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: m = 50 g ZnSO4
Explanation: First is convert the moles of Zn to the moles of ZnSO4 by having their mole ratio which is 2:2 based from the balanced equation. Next is convert the moles of ZnSO4 to mass using its molar mass.
0.311 mole Zn x 2 moles ZnSO4 / 2 moles Zn
= 0.311 moles ZnSO4
0.311 moles ZnSO4 x 161 g ZnSO4 / 1 mole ZnSO4
= 50 ZnSO4
Balanced chemical equation:
2 C2H2 + 5 O2 = 4 CO2 + 2 H2O
2 moles C2H2 ---------------- 5 moles O2
moles C2H2 ------------------ 84 moles O2
moles C2H2 = 84 * 2 / 5
molesC2H2 = 168 / 5 => 33.6 moles of C2H2
It’s one... electron affinity
Their valence electrons don't give them the ability to bond, they are also not in need to any electrons, since they won't get an octet soon.
