How many units of AgNO₃ are present in 1.36g of this compound? A)3.84 X 10^-22 B)4.82 X 10 ²¹ C)1.36 X 10²³ D)7.52 X 10²⁵

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Quizplus · Accepted Answer

To solve this problem, we need to use stoichiometry and molar mass. First, we need to find the molar mass of AgNO₃: Ag: 1 atom x 107.9 g/mol = 107.9 g/mol N: 1 atom x 14.0 g/mol = 14.0 g/mol O: 2 atoms x 16.0 g/mol = 32.0 g/mol S: 2 atoms x 32.1 g/mol = 64.2 g/mol Total molar mass = 107.9 + 14.0 + 32.0 + 64.2 = 218.1 g/mol Next, we can set up a conversion factor to find the number of moles present in 1.36 g of AgNO₃: 1.36 g AgNO₃ x (1 mol / 218.1 g) = 0.00624 mol AgNO₃ Finally, we can use the coefficients from the balanced chemical equation for AgNO₃ to find the number of units present: 1 mol AgNO₃ = 4.82 x 10²¹ units AgNO₃ 0.00624 mol AgNO₃ x (4.82 x 10²¹ units / 1 mol) = 3.00 x 10²² units AgNO₃ Therefore, the answer is B, 4.82 x 10²¹.

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