Question

Question 8 part c, and question 9?? help?

Answer 1

8)

a.
p = 77
e = 77
n = 114

b.
Ir-191 has 114 neutrons, whereas, Ir-193 has 116 neutrons.

c.
Let the Percentage of Ir-191 in a sample be 'x', then, the Percentage of Ir-193 is (100 - x).

`191.22 \: = \: (191 \: * \: x \: + \: 193 \: * \: (100 \: - \: x))/(x \: + \: 100 \: - \: x)`

`191.22 \: = \: (191x \: + 19300 \: - \: 193x)/(100)`

`19122 \: = \: - 2x \: + \: 19300`

`2x \: = \: 19300 \: - \: 19122`

`2x \: = \: 178`

`x \: = \: 89 \%`

`100 \: - \: x \: = \: 11\%`
Therefore, in a naturally occurring Iridium sample, the amount of Ir-191 present is 89%, and the amount of Ir-193 present is 11%.



9)
False, Ir-191 is a common isotope having more no. of neutrons (114) than protons (77).