Because I (iodide) is better leaving group than Cl, so it will leave when this molecule is reacted with strong base (sodium tert-butyl oxide) giving the elimination product provided in picture<span />
When adjusted for any changes in δh and δs with temperature, the standard free energy change δg∘t at 2400 k is equal to 1.22×105j/mol, then the equilibrium constant at 2400 k is 2.21×10−3. The answer to the statement is 2.21×10−3.
Answer:- 14.9 M
Solution:- Given commercial sample of ammonia is 28% by mass. Let's say we have 100 grams of the sample. Then mass of ammonia would be 28 grams.
Density of the solution is given as 0.90 grams per mL.
From the mass and density we could calculate the volume of the solution as:

= 111 mL
Let's convert the volume from mL to L as molarity is moles of solute per liter of solution.
= 0.111 L
Now, we convert grams of ammonia to moles on dividing the grams by molar mass. Molar mass of ammonia is 17 gram per mole.

= 1.65 mole
To calculate the molarity we divide the moles of ammonia by the liters of solution:

= 14.9 M
So, the molarity of the given commercial sample of ammonia is 14.9 M.
Temperature can change a reaction rate because adding or taking away heat means energy is being added or taken away. When energy is added, the particles speed up, so there is a greater chance of the reactants colliding to form the products, which increases the reaction rate. When energy is taken away, the particles more slower, so they don't collide as easily, which slows down the reaction rate.
Therefore, the answer is D.
A is the answer
Hope it helps :)