In 1 mole of magnesium there are Avogadro's number of atoms are present.
Avogadro's number = 6.023 x 10²³
1 mole = 6.023 x 10²³ atoms
3.75 moles = 3.75 x 6.023 x 10²³
=2.26 x 10²⁴
So, in 3.75 moles of magnesium there are 2.26 x 10²⁴ atoms of magnesium are present.
Answer:
lead ii nitrate is the answer
Given :
Moles of Na : 1.06
Moles of C : 0.528
Moles of O : 1.59
To Find :
The empirical formula of the compound.
Solution :
Dividing moles of each atom with the smallest one i.e 0.528 .
So,
Na : 1.06/0.528 = 2.007 ≈ 2
C : 0.528/0.528 = 1
O : 1.59/0.528 = 3.011 ≈ 3
Rounding all them to nearest integer, we will get the number of each atom in the empirical formula.
So, empirical formula is
.
Hence, this is the required solution.