Oblicz zawartość procentową dwóch izotopów boru: 10B i 11B. Wiedząc, że średnia masa atomowa boru wynosi 10,814 u.
¹⁰B
A₁=10
%x₁=?
¹¹B
A₂=11
%x₂=?
mat=10,814 [u]
%x₁-x
%x₂-100-x
mat=A₁*%x₁+A₂*%x₂ / 100%
10,814=10*x+11*(100-x) / 100% /*100
1081,4=10x+1100-11x
1081,4-1100=-x
-18,6=-x /*(-1)
x=18,6
100-18,6=81,4
¹⁰B=18,6%
¹¹B=81,4%
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¹⁰B
A₁=10
%x₁=?
¹¹B
A₂=11
%x₂=?
mat=10,814 [u]
%x₁-x
%x₂-100-x
mat=A₁*%x₁+A₂*%x₂ / 100%
10,814=10*x+11*(100-x) / 100% /*100
1081,4=10x+1100-11x
1081,4-1100=-x
-18,6=-x /*(-1)
x=18,6
100-18,6=81,4
¹⁰B=18,6%
¹¹B=81,4%