Tes Ekstensi mhchem (Rumus KIMIA di MathJax)

 <!--Komen: awas diatas baris ini ada script MathJax lho-->

Berikut ini adalah script dasar MathJax:

<script type="text/x-mathjax-config">
MathJax.Hub.Config({
  tex2jax: {inlineMath: [['$','$'], ['\\(','\\)']]}
});
</script>

<script type="text/javascript" async
  src="https://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-MML-AM_CHTML">
</script>

Berikut ini  adalah script ekstensi mhchem:

TeX: {
  extensions: ["mhchem.js"]
}

Sehingga pada script seluruhnya menjadi:

<script type="text/x-mathjax-config">
MathJax.Hub.Config({
  tex2jax: {inlineMath: [['$','$'], ['\\(','\\)']]}
});

MathJax.Hub.Config({
  TeX: {
    extensions: ["mhchem.js"]
}
});
</script>

<script type="text/javascript" async
  src="https://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-MML-AM_CHTML">
</script>

atau bisa juga ditulis seperti ini:

<script type="text/x-mathjax-config">
MathJax.Hub.Config({
  tex2jax: {
             inlineMath: [['$','$'], ['\\(','\\)']]
           },
  TeX: {
         extensions: ["mhchem.js"]
       }
});
</script>

<script type="text/javascript" async
  src="https://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-MML-AM_CHTML">
</script>

TES RUMUS KIMIA

\ce{C6H5-CHO}
$\ce{C6H5-CHO}$


\ce{SO4^2- + Ba^2+ -> BaSO4 v}
$\ce{SO4^2- + Ba^2+ -> BaSO4 v}$


\ce{2HBr + H2SO4 → Br2 + SO2 + 2H2O}
$\ce{2HBr + H2SO4 → Br2 + SO2 + 2H2O}$

\ce{Al3(aq) + 3H2O(l) ⇄ Al(OH)3(s) + 3H+(aq)}
$\ce{Al3(aq) + 3H2O(l) ⇄ Al(OH)3(s) + 3H+(aq)}$


\ce{Kc} = {\displaystyle\frac{\Big[ \ce{Al(OH)3} \Big] \Big[ \ce{H+} \Big]^3}{\Big[ \ce{Al^3+} \Big] \Big[ \ce{H2O} \Big]}}
$\ce{Kc} = {\displaystyle\frac{\Big[ \ce{Al(OH)3} \Big] \Big[ \ce{H+} \Big]^3}{\Big[ \ce{Al^3+} \Big] \Big[ \ce{H2O} \Big]}}$


\ce{Kc} = {\displaystyle{{\Big[ \ce{Al(OH)3} \Big] \Big[ \ce{H+} \Big]^3} \over {\Big[ \ce{Al^3+} \Big] \Big[ \ce{H2O} \Big]}}}
$\ce{Kc} = {\displaystyle{{\Big[ \ce{Al(OH)3} \Big] \Big[ \ce{H+} \Big]^3} \over {\Big[ \ce{Al^3+} \Big] \Big[ \ce{H2O} \Big]}}}$


\ce{Kc} = {\displaystyle{{\Big[ \ce{H+} \Big]^3} \over {\Big[ \ce{Al^3+} \Big] \Big[ \ce{H2O} \Big]^3}}}
$\ce{Kc} = {\displaystyle{{\Big[ \ce{H+} \Big]^3} \over {\Big[ \ce{Al^3+} \Big] \Big[ \ce{H2O} \Big]^3}}}$


\ce{Kc} = {\displaystyle{{\Big[ \ce{A} \Big] \Big[ \ce{B} \Big]} \over {\Big[ \ce{C} \Big] \Big[ \ce{D} \Big]}}}
$\ce{Kc} = {\displaystyle{{\Big[ \ce{A} \Big] \Big[ \ce{B} \Big]} \over {\Big[ \ce{C} \Big] \Big[ \ce{D} \Big]}}}$

\ce{Kc} = {\displaystyle{{\Big[ \ce{Al^3+} \Big] \Big[ \ce{H2O} \Big]} \over {\Big[ \ce{Al(OH)3} \Big] \Big[ \ce{H+} \Big]^3}}}
$\ce{Kc} = {\displaystyle{{\Big[ \ce{Al^3+} \Big] \Big[ \ce{H2O} \Big]} \over {\Big[ \ce{Al(OH)3} \Big] \Big[ \ce{H+} \Big]^3}}}$


\ce{Kc} = {\displaystyle{{\Big[ \ce{H+} \Big]^3} \over {\Big[ \ce{Al^3+} \Big]}}}
$\ce{Kc} = {\displaystyle{{\Big[ \ce{H+} \Big]^3} \over {\Big[ \ce{Al^3+} \Big]}}}$


\ce{Kc} = {\displaystyle\frac{\Big[ \ce{Al(OH)3} \Big]}{\Big[ \ce{H2} \Big]}}
$\ce{Kc} = {\displaystyle\frac{\Big[ \ce{Al(OH)3} \Big]}{\Big[ \ce{H2} \Big]}}$

PART 2 >>>