A powder is used to remove a dissolved pollutant from water. The powder is made of cube-shaped particles.
In test P, the particles have sides of length 18 nm18\,\text{nm}18nm. A fixed volume of the powder removes 2.5 mg of pollutant in 60 s.
In test Q, the same volume of the powder is used, but the particles have sides of length 3.0 nm3.0\,\text{nm}3.0nm.
For this powder, the mass of pollutant removed in 60 s is directly proportional to the total surface area of the particles.
Calculate the mass of pollutant predicted to be removed in test Q in 60 s.
Explain why the smaller particles produce this change.
122 exam-style questions on OCR GCSE Chemistry 2.3 Properties of materials, covering 2.3.1 Carbon bonding and the variety of organic compounds, 2.3.2 Diamond, graphite, fullerenes and graphene, 2.3.3 Bond strength, changes of state and predicting states, 2.3.4 Bulk properties and types of bonding, and 2.3.5 Nanoparticles: scale, surface area, uses and risks. Each one has a worked solution and a mark scheme showing where the marks go.