When 2 grams of powdered lead (IV) oxide was added to 100 cm3 of hydrogen peroxide, water and oxygen were produced. Lead (IV) ox
ide 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 its large crystals
Explanation:
Powdered lead (IV) oxide has larger surface area than its crystal form. Surface area increases the rate of the reaction. Larger the surface larger is the effective collision thereby faster reaction completion.
Using large of crystals of lead (IV) oxide is likely to decrease the rate of formation of the products.
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.
The two notations that represent isotopes of the same element is the one that represented in option 1 The lower number is the number of protons while the upper number is the atomic weight
