An aluminum cylinder in a physics lab had a mass of 29.7 grams. How many aluminum atoms are present in this cylinder?
___________ × 1023 Al atoms. Do NOT enter the unit and report your final answer with 3 SFs.
a mol, Avogadro's number , of Al atoms has mass of 26.98 grams
so
(29.7 / 26.98) 6.023 * 10^23
So i plugged into that calculator and got 6.630211268e23
Is that sound about right?
yes, but 3 SF you said
6.63 * 10^23
To determine the number of aluminum atoms present in the cylinder, we need to use the concept of moles and Avogadro's number.
First, we need to calculate the number of moles of aluminum in the cylinder. We can do this using the molar mass of aluminum, which is 26.98 grams/mol. The formula to calculate the number of moles is:
moles = mass / molar mass
Substituting the given values:
moles = 29.7 g / 26.98 g/mol
moles ≈ 1.1003 mol
Next, we need to convert the number of moles to atoms using Avogadro's number, which is approximately 6.022 × 10^23 atoms/mol.
number of atoms = moles × Avogadro's number
Substituting the calculated value of moles:
number of atoms = 1.1003 mol × 6.022 × 10^23 atoms/mol
number of atoms ≈ 6.625 × 10^23 atoms
Therefore, the number of aluminum atoms present in the cylinder is approximately 6.625 × 10^23 atoms.