To find the mass of the chronium-51 sample that remains after
Recall the half-life formula:
$color(blue)(|bar(ul(color(white)(a/a)A=A_0(1/2)^(t/h)color(white)(a/a)|)))$ where:
$A=$ final amount
$A_0=$ initial amount
$t=$ time elapsed
$h=$ time for half of initial amount to remain after decay (also known as half-life)
Plugging in our known values,
$A=28.0g(1/2)^((84color(white)(i)"days")/(28color(white)(i)"days"))$
$A=color(green)(|bar(ul(color(white)(a/a)color(black)(3.50g)color(white)(a/a)|)))$