MgCl2 = 1Mg + 2Cl = 1(24.3) + 2(35.45) = 95.2g/1mole
7.50moles MgCl2 x 95.2g MgCl2 = 714g MgCl2
Answer:
1.12 × 10⁻⁴ M
Explanation:
Step 1: Write the reaction for the solution of Mg(OH)₂
Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)
Step 2: Make an ICE chart
We can relate the solubility product constant (Ksp) with the solubility (S) through an ICE chart.
Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)
I 0 0
C +S +2S
E S 2S
The solubility product constant is:
Ksp = 5.61 × 10⁻¹² = [Mg²⁺] × [OH⁻]² = S × (2S)² = 4S³
S = 1.12 × 10⁻⁴ M
Hey there !
Molecular Weight: 262.86 g/mol
Volume in liters : 1,000 mL => 1,000 / 1000 => 1 L
Number of moles of solution :
number of moles = volume x molarity
number of moles = 1 x 0.3 => 0.3 moles
Therefore:
1 mole Mg₃(PO₄)₂ ------------------- 262.86 g
0.3 moles ----------------------------- mass of Mg₃(PO₄)₂ ??
mass of Mg₃(PO₄)₂ = 0.3 x 262.86 / 1
mass of Mg(PO₄)₂ => 78.858 g
Hope this helps!
Answer: 448 g of 
will be required to completely react with 784g moles of CO(g) during this reaction.
Explanation:
To calculate the moles :
The balanced chemical equation is:
According to stoichiometry :
2 moles of 
require = 1 mole of
Thus 28 moles of will require=
of
Mass of 
Thus 448g of will be required to completely react with 784g moles of CO(g) during this reaction.
carbon dioxide, that is what I found
