Answer:
hola soy jess, tu respuesta esta aqui
¿cuantos moles de CO2 se requiere para reaccionar 2 moles de Ba(OH)2
2 mol Ba(OH)₂ × \frac{1molCO_{2} }{1molBa (OH)_{2}}
1molBa(OH)
2
1molCO
2
= 2 moles CO₂
Explanation:
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Answer:
The equilibrium shifts to produce more reactants.
Explanation:
According to the Le- Chatelier principle,
At equilibrium state when stress is applied to the system, the system will behave in such a way to nullify the stress.
The equilibrium can be disturb,
By changing the concentration
By changing the volume
By changing the pressure
By changing the temperature
Consider the following chemical reaction.
Chemical reaction:
2SO₂ + O₂ ⇄ 2SO₃
In this reaction the equilibrium is disturb by increasing the concentration of Product.
When the concentration of product is increased the system will proceed in backward direction in order to regain the equilibrium. Because when product concentration is high it means reaction is not on equilibrium state. As the concentration of SO₃ increased the reaction proceed in backward direction to regain the equilibrium state and more reactant is formed.
Answer:
salt bridge balances the charge when electrons move from one half cell to another half cell.
Explanation:
Explanation: A salt bridge balances the charge when electrons move from one half cell to another half cell. During this process the salt bridge uses its electrolyte solution which further helps in balancing charges in both the half cells. ... Therefore, for each electrochemical cell a new salt bridge is used.
Answer:
Rate of formation of SO₃ ![[\frac{d[SO_{3}] }{dt}]](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bd%5BSO_%7B3%7D%5D%20%7D%7Bdt%7D%5D)
= 7.28 x 10⁻³ M/s
Explanation:
According to equation 2 SO₂(g) + O₂(g) → 2 SO₃(g)
Rate of disappearance of reactants = rate of appearance of products
⇒ ![-\frac{1}{2} \frac{d[SO_{2} ]}{dt} = -\frac{d[O_{2} ]}{dt}=\frac{1}{2} \frac{d[SO_{3} ]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D%20%5Cfrac%7Bd%5BSO_%7B2%7D%20%5D%7D%7Bdt%7D%20%3D%20-%5Cfrac%7Bd%5BO_%7B2%7D%20%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B2%7D%20%5Cfrac%7Bd%5BSO_%7B3%7D%20%5D%7D%7Bdt%7D)
-----------------------------(1)
Given that the rate of disappearance of oxygen = ![-\frac{d[O_{2} ]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BO_%7B2%7D%20%5D%7D%7Bdt%7D)
= 3.64 x 10⁻³ M/s
So the rate of formation of SO₃ = ?
from equation (1) we can write
![\frac{d[SO_{3}] }{dt} = 2 [-\frac{d[O_{2}] }{dt} ]](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BSO_%7B3%7D%5D%20%7D%7Bdt%7D%20%3D%202%20%5B-%5Cfrac%7Bd%5BO_%7B2%7D%5D%20%7D%7Bdt%7D%20%5D)
⇒ ![\frac{d[SO_{3}] }{dt}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BSO_%7B3%7D%5D%20%7D%7Bdt%7D)
= 2 x 3.64 x 10⁻³ M/s
⇒ = 7.28 x 10⁻³ M/s
∴ So the rate of formation of SO₃ = 7.28 x 10⁻³ M/s
