Answer:
That the isotope H-1 is the most abundant in nature.
Explanation:
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In this case, since the average atomic mass of an element is computed considering the mass of each isotope and the percent abundance each, for hydrogen we would set up something like this:

Moreover, since the isotope notation H-1 and H-2 means that the atomic mass of H-1 is 1 amu, that of H-2 is 2 amu and the average one is 1.0079 amu, we can infer that the most of the hydrogen in nature is H-1 as the most of it composes the average hydrogen atom.
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Answer:
B = - 0.0326 dm³/mol
Explanation:
virial eq until second term:
∴ P = 10 bar * (atm/ 1.01325 bar) = 9.869 atm
∴ T = 200°C = 473 K
∴ Vm = 3.90 dm³/mol
∴ R = 0.08206 dm³.atm/K.mol
⇒ PVm / RT = 1 + B/Vm
⇒ ((9.869 atm)*(3.90 dm³/mol)) / ((0.08206 dm³.atm/mol.K)*(473K)) = 1 + B/Vm
⇒ 0.99164 = 1 + B/Vm
⇒ B/Vm = - 8.357 E-3
⇒ B = (3.90 dm³/mol)*( - 8.357 E-3 )
⇒ B = - 0.0326 dm³/mol
The Best Answer: 1 - (.47+.23) = 0.30
If Ne has a mole fraction of 0.47 (or 47/100) and Ar is 0.23, then H2(or He) has a mole fraction of 0.30
This means the gas mixture is 30/100 H2(or He).
7.85 x 0.30 = 2.355 atm
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