The rubidium isotope 87rb is a β emitter that has a half-life of 4.9 ✕ 1010 y that decays into 87sr. it is used to determine the age of rocks and fossils. rocks containing the fossils of early animal have a ratio of 87sr to 87rb of 0.0040. assuming that there was no 87sr present when the rocks were formed, calculate the age of the fossils.

t = 39.04 1010 year

Explanation:

This is a nuclear disintegration exercise that is governed by the equation.

     N = N0 e (-lam t)

The average life time is related to nuclear activity

      T ½ = ln 2 / lam

Let's use these two equations for exercise, let's start by finding nuclear activity

     Lam = ln 2 / T ½

     Lam = ln 2 / 4.9 10 10

    Lam = 0.14146 10-10 y-1

They tell us that the relationship atoms

      No / N = 0.0040

Let's look

       No / N = 1/0040

       N/No = 250                                                                                                                                                                                

Let's calculate the time

       (-lam t) = ln (N / No)

,        t = - 1 / lam ln (n / No)

        t = - 1 / 0.14146 10-10 ln (250)

        t = 39.04 1010 year

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  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