Question

Относительная молекулярная масса наибольшая у вещества с формулой а)Al²O⁴б)Fe²O³в)BaO

Answer 1

Mr(Al2O3) = 27*2 + 16*3 = 102

Mr(Fe2O3) = 56*2 + 16*3 = 160

Mr(BaO) = 137 + 16 = 153

Наибольшая у Fe2O3

Answer 2

Relative Molecular Mass of Substances

The relative molecular mass of a substance is calculated by adding up the atomic masses of all the atoms in a molecule. Let's calculate the relative molecular masses for the given substances:

a) Al₂O₄ The relative molecular mass of aluminum oxide (Al₂O₄) can be calculated as follows: - Atomic mass of aluminum (Al) = 26.98 g/mol - Atomic mass of oxygen (O) = 16.00 g/mol - Relative molecular mass of Al₂O₄ = (2 * 26.98) + (4 * 16.00) = 53.96 + 64.00 = 117.96 g/mol

b) Fe₂O₃ The relative molecular mass of iron(III) oxide (Fe₂O₃) can be calculated as follows: - Atomic mass of iron (Fe) = 55.85 g/mol - Atomic mass of oxygen (O) = 16.00 g/mol - Relative molecular mass of Fe₂O₃ = (2 * 55.85) + (3 * 16.00) = 111.70 + 48.00 = 159.70 g/mol

c) BaO The relative molecular mass of barium oxide (BaO) can be calculated as follows: - Atomic mass of barium (Ba) = 137.33 g/mol - Atomic mass of oxygen (O) = 16.00 g/mol - Relative molecular mass of BaO = 137.33 + 16.00 = 153.33 g/mol

Conclusion

The substance with the highest relative molecular mass among the given options is Fe₂O₃ with a relative molecular mass of 159.70 g/mol.