What volume of aluminium has the same number of atoms as 8.0cm^3 of mercury?
figure out the mass of the mercury..
mass=volumemercury/densitymercury.
then the moles of mercury..
molesHg=massHg/molmassHg
then you want the same moles of al
molesAl=molesHg=massHg/molmassHg
then
volume al=massAl/densityAl=molesAl*molmassAl/densitAl
So if I can put all that togetheer..
volumsAl=molmassAl*massHg/molmassHg*1/densityAl
=massHg (molmassAl/molmassHg)*1/dsnisty Al
To compare the volume of aluminium and mercury in terms of the number of atoms, we need to calculate the number of atoms in 8.0 cm³ of mercury, and then determine the volume of aluminium that contains the same number of atoms.
Step 1: Calculate the number of atoms in 8.0 cm³ of mercury.
The molar volume of a substance at standard temperature and pressure (STP) is 22.4 liters (or 22.4 x 10⁻³ m³) and contains Avogadro's number (6.022 x 10²³) of atoms.
The molar mass of mercury (Hg) is approximately 200.59 g/mol.
First, we can calculate the number of moles of Hg in 8.0 cm³:
V(Hg) = 8.0 cm³ = 8.0 x 10⁻⁶ m³ (since 1 cm³ = 10⁻⁶ m³)
Moles(Hg) = Volume(Hg) / Molar Volume(Hg)
= (8.0 x 10⁻⁶ m³) / (22.4 x 10⁻³ m³/mol)
= 3.57 x 10⁻⁴ mol
Next, we can calculate the number of atoms in 8.0 cm³ of mercury using Avogadro's number:
Number of Atoms(Hg) = Moles(Hg) x Avogadro's Number
= (3.57 x 10⁻⁴ mol) x (6.022 x 10²³ atoms/mol)
= 2.15 x 10¹⁹ atoms
Step 2: Determine the volume of aluminium with the same number of atoms.
The molar mass of aluminium (Al) is approximately 26.98 g/mol.
Since the number of atoms is directly proportional to the number of moles, we can use the molar mass of aluminium to find the volume of aluminium.
Number of Atoms(Al) = Number of Atoms(Hg)
Moles(Al) = Number of Atoms(Al) / Avogadro's Number
= 2.15 x 10¹⁹ atoms / (6.022 x 10²³ atoms/mol)
= 3.57 x 10⁻⁵ mol
Finally, we can find the volume of aluminium using the molar volume at STP:
Volume(Al) = Moles(Al) x Molar Volume(Al)
= (3.57 x 10⁻⁵ mol) x (22.4 x 10⁻³ m³/mol)
= 0.799 cm³ (Convert to cm³ for consistency)
Therefore, the volume of aluminium that contains the same number of atoms as 8.0 cm³ of mercury is approximately 0.799 cm³.
To find the volume of aluminum that has the same number of atoms as 8.0 cm^3 of mercury, we need to compare the densities and atomic mass of both elements.
Step 1: Determine the density of mercury.
The density of mercury is 13.6 g/cm^3. This means that 1 cm^3 of mercury has a mass of 13.6 grams.
Step 2: Calculate the mass of 8.0 cm^3 of mercury.
Using the density information, the mass of 8.0 cm^3 of mercury can be calculated as follows:
Mass of mercury = Volume × Density
Mass of mercury = 8.0 cm^3 × 13.6 g/cm^3 = 108.8 grams
Step 3: Determine the atomic mass of aluminum.
The atomic mass of aluminum is 26.98 g/mol, which means that one mole of aluminum atoms weighs 26.98 grams.
Step 4: Calculate the number of moles of aluminum.
To find the number of moles of aluminum, we need to divide the mass of mercury (108.8 grams) by the atomic mass of aluminum (26.98 g/mol):
Number of moles of aluminum = Mass of aluminum / Atomic mass of aluminum
Number of moles of aluminum = 108.8 g / 26.98 g/mol = 4.026 moles
Step 5: Calculate the volume of aluminum.
Finally, to find the volume of aluminum with the same number of atoms as 8.0 cm^3 of mercury, we can use Avogadro's number. One mole of any substance contains 6.022 × 10^23 atoms or molecules.
Volume of aluminum = Number of moles of aluminum × (1 mole / Avogadro's number)
Volume of aluminum = 4.026 moles × (1 mol / 6.022 × 10^23) ≈ 6.69 × 10^-24 cm^3
Therefore, the volume of aluminum that has the same number of atoms as 8.0 cm^3 of mercury is approximately 6.69 × 10^-24 cm^3.