How do you convert 2.750 mmHg to atmospheres?

$2.750 \cdot m m \cdot H g \equiv 3.618 \times {10}^{-} 3 \cdot a t m$
We know that one atmosphere's pressure will support a column of mercury $760 \cdot m m \cdot H g$ high.
And thus a pressure of, $2.750 \cdot m m \cdot H g \equiv \frac{2.750 \cdot m m \cdot H g}{760 \cdot m m \cdot H g \cdot a t {m}^{-} 1} = 3.618 \times {10}^{-} 3 \cdot a t m$