for questions 8 - 12, balance the nuclear equation, then indicate if the equation represents a fission or a…

Explanation:

Step1: Recall nuclear - equation rules

In a nuclear equation, the sum of mass numbers and atomic numbers on both sides must be equal.

Step2: Balance equation 8

For ({91}^{231}Pa\rightarrow{77}^{192}Ir + X). Mass - number of (X): (231-192 = 39). Atomic - number of (X): (91 - 77=14). So (X = _{14}^{39}Si). This is a fission reaction as a heavy nucleus ((Pa)) splits into lighter nuclei ((Ir) and (Si)).

Step3: Balance equation 9

For ({2}^{4}He+{1}^{2}H\rightarrow X). Mass - number of (X): (4 + 2=6). Atomic - number of (X): (2+1 = 3). So (X=_{3}^{6}Li). This is a fusion reaction as lighter nuclei combine to form a heavier nucleus.

Step4: Balance equation 10

For ({4}^{9}Be+{6}^{12}C\rightarrow X). Mass - number of (X): (9 + 12=21). Atomic - number of (X): (4+6 = 10). So (X = _{10}^{21}Ne). This is a fusion reaction.

Step5: Balance equation 11

For ({92}^{238}U+{0}^{1}n\rightarrow_{52}^{128}Te+X). Mass - number of (X): (238 + 1-128=111). Atomic - number of (X): (92+0 - 52 = 40). So (X=_{40}^{111}Zr). This is a fission reaction.

Step6: Balance equation 12

For ({105}^{262}Db\rightarrow{42}^{96}Mo+{14}^{28}Si+X). Mass - number of (X): (262-(96 + 28)=138). Atomic - number of (X): (105-(42 + 14)=49). So (X={49}^{138}In). This is a fission reaction.

Answer:

8. ({91}^{231}Pa\rightarrow{77}^{192}Ir+_{14}^{39}Si), Fission
9. ({2}^{4}He+{1}^{2}H\rightarrow_{3}^{6}Li), Fusion
10. ({4}^{9}Be+{6}^{12}C\rightarrow_{10}^{21}Ne), Fusion
11. ({92}^{238}U+{0}^{1}n\rightarrow_{52}^{128}Te+_{40}^{111}Zr), Fission
12. ({105}^{262}Db\rightarrow{42}^{96}Mo+{14}^{28}Si+{49}^{138}In), Fission