Iodide ion is oxidized by acidified dichromate ions as shown in this equation.

$\begin{array}{l}{\mathrm{Cr}}_2{{\mathrm{O}}_7}^{2-}\left(aq\right)+9{\mathrm{I}}^{-}\left(aq\right)+14{\mathrm{H}}^{+}\left(aq\right)⟶2{\mathrm{Cr}}^{3+}\left(aq\right)+3{{\mathrm{I}}_3}^{-}\left(aq\right)+7{\mathrm{H}}_2\mathrm{O}\left(l\right)\\ \end{array}$

These data were obtained when the reaction was studied at a constant pH.

Exp.	$\left[{\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}\right],\mathrm{M}$	$\left[{\mathrm{I}}^{-}\right],\mathrm{M}$	Rate, $\mathrm{M}.{\mathrm{s}}^{-1}$
1	0.0040	0.010	0.00050
2	0.0080	0.010	0.0010
3	0.0120	0.020	0.0060

What is the order of the reaction with respect to ${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}\left(\mathrm{aq}\right)$ and ${\mathrm{I}}^{-}\left(\mathrm{aq}\right)$?