Question

How many atoms are in 29.79 g of copper (Cu)?

Answer 1

2.80×10^23atoms

Step-by-step explanation:

1mol of any element contains 6.02×10^23 atoms (Avogadro's constant)

1mole of copper is 64g

64g contain 6.02×10^23 atoms

29.78g will contain X atoms

• cross multiply

29.79g × 6.02×10^23atoms= 64g×Xatom

• divide both by 64g

X= 29.79g×6.02×10^23atoms/64g

X=2.80×10^23atoms