When the concentrations of CO2 and H2CO3 are both horizontal lines then the rate of the forward reaction is the same as the rate of the reverse reaction.
<h3>What is rate of reaction?</h3>
The term rate of reaction refers to how fast or slow a reaction proceeds. Recall that the rate of reaction is measured from the rate of disappearance of reactants or the rate of appearance of products.
When the [CO2] and [H2CO3 ] are both horizontal lines, the rate of the forward reaction is the same as the rate of the reverse reaction.
Let us recall that the reaction is reversible hence addition of H2CO3 will increase the concentration of H2CO3, the reverse reaction would be favored.
Learn more about rate of reaction: brainly.com/question/8592296
The right answer is D - mass.
Answer:
The concentration of I at equilibrium = 3.3166×10⁻² M
Explanation:
For the equilibrium reaction,
I₂ (g) ⇄ 2I (g)
The expression for Kc for the reaction is:
![K_c=\frac {\left[I_{Equilibrium} \right]^2}{\left[I_2_{Equilibrium} \right]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%20%7B%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%7D%7B%5Cleft%5BI_2_%7BEquilibrium%7D%20%5Cright%5D%7D)
Given:
= 0.10 M
Kc = 0.011
Applying in the above formula to find the equilibrium concentration of I as:
![0.011=\frac {\left[I_{Equilibrium} \right]^2}{0.10}](https://tex.z-dn.net/?f=0.011%3D%5Cfrac%20%7B%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%7D%7B0.10%7D)
So,
![\left[I_{Equilibrium} \right]^2=0.011\times 0.10](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%3D0.011%5Ctimes%200.10)
![\left[I_{Equilibrium} \right]^2=0.0011](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%3D0.0011)
![\left[I_{Equilibrium} \right]=3.3166\times 10^{-2}\ M](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%3D3.3166%5Ctimes%2010%5E%7B-2%7D%5C%20M)
<u>Thus, The concentration of I at equilibrium = 3.3166×10⁻² M</u>
Answer:
An element
Explanation:
Elements are not able to be separated, but everything else can, whether that be physically or chemically.