Question

What volume of aluminum has the same number of atoms as 11 cm3 of mercury?

Answer 1

The volume of aluminum that has the same number of atoms as 11 cm³ of mercury is approximately 55.4 cm³.

Step-by-step explanation:

The volume of aluminum that has the same number of atoms as 11 cm³ of mercury can be calculated using the mass-to-volume ratio and density of aluminum. Given that 1 cm³ of aluminum has a mass of 2.70 grams and the density of aluminum is 2.70 g/cm³, we can use the mass-to-volume ratio to find the mass of aluminum equivalent to 11 cm³ of mercury, and then convert that mass back to volume using the density.

First, calculate the mass of 11 cm³ of mercury using its density, which is 13.6 g/cm³. The mass of 11 cm³ of mercury is: 11 cm³ x 13.6 g/cm³ = 149.6 grams.

Next, use the mass-to-volume ratio for aluminum to find the equivalent volume of aluminum. The mass-to-volume ratio of aluminum is 2.70 g/cm³. So, the volume of aluminum with a mass of 149.6 grams can be calculated by dividing the mass by the mass-to-volume ratio: 149.6 g / 2.70 g/cm³ = 55.4 cm³.

Therefore, the volume of aluminum that has the same number of atoms as 11 cm³ of mercury is approximately 55.4 cm³.

Answer 2

To find the volume of aluminum that has the same number of atoms as 11 cm³ of mercury, you need to use the molar mass and Avogadro's number. First, calculate the number of moles of mercury in 11 cm³ using its density. Then, divide the mass by the molar mass of mercury. Next, find the volume of aluminum that has the same number of atoms as the calculated number of moles of mercury. Finally, divide the number of atoms by the density of aluminum.

Step-by-step explanation:

The question asks for the volume of aluminum that has the same number of atoms as 11 cm³ of mercury. To solve this problem, we need to use the molar mass and Avogadro's number. The molar mass of aluminum is 26.98 g/mol and Avogadro's number is 6.022 × 10^23 atoms/mol.

First, we calculate the number of moles of mercury in 11 cm³ using its density. The density of mercury is 13.6 g/cm³, so the mass of 11 cm³ of mercury is 13.6 g/cm³ × 11 cm³ = 149.6 g. Then, we divide the mass by the molar mass of mercury (200.59 g/mol) to get the number of moles. This gives us 149.6 g / 200.59 g/mol = 0.746 mol.

Next, we need to find the volume of aluminum that has the same number of atoms as 0.746 mol of mercury. We multiply the number of moles by Avogadro's number to get the number of atoms: 0.746 mol × 6.022 × 10^23 atoms/mol = 4.49 × 10^23 atoms. Finally, we divide the number of atoms by the density of aluminum (2.70 g/cm³) to find the volume: (4.49 × 10^23 atoms) / (2.70 g/cm³) = 1.66 × 10^23 cm³.