At 100°C and 1 atm, if the density of liquid water is 1g cm-3 and that of water vapor is 0.0006 g cm-3, then the volume occupied by water molecules in 1 liter of steam at that temperature is:
Correct answer: C. 0.6 cm3
- A. 3.6 cm3
- B. 60 cm3
- C. 0.6 cm3
- D. 0.06 cm3
Explanation
At 100°C and 1 atm, the volume occupied by water molecules in 1 liter of steam is 0.6 cm³. This can be calculated using the following formula: Volume of water molecules in steam = (Density of liquid water / Density of water vapor) * Volume of steam Volume of water molecules in steam = (1 g cm-3 / 0.0006 g cm-3) * 1 liter The volume of water molecules in steam = 0.6 cm³ It is important to note that the density of water vapor is much lower than the density of liquid water. This is because water vapor molecules are much further apart than liquid water molecules. Therefore, the volume occupied by water molecules in 1 liter of steam is much smaller than the volume occupied by liquid water in 1 liter of water.
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Mole calculations connect mass, number of particles and Avogadro's number, while balanced equations provide the mole ratios used in stoichiometry. Questions cover limiting and excess reactants, theoretical yield and percentage yield, including identifying which reactant is consumed first and comparing the actual product with the maximum possible product.
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