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likoan [24]
3 years ago
7

The average atomic masses of some elements may vary, depending upon the sources of their ores. Naturally occurring boron consist

s of two isotopes with accurately known masses (10B, 10.0129 amu and 11B, 11.0931 amu). The actual atomic mass of boron can vary from 10.807 to 10.819, depending on whether the mineral source is from Turkey or the United States. Calculate the percent abundances leading to the two values of the average atomic masses of boron from these two countries.
Chemistry
2 answers:
vodomira [7]3 years ago
7 0

Explanation:

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i

Let the fractional abundance of B-10 isotope be 'x'. So, fractional abundance of B-11 isotope will be '1 - x'

For B-10 isotope:

Mass of B-10 isotope = 10.0129 amu

For Li-7 isotope:

Mass of B-11 isotope = 11.0931 amu

1) Average atomic mass of boron = 10.807 amu

10.807 amu=x\times 10.0129 amu+(1-x)\times 11.0931 amu

x= 0.2648

The  percent abundances B-10 isotope = 0.2648 × 100 =26.48%

The  percent abundances B-11 isotope = 100% - 26.48% =73.51%

2) Average atomic mass of boron = 10.819 amu

10.819 amu=x\times 10.0129 amu+(1-x)\times 11.0931 amu

x= 0.2537

The  percent abundances B-10 isotope = 0.2537 × 100 =25.37%

The  percent abundances B-11 isotope = 100% - 25.37% =74.63%

frez [133]3 years ago
6 0
Sorry I don’t know but good luck!
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So, let's replace m_1^{N_2} in the energy balance:

m_C*Cp*(T_1^C-T_2^C)=m_{2,liq}^{N_2}u_{2,liq}^{N_2}+m_{2,vap}^{N_2}u_{2,vap}^{N_2}-m_{2,liq}^{N_2}u_1^{N_2}-m_{2,vap}^{N_2}u_1^{N_2}

So, as you can see, the term m_{2,liq}^{N_2}u_{2,liq}^{N_2} disappear because u_{2,liq}^{N_2}=u_{1,liq}^{N_2} (The specific energy in the liquid is the same because the temperature does not change).

m_C*Cp*(T_1^C-T_2^C)=m_{2,vap}^{N_2}u_{2,vap}^{N_2}-m_{2,vap}^{N_2}u_1^{N_2}

m_C*Cp*(T_1^C-T_2^C)=m_{2,vap}^{N_2}(u_{2,vap}^{N_2}-u_1^{N_2})

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