Balance the nuclear reaction equation for the beta minus decay of nickel-63 by completing the missing numbers. Superscript question mark subscript 28 upper N i right arrow superscript 63 subscript question mark upper C u plus superscript 0 subscript 1 e.

The superscript of nickel is
.
The subscript of copper is
.

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Explanation:

Symbiology of nuclear isotopes can be written in one of two ways. That is, for the symbiology Ni-63, the '63' represents the nuclear isotopic mass and represents the sum of the number of neutrons and protons. When expressed in subscript and superscript notation Ni-63 becomes ₂₈Ni⁶³ where '28' is the atomic number and represents the number of protons (p⁺) in the nucleus and the number '63', as mentioned above, represents the number of neutrons and protons. Thus, for any isotope the number of protons can NOT change or the element changes but the number of neutrons can change giving the various isotopes of a given element. Example: Oxygen exists in 3 primary isotopes, O-16, O-17 & O-18. Since the atomic number of oxygen is always 8 (i.e., 8 protons), writing it into the symbiology is optional => ₈O¹⁶, ₈O¹⁷ & ₈O¹⁸. However, the number of neutrons can change and is always listed in the symbiology by X-(mass no.) or ₐXᵇ, where a = # of protons (atomic no.) and b = #protons + #neutrons.  

For determining the numbers of protons and neutrons, remember the large number is always a superscript (p⁺ + n°) and the smaller number always the subscript (p⁺). So, number of neutrons, therefore, can be easily be determined by 'superscript - subscript' = (p⁺ + n°) - (p⁺) = number of n°.

For nuclear equations, the mass and charge must balance for reactants and products. That is, ∑reactant superscripts = ∑product superscripts and ∑reactant subscripts = ∑product subscripts.

Given Ni-63 as decaying by β⁻ emissions (high energy electrons => ₋₁e°), one writes ...

₂₈Ni⁶³ => ₐXᵇ + ₋₁e°  =>  ₂₈Ni⁶³ => ₂₉Cu⁶³ + ₋₁e°

Note: ∑reactant superscripts (63) = ∑product superscripts (63 + 0) and ∑reactant subscripts (28) = ∑product subscripts [(29 + (-1)].

Hope this helps. Doc :-)

63 and 29

Explanation:

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2

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