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2 years ago

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Boron has an average atomic mass of 10.81. One isotope of boron has a mass of 10.012938 and a relative abundance of 19.80 percen

t . The other isotope has a relative abundance of 80.20 percent what is the mass of that isotope? Report two decimal places

1 answer:

slega [8]2 years ago

7 0

Answer: The atomic mass of the second isotope of boron (B) is 11.01 amu.

Explanation:

The average atomic mass is can be calculated by the summation of the product of the atomic mass of each isotope multiplied by its abundance (abundance % / 100.0).

Boron has two isotopes:

1) At.mass = 10.012938 amu ≅ 10.013 amu and its abundance = 0.198

2) At.mass = ??? amu and its abundance = 0.802

Average atomic mass of boron = 10.81 amu.

The average atomic mass of boron = (10.013 amu)*(0.198) + (atomic mass of the second isotope amu)*(0.802) = 10.81 amu

10.81 = 1.9825 + (0.802 x atomic mass of the second isotope)

(0.802 x atomic mass of the second isotope) = 10.81 - 1.9825 = 8.8275.

Atomic mass of the second isotope = 8.8275 / 0.802 = 11.0068 ≅ 11.01 amu.

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Answer:

Explanation:

Given that:

the temperature $T_1$ = 250 °C= ( 250+ 273.15 ) K = 523.15 K

Pressure = 1800 kPa

a)

The truncated viral equation is expressed as:

$\frac{PV}{RT} = 1 + \frac{B}{V} + \frac{C}{V^2}$

where; B = - $152.5 \ cm^3 /mol$ C = -5800 $cm^6/mol^2$

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Plugging all our values; we have

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Multiplying through with V² ; we have

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$T__{\gamma}} = \frac{523.15}{647.1}$

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V = 0.1249 × 10³ × 18.015

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Z = $\frac{PV}{RT}$

Z = $\frac{1800*10^3 *0.1249}{729.77*523.15}$

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