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To find the number of molecules in 347 grams of CaCl₂, first calculate the molar mass of CaCl₂, which is approximately 110.98 g/mol (40.08 g/mol for Ca and 35.45 g/mol for each Cl atom). Then, divide the mass by the molar mass: 347 g ÷ 110.98 g/mol ≈ 3.13 moles. Finally, multiply the number of moles by Avogadro's number (approximately 6.022 x 10²³ molecules/mol): 3.13 moles × 6.022 x 10²³ molecules/mol ≈ 1.89 x 10²⁴ molecules.

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How many total ions are present in 347g of CaCl2?

To find the total ions in 347 g of CaCl₂, first calculate the number of moles of CaCl₂ using its molar mass, which is approximately 110.98 g/mol. Dividing 347 g by the molar mass gives about 3.13 moles of CaCl₂. Each formula unit of CaCl₂ dissociates into one calcium ion (Ca²⁺) and two chloride ions (Cl⁻), totaling three ions per formula unit. Therefore, the total number of ions is 3.13 moles × 3 ions/mole = approximately 9.39 moles of ions, or about 5.65 × 10²⁴ ions.


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