How much of a 5.0 g sample of strontium-90 will remain after 87 years?

Since radioacitive decay fallows first order kinetics you can use the equation lnA(t)=-kt+lnA₀.  
A₀=the initial amount of sample
A(t)=the amount of sample after t years
t=time
k=rate constant

First you need to solve for the rate constant (k) to get the equation
-((lnA(t)-lnA(0))/t=k which can simplify into k=-ln(A(t)/A₀)/t which can then be simplified further to ln(A₀/A(t))/t=k.   I think the half life of strontium-90 is 28.8 years which means that when t=28.8 years, A(t) is going to equal 1/2 of A₀ and you can make the equation k=ln2/28.8 to get k=0.02408.

Now you need to solve for A(t) in the equation ln(A(t))=-kt+lnA₀ to get A(t)=e^(-kt+lnA₀) which simplifies to A(t)=A₀e^-kt.  When you plug t=87, A₀=5g, and k=0.02408 into that equation you should get 0.616g.

Therefore after 87 years you will have 0.616g of strontium-90 if you started with a 5g sample.

I hope this helps.  Let me know in the comments if anything is unclear.

The correct name for n3-ion should be azide.

Ibelieve y y even though i've only taken chemistry for one year