Answer:
d. [HI] > [H2]
Explanation:
The explanation at equilibrium is shown below:-
Data provided 
Initial concentration - - 
= 0.280 M
At equilibrium x x 0.280 - 2x
After solve the above equation we will get
x = 0.0282 M
Therefore at equilibrium
![[H_2] = [I_2] = x = 0.0282M\\\\](https://tex.z-dn.net/?f=%5BH_2%5D%20%3D%20%5BI_2%5D%20%3D%20x%20%3D%200.0282M%5C%5C%5C%5C)
![[HI] = 0.280 - 2x = 0.2236 M](https://tex.z-dn.net/?f=%5BHI%5D%20%3D%200.280%20-%202x%20%3D%200.2236%20M)
Hence, the correct option is d.
<span>We can solve this problem by assuming that the decay of
cyclopropane follows a 1st order rate of reaction. So that the
equation for decay follows the expression:</span>
A = Ao e^(- k t)
Where,
A = amount remaining at
time t = unknown (what to solve for) <span>
Ao = amount at time zero = 0.00560
M </span><span>
<span>k = rate constant
t = time = 1.50 hours or 5400 s </span></span>
The rate constant should
be given in the problem which I think you forgot to include. For the sake of
calculation, I will assume a rate constant which I found in other sources:
k = 5.29× 10^–4 s–1 (plug in the correct k value)
<span>Plugging in the values
in the 1st equation:</span>
A = 0.00560 M * e^(-5.29 × 10^–4 s–1 * 5400 s )
A = 3.218 <span>× 10^–4 M (simplify
as necessary)</span>
4.7e^6 is how I would write it. You simply count in from the right side where the decimal is until you reach the point just before the 4.
Well, if u had a spilled liquid in there (we'll simply go with water) and you had the freezer at a cold temperature it would change (like,icycles on trees when it's snowing)
