Question

Mg(OH)2M=%O=God help me​

Answer 1

Ответ:

M(Mg(OH)2) = 24+ (16+1)*2)  = 58 g/mol

W(O) = Ar(O)*n / M(Mg(OH)2) * 100%

W(O) = 16*2 / 58 *100% = 55.17%

Объяснение:

Answer 2

I believe you are asking for the molecular weight (M) and percentage composition of oxygen (%O) in Mg(OH)2.

To calculate the molecular weight of Mg(OH)2, you need to add up the atomic weights of each element in the compound. The atomic weights for Mg, O, and H are 24.31, 15.99, and 1.01, respectively. Mg(OH)2 has one Mg atom, two O atoms, and two H atoms.

Therefore, the molecular weight (M) of Mg(OH)2 is:

M = (1 x 24.31) + (2 x 15.99) + (2 x 1.01) = 58.33 g/mol

To calculate the percentage composition of oxygen (%O) in Mg(OH)2, you need to divide the total atomic weight of oxygen by the total molecular weight of the compound and then multiply by 100.

The atomic weight of oxygen in Mg(OH)2 is 2 x 15.99 = 31.98 g/mol.

Therefore, the percentage composition of oxygen in Mg(OH)2 is:

%O = (31.98 g/mol / 58.33 g/mol) x 100% = 54.94%

So the molecular weight of Mg(OH)2 is 58.33 g/mol, and the percentage composition of oxygen is 54.94%.