Question

Alguém pra me ajudar na 1 e na 3?

Answer 1

1)

$\frac{1mol}{1,32mol} = \frac{6.10^{23}moleculas}{x} \\ \\ x = 7,92.10^{23}moleculas$


3)

CaCO₃ = 40,08 + 12 + 3.15,99 = 62,08 + 47,97 = 110,05g/mol

$\frac{1mol}{0,1mol} = \frac{110,05g}{x} \\ \\ x = 11,005g CaCO_3$