The molecular weight of Mg(OH)2 : 58 g/mol
<h3>Further explanation</h3>
Given
Mg(OH)2 compound
Required
The molecular weight
Solution
Relative atomic mass (Ar) of element : the average atomic mass of its isotopes
Relative molecular weight (M) : The sum of the relative atomic mass of Ar
M AxBy = (x.Ar A + y. Ar B)
So for Mg(OH)2 :
= Ar Mg + 2 x Ar O + 2 x Ar H
= 24 g/mol + 2 x 16 g/mol + 2 x 1 g/mol
= 24 + 32 + 2
= 58 g/mol
Answer:I think it could be A or B but I would choose A.
Explanation:
Answer: -
Concentration of PbI₂ = 1.5 x 10⁻³ M
PbI₂ dissociates in water as
PbI₂ ⇄ Pb²⁺ + 2 I⁻
So PbI₂ releases two times the amount of I⁻ as it's own concentration when saturated.
Thus the molar concentration of iodide ion in a saturated PbI₂ solution = [ I⁻] =
= 1.5 x 10⁻³ x 2 M
= 3 x 10⁻³ M
PbI₂ releases the same amount of Pb²⁺ as it's own concentration when saturated.
[Pb²⁺] = 1.5 x 10⁻³ M
So solubility product for PbI₂
Ksp = [Pb²⁺] x [ I⁻]²
=1.5 x 10⁻³ x (3 x 10⁻³)²
= 4.5 x 10⁻⁹
Answer:
i think the answer is C , atom mass increases with temperature
Electrolysis and heat can separate a compound into its elements.
I hope this helped you!
