A1.00 * 10^-6 -g sample of nobelium, 254/102 no, has a half-life of 55 seconds after it is formed. what is the percentage of 254/102 no remaining at the following times? a) 5.0 min after it formsb) 1.0 h after it forms

2.2 % and 0 %

Explanation:

The equation we will be using to solve this question is:

N/N₀  = e⁻λ t

where  N₀ : Number of paricles at t= 0

            N=  Number of particles after time t

             λ= Radioactive decay constant

             e= Euler´s constant

We are not given λ , but it can be determined from the half life with the equation:

λ = 0.693 / t 1/2 where t 1/2 is the half-life

Substituting our values:

λ = 0.693 / 55 s = 0.0126/s

a) For t = 5 min = 300 s

N / N₀ = e^-(0.0126/s x 300 s) = e^-3.8 = 0.022 = 2.2 %

b) For t = 1 hr = 3600 s

N / N₀ =  e^-(0.0126/s x 3600 s) = 2.9 x 10 ⁻²⁰ = 0 % (For all practical purposes)

it's filtration, but this is an assumption so my apologies if i got this

i think its 5049.

i hope this !