No porque plantas estan vivas pero no se mueven solo.
B. Gain information about the current state of the environment and ways to improve it.
Answer:
![K_2=\frac{[NOBr]^4_{eq}}{[NO]^4_{eq}[Br]^2_{eq}}](https://tex.z-dn.net/?f=K_2%3D%5Cfrac%7B%5BNOBr%5D%5E4_%7Beq%7D%7D%7B%5BNO%5D%5E4_%7Beq%7D%5BBr%5D%5E2_%7Beq%7D%7D)
Explanation:
Hello,
In this case, for the equilibrium condition, the equilibrium constant is defined via the law of mass action, which states that the division between the concentrations of the products over the concentration of the reactants at equilibrium equals the equilibrium constant, for the given reaction:

The suitable equilibrium constant turns out:
![K_2=\frac{[NOBr]^4_{eq}}{[NO]^4_{eq}[Br]^2_{eq}}](https://tex.z-dn.net/?f=K_2%3D%5Cfrac%7B%5BNOBr%5D%5E4_%7Beq%7D%7D%7B%5BNO%5D%5E4_%7Beq%7D%5BBr%5D%5E2_%7Beq%7D%7D)
Or in terms of the initial equilibrium constant:

Since the second reaction is a doubled version of the first one.
Best regards.
<h3>
Answer:</h3>
5.6 Liters
<h3>
Explanation:</h3>
- N.T.P. refers to the standard temperature and pressure (S.T.P).
We need to know that;
- One mole of a gas occupies a volume of 22.4 liters at N.T.P.
In this case;
We have 11 g of CO₂
But, 1 mole of CO₂ occupies 22.4 l at N.T.P.
1 mole of CO₂ = 44 g
Therefore;
44 g of CO₂ = 22.4 liters
What about 11 g ?
= (11 g × 22.4 l)÷ 44 g
= 5.6 l
Therefore, 11 g of CO₂ will occupy a volume of 5.6 liters at N.T.P.