Answer:
9 × 10⁻³ mol·L⁻¹s⁻¹
Explanation:
Data:
k = 1 × 10⁻³ L·mol⁻¹s⁻¹
[A] = 3 mol·L⁻¹
Calculation:
rate = k[A]² = 1 × 10⁻³ L·mol⁻¹s⁻¹ × (3 mol·L⁻¹)² = 9 × 10⁻³ mol·L⁻¹s⁻¹
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 />
Od. 0-0 does not pair of the Molecules bid it’s polar bond due to certain pairs of molecules not being present within its presentations
Answer:
2
Explanation:
Tried out 1, but couldn't get whole numbers on the right side then, so went up to 2, worked
all four numbrrs: 2, 7, 4, 6
The 7 for oxygen got adjusted in the last step of the thinking, because it's the simplest to adjust.
since there are different prime numbers (or rather numbers that don't share prime factors) in the set, the numbers can't be revived by the same number and still give whole numbers as results (they can't be smaller)
