To solve this problem, we have to use the half-life formula, where A represents the remaining amount, P is the initial amount, t is the time that has passed, and h is the substance's half-life:
[tex]A = P( \frac{1}{2} )^{\frac{t}{h} } [/tex]
This is what it looks like with the values from this scenario plugged in:
[tex]A = 80( \frac{1}{2} )^{\frac{140}{35} } \\ A = 5[/tex]
There are 5 mL of the 80 mL Bromine sample left after 140 hours.