar mass of the compound}}[/tex]

Moles of Calcium iodide :

Moles of [tex]CaI_2=\frac{12.75 g}{293.8 g/mol}=0.0433 moles[/tex]

number of molecules of [tex]CaI_2[/tex] in 0.0433 moles = [tex]N_A\times \text{number of moles}=6.022\times 10^{23}\times 0.0433=2.60\times 10^{22} molecules [/tex]

In one molecule of calcium iodide there are two iodine atoms, then number of iodine atoms in [tex]2.60\times 10^{22} molecules [/tex] of [tex]CaI_2[/tex]

[tex]2\times 2.60\times 10^{22} atoms=5.2\times 10^{22} atoms[/tex]

Hence, there are [tex]5.2\times 10^{22} atoms[/tex] of iodine are present in 12.75 grams of [tex]CaI_2[/tex].

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