Magnesium iodide = MgI₂
mass of Mg = 24.3g
mass of I = 126.9g
mass of MgI₂ = 24.3 + 2*126.9 = 278.1g = 1 mole
in 5.36x10⁻⁴ mole of MgI₂ ---------------- x g of Mg
in 1 mole of MgI₂ ------------------------------ 1 mole of Mg
x = 5.36x10⁻⁴ moles of Mg = 0.000536 moles of Mg
answer: we've 0.000536 moles of Mg (magnesium ions) in 5.36x10⁻⁴ moles of MgI₂
Answer:- 2,1
Explanations:- Water is 
and from it's formula it is clear that the ratio of H to O moles or atoms is 2:1 means two hydrogen atoms for each oxygen atom. No matter what amount of water we have, the ratio of H atoms to O atoms is always same.
Hence, the answer is 2,1.
Answer:
the enantiomeric excess of the mixture is 40%
Explanation:
The computation of the enantiomeric excess of the mixture is shown below:
As we know that
Hence, the enantiomeric excess of the mixture is 40%
