The half-life for the process 238u→206pb is 4.5×109 yr. a mineral sample contains 61.0 mg of 238u and 15.5 mg of 206pb. part a what is the age of the mineral

Answer : The age of the mineral is,

Explanation :

First we have to calculate the rate constant, we use the formula :

Now we have to calculate the time.

Expression for rate law for first order kinetics is given by:

where,

k = rate constant

t = time passed by the sample  = ?

a = initial amount of the reactant  = 61.0 mg

a - x = amount left after decay process = 15.5 mg

Now put all the given values in above equation, we get

Therefore, the age of the mineral is,

so what is the problem with your lab report and whats the subject of the lab report? i will you

a pan containing water is placed in a positio n where it is exposed to a 10 mile per hour wind. the pan contains a uniform depth of 1.27 cm of water, is quite wide, and has a length of 2.44 m in the wind direction. the water in the pan is at a constant temperature of 15.5 °c.