How many grams of sulphur contain 3.0×1021 atoms​

There are 0.16 g of sulphur in 3.0 × 10²¹ atoms.

Explanation:

One mole of a substance contains 6.02 × 10²³ particles

One mole of sulphur atom contains 6.02 × 10²³ atoms

Number of moles of sulphur present in 3.0 × 10²¹ atoms = 3.0 ×10²¹/6.02 10²³

Number of moles of sulphur atom = 0.00498 moles of sulphur atom

The mole of a substance is related to its mass by a quantity known as molar mass which is defined as the mass of one mole of that substance.

Molar mass of sulphur atom = 32 grams per mole (g/mol)

Mass of 0.00498 moles of sulphur atom = 0.00498 moles × 32 g/mol

Mass of sulphur = 0.16 g

Therefore, there are 0.16 g of sulphur in 3.0 × 10²¹ atoms.

> b takes place in the gas phase in an isothermal tubular reactor. the feed is one mole of a per one mole of c, an inert. the entering temperature and pressure are 427°c and 10 atm, respectively. the gas constant r = 0.08206 atm · l/mol · k.