The Aqua Regia Test: Test-suite

The photo below [1] shows freshly prepared aqua regia to remove metal salt deposits.

Aqua regia was first discovered around the year 800 by جابر ابن حيان who mixed common salt with vitriol.

The following chemical equations explains how aqua regia dissolves gold :

$\{\begin{array}{c}\mathrm{Au}\left(s\right)+3{N{O}_3}^{-}\left(\mathrm{aq}\right)+6H^{+}\left(\mathrm{aq}\right)\to {\mathrm{Au}}^{3+}\left(\mathrm{aq}\right)+3N{O}_2\left(g\right)+3H_{2}O\left(l\right)\\ {\mathrm{Au}}^{3+}\left(\mathrm{aq}\right)+4{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)\to {\mathrm{Au}{\mathrm{Cl}}_4}^{-}\left(\mathrm{aq}\right)\end{array}$

A blue circle A red rectangle A green triangle

In the paragraph below, a gold prospector enumerates the objects he owns [2]:

À la hâte, je ramasse les quelques objets qui sont ma trace dans ce monde, ma couverture kaki, mon sac de soldat, et mes outils d’orpailleur, batée, tamis, flacon d’eau régale.

The bottle of "Aqua Regia" enables gold seekers to check whether a metal is gold, using the "Acid Test".

The generalized continuum hypothesis (GCH) is the statement that for all cardinal $\kappa$, $2^{\kappa }={\kappa }^{+}$ for all ordinal α, $2^{{\aleph }_{\alpha }}={\aleph }_{\alpha +1}$ . Gödel proved that given a model of ZF, one can build a class $L$ which is a model of ZFC and satisfies the generalized continum hypothesis. Hence the consistency of ZF implies the one of ZFC + GCH.

Animated construction of a smiley This SVG image initially represents some black, red and orange shapes. Then, this shapes a removed by rotation, translation, scale etc At the end, you get a yellow smiley with green eyes, similar to theone of Acid Test2.

A black ball $V=\frac{4}{3}\pi R^3$ A black square

Here is an example of a ruby annotation [3] :

民政局みんせいきょくガバメント・セクシヨン

The probability density function of the normal distribution is $\frac{1}{\sigma \sqrt{2\pi }}e^{-\left(\frac{1}{2}\right)\left(\frac{x-\mu }{\sigma }\right)^2}$

We consider two crashes :

• A car hits a wall at the speed $v$.
• The same car falls from a height $h$.

How must we choose the height $h$ so that the two crashes are equivalent? We can reason using the concept of energy. In the first case, the kinetic energy of the car is always $E_k=\frac{1}{2}m{v}^2$. In the second one, the kinetic energy of the car when the impact occurs is its initial potential energy $E_p=mgh$. Hence we get : $h=\frac{v^2}{2g}$. If we take $g=9.8m.s^{-2}$ and $v=30m.s^{-1}=108\mathrm{km}.h^{-1}=67\mathrm{mph}$ , the initial height must be : $h=46m=151\mathrm{ft}$.

A wall (grey square) at the left bottom corner. A red car at the bottom, going from the left to right where ithit the wall. $E_k=\frac{1}{2}m{v}^2$ A red car on the right, falling on the wall. $E_p=mgh$