This is a difficult task because zinc is much more active than copper and could hardly be passivated. ... The sur- face immediately turns white (the color of copper(I) iodide) and the yellow-brown color of iodine quickly fades. Rinse the coin with water, brighten it with polish and cloth and begin the whole process again.N
Answer:
- Acetic acid (CH₃COOH) and hydronium ion (H₃O⁺)
Explanation:
Hello,
In this case, based on the acid-base theory which states that acids are known as H⁺ donors, if we consider the direct reaction:
It is clear that the acetic acid is the first H⁺ donor as it losses one H⁺ to turn into the acetate ion. Moreover, if we consider the inverse reaction:
It is also clear that the hydronium ion is the second H⁺ donor as it losses one H⁺ to turn into water.
Best regards.
Use the question marck Moles of CO2
The the giving = 0.624 mol O2
Find the CF faction = 1 mole= 32.00 of O2
O= 2x16.00= 32.00amu ( writte this in the cf fraction)
SET UP THE CHART
Always start with the giving
0.624 mol O2 / 1mol of CO2
___________ / _____________ = Cancel the queal ( O2)
/ 32.00c O2
/
/
Multiply the top and divide by the bottom
0.624 mol CO x 1mol CO2 = 0.624 divide by 32.00 O2 =0.0195
You should look at the giving number ( how many num u gor ever there)
Ur answer should have the same # as ur givin so
= 0.0195
= .0195 mol of CO2
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.
Answer:
cm3 = 2500.0 g / 10.5 g/cm3 = 238 cm3