Answer : The percentage reduction in intensity is 79.80 %
Explanation :
Using Beer-Lambert's law :



where,
A = absorbance of solution
C = concentration of solution = 
l = path length = 2.5 mm = 0.25 cm
= incident light
= transmitted light
= molar absorptivity coefficient = 
Now put all the given values in the above formula, we get:



If we consider
= 100
then, 
Here 'I' intensity of transmitted light = 20.198
Thus, the intensity of absorbed light
= 100 - 20.198 = 79.80
Now we have to calculate the percentage reduction in intensity.


Therefore, the percentage reduction in intensity is 79.80 %
Hello,
Here is your answer:
The proper answer to this question is that "e<span>ach sub-level electron type has a unique path where it will likely to be found".
If you need anymore help feel free to ask me!
Hope this helps!</span>
Answer:
+1
Explanation:
Na₂O₂
NOTE: the oxidation number of oxygen is always –2 except in peroxides where it is –1.
Thus, we can obtain the oxidation number of sodium (Na) in Na₂O₂ as illustrated below:
Na₂O₂ = 0 (oxidation number of ground state compound is zero)
2Na + 2O = 0
O = –1
2Na + 2(–1) = 0
2Na – 2 = 0
Collect like terms
2Na = 0 + 2
2Na = 2
Divide both side by 2
Na = 2/2
Na = +1
Thus, the oxidation number of sodium (Na) in Na₂O₂ is +1
Once the student has finished his experiment, additional steps he may take are to analyze the data gathered and draw a conclusion, then publish his findings to be seen by his peers (other students).
<h3 /><h3>What comes after the experiment?</h3>
- Once an experiment is concluded, one must go over the data gathered.
- By this, we mean to analyze the information obtained through the experiment.
- This allows the student to draw a conclusion backed by evidence.
- Once complete, it is wise to share and discuss the findings with peers.
Therefore, we can confirm that once the experiment is completed, the most common action that follows is to draw a conclusion based on the analysis of the data gathered during the experiment, then discuss the findings with peers.
To learn more about experiments visit:
brainly.com/question/9199868?referrer=searchResults
Answer:1.
Explanation: This reaction is catalyzes by pyruvate dehydrogenase. Pyruvate being the end product of glycolysis has many fates after glycolysis,one of which is to enter the TCA(Tricarboxylic acid cycle) cycle. It is first converted to actetate by the action of pyruvate dehydrogenase. This enzyme converts pyruvate to acetate releasing CO2 and NADH because this oxidative decarboxylation of pyruvate is coupled with reduction of NAD+ which can feed into the electron transport chain.