Could you please explain in detail how to do these questions? Thank you in advance.
Consider reaction, Z A X 89 223 Ac + 24 He
For balanced nuclear reaction, total Z on L.H.S = Total Z on R.H.S
& Total A on L.H.S = Total A on R.H.S
On equating atomic numbers on both sides, we get Z = 89 + 2
i e Z = 91
On equating mass numbers on both sides, we get A =223 + 4 = 227
Element with atomic number 91 is Pa. Hence, balanced nuclear equation is 91 227 Pa 89 223 Ac + 24 He
b) Consider reaction,20 41 Ca + Z A X 19 41 K
On equating atomic numbers on both sides, we get 20 + Z = 19
Z = 19 -20 = -1
On equating mass numbers on both sides, we get 41 + A = 41
A = 41 - 41 = 0
Hence, Z A X = -10 e
Hence, balanced nuclear equation is 20 41 Ca + -1 0 e 19 41 K
C)
79 201 Au 80201Hg + Z A X
On equating atomic numbers on both sides, we get 79 = 80 + Z
Z = 79 - 80 = -1
On equating mass numbers on both sides, we get 201 = 201 + A
A = 201 - 201 = 0
Hence, Z A X = -10 e
Hence, balanced nuclear equation is 79 201 Au 80201Hg + -1 0 e
d)
63150 Eu Z A X + +10
On equating atomic numbers on both sides, we get 63 = Z + 1
Z = 63 -1 = 62
On equating mass numbers on both sides, we get 150 = A + 0
A = 150
Element with atomic number 62 is Sm.
Hence, balanced nuclear equation is 63150 Eu 62150Sm + +10
Question 12
Boran-10 contain 5 protons and 5 neutrons.
Mass defect =[ [ 5 ( 1.007825 ) + 5 ( 1.008665 ) ] - 10.012937 ] amu
Mass defect = [10.08245 - 10.012937 ] g / mol = 0.069513 amu
We have relation , 1 amu = 1.6605 10 -27 kg
We can calculate B.E from the mass defect using the mass-energy equation.E = m c 2
E = [0.069513 amu ( 1.6605 10 -27 kg / 1 amu ) ]( 2.998 10 8 m /s ) 2
E = 1.037 10 -11 J per nucleus.
In 1 mole B , there are 6.022 10 23 nucleii.
Therefore, B,E in J / mol = ( 1.037 10 -11 J / nucleus ) ( 6.022 10 23 nucleus ) = 6.248 10 12 J / mol
ANSWER : B.E in J / mol = 6.248 10 12 J / mol
Could you please explain in detail how to do these questions? Thank you in advance. 11....
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Can anyone explain how to do this type of problem? I'll give you a thumbs up if it's the correct answer!! Thanks in advance :) Question 8 0.38 pts Chemistrium is an essential component in science knowledge. Calculate the binding energy per nucleon for Chemistrium (mass of Chemistrium = 100.7654 amu). Chemistrium has 51 protons and 50 neutrons. Enter to 4 decimal places. mass of 1H = 1.007825 amu, mass of neutron = 1.008665 amu
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Please fill in the empty shells in excel and include how you calculated each of the cell values in excel. Mass of atomic constituents (MeV Mass Defect [MeV Mass Number Atomic Mass Mass of neutrons Mass of protons & electrons [Mev Mass defect per proton+neutron, aka Binding Energy per Nucleon [MeV Nuclide Atomic Mass |MeV] [amu [MeV) H-1 1 H-2 2 H-3 3 Не3 1.007825 938.79100 2,014101 3.016049 1 2 1876.12300 2809.43173 2 1 3.016029 2809.41292 He-4 2 7 4.002603...
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