When 100 photons of light pass through a sample and 64 photons are detected after the passage of light, the number of photons transmitted through the sample is 64.
This is based on the methods of calculating the absorbance of light, which is depicted as the higher the amount of light transmission, the lower the amount of light absorbed.
Thus, when 64 photons of light in 100photons are detected, 64 photons are transmitted, and therefore, the number of photons absorbed is 36.
Hence, hypothetically, if 100 photons of light are transmitted, 0 photons of light will be absorbed.
Therefore, in this case, it is concluded that the correct answer is 64 photos.
Option (1) Br– is the catalyst, and the reaction follows a faster pathway with Br– than without
Explanation:
Let us consider the equation below:
Step 1:
H2O2(aq) + Br–(aq) → H2O(l) + BrO–(aq)
Step 2:
BrO–(aq) + H2O2(aq) → H2O(l) + O2(g) + Br–(aq)
From the above equation, we can see that Br– is unchanged.
This implies that Br– is the catalyst as catalyst does not take part in a chemical reaction but they create an alternate pathway to lower the activation energy in order for the reaction to proceed at a much faster rate to arrive at the products.
In this case, your facing a problem in which a carboxylic acid is produced starting by an alcohol. More specifically, cyclopentanol must react with phosphorous tribromide in order to yield bromocyclopentane which is more likely to produce a carboxylic acid, therefore, reagent A is PBr₃.
On the other hand, by means of the production of the specified product, bromocyclopentane must react with carbon dioxide and magnesium in diethyl ether in acidic media to promote the production of the cyclopentanoic acid via the grignard reaction (substitution of the bromine by the carboxyle group), therefore, reagent B is Mg in Et₂O.