Nuclear scientists have synthesized approximately 1600 nuclei not known in nature. more might be discovered with heavy-ion bombardment using high-energy particle accelerators. complete and balance the following reactions, which involve heavy-ion bombardments.

I shall assume the following equations are the ones of interest as none was provided:

(a)63 Li + 56 28Ni →?

(b)40 20Ca + 248 96Cm→147 62Sm + ?

(c)88 38Sr + 84 36Kr→116 46Pd + ?

(d)40 20Ca + 238 92U→70 30Zn + 4 10 n + 2?

I shall also assume in the following calculations that the mass number of the unknown products is 'a' and the atomic number of the unknown products is 'b'.

In balancing nuclear equations, we observe the following rules:

I.The total number of protons and neutrons in the products and reactants must be the same ( conservation of mass number, A ).

II.The total number of nuclear charges in the products and reactants must be the same ( conservation of atomic number, Z ).

For (a),

63 Li + 56 28Ni→?

Add mass number, A, of reactants = 6+56 = 62

Add atomic number, Z, of reactants = 3+28 = 31

A of reactants = A of unknown product = 62

and, Z of reactants = Z of unknown product = 31

Hence, unknown product is the element with Z as 31 = Ga

Balanced equation becomes:

63 Li + 56 28Ni → 62 31Ga

For (b),

40 20Ca + 248 96Cm→147 62Sm + ?

A of reactants = 40+248 = 288

A of products = 147 + 'a'

But A of reactants = A of products

288 = 147 + 'a'

‘a’ = 288 – 147 = 141

Also,

Z of reactants = 20 + 96 = 116

Z of products = 62 + 'b'

Z of reactants = Z of products

116 = 62 + 'b'

'b' = 116 – 62 = 54

So, unknown product is the element with Z = 54 = Xe

Balanced equation is:

40 20Ca + 248 96Cm→147 62Sm + 141 54Xe

For (c),

88 38Sr + 84 36Kr→116 46Pd + ?

A of reactants = 88 + 84 = 172

A of products = 116 + 'a'

But A of reactants = A of products

172 = 116 + 'a'

‘a’ = 172 – 116 = 56

Also,

Z of reactants = 38 + 36 = 74

Z of products = 46 + 'b'

Z of reactants = Z of products

74 = 46 + 'b'

'b' = 74 – 46 = 28

So, unknown product is the element with Z = 28 = Ni

Balanced equation is:

88 38Sr + 84 36Kr→116 46Pd + 56 28Ni

For (d),

40 20Ca + 238 92U→70 30Zn + 4 10 n + 2?

A of reactants = 40 + 238 = 278

A of products = 70 + (4 x 1) + (2 x 'a') = 70 + 4 + 2a = 74 + 2a

But, A of reactants = A of products

278 = 74 + 2a

2a = 278 – 74

2a = 204

'a' = 204/2 = 102

Also,

Z of reactants = 20 + 92 = 112

Z of products = 30 + (4 x 0) + (2 x b) = 30 + 0 + 2b = 30 + 2b

Z of reactants = Z of products

112 = 30 + 2b

2b = 112 – 30

2b = 82

‘b’ = 82/2 = 41

Therefore, unknown product is element with Z = 41 =Nb

Balanced equation is:

40 20Ca + 238 92U→70 30Zn + 4 10 n + 2 102 41Nb

  the   balanced   equation   for   reaction of   ca metal   with   o2 to   produce   cao   is as below

2 ca+ o2→ 2 cao2

  explanation

the reaction is   balanced     since the   number of atoms   in reactant side is   equal to   number of atoms   in   product side.   for   example   they   are 2   atom of   ca in   reactant   side   and   2   atoms of ca   in   the product side.

  the   process   is a   meet the   requirement   of   redox   reaction because :     calcium   is   oxidized   from   oxidation state 0   to 2+   while   oxygen   is reduced from   oxidation   state   0   to 2-