A sample of a compound was found that has 40.25g of (C) Carbon and 59.75g of (O) Oxygen. What is the empirical formula for this sample.

CO

Explanation:

From the question given above, the following data were obtained:

Mass of Carbon (C) = 40.25 g

Mass of oxygen (O) = 59.75 g

Empirical formula =?

The empirical formula of the compound can be obtained as follow:

C = 40.25 g

O = 59.75 g

Divide by their molar mass

C = 40.25 / 12 = 3.354

O = 59.75 / 16 = 3.734

Divide by the smallest

C = 3.354 / 3.354 = 1

O = 3.734 / 3.354 = 1

The empirical formula of the compound is CO

8 m/s because for him to get the same amount of energy as ray is putting out, bobby, who weighs half as much as ray, needs to run twice as fast.