Answer is: the average atomic mass 217.606 amu.
Ar₁= 203.973 amu; the average atomic mass of isotope.
Ar₂ = 205.9745 amu.
Ar₃ = 206.9745 amu.
Ar₄ = 207.9766 amu.
ω₁ = 1.40% = 0.014; mass percentage of isotope.
ω₂ = 24.10% = 0.241.
ω₃ = 22.10% = 0.221.
ω₄ = 57.40% = 0.574.
Ar = Ar₁ · ω₁+ Ar₂ · ω₂ + Ar₃ · ω₃ + Ar₄ · ω₄.
Ar = 203.973 amu · 0.014 + 205.9745 amu · 0.241 + 206.9745 amu · 0.221 + 207.9766 amu · 0.574.
Ar = 2.855 amu + 49.632 amu + 45.741 amu + 119.378 amu.
Ar = 217.606 amu.
But abundance of isotopes is greater than 100%.
It should be lead, with the fourth isotope weighs 207.9766 amu and an abundance of 52.40.
Answer:
Thus, the radius of the helium atom in nanometers is - 0.031 nm
Explanation:
Given that:-
The radius of the helium atom = 31 pm
Considering the conversion of length in pm to the length in nm as:-
1 pm = 0.001 nm
So,
Applying the above conversion factor in the radius of helium atom as:-
Radius =
nm = 0.031 nm
<u>Thus, the radius of the helium atom in nanometers is - 0.031 nm</u>
We have to know the molarity of solution obtained when 5.71 g of Na₂CO₃.10 H₂O is dissolved in water and made up to 250 cm³ solution.
The molarity of solution obtained when 5.71 g of sodium carbonate-10-water (Na₂CO₃.10 H₂O) is dissolved in water and made up to 250.0 cm^3 solutionis: (A) 0.08 mol dm⁻³
The molarit y of solution means the number of moles of solute present in one litre of solution. Here solute is Na₂CO₃.10 H₂O and solvent is water. Volume of solution is 250 cm³.
Molar mass of Na₂CO₃.10 H₂O is 286 grams which means mass of one mole of Na₂CO₃.10 H₂O is 286 grams.
5.71 grams of Na₂CO₃.10 H₂O is equal to
= 0.0199 moles of Na₂CO₃.10 H₂O. So, 0.0199 moles of Na₂CO₃.10 H₂O present in 250 cm³ volume of solution.
Hence, number of moles of Na₂CO₃.10 H₂O present in one litre (equal to 1000 cm³) of solution is
= 0.0796 moles. So, the molarity of the solution is 0.0796 mol/dm³ ≅ 0.08 mol/dm³
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