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Artist 52 [7]
2 years ago
7

Given the atomic radius of xenon, 1.3 ?, and knowing that a sphere has a volume of 4?r3/3, calculate the fraction of space that

Xe atoms occupy in a sample of xenon at STP.
Chemistry
1 answer:
34kurt2 years ago
7 0
R = 1.3 A° = 1.3 × 10^(-10) m
v \:  =  \:  \frac{4}{3} \pi {r}^{3}
v \:  =  \:  \frac{4}{3}  \times 3.1415 \times  {(1.3 \times  {10}^{ - 10} )}^{3}
V = 9.20277208×10^(−30) m.
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Answer:

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% of N in the compound = (mass of N2/mass of compound) × 100 = (44.52/8750) × 100 = 5.088×10^-3 × 100 = 0.5088

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Lithium has two stable isotopes with masses of 6.01512 amu and 7.01600 amu. The average molar mass of Li is 6.941 amu. What is t
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Answer :  The percent abundance of Li isotope-1 and Li isotope-2 is, 6.94 % and 93.1 % respectively.

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

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   .....(1)

Let the fractional abundance of Li isotope-1 be 'x' and the fractional abundance of Li isotope-2 will be '100-x'

For Li isotope-1 :

Mass of Li isotope-1 = 6.01512 amu

Fractional abundance of Li isotope-1 = x

For Li isotope-2 :

Mass of Li isotope-2 = 7.01600 amu

Fractional abundance of Li isotope-2 = 100-x

Average atomic mass of Li = 6.941 amu

Putting values in equation 1, we get:

6.941=[(6.01512\times x)+(7.01600\times (100-x))]

By solving the term 'x', we get:

x=694.048

Percent abundance of Li isotope-1 = \frac{694.048}{100}=6.94\%

Percent abundance of Li isotope-2 = 100 - x = 100-6.94 = 93.1 %

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