Final answer:
There are approximately 3.27 x 10¹¹ bromide anions in 0.500 g of MgBr₂, calculated by using the molar mass of MgBr₂ and Avogadro's number.
Step-by-step explanation:
To determine how many anions are in 0.500 g of MgBr₂, we first need to calculate the number of moles of MgBr₂. The molar mass of MgBr₂ is the sum of the atomic masses of magnesium (24.305 g/mol) and two bromine atoms (2 x 79.904 g/mol), which equals 184.113 g/mol. With this, we can now calculate the moles of MgBr₂ in 0.500 g:
0.500 g MgBr₂ × (1 mol MgBr₂ / 184.113 g MgBr₂) = 0.002715 mol MgBr₂
Because each molecule of MgBr₂ contains two bromide (Br⁻) ions, we multiply the moles of MgBr₂ by 2 to get the number of moles of Br⁻ ions:
0.002715 mol MgBr₂ × 2 Br⁻/mol MgBr₂ = 0.005430 mol Br⁻
To find the number of Br⁻ anions, we convert moles to anions using Avogadro's number (6.022 × 10¹⁺ anions/mol):
0.005430 mol Br⁻ × (6.022 × 10¹⁺ anions/mol) = 3.27 × 10¹¹ anions
Therefore, there are approximately 3.27 × 10¹¹ bromide anions in 0.500 g of MgBr₂.