13 of 13
extension with allignment

Align at point:
Define for easier typsetting \newcommand{\d}[2]{#1.&\hspace{-1em}#2}
and use in table 1.23 as \d{1}{23}

$$\newcommand{\d}[2]{#1.&\hspace{-1em}#2} \begin{array}{lrl} \hline \text{Beverage} & \text{Nomnom}&\hspace{-1em}\text{-factor} \\ \hline \text{Beer} & 100.&\hspace{-1em}000 \\ \text{Wine} & 23.&\hspace{-1em}4567\\ \text{Cola stuff} & 1.&\hspace{-1em}23\\ \text{Coffee} & 66.&\hspace{-1em}6\\ \text{Tea (black)}&\d{0}{3}\\ \hline \end{array} $$


The mhchem package is not enabled on StackPrinter? Testcode:

$$\ce{2H2 + O2 -> 2H2O}$$

$$\ce{2H2 + O2 -> 2H2O}$$

Enable mhchemmanually via $\require{mhchem}$ $\require{mhchem}$ and try again: $$\ce{2H2 + O2 -> 2H2O}$$


$\mathcal{M}_{ath}\mathrm{J^{a}X}\neq\LaTeX$


Operation Broken Arrow

$$\require{cancel}\ce{\bcancel{->}\cancel{->}\xcancel{->}}$$ $$%\require{Extpfeil}\Newextarrow{\xrightharpoonup}{5,10}{0x21C0}\xrightharpoonup{\not{}}$$ $$\nrightarrow$$ $$\ce{\rlap{~~/\!/}->}$$ $$\def\nnot{\color{red}{\rlap{~~/\!/}}}$$ $$\ce{\nnot->T[water]}$$


$$\begin{multline} E^\ominus(\ce{Fe/Fe^{3+}}) - \frac13\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe}}{\ce{Fe^{3+}}} = \\ a\cdot\left(E^\ominus(\ce{Fe/Fe^{2+}}) - \frac12\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe}}{\ce{Fe^{2+}}}\right) \\+ b\cdot\left(E^\ominus(\ce{Fe^{2+}/Fe^{3+}}) - \frac11\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe^{2+}}}{\ce{Fe^{3+}}}\right) \end{multline}$$

Heureka! Now I know how to use \begin{multline}...\end{multline} (enclose in $$).


Let us add some colour by \color{\red}{...}:

$$\ce{\color{\red}{H}-\color{\green}{OH} + H2O <=>\color{\red}{H+}OH2 + \color{\green}{{}^{-}OH}} $$

  • red $\ce{\color{\red}{H}}$
  • navy $\ce{\color{\navy}{H}}$
  • green $\ce{\color{\green}{H}}$
  • orange $\ce{\color{\orange}{H}}$
  • pink $\ce{\color{\pink}{H}}$
  • purple $\ce{\color{\purple}{H}}$
  • yellow $\ce{\color{\yellow}{H}}$
  • white $\ce{\color{\white}{H}}$
  • turquoise $\ce{\color{\turquoise}{H}}$
  • but no blue $\ce{\color{\blue}{H}}$ - Why?

$$\ce {4\overset{+I}{K}_2\overset{+VI}{Cr}_2\overset{-II}{O}_7 -> 4\overset{+I}{K}_2\overset{+VI}{Cr}\overset{-II}{O}_4 + 2\overset{+III}{Cr}_2\overset{-II}{O}_3 + 3\overset{\pm0}{O}_2}$$


