使用 Papeeria — — 第二部分根本没有编译

Question

解决！

在我的文档中，有两个部分。文档结尾位于第二部分之后（我检查过了）。但是，第二部分根本没有显示在我编译的 PDF 上。

编辑：问题似乎出在分页上？每次我尝试将第 2 部分拉到新页面上时，它都不会编译该页面。

目前总体架构如下：

\begin{document}

\maketitle

\section{Problem 1}

bla bla bla

\section{Problem 2}


 When working with reactions, it is often necessary to work with Gibbs energies, equilibrium constants, and concentrations. The solubility reaction plutonium oxide, shown below, 
 has a $\Delta G$ of 24.0 $\frac{kJ}{mol}$. First, we will explore this value with an uncertainty of $\pm 7$, and then we an uncertainty of $\pm 1$.

 $\mathbf{PuO_{2}(OH) (s) \rightarrow {PuO_{2}(OH) (aq)}}$

 $\Delta G = -RT ln(K_{eq})$

 $K_{eq} = \frac{[{PuO_{2}(OH) (s)]}{[{PuO_{2}(OH)(aq)]}$

 Given that the uncertainty in Gibbs energy is Gaussian, there is a range of numbers that should be plugged into the equation to find the range of $K_{eq}$ values. This can be done
 in a computer by finding random numbers and having it add and subtract those from $\Delta G$ to create a Gaussian distribution. The computer can then plug those values into the function to create a curve representing $K_eq$.

 Here is that curve given a Gibbs free energy of $24.0 \pm 7 frac{kJ}{mol}$.

 \begin{figure}[h!]
      \includegraphics[width=\linewidth]{analyticalhomework1plusminus7.jpg}
 \end{figure}

 Here is that curve given a Gibbs free energy of $24.0 \pm 1 frac{kJ}{mol}$.

 \begin{figure}[h!]
      \includegraphics[width=\linewidth]{analyticalhomework1plusminus1.jpg}
 \end{figure}

 Given that the solid is not in the solution, its concentration can be safely ignored, making the calculation of the concentration fairly straightforward. 

 Here is the concentration of the aqueous plutonium oxide given a Gibbs free energy of $24.0 \pm 7 frac{kJ}{mol}$.

 \begin{figure}[h!]
      \includegraphics[width=\linewidth]{analyticalhomework1pm7conc.jpg}
 \end{figure}

 Here is the concentration given a Gibbs free energy of $24.0 \pm 1 frac{kJ}{mol}$.

 \begin{figure}[h!]
      \includegraphics[width=\linewidth]{analyticalhomework1pm1conc.jpg}
 \end{figure}

 Clearly, even a fairly small uncertainty in the Gibbs free energy makes the $K_{eq}$ and the concentration of the product fairly difficult to identify.


\end{document}