Answer:
All answers attached in the pictures above.
Explanation:
first of all open the menu
<span>Answer is: pH of solution of sodium cyanide is 11.3.
Chemical reaction 1: NaCN(aq) → CN</span>⁻(aq)
+ Na⁺<span>(aq).
Chemical reaction 2: CN</span>⁻ +
H₂O(l) ⇄ HCN(aq) + OH⁻<span>(aq).
c(NaCN) = c(CN</span>⁻<span>)
= 0.021 M.
Ka(HCN) = 4.9·10</span>⁻¹⁰<span>.
Kb(CN</span>⁻) = 10⁻¹⁴ ÷
4.9·10⁻¹⁰ = 2.04·10⁻⁵<span>.
Kb = [HCN] · [OH</span>⁻]
/ [CN⁻<span>].
[HCN] · [OH</span>⁻<span>] =
x.
[CN</span>⁻<span>] = 0.021 M - x..
2.04·10</span>⁻⁵<span> = x² / (0.021 M
- x).
Solve quadratic equation: x = [OH</span>⁻<span>] = 0.00198 M.
pOH = -log(0.00198 M) = 2.70.
pH = 14 - 2.70 = 11.3.</span>
Water. All organisms that depend on oxygen need water to live.
Answer:
Initial volume of the container (V1) = 1.27 L (Approx)
Explanation:
Given:
Number of mol (n1) = 5.67 x 10⁻²
Number of mol (n2) = (5.67 +2.95) x 10⁻² = 8.62 x 10⁻²
New volume (V2) = 1.93 L
Find:
Initial volume of the container (V1)
Computation:
Using Avogadro's law
V1 / n1 = V2 / n2
V1 / 5.67 x 10⁻² = 1.93 / 8.62 x 10⁻²
V1 = 10.9431 / 8.62
Initial volume of the container (V1) = 1.2695
Initial volume of the container (V1) = 1.27 L (Approx)
