Answer:
Explanation:
The balanced chemical equation is Cl2 + 2KBr → 2KCl + Br2. The amount of bromine is calculated as follows:
11.0 g KBr (1mol KBr/119.002g KBr * 1mol Br2/2mol KBr * 159.808g Br2/1mol Br2= 7.39 g Br2.)
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3.08 is the pOH of a 0.175 M aqueous solution of .
0.215% is the per cent ionization of a 0.325 M aqueous solution of
pH is a measure of how acidic/basic water is.
A)
→
Kb = 4.5 x
Kb = {concentration of (NH₄⁺) x concentration of (OH⁻)} ÷ concentration of (NH₃).
concentration of (NH₄⁺) = concentration of (OH⁻) = x.
x² = Kb x concentration of (NH₃)
x² = 4.5 × 10⁻⁶ × 0.175 = 7.0 × 10⁻⁷.
x = concentration of (OH⁻) = √(7.0 × 10⁻⁷)
= 8.367 × 10⁻⁴
pOH = -log(c(OH⁻))
=- log ( 8.367 × 10⁻⁴)
= 3.08
B)
Chemical reaction: NX₃ + H₂O ⇄ NX₃H⁺ + OH⁻.
Concentration of (NX₃) = 0.325 M.
Kb = 4.5 x 10⁻⁶.
[NX₃H⁺] = [OH⁻] = x.
[NX₃] = 0.325 M - x.
Kb = [NX₃H⁺] x [OH⁻] ÷ [NX₃].
4.5 x 10⁻⁶ = x² ÷ (0.325 M - x).
x = 0.0007 M.
Per cent of ionization:
α = 0. 0007 M ÷ 0. 325 M x 100%
= 0.215%.
Hence,
3.08 is the pOH of a 0.175 M aqueous solution of .
0.215% is the per cent ionization of a 0.325 M aqueous solution of
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