Answer is (1) - no reaction.
<em>Explanation;
</em>
Some of you may think this reaction as a single replacement reaction which gives NaBr + F₂ as products.
But, according to the reactivity of the halogens, reactivity decreases from up to bottom of the group. F is placed above Br. Hence, F is more reactive than Br. Hence, Br can't replace F.
I dont know what subject is this
Answer:
It would move either left or right
Explanation: Taking assumption that,
Fructose + ATP fructose - 6 - phosphate + ADP (The standard free energy of hydrolysis for fructose-6-phosphate is - 15.9 kJ/mol.) 3 - phosphoglycerate + ATP 1,3 - bisphosphoglycerate + ADP (The standard free energy of hydrolysis for 1,3-bisphosphoglycerate is - 4 9.3 kJ/mol.) pyruvate + ATP phosphoenolpyruvate + ADP (The standard free energy of hydrolysis for phosphoenolpyruvate -is -61.9 kJ/mol.)
This is what i got The KB expression for aniline c6h5nh2 is: For C6H5NH2 + H2O >< C6H5NH3+ <span>OH-Kb = 4.3 x (10 ^ -10) = [C6H5NH3+][OH-] / [C6H5NH2]
hope this helps:)
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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.
