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grin007 [14]
3 years ago
9

A newly discovered element has two isotopes. One has an atomic weight of 120.9038 amu with 57.25% abundance. The other has an at

omic weight of 122.8831 amu. What is the atomic weight of the element? 1. 123.45 amu 2. 121.54 amu 3. 121.17 amu 4. 121.75 amu 5. 122.15 amu 6. 122.38 amu
Physics
1 answer:
Katarina [22]3 years ago
6 0

Answer : The atomic weight of the element is, 121.75 amu

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

As we are given that,

Mass of isotope 1 = 120.9038 amu

Percentage abundance of isotope 1 = 57.25 %

Fractional abundance of isotope 1 = 0.5725

Mass of isotope 2 = 122.8831 amu

Percentage abundance of isotope 2 = 100 - 57.25 = 42.75 %

Fractional abundance of isotope 2 = 0.4275

Now put all the given values in above formula, we get:

\text{Average atomic mass of element}=\sum[(120.9038\times 0.5725)+(122.8831\times 0.4275)]

\text{Average atomic mass of element}=121.75amu

Therefore, the atomic weight of the element is, 121.75 amu

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Answer:

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Explanation:

Here is the complete question

The volume electric charge density of a solid sphere is given by the following equation: ρ = (0.2 mC/m⁵)r²The variable r denotes the distance from the center of the sphere, in spherical coordinates. What is the net electric charge (in μC) of the sphere if the radius of the sphere is 0.5 m?

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The total charge on the sphere Q = ∫∫∫ρdV where ρ = volume charge density = 0.2r² and dV = volume element in spherical coordinates = r²sinθdθdrdΦ

So,  Q =  ∫∫∫ρdV

Q =  ∫∫∫ρr²sinθdθdrdΦ

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So, Q =  ∫∫∫0.2r⁴sinθdθdrdΦ

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Q =  ∫∫0.2r⁴-[- 2]drdΦ

Q =  ∫∫0.2r⁴(2)drdΦ

Q =  ∫∫0.4r⁴drdΦ

Q =  ∫0.4r⁴dr∫dΦ

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Q = 0.0628 mC

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Q = 62.8 × 10⁻³ × 10⁻³ C

Q = 62.8 × 10⁻⁶ C

Q = 62.8 μC

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