Loaded dice with an evenly spread total

Two six-sided dice, faces numbered 1 to 6, may each be loaded in any way you like (independently). Can they be weighted so that every total from 2 to 12 is equally likely?

No. Uniform totals force $p_1q_1=p_6q_6=\tfrac1{11}$. But a total of 7 includes the outcomes $(1,6)$ and $(6,1)$, so $\tfrac1{11}\ge p_1q_6+p_6q_1=\tfrac1{11}\left(\tfrac{p_1}{p_6}+\tfrac{p_6}{p_1}\right)$, requiring $x+\tfrac1x\le1$ for some $x>0$, which is impossible since $x+\tfrac1x\ge2$.