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refactor: lint documentation source markdown

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docs/academic/chemistry/problems/03-02.md
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@ -5,6 +5,7 @@
$$ $$
2 \mathrm{~N}_2 \mathrm{O}_5(g) \rightarrow 4 \mathrm{NO}_2(g)+\mathrm{O}_2(g) 2 \mathrm{~N}_2 \mathrm{O}_5(g) \rightarrow 4 \mathrm{NO}_2(g)+\mathrm{O}_2(g)
$$ $$
The decomposition of $\mathrm{N}_2 \mathrm{O}_5(g)$ is represented by the equation above. A sample of $\mathrm{N}_2 \mathrm{O}_5(g)$ is monitored as it decomposes, and the concentration of $\mathrm{N}_2 \mathrm{O}_5$ as a function of time is recorded. The results are shown in the table below. The decomposition of $\mathrm{N}_2 \mathrm{O}_5(g)$ is represented by the equation above. A sample of $\mathrm{N}_2 \mathrm{O}_5(g)$ is monitored as it decomposes, and the concentration of $\mathrm{N}_2 \mathrm{O}_5$ as a function of time is recorded. The results are shown in the table below.
| Time (s) | $\mathrm{[N_2O_5]}$ | | Time (s) | $\mathrm{[N_2O_5]}$ |
@ -21,23 +22,30 @@ Calculate the average rate of the reaction between $50.0$ and $75.0$ seconds.
$$ $$
a \mathrm{~A}+b \mathrm{~B} \rightarrow c \mathrm{C}+d \mathrm{D} a \mathrm{~A}+b \mathrm{~B} \rightarrow c \mathrm{C}+d \mathrm{D}
$$ $$
the rate of reaction is defined as the rate of reaction is defined as
$$ $$
\color{cyan}\text { rate }=-\frac{1}{a} \frac{\Delta[\mathrm{A}]}{\Delta t}=-\frac{1}{b} \frac{\Delta[\mathrm{B}]}{\Delta t}=+\frac{1}{c} \frac{\Delta[\mathrm{C}]}{\Delta t}=+\frac{1}{d} \frac{\Delta[\mathrm{D}]}{\Delta t} \color{cyan}\text { rate }=-\frac{1}{a} \frac{\Delta[\mathrm{A}]}{\Delta t}=-\frac{1}{b} \frac{\Delta[\mathrm{B}]}{\Delta t}=+\frac{1}{c} \frac{\Delta[\mathrm{C}]}{\Delta t}=+\frac{1}{d} \frac{\Delta[\mathrm{D}]}{\Delta t}
$$ $$
Notice that the rate of change in concentration of each species is divided by its coefficient from the balanced chemical equation ( $a$, $b, c$, or $d$ ). This ensures that the calculated reaction rate is the same no matter which reactant or product is monitored for changes in concentration. Notice that the rate of change in concentration of each species is divided by its coefficient from the balanced chemical equation ( $a$, $b, c$, or $d$ ). This ensures that the calculated reaction rate is the same no matter which reactant or product is monitored for changes in concentration.
In this case, the monitored species was $\mathrm{N}_2 \mathrm{O}_5$. With that in mind, let's write the reaction rate in terms of the rate of change in concentration of $\mathrm{N}_2 \mathrm{O}_5$ : In this case, the monitored species was $\mathrm{N}_2 \mathrm{O}_5$. With that in mind, let's write the reaction rate in terms of the rate of change in concentration of $\mathrm{N}_2 \mathrm{O}_5$ :
$$ $$
\text { rate }=-\frac{1}{2} \frac{\Delta\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\Delta t} \text { rate }=-\frac{1}{2} \frac{\Delta\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\Delta t}
$$ $$
Since the coefficient for $\mathrm{N}_2 \mathrm{O}_5$ in the balanced equation is 2 , we divided the rate of change in concentration of $\mathrm{N}_2 \mathrm{O}_5$ by 2 . Since the coefficient for $\mathrm{N}_2 \mathrm{O}_5$ in the balanced equation is 2 , we divided the rate of change in concentration of $\mathrm{N}_2 \mathrm{O}_5$ by 2 .
Additionally, since $\mathrm{N}_2 \mathrm{O}_5$ is being consumed in the reaction, we included a negative sign in front of the expression. Additionally, since $\mathrm{N}_2 \mathrm{O}_5$ is being consumed in the reaction, we included a negative sign in front of the expression.
Now, let's plug in the information from the table to calculate the average reaction rate between $50.0$ and $75.0$ seconds: Now, let's plug in the information from the table to calculate the average reaction rate between $50.0$ and $75.0$ seconds:
$$ $$
\begin{aligned} \begin{aligned}
\text { rate } & =-\frac{1}{2} \frac{(0.515 M-0.642 M)}{(75.0 \mathrm{~s}-50.0 \mathrm{~s})} \\ \text { rate } & =-\frac{1}{2} \frac{(0.515 M-0.642 M)}{(75.0 \mathrm{~s}-50.0 \mathrm{~s})} \\
& =2.54 \times 10^{-3} M \mathrm{~s}^{-1} & =2.54 \times 10^{-3} M \mathrm{~s}^{-1}
\end{aligned} \end{aligned}
$$ $$
So, the average rate of the reaction between $50.0$ and $75.0$ seconds is $\bold{2.54 \times 10^{-3} \mathrm{M} \mathrm{s}^{-1}}$.
So, the average rate of the reaction between $50.0$ and $75.0$ seconds is $\bold{2.54 \times 10^{-3} \mathrm{M} \mathrm{s}^{-1}}$.