In a typical double displacement reaction, you would have a total of two products (AB + CD —> AD + BC).
P₄ + 10Cl₂ ---> 4PCl₅
stoichiometry of P₄ to PCl₅ is 1:4
number of moles of P₄ reacted - 28.0 g / 124 g/mol = 0.22 mol
Cl₂ is in excess therefore P₄ is the limiting reactant, amount of product formed depends on amount of limiting reactant present
according to molar ratio of 1:4
number of PCl₅ moles formed -0.22 mol x 4 = 0.88 mol
0.88 mol of PCl₅ is formed
Answer:
High levels of moisture are contained within an air mass.
Tell me if I'm correct, plz!
Explanation:
<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>
