Answer:
1.8 × 10⁻⁴ mol M/s
Explanation:
Step 1: Write the balanced reaction
2 Br⁻ ⇒ Br₂
Step 2: Establish the appropriate molar ratio
The molar ratio of Br⁻ to Br₂ is 2:1.
Step 3: Calculate the rate of appearance of Br₂
The rate of disappearance of Br⁻ at some moment in time was determined to be 3.5 × 10⁻⁴ M/s. The rate of appearance of Br₂ is:
3.5 × 10⁻⁴ mol Br⁻/L.s × (1 mol Br₂/2 mol Br⁻) = 1.8 × 10⁻⁴ mol Br₂/L.s
Answer:

Explanation:
First you should calculate the volume of a big sphere,so:



Then you calculate the volume of a small spehre, so:



Finally you subtract the two quantities:


Answer: The volume of the sample after the reaction takes place is 29.25 L.
Explanation:
The given reaction equation is as follows.

So, moles of product formed are calculated as follows.
Hence, the given data is as follows.
= 0.17 mol,
= 0.255 mol
= 19.5 L, 
As the temperature and pressure are constant. Hence, formula used to calculate the volume of sample after the reaction is as follows.

Substitute the values into above formula as follows.

Thus, we can conclude that the volume of the sample after the reaction takes place is 29.25 L.
Is this a multi choice question?????????????????????