<span>Hydrogen fusion generates the energy for proton - proton chains and the carbon nitrogen oxygen cycle. It is the nuclear fusion of 4 protons to form a helium 4 nucleus.</span>
<u>Answer:</u> The atomic weight of the second isotope is 64.81 amu.
<u>Explanation:</u>
Average atomic mass of an element is defined as the sum of atomic masses of each isotope each multiplied by their natural fractional abundance
Formula used to calculate average atomic mass follows:
.....(1)
We are given:
Let the mass of isotope 2 be 'x'
Mass of isotope 1 = 62.9 amu
Percentage abundance of isotope 1 = 69.1 %
Fractional abundance of isotope 1 = 0.691
Mass of isotope 2 = 'x'
Percentage abundance of isotope 2 = 30.9%
Fractional abundance of isotope 2 = 0.309
Average atomic mass of copper = 63.5 amu
Putting values in equation 1, we get:
![\text{Average atomic mass of copper}=[(62.9\times 0.691)+(x\times 0.309)]](https://tex.z-dn.net/?f=%5Ctext%7BAverage%20atomic%20mass%20of%20copper%7D%3D%5B%2862.9%5Ctimes%200.691%29%2B%28x%5Ctimes%200.309%29%5D)

Hence, the atomic weight of second isotope will be 64.81 amu.
We are already given with the mass of the Xe and it is 5.08 g. We can calculate for the mass of the fluorine in the compound by subtracting the mass of xenon from the mass of the compound.
mass of Xenon (Xe) = 5.08 g
mass of Fluorine (F) = 9.49 g - 5.08 g = 4.41 g
Determine the number of moles of each of the element in the compound.
moles of Xenon (Xe) = (5.08 g)(1 mol Xe / 131.29 g of Xe) = 0.0387 mols of Xe
moles of Fluorine (F) = (4.41 g)(1 mol F/ 19 g of F) = 0.232 mols of F
The empirical formula is therefore,
Xe(0.0387)F(0.232)
Dividing the numerical coefficient by the lesser number.
<em> XeF₆</em>
Answer:
38.7 g is the answer to your question. Your welcome:)