Question

57.7 g Ni contains how many atoms?

Answer 1

58.69 g if Ni contains 6.023×10^23atoms

1g of Ni contains 6.023×10^23/58.69 atoms

57.7 g of Ni contains (6.023×10^23/58.69)×57.7=5.921×10^23atoms

Answer 2

$5.92\times 10^{23}$ nickel atoms are there in 57.7 g nickel

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{57.7g}{58.7g/mol}=0.983moles$

1 mole of nickel $(Ni)$ contains = $6.023\times 10^{23}$ nickel atoms

0.983 moles of nickel $(Ni)$ contain = $\frac{6.023\times 10^{23}}{1}\times 0.983=5.92\times 10^{23}$ nickel atoms

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