an aluminum cylinder is 10.0 cm in length and has a radius of .25 cm. if the mass of a single atom of aluminum is 4.48X10^-23 calculate the number of aluminum atoms present in the cylinder. the density of aluminum is 2.70 g/cm^3
Please answer quickly.
g/cm^3 * cm^3 * atoms/g = atoms
or,
g/cm^3 * cm^3 ÷ g/atom = atoms
2.70 * (pi * .25^2 * 10) / (4.48*10^-23) = 1.18 * 10^23
An aluminum cylinder is 10.0 cm in length and has a radius of 0.25 cm.
If the mass of a single Al atom is 4.48 × 10−23g, calculate the number of
Al atoms present in the cylinder. The density of aluminum is 2.70 g/cm
To find the number of aluminum atoms present in the cylinder, we need to calculate the volume of the cylinder and then convert it to moles using the mass and molar mass of aluminum. Finally, we can determine the number of atoms using Avogadro's number.
Step 1: Calculate the volume of the cylinder using the formula V = πr^2h, where r is the radius and h is the height.
Given:
Radius (r) = 0.25 cm
Height (h) = 10.0 cm
V = π(0.25 cm)^2(10.0 cm)
V ≈ 1.9635 cm^3 (rounded to four decimal places)
Step 2: Convert the volume of the cylinder to mass using the density of aluminum.
Given:
Density (ρ) = 2.70 g/cm^3
Mass (m) = ρV
m = (2.70 g/cm^3)(1.9635 cm^3)
m ≈ 5.304 g (rounded to four decimal places)
Step 3: Convert the mass of aluminum to moles using the molar mass.
Given:
Mass of a single aluminum atom = 4.48 × 10^-23 g
Molar mass of aluminum (M) = 26.98 g/mol
moles (n) = m/M
n = (5.304 g)/(26.98 g/mol)
n ≈ 0.1964 mol (rounded to four decimal places)
Step 4: Convert moles to the number of atoms using Avogadro's number.
Avogadro's number (N_A) = 6.022 × 10^23 atoms/mol
Number of atoms (N) = n × N_A
N ≈ (0.1964 mol)(6.022 × 10^23 atoms/mol)
N ≈ 1.181 × 10^23 atoms
Therefore, the number of aluminum atoms present in the cylinder is approximately 1.181 × 10^23 atoms.
To calculate the number of aluminum atoms present in the cylinder, we need to determine the volume and mass of the cylinder and then use the given mass of a single aluminum atom to find the total number of atoms.
1. Firstly, we need to calculate the volume of the cylinder using the formula for the volume of a cylinder: V = πr^2h, where r is the radius and h is the height (length) of the cylinder.
V = π(0.25 cm)^2 (10.0 cm) ≈ 19.63 cm^3
2. Next, we can calculate the mass of the cylinder using the formula: mass = density × volume.
mass = 2.70 g/cm^3 × 19.63 cm^3 ≈ 52.99 g
3. Now, we need to convert the mass of the cylinder to the number of moles of aluminum using the molar mass of aluminum, which is approximately 26.98 g/mol.
moles = mass / molar mass
moles = 52.99 g / 26.98 g/mol ≈ 1.963 mol
4. Finally, we can calculate the number of aluminum atoms by multiplying the number of moles by Avogadro's number, which is approximately 6.022 × 10^23 atoms/mol.
number of atoms = moles × Avogadro's number
number of atoms = 1.963 mol × 6.022 × 10^23 atoms/mol ≈ 1.180 × 10^24 atoms
Therefore, there are approximately 1.180 × 10^24 aluminum atoms present in the cylinder.