Answer:
The answer to your question is: 8.82 x 10 ²⁴ atoms of oxygen
Explanation:
Data
Cube measuring : 2.805in on each edge
density = 2.7 g/cm3
# of oxygen atoms in the cube = ?
Process
Volume of the cube = (2.805)³ = 22.07 in³
Convert in³ to cm³ 1 in³ ------------------- 16.39 cm³
22.07 in³ -------------------- x
x = (22.07 x 16.39) / 1 = 361.72 cm³
Density = mass / volume
Mass = density x volume
Mass = (2.7)(361.72) = 976.66 g
Atomic mass CaCO3 = 40 + 12 + (16 x 3) = 100 g
100 g of CaCO3 -------------------- 48 g of O2
976.66 g of CaCO3 --------------- x
x = (976.66 x 48) / 100 = 468.8 g of O2
32 g of O2 ---------------------- 1 mol
468.8 g of O2 -------------------- x
x = (468.8 x 1) / 32 = 14.65 mol
1 mol ----------------------- 6.023 x 10 ²³ atoms
14.65 mol ------------------- x
x = (14.65 x 6.023 x 10 ²³) / 1 = 8.82 x 10 ²⁴ atoms of oxygen
