The study strategy Lauren is using in spreading her study sessions over a period of time is pacing, which helps the student develop a schedule focused on their own study pace.
<h3>Pacing Study Sessions</h3>
This study strategy of distributing the study into short sessions rather than studying the entire content through one long session is more effective in retaining content and learning.
What happens is that Lauren is using mass repetition processing, which can be compared to a longitudinal wave in physics, with spaces in between, concentrating the initial review close to the proof to ensure retention and avoid forgetting.
Through pacing, Lauren achieves greater motivation to carry out her studies in a concentrated and focused way, helping her to retain and preserve knowledge.
The correct answer is:
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Answer:

Explanation:
Hello,
In this case, by knowing the given reference reactions, one could rearrange them as follows:


Subsequently, to obtain the main reaction, we add the aforementioned reference rearranged reactions as shown below (just as reference):

Consequently, the equilibrium constant is computed as:
![Kp=\frac{[N_2][O_2]}{[NO]^2} * \frac{[NO_2]^2}{[N_2][O_2]^2} =Kp_2*Kp_3=4.35x10^{18}*7.056x10^{-13}=3.07x10^6](https://tex.z-dn.net/?f=Kp%3D%5Cfrac%7B%5BN_2%5D%5BO_2%5D%7D%7B%5BNO%5D%5E2%7D%20%2A%20%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BN_2%5D%5BO_2%5D%5E2%7D%20%3DKp_2%2AKp_3%3D4.35x10%5E%7B18%7D%2A7.056x10%5E%7B-13%7D%3D3.07x10%5E6)
Best regards.
The given equation from the problem above is already balance,
N2O5 ---> 2NO2 + 0.5O2
Since, in every mole of N2O5 consumed, 2 moles of NO2 are formed, we can answer the problem by multiplying the given rate, 7.81 mol/L.s with the ratio.
(7.81 mol/L.s) x (2 moles NO2 formed/ 1 mole of N2O5 consumed)
= 15.62 mol/L.s
The answer is the rate of formation of NO2 is approximately 15.62 mol/L.s.