Question

The mass defect for the following reaction, 1 over 1 H plus 1 over 0 n yields 2 over 1 H, is 0.00239 amu. (1 amu= 1.66 × 10-27 kg) How much energy will this reaction provide? [(1 J = 1 kg times m squared over s squared ) and the speed of light = 3.00 × 108 m over s ]


1.19 × 10-22 J


1.32 × 10-38 J


3.57 ×10-13 J


4.41 × 1022 J
@Abmon98

Answer 1

1 amu=1.66*10^-27 Kg
0.00239 amu= x Kg
Solve for x and use e=mc^2

Answer 2

so I do 0.00239*1.66^-27= 3.00*108 @Abmon98

Answer 3

that's wrong right @Abmon98

Answer 4

no that's roght your changing units from amu to kg 1.66*10^-27*0.00239=0.0000000000000000000000000000039674
e=mc^2
e=0.0000000000000000000000000000039674*(3.0*10^8)^2=0.000000000000357066

Answer 5

so its C

Answer 6

ok lol that was lot of zeros lol @Abmon98

Answer 7

thank you @Abmon98

Answer 8

yes hahah your welcome :)