sab three isotopes x 20 X 21 and 22 the percentage abundance of x20 is 90% and its average atomic mass of the element is 20.11 the percentage abundance of x21 should be
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21
Percentage of x20 = 90%
The remaining percentage = 100 - 90 = 10%
Let
Percentage of x21 = y%
Percentage of x22 = (10-y)%
Average atomic mass = [20x90 + 21y + 22(10-y)]/100 = 20.11
1800 + 21y + 220 - 22y = 2011
21y - 22y = 2011 - 1800 - 220
-y = -9
y = 9
The percentage of y21 = 9%
The remaining percentage = 100 - 90 = 10%
Let
Percentage of x21 = y%
Percentage of x22 = (10-y)%
Average atomic mass = [20x90 + 21y + 22(10-y)]/100 = 20.11
1800 + 21y + 220 - 22y = 2011
21y - 22y = 2011 - 1800 - 220
-y = -9
y = 9
The percentage of y21 = 9%
Answered by
6
Answer:
Explanation:
Percentage of x20 = 90%
The remaining percentage = 100 - 90 = 10%
Let
Percentage of x21 = y%
Percentage of x22 = (10-y)%
Average atomic mass = [20x90 + 21y + 22(10-y)]/100 = 20.11
1800 + 21y + 220 - 22y = 2011
21y - 22y = 2011 - 1800 - 220
-y = -9
y = 9
The percentage of y21 = 9%
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