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>
<h3><em><u>ᎪꪀsωꫀᏒ</u></em></h3>
even no = 3/6 = 1/2
no. less than 5 = 4/6 = 2/3
The least net force applied : Car 3(12 N)
<h3>Further explanation </h3>
Newton's 2nd law explains that the acceleration produced by the resultant force on an object is proportional and in line with the resultant force and inversely proportional to the mass of the object
∑F = m. a
Car 1 ⇒m=0.5 kg, a=36 m/s²

Car 2⇒m=0.8 kg, a=50 m/s²

Car 3⇒m=0.6, a=20 m/s²

Car 4⇒m=1, a=19~m/s²
