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Elis [28]
3 years ago
9

Determine the missing species:

Chemistry
1 answer:
Galina-37 [17]3 years ago
5 0

Answer: 4. ^{0}_{-1}\textrm{e} and ^{0}_{1}\textrm{e}

Explanation:

a) The given reaction is ^{41}_{20}\textrm{Ca}+^{x}_{y}\textrm{X}\rightarrow ^{41}_{19}\textrm {K}

As the mass on both reactant and product side must be equal:

41+x=41

x=0

As the atomic number on both reactant and product side must be equal:

20+y=19

y=-1

^{41}_{20}\textrm{Ca}+^{0}_{-1}\textrm{e}\rightarrow ^{41}_{19}\textrm {K}

b) ^{15}_{8}\textrm{O}\rightarrow ^{15}_{7}\textrm{N}+^{x}_{y}\textrm{X}

Total mass on reactant side = total mass on product side

15 =15 + x

x = 0

Total atomic number on reactant side = total atomic number on product side

8 = 7 + y

y = 1

^{15}_{8}\textrm{O}\rightarrow ^{15}_{7}\textrm{N}+^{0}_{1}\textrm{e}

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Hello,

I think what the question is requesting is to calculate the energy difference when an excited electron drops from N = 15 to N = 5

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