A helium atom weighs 6.65 X 10 to the negative 24 grams. How many helium atoms are contained in a 1.0 g sample?
to get the number of He atoms, we dive the given mass to the mass of each He:
1.0 / (6.65 x 10^(-24)) = 1.50376 x 10^23 He atoms
hope this helps~ :)
To find the number of helium atoms contained in a 1.0 gram sample, we need to use the atomic mass of helium and Avogadro's number.
Step 1: Determine the number of moles of helium in the sample using its atomic mass.
The atomic mass of helium (He) is approximately 4 grams/mol.
Number of moles = mass of sample (in grams) / atomic mass
Number of moles = 1.0 g / 4 g/mol
Number of moles = 0.25 mol
Step 2: Use Avogadro's number to calculate the number of helium atoms.
Avogadro's number is approximately 6.022 x 10^23 atoms/mol.
Number of atoms = number of moles x Avogadro's number
Number of atoms = 0.25 mol x 6.022 x 10^23 atoms/mol
Calculating the result:
Number of atoms = 0.25 x 6.022 x 10^23
Number of atoms = 1.5055 x 10^23
Therefore, there are approximately 1.5055 x 10^23 helium atoms contained in a 1.0 gram sample.
To determine the number of helium atoms in a given mass, we need to use the concept of molar mass and Avogadro's constant.
Molar mass is the mass of one mole of a substance, expressed in grams. For helium, the molar mass is approximately 4 grams per mole.
Avogadro's constant is the number of particles (atoms, molecules, ions) in one mole of a substance, which is approximately 6.022 × 10^23 particles per mole.
Now, let's use these values to find the number of helium atoms in a 1.0 g sample:
1. Determine the number of moles of helium atoms in the 1.0 g sample.
Number of moles = given mass / molar mass
Number of moles = 1.0 g / 4 g/mol
Number of moles = 0.25 mol
2. Use Avogadro's constant to calculate the number of helium atoms.
Number of atoms = number of moles * Avogadro's constant
Number of atoms = 0.25 mol * 6.022 × 10^23 atoms/mol
Number of atoms = 1.5055 × 10^23 atoms
Therefore, there are approximately 1.5055 × 10^23 helium atoms in a 1.0 g sample.