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Explanation:
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Based on the data provided, there are 25 g of calcium carbonate in 1.505 × 10^23 atoms.
<h3>What is the moles of calcium carbonate in 1.505 × 10^23 atoms of calcium carbonate?</h3>
The mole of a substance can be calculated as follows:
- Moles of substance = number of particles/6.02 × 10^23
Moles of calcium carbonate = 1.505 × 10^23/6.02 × 10^23
Moles of calcium carbonate = 0.25 moles
The mass of calcium carbonate in 0.25 moles is calculated as follows:
- mass = moles × molar mass
molar mass of a calcium carbonate = 100 g/mol
mass of calcium carbonate = 0.25 × 100 = 25 g.
Therefore, there are 25 g of calcium carbonate in 1.505 × 10^23 atoms.
Learn more about molar mass and mass at: brainly.com/question/15476873
Explanation:
The boiling point of a substance is 293°C.
The boiling point is the temperature at which the vapor pressure of a liquid becomes equal to the atmospheric pressure.
So, at the boiling point:
The vapor pressure =atmospheric pressure = 1 atm.
To get the vapor pressure of the substance at 220°C is definitely less than 1 atm.
Since, as the temperature increases, the vapor pressure of the liquid increases.