Na₂S(aq) + Cd(NO₃)₂(aq) = CdS(s) + 2NaNO₃(aq)
v=25.00 mL
c=0.0100 mmol/mL
M(Na₂S)=78.046 mg/mmol
n(Na₂S)=n{Cd(NO₃)₂}=cv
m(Na₂S)=M(Na₂S)n(Na₂S)=M(Na₂S)cv
m(Na₂S)=78.046*0.0100*25.00≈19.5 mg
Mass atomic of Ne=20.18 u
Therefore:
molar mass=20.18 g/1 mol
1 mole=6.022*10²³ particles (atoms or molecules)
Then: 6.022*10²³ atoms are contained in 20.18g
Now, We can solve this problem by the three rule.
6.022*10²³ atoms-------------------20.18 g
x------------------------------------------32 g
x=(6.022*10²³ atoms * 32 g)/20.18 g=9.55*10²³ atoms.
Answer: 9.55*10²³ Ne atoms are contained in 32 g of the element.
This separation technique is a 4-step procedure. First, add H₂SO₄ to the solution. Because of common ion effect, BaSO₄ will not react, only Mg(OH)₂.
Mg(OH)₂ + H₂SO₄ → MgSO₄ + 2 H₂O
The aqueous solution will now contain MgSO₄ and BaSO₄. Unlike BaSO₄, MgSO₄ is soluble in water. So, you filter out the solution. You can set aside the BaSO₄ on the filter paper. To retrieve Mg(OH)₂, add NaOH.
MgSO₄ + 2 NaOH = Mg(OH)₂ + Na₂SO₄
Na₂SO₄ is soluble in water, while Mg(OH)₂ is not. Filter this solution again. The Mg(OH)₂ is retrieved in solid form on the filter paper.
