a 182.4g sample of germanium-66 is left undisturbed for 22.5 hours. At the end of that period, only 5.70g remain. What is the half-life of this material?

Approximately .

Explanation:

Calculate the ratio between the mass of this sample after and the initial mass:

.

Let denote the number of half-lives in that (where might not necessarily be an integer.) The mass of the sample is supposed to become the previous quantity after each half-life. Therefore, if the initial mass of the sample is (for example,) the mass of the sample after half-lives would be . Regardless of the initial mass, the ratio between the mass of the sample after half-lives and the initial mass should be .

For this question:

.

Take the natural logarithm of both sides of this equation to solve for :

.

.

.

In other words, there are half-lives of this sample in . If the length of each half-life is constant, that length should be .

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml