Asample originally contained 1.28 g of a radioisotope. it now contains 1.12 g of its daughter isotope. how many half-lives have passed since the sample originally formed?

3 half lives

Solution :

Formula used :

where,

a = amount of reactant left after n-half lives = initial amount - amount of daughter isotope= (1.28-1.12)= 0.16 g

= Initial amount of the reactant = 1.28 g

n = number of half lives = ?

Putting values in above equation, we get:

Therefore,  3 half-lives have passed since the sample originally formed.

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"

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