1.

Calculate mass defect in the following reaction ._(1)H^(2) + ._(1)H^(3) rarr ._(2) He^(4) + ._(0)n^(1) (Given : mass H^(2) = 2.014, H^(3) = 3.016, He = 4.004, n = 1.008amu)

Answer»

0.018 AMU
0.18 amu
0.0018 amu
1.8 amu

Solution :MASS LOSS = mass of REACTANT - mass of PRODUCT
`= (2.014 + 3.016) - (4.004 + 1.008)`
`= 5.030 - 5.012 = 0.018` amu


Discussion

No Comment Found

Related InterviewSolutions