Calculate the mass defect and nuclear binding energy per nucleon for each of the nuclides indicated below.
Part A
Ca-40 (atomic mass = 39.96259 amu)
Part B
Binding Energy per nucleon for above problem
Part C
Ti-48 (atomic mass = 47.947947 amu)
Part D
Binding Energy per nucleon for above problem
Part E
Ag-107 (atomic mass = 106.905092 amu)
Part F
Binding Energy per nucleon for above problem

8 carbon atoms forming single bonds with 18 hydrogen atoms; a triple bond exists between the third and fourth carbon atom from the right

explanation:

imma come back to you hold on

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2