$\Tiny1\scriptsize2 \small3 \normalsize4 \large5 \Large6 \LARGE7 \huge8 \Huge9$


$$\scriptstyle\Tiny \begin{array}{l|cccccccccccccccccc} \text{Gr.} &1 &2 &3 &4 &5 &6 &7 &8 &9 &10 &11 &12 &13 &14 &15 &16 &17 &18\\ \text{Per.}\\\hline 1& \ce{^1H} & &&&&&&&&&&&&&&&&\ce{^2He}\\ 2& \ce{^3Li}& \ce{^4Be}& &&&&&&&&&&\ce{^5B}& \ce{^6C}& \ce{^7N}& \ce{^8O}& \ce{^9F}& \ce{^{10}Ne}\\ 3& \ce{^{11}Na}& \ce{^{12}Mg}& &&&&&&&&&&\ce{^{13}Al}& \ce{^{14}Si}& \ce{^{15}P}& \ce{^{16}S}& \ce{^{17}Cl}& \ce{^{18}Ar}\\ 4& \ce{^{19}K}& \ce{^{20}Ca}& \ce{^{21}Sc}& \ce{^{22}Ti}& \ce{^{23}V}& \ce{^{24}Cr}& \ce{^{25}Mn}& \ce{^{26}Fe}& \ce{^{27}Co}& \ce{^{28}Ni}& \ce{^{29}Cu}& \ce{^{30}Zn}& \ce{^{31}Ga}& \ce{^{32}Ge}& \ce{^{33}As}& \ce{^{34}Se}& \ce{^{35}Br}& \ce{^{36}Kr}\\ 5& \ce{^{37}Rb}& \ce{^{38}Sr}& \ce{^{39}Y}& \ce{^{40}Zr}& \ce{^{41}Nb}& \ce{^{42}Mo}& \ce{^{43}Tc}& \ce{^{44}Ru}& \ce{^{45}Rh}& \ce{^{46}Pd}& \ce{^{47}Ag}& \ce{^{48}Cd}& \ce{^{49}In}& \ce{^{50}Sn}& \ce{^{51}Sb}& \ce{^{52}Te}& \ce{^{53}I}& \ce{^{54}Xe}\\ 6&\ce{^{55}Cs}& \ce{^{56}Ba}& \text{*}& \ce{^{72}Hf}& \ce{^{73}Ta}& \ce{^{74}W}& \ce{^{75}Re}& \ce{^{76}Os}& \ce{^{77}Ir}& \ce{^{78}Pt}& \ce{^{79}Au}& \ce{^{80}Hg}& \ce{^{81}Tl}& \ce{^{82}Pb}& \ce{^{83}Bi}& \ce{^{84}Po}& \ce{^{85}At}& \ce{^{86}Rn}\\ 7& \ce{^{87}Fr}& \ce{^{88}Ra}& \text{**}& \ce{^{104}Rf}& \ce{^{105}Db}& \ce{^{106}Sg}& \ce{^{107}Bh}& \ce{^{108}Hs}& \ce{^{109}Mt}& \ce{^{110}Ds}& \ce{^{111}Rg}& \ce{^{112}Cn}& \ce{^{113}Uut}& \ce{^{114}Fl}& \ce{^{115}Uup}& \ce{^{116}Lv}& \ce{^{117}Uus}& \ce{^{118}Uuo}\\\hline \end{array}\\ \scriptstyle\Tiny \begin{array}{lc} \text{* Lanthanoide}& \ce{^{57}La}& \ce{^{58}Ce}& \ce{^{59}Pr}& \ce{^{60}Nd}& \ce{^{61}Pm}& \ce{^{62}Sm}& \ce{^{63}Eu}& \ce{^{64}Gd}& \ce{^{65}Tb}& \ce{^{66}Dy}& \ce{^{67}Ho}& \ce{^{68}Er}& \ce{^{69}Tm}& \ce{^{70}Yb}& \ce{^{71}Lu}\\ \text{** Actinoide}& \ce{^{89}Ac}& \ce{^{90}Th}& \ce{^{91}Pa}& \ce{^{92}U}& \ce{^{93}Np}& \ce{^{94}Pu}& \ce{^{95}Am}& \ce{^{96}Cm}& \ce{^{97}Bk}& \ce{^{98}Cf}& \ce{^{99}Es}& \ce{^{100}Fm}& \ce{^{101}Md}& \ce{^{102}No}& \ce{^{103}Lr} \\ \end{array}$$


Can I haz two labels in one alignment? \begin{align} A&=B\tag{1}\\ B&=C\tag{2}\\\hline A&=C\tag{1 = 2} \end{align}

Yes, we can!

Escape math in tags \tag{$\alpha$} \begin{align} A&=B\tag{$\alpha$}\\ B&=C\tag{$\beta$}\\\hline A&=C\tag{$\alpha = \beta$} \end{align}