Answer:
8510 mol
Explanation:you must divide both of the molar masses into each other
Answer:
SO₄²⁻(aq) + Ba²⁺(aq) ⇒ BaSO₄(s)
Explanation:
Let's consider the molecular equation that occurs when aqueous solutions of ammonium sulfate and barium nitrate are mixed.
(NH₄)₂SO₄(aq) + Ba(NO₃)₂(aq) ⇒ BaSO₄(s) + 2 NH₄NO₃(aq)
The complete ionic equation includes all the ions and insoluble species.
2 NH₄⁺(aq) + SO₄²⁻(aq) + Ba²⁺(aq) + 2 NO₃⁻(aq) ⇒ BaSO₄(s) + 2 NH₄⁺(aq) + 2 NO₃⁻(aq)
The net ionic equation includes only the ions that participate in the reaction and the insoluble species.
SO₄²⁻(aq) + Ba²⁺(aq) ⇒ BaSO₄(s)
Answer:
1. 35 mg of H₃PO₄
2. 27 mol AlF₃; 82 mol F⁻
3. 300 mL of stock solution.
Explanation:
1. Preparing a solution of known molar concentration
Data:
V = 80 mL
c = 4.5 × 10⁻³ mol·L⁻¹
Calculations:
(a) Moles of H₃PO₄
Molar concentration = moles of solute/litres of solution
c = n/V
n = Vc = 0.080L × (4.5 × 10⁻³ mol/1 L) = 3.60 × 10⁻⁴ mol
(b) Mass of H₃PO₄
moles = mass/molar mass
n = m/MM
m = n × MM = 3.60 × 10⁻⁴ mol × (98 g/1 mol) = 0.035 g = 35 mg
(c) Procedure
Dissolve 35 mg of solid H₃PO₄ in enough water to make 80 mL of solution,
2. Moles of solute.
Data:
V = 4900 mL
c = 5.6 mol·L⁻¹
Calculations:
Moles of AlF₃ = cV = 4.9 L AlF₃ × (5.6 mol AlF₃/1L AlF₃) = 27 mol AlF₃
Moles of F⁻ = 27 mol AlF₃ × (3 mol F⁻/1 mol AlF₃) = 82 mol F⁻.
3. Dilution calculation
Data:
V₁= 750 mL; c₁ = 0.80 mol·L⁻¹
V₂ = ? ; c₂ = 2.0 mol·L⁻¹
Calculation:
V₁c₁ = V₂c₂
V₂ = V₁ × c₁/c₂ = 750 mL × (0.80/2.0) = 300 mL
Procedure:
Measure out 300 mL of stock solution. Then add 500 mL of water.
