How many atoms are in 5.7 g of helium?

The number of atoms in 5.7 grams of helium can be found by dividing the mass of the helium by its atomic weight.

The atomic weight of helium is 4.0026 g/mol, which means that one mole of helium has a mass of 4.0026 grams.

To find the number of atoms in 5.7 grams of helium, we need to convert the mass of the helium to moles and then multiply by Avogadro's number (6.022 x 10^23 atoms/mol). The conversion can be done using the following equation:

5.7 g He / 4.0026 g/mol = 1.424 moles He

Then, we multiply the number of moles by Avogadro's number to find the number of atoms:

1.424 moles He * (6.022 x 10^23 atoms/mol) = 8.53 x 10^23 atoms

Therefore, there are approximately 8.53 x 10^23 atoms in 5.7 grams of helium.