Answer is: pH of solution is 5,17.
Kb(NH₃) = 1,8·10⁻⁵.
c(NH₄Cl) = 0,084 M = 0,084 mol/L.
Chemical reaction: NH₄⁺ + H₂O → NH₃ + H₃O⁺.
Ka · Kb = 10⁻¹⁴.
Ka(NH₄⁺) = 10⁻¹⁴ ÷ 1,8·10⁻⁵.
Ka(NH₄⁺) = 5,55·10⁻¹⁰.
[H₃O⁺] = [NH₃] = x.
Ka(NH₄⁺) = [H₃O⁺] · [NH₃] ÷ [NH₄⁺].
5,55·10⁻¹⁰ = x² ÷ (0,084 M - x).
Solve quadratic equation: x = [H₃O⁺] = 6,8·10⁻⁶ M.
pH = -log[H₃O⁺].
pH = -log(6,8·10⁻⁶ M) = 5,17.
From the periodic chart the only element between F and Na is Neon. Neon has one more valence electron 8 than fluorine 7, and is one energy level shy of Na.<span>
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Respuesta:
2 FeCl₃(aq) + 3 Na₂CO₃(s) ⇒ Fe₂(CO₃)₃(s) + 6 NaCl(aq)
Explicación:
Consideremos la ecuación no balanceada que ocurre cuando cloruro férrico acuoso reacciona con carbonato de sodio sólido para formar carbonato férrico sólido y cloruro de sodio acuoso. Esta es una reacción de doble desplazamiento.
FeCl₃(aq) + Na₂CO₃(s) ⇒ Fe₂(CO₃)₃(s) + NaCl(aq)
Vamos a usar el método de tanteo. Empezaremos balanceando los átomos de C, multiplicando Na₂CO₃ por 3.
FeCl₃(aq) + 3 Na₂CO₃(s) ⇒ Fe₂(CO₃)₃(s) + NaCl(aq)
Luego, balancearemos los átomos de Fe, multiplicando FeCl₃ por 2.
2 FeCl₃(aq) + 3 Na₂CO₃(s) ⇒ Fe₂(CO₃)₃(s) + NaCl(aq)
Finalmente, obtendremos la ecuación balanceada, multiplicando NaCl por 6.
2 FeCl₃(aq) + 3 Na₂CO₃(s) ⇒ Fe₂(CO₃)₃(s) + 6 NaCl(aq)
Answer:

Explanation:
10.11 g sample of
contains 22.34%
by mass
According to the law of constant composition the if one sample of
has 22.34% of
by mass then any other sample of
will have the same percentage of the amount of
.
For a sample of 7.09 g we have

The mass of sodium in the required sample is
.