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).
[tex]191.22 \: = \: \frac{191 \: \times \: x \: + \: 193 \: \times \: (100 \: - \: x)}{x \: + \: 100 \: - \: x} [/tex]
[tex]191.22 \: = \: \frac{191x \: + 19300 \: - \: 193x}{100} [/tex]
[tex]19122 \: = \: - 2x \: + \: 19300[/tex]
[tex]2x \: = \: 19300 \: - \: 19122[/tex]
[tex]2x \: = \: 178[/tex]
[tex]x \: = \: 89 \%[/tex]
[tex]100 \: - \: x \: = \: 11\%[/tex]
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).
