When 2 grams of powdered lead (iv) oxide was added to 100 cm3 of hydrogen peroxide, water and oxygen were produced. lead (iv) oxide was not used up in the reaction. based on the information, which of the following is likely to decrease the rate of formation of the products? replacing the powdered lead oxide with its large crystals removing lead (iv) oxide from the reaction mixture using 50 cm3 of hydrogen peroxide using 1.0 gram of lead (iv) oxide

Replacing the powdered lead oxide with large crystals

Explanation:

The large crystals have less surface area exposed to the other reagents than the powdered  lead oxide. High surface area leads to a high rate of reaction thus the products are formed faster, while a low surface area leads to a lower rate of reaction since the reagent is less exposed to the other reagents.

i don't see a text, but i am gonna assume this *orbit* the curved path, usually elliptical, described by a planet, satellite, spaceship, etc., around a celestial body, as the sun.