To find the mass of rhodium that contains the same number of atoms as 24 g of zirconium, we need to compare their molar masses.
First, we need to determine the number of zirconium atoms in 24 g of zirconium. We can use Avogadro's number, which states that 1 mole of any substance contains 6.022 × 10^23 atoms.
The molar mass of zirconium (Zr) is 91.22 g/mol. This means that 1 mole of zirconium weighs 91.22 grams.
To calculate the number of moles of zirconium in 24 g, we divide the mass by the molar mass:
24 g Zr / 91.22 g/mol = 0.263 moles Zr
Now, we need to find the molar mass of rhodium (Rh). The molar mass of rhodium is 102.91 g/mol.
To find the mass of rhodium equivalent to 0.263 moles, we multiply the number of moles by the molar mass:
0.263 moles Rh * 102.91 g/mol = 27.04 g Rh
Therefore, the mass of rhodium that contains the same number of atoms as there are zirconium atoms in 24 g of zirconium is 27.04 g.
What mass of rhodium contains as many
atoms as there are zirconium atoms in 24 g of
zirconium?
Answer in units of g.
3 answers
How many formula units are in 5.5 g of
alumina (Al2O3)?
Answer in units of formula units.
alumina (Al2O3)?
Answer in units of formula units.
To determine the number of formula units in 5.5 g of alumina (Al2O3), we need to use the molar mass of the compound.
The molar mass of Al2O3 can be calculated by summing the atomic masses of aluminum (Al) and oxygen (O):
Atomic mass Al = 26.98 g/mol
Atomic mass O = 16.00 g/mol
Molar mass Al2O3 = [(2 x 26.98) + (3 x 16.00)] g/mol
Molar mass Al2O3 = 101.96 g/mol
Now, we can use the molar mass to convert grams of Al2O3 to moles:
5.5 g Al2O3 / 101.96 g/mol = 0.0539 moles Al2O3
According to Avogadro's number, 1 mole of any substance contains 6.022 × 10^23 formula units.
To find the number of formula units in 0.0539 moles, we multiply by Avogadro's number:
0.0539 moles Al2O3 x (6.022 × 10^23 formula units/mole) = 3.24 × 10^22 formula units
Therefore, there are approximately 3.24 × 10^22 formula units in 5.5 g of alumina (Al2O3).
The molar mass of Al2O3 can be calculated by summing the atomic masses of aluminum (Al) and oxygen (O):
Atomic mass Al = 26.98 g/mol
Atomic mass O = 16.00 g/mol
Molar mass Al2O3 = [(2 x 26.98) + (3 x 16.00)] g/mol
Molar mass Al2O3 = 101.96 g/mol
Now, we can use the molar mass to convert grams of Al2O3 to moles:
5.5 g Al2O3 / 101.96 g/mol = 0.0539 moles Al2O3
According to Avogadro's number, 1 mole of any substance contains 6.022 × 10^23 formula units.
To find the number of formula units in 0.0539 moles, we multiply by Avogadro's number:
0.0539 moles Al2O3 x (6.022 × 10^23 formula units/mole) = 3.24 × 10^22 formula units
Therefore, there are approximately 3.24 × 10^22 formula units in 5.5 g of alumina (Al2O3).