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3 years ago

Consider a sample of calcium carbonate in the form of a cube measuring 2.805 in. on each edge.

Chemistry

1 answer:

Naya [18.7K]3 years ago

3 0

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

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soldier1979 [14.2K]

Answer:

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Explanation:

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Answer:

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Explanation:

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