Respuesta :
Answer:
10.13086%
Explanation:
129y + 132(1-y) = 129.91
129y + 132 - 132y = 129.91
-3y = -2.09
y = 0.69666 (5 s.f.)
% abundance of 132X = 100% - (129×0.69666)%
= 100% - 89.86914%
= 10.13086%
= 10.1% (3 significant figures)
*y is just a variable for the equation
The percentage abundance of ¹³²X is 30.33%
Let Isotope ¹³²X be A
Let Isotope ¹²⁹X be B
From the question given above, the following data were obtained:
For A (Isotope ¹³²X):
Mass of A = 132
Abundance of A (A%) =?
For B (Isotope ¹²⁹X):
Mass of B = 129
Abundance of B (B%) = 100 – A%
Atomic mass of X = 129.91 amu
The abundance of ¹³²X can be obtained as follow:
Atomic mass = [(mass of A × A%)/100] + [(mass of B × B%)/100]
129.91 = [(132 × A%)/100] + [(129 × (100 –A%))/100]
129.91 = 1.32A% + 129 – 1.29A%
Collect like terms
129.91 – 129 = 1.32A% – 1.29A%
0.91 = 0.03A%
Divide both side by 0.03
A% = 0.91 / 0.03
A% = 30.33%
Therefore, the abundance of ¹³²X is 30.33%
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