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.
Answer:
For the most part, non-metals (excluding Nobel gases) are the most likely to form covalent bonds. Pure covalent bonds are formed between atoms with the same electronegativity, ie. they are trying to hold on to the electrons in the bond with the same strength.
Answer:
Ba(ClO₃)₂ → BaCl₂ + 3 O₂
Explanation:
When exposed to heat, barium chlorate (Ba(ClO₃)₂ breaks down into an inorganic compound (Barium chloride - BaCl₂) and a molecule (Oxygen - O₂).
