新的 latex 用户，如果有人能仔细检查我的表单我将不胜感激

Question

\documentclass[12pt]{report}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc} 
\usepackage{graphicx}
\usepackage{pdfpages}
\usepackage[margin=1in]{geometry}
\usepackage{microtype}
\usepackage{amsmath}
\usepackage{amssymb}

\usepackage{siunitx} 
\usepackage{mhchem}  
\begin{document}

\begin{enumerate}
\item {Find the actual (undiluted) concentration of vinegar mol/L.}

\[  C_{NaOH} = \SI{0.120}{mol/L}  , \ V_{NaOH} = \SI{0.00774}{L} \ , \ V_{\ce{CH3COOH}} = \SI{0.01}{L} \ , C_{\ce{CH3COOH}} = \ ? \]

\[ n_{NaOH} = c \cdot V\]
\[ n_{NaOH} = (\SI{0.120}{mol/L})(\SI{0.00774}{L})\] 
\[ n_{NaOH} = \SI{0.0009288}{mol}\]

\[ n_{CH_{3}COOH} = n_{NaOH} = \SI{0.0009288}{mol} \ (1:1 \ ratio) \]

\[ C_{\ce{CH3COOH}} = \frac{n}{V}\]
\[ C_{\ce{CH3COOH}} = \frac{\SI{0.0009288}{mol}}{\SI{0.01}{L}}\]

\[ C_{\ce{CH3COOH}} = \SI{0.00928}{mol/L}\]

$\therefore$ The actual (undiluted) concentration of vinegar is \SI{0.1}{mol/L}.
\end{enumerate}

\end{document}

整合了我第一篇文章的反馈为什么列表中的第二项要进一步缩进只是想再检查一下我的表格:)

Answer 1

这里有一个例子，你知道我在说什么。你可能想在方程式之间添加一些空格，...，但这是我想到的结果。

\documentclass[12pt]{report}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc} 
\usepackage{graphicx}
\usepackage{pdfpages}
\usepackage[margin=1in]{geometry}
\usepackage[onehalfspacing]{setspace}
\usepackage{microtype}
\usepackage{amsmath}
\usepackage{amssymb}

\usepackage{siunitx} 
\usepackage{mhchem}  

\begin{document}
    
\begin{enumerate}
    \item {Find the actual (undiluted) concentration of vinegar mol/L.}
        
        \begin{itemize}
            \item $c_{\ce{NaOH}} = \SI{0.120}{mol/L}$
            \item $V_{\ce{NaOH}} = \SI{0.00774}{L}$
            \item $V_{\ce{CH3COOH}} = \SI{0.01}{L}$
            \item $c_{\ce{CH3COOH}} = \ ?$
        \end{itemize}
        
        \begin{align*}
            n_{\ce{NaOH}} &= c \cdot V \\
            n_{\ce{NaOH}} &= \SI{0.120}{mol/L} \cdot \SI{0.00774}{L} \\
            n_{\ce{NaOH}} &= \SI{0.0009288}{mol} \\
            \frac{n_{\ce{CH3COOH}}}{n_{\ce{NaOH}}} &= \frac{1}{1} \Leftrightarrow n_{\ce{CH3COOH}} = \SI{0.0009288}{mol} \\
            c_{\ce{CH3COOH}} &= \frac{n}{V} \\
            c_{\ce{CH3COOH}} &= \frac{\SI{0.0009288}{mol}}{\SI{0.01}{L}} \\
            c_{\ce{CH3COOH}} &= \SI{0.0928}{mol/L} \\
        \end{align*} 
    
    $\therefore$ The actual (undiluted) concentration of vinegar is \SI{0.1}{mol/L}.
\end{enumerate}

\end{document}