Consider a sample of calcium carbonate in the form of a cube measuring 2.805 in. on each edge.
if the sample has a density of 2.70 g/cm3 , how many oxygen atoms does it contain?

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          

what is the volume occupied by 40 grams of argon gas (ar) at standard conditions? what volume, in liters, is occupied by 1.5 x 1023 

"the periodic table is arranged in order of atomic number."