Em 150 gramas de alumínio, quantos átomos deste elemento estão presentes? dados: mm(ai) = 27 gmol-¹ número de avogadro = 6,02 x 10²³

The question in English:
In 150 grams of aluminum, how many atoms of this element are present? 

Data: MM (AI) = 27 g mol-1 

Number of avogadro = 6.02 x 10²³



Moles (mol) = mass (g) / molar mass (g/mol)

Mass of Al is given as 150 g.
Hence, moles of Al = 150 g / 27 g/mol  = 5.56 mol

According to the Avogadro number, 1 mole of substance has 6.02 x 10²³ of particles.

Hence, 1 mole of Al = 6.02 x 10²³ of Al atoms
Then, atoms in 5.56 mol of Al = 6.02 x 10²³ mol⁻¹ x 5.56 mol
                                                = 3.34712 x 10²⁴

Hence, 150 g of  Al have 3.34712 x 10²⁴ of atoms.

Its a hope this

Answerit curves up and levels out.explanationwe assume that the container contains absolutely no ammonia to begin with as it states that "nitrogen gas and hydrogen gas" react to "produce ammonia."at dynamic equilibrium, the product and reactant concentrations are equal.this means that, starting from a concentration of zero, [nh₃] must curve "up" from zero and then level out to a constant concentration.