How much will be left after 105 days?

`105/35 = 3`

in `105` days, `3` half-lives will have passed.

`1` half-life is the time taken for a radioactive substance to halve in mass.

after `3` half-lives, the mass will have halved `3` times.

after `1` half-life, `(1/2)^1` of the original substance is left.

after `3` half-lives, `(1/2)^3` of the original substance is left.

`(1/2)^3 = 1/(2^3) = 1/8`

after `3` half-lives, `1/8` of the original mass of niobium is left.

the original mass given is `816g`.

`1/8 * 816 = 816/8`

`=102`

`(816g)/8 = 102g`

after `3` half-lives, `102g` of niobium will be left.

The half-life of an isotope is the amount of time it takes for half of the mass of an element to decay.
If the half-life of a radioactive kind of niobium is 35 days, it will decay to `(816g)*1/2=408g` in 35 days. After 105 days, the period of its half-life will have occurred `105/35=3` times, so it will decay to `(816g)*1/2*1/2*1/2=(816g)*(1/2)^3=102g`

Consider that radioactive decay follows first order kinetics. Moreover, assume that the concentration for the purposes of this calculation is mass.

Recall,

`ln[A]_"t" = -kt + ln[A]_0`, and by extension,

`t_(1/2) = ln(2)/k`

If `t_(1/2) = 35"d"`, then,

`35"d" = ln(2)/k = > k approx 1.98*10^-2"d"^-1`

Hence,

`[A]_"t" = e^(-kt + ln[A]_0) approx 102"g"`

of niobium are left after that time interval.