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777dan777 [17]
3 years ago
8

En un recipiente cerrado y rígido se introdujo una mezcla gaseosa a cierta temperatura y las presiones parciales de cada gas son

: p(F2) = 2,00 atm, p(BrF) = 1,50 atm y p(BrF3) = 0,0150 atm. A la temperatura que se preparó la mezcla tiene lugar la reacción representada por:
BrF3 (g) ⇌ BrF(g) + F2(g) Kp(T) = 64,0

Elegir la afirmación correcta.

Seleccione una:
Qp > Kp, por lo tanto, las presiones parciales de BrF(g) y F2(g) aumentan hasta alcanzar el equilibrio.
Qp < Kp, por lo tanto, la presión parcial de BrF3(g) disminuye hasta alcanzar el equilibrio.
Qp = Kp, por lo tanto, las presiones parciales de BrF3(g), BrF(g) y F2(g) no cambian.
Qp < Kp, por lo tanto, las presiones parciales de BrF(g) y F2(g) disminuyen hasta alcanzar el equilibrio.
Qp > Kp, por lo tanto, la presión parcial de BrF3(g) aumenta hasta alcanzar el equilibrio.
Chemistry
1 answer:
valkas [14]3 years ago
7 0

Answer:

Qp > Kp, por lo tanto, la presión parcial de BrF₃(g) aumenta hasta alcanzar el equilibrio.

Explanation:

Paso 1: Escribir la ecuación balanceada

BrF₃ (g) ⇌ BrF(g) + F₂(g)      Kp(T) = 64,0

Paso 2: Calcular el cociente de reacción (Qp)

Qp = pBrF × pF₂ / pBrF₃

Qp = 1,50 × 2,00 / 0,0150 = 200

Paso 3: Sacar una conclusión

Dado que Qp > Kp, la reacción se desplazará hacia la izquierda para alcanzar el equilibrio, es decir, la presión parcial de BrF₃(g) aumenta hasta alcanzar el equilibrio.

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What is the mass in grams of 1.00 x 10 24 atoms of Mn?
Alex17521 [72]
<h3>Answer:</h3>

91.2 g Mn

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
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<u>Chemistry</u>

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<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 1.00 × 10²⁴ atoms Mn

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Avogadro's Numer

[PT] Molar Mass of Mn - 54.94 g/mol

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                      \displaystyle 1.00 \cdot 10^{24} \ atoms \ Mn(\frac{1 \ mol \ Mn}{6.022 \cdot 10^{23} \ atoms \ Mn})(\frac{54.94 \ g \ Mn}{1 \ mol \ Mn})
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<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

91.2321 g Mn ≈ 91.2 g Mn

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