Two samples of the same material are cut into cubes. The cubes in one sample have sides of 4 mm, and the cubes in the other sample have sides of 10 mm. Both samples are placed in a substance that reacts with the material.
Use the equations:
surface area of a cube = 6×(side length)26 \times (\text{side length})^{2}6×(side length)2
volume of a cube = (side length)3(\text{side length})^{3}(side length)3
surface area to volume ratio = surface area ÷ \div\,÷ volume
A technician claims that the 10 mm cubes expose a greater fraction of their material to the reacting substance.
Calculate the surface area to volume ratio of a 4 mm cube and of a 10 mm cube. Give your answers in mm−1\text{mm}^{-1}mm−1.
Explain whether the technician's claim is correct.
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.