calculate the % of the naturally occuring isotopes 35Cl and 37Cl that accounts for the atomic mass of chlorine taken as 35.5u with solutions please thanku
Answers
Answered by
0
let the % of 35Cl is x%
let the % of 37Cl is (100-x)%
given atomic mass of Cl is 35.5u
so, [35(x)+37(100-x)]/100 = 35.5
35x+3700-37x = 3550
-2x = 3550-3700
2x = 150
x = 75
100-x = 100-75 = 25
therefore, % of 35Cl is 75% in nature and % of 37Cl is 25% in nature
let the % of 37Cl is (100-x)%
given atomic mass of Cl is 35.5u
so, [35(x)+37(100-x)]/100 = 35.5
35x+3700-37x = 3550
-2x = 3550-3700
2x = 150
x = 75
100-x = 100-75 = 25
therefore, % of 35Cl is 75% in nature and % of 37Cl is 25% in nature
honeysvb:
plz mark as brainliest if this helps you
Similar questions