andrea1375
contestada

If 1 mole of aluminum contains 6.02 ×10^23 atoms aluminum. how many atoms are contained in 0.9g of aluminum (Al=27)​

Respuesta :

Sueraiuka

Answer:

Explanation:

Hi there!

We have,

1 mole of Al = 6.02*[tex]10^{23}[/tex] atoms

1 mole of Al = 27 g

Now, from the above relation;

27 g = 6.02*[tex]10^{23}[/tex] atoms

1 g = [tex]\frac{6.02*10^{23} }{27}[/tex] atoms

or, 0.9 g of Al =  [tex]\frac{6.02*10^{23} }{27}[/tex]*0.9 atoms

                       = 2.0066*[tex]10^{22}[/tex] atoms

Therefore,  0.9 g of Al contains 2.0066*[tex]10^{22}[/tex] atoms.

Hope it helps!