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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