Question #310036

The number of atoms of chlorine present in 5.85g of Nacl is ?( Na =23 Cl=35.5)(Avogadro's number=6.02*10


Expert's answer

Answer

The number of atoms of chloride present in 5.85g of NaCl is

5.85∗6.02∗1023=35.22∗1023atoms5.85 *6.02 * 10 ^{23} = 35.22*10^{23}atoms



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Canada #340153 on Dec 2023