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Y2O3 is the molecular formula. Yttrium (III) oxide is also called yttria. It is a white substance and is air-stable. The usual application for this compound is a starting material for inorganic compounds and in material science. IT is insoluble in water and has a high melting point.
Answer is: <span>the mass of the excess reactant (ethane) leftover is 90.135 grams.
</span>Chemical reaction: 2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O<span>(g).
m(</span>C₂H₆) = 152 g.
n(C₂H₆) = m(C₂H₆) ÷ M(C₂H₆).
n(C₂H₆) = 152 g ÷ 30 g/mol.
n(C₂H₆) = 5.067 mol.
m(O₂) = 231 g.
n(O₂) = 231 g ÷ 32 g/mol.
n(O₂) = 7.218 mol; limiting reactant.
From chemical reaction: n(O₂) : n(C₂H₆) = 7 : 2.
n(C₂H₆) = 2 · 7.218 mol ÷ 7.
n(C₂H₆) = 2.0625mol.
Δn(C₂H₆) = 5.067 mol - 2.0625 mol.
Δn(C₂H₆) = 3.0045 mol.
Δm(C₂H₆) = 3.0045 mol · 30 g/mol = 90.135 g.
Answer:
10.10
Explanation:
Step 1: Write the basic dissociation reaction for pyridine
C₅H₅N(aq) + H₂O(l) ⇌ C₅H₅NH⁺(aq) + OH⁻(aq) Kb = 1.9 × 10⁻⁹
Step 2: Calculate [OH⁻]
For a weak base, we will use the following expression.
[OH⁻] = √(Cb × Kb) = √(9.2 × 1.9 × 10⁻⁹) = 1.3 × 10⁻⁴ M
Step 3: Calculate pOH
We will use the definition of pOH.
pOH = -log [OH⁻] = -log 1.3 × 10⁻⁴ = 3.9
Step 4: Calculate pH
We will use the following expression.
pH = 14 - pOH = 14 - 3.9 = 10.10
