naturally occurring boron consists of two isotopes whose atomic masses are 10u and 11u.the atomic mass of natural boron is 10.80 calculate the percentage abundance of each isotope in natural boron pls tell asap
Answers
Answer:
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Explanation:
The atomic mass of boron is 10.81 u.
And 10.81 u is a lot closer to 11u than it is to 10u, so there must be more of boron-11.
To convince you fully, we can also do a simple calculation to find the exact proportion of boron-11 using the following formula:
(
10
u
)
(
x
)
+
(
11
u
)
(
1
−
x
)
100
%
=
10.81
u
Where u is the unit for atomic mass and x is the proportion of boron-10 out of the total boron abundance which is 100%.
Solving for x we get:
11
u
−
u
x
=
10.81
u
0.19
u
=
u
x
x
=
0.19
1
−
x
=
0.81
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Explanation:
The atomic mass of boron is 10.81 u.
And 10.81 u is a lot closer to 11u than it is to 10u, so there must be more of boron-11.
To convince you fully, we can also do a simple calculation to find the exact proportion of boron-11 using the following formula:
(10u)(x)+(11u)(1−x)100%=10.80u
Where u is the unit for atomic mass and x is the proportion of boron-10 out of the total boron abundance which is 100%.
Solving for x we get:
11u−ux=10.81u
0.19u = ux x = 0.19
1x=0.81
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