How many cations are in 0.500 g of MgBr₂?

Question

Quizplus · Accepted Answer

First, we need to determine the molar mass of MgBr₂: Mg: 24.31 g/mol Br: 79.90 g/mol S: 32.06 g/mol Total molar mass: 136.27 g/mol Next, we can use this molar mass to convert 0.500 g to moles: 0.500 g / 136.27 g/mol = 0.00367 mol MgBr₂ Finally, we can use the stoichiometry of MgBr₂ to determine the number of cations present. Since MgBr₂ has one magnesium ion and one sulfate ion: 0.00367 mol MgBr₂ × 1 mol Mg2+ / 1 mol MgBr₂ = 0.00367 mol Mg2+ Therefore, there are 0.00367 moles (or 1.64 × 10²¹) of anions and 0.00367 moles (or 1) of cations present in 0.500 g of MgBr₂.

How Many Cations Are in 0