Answer to Question #98997 in General Chemistry for Manilyn Atractivo

Question

Complete the following nuclear reaction:
(_4^9)Be + (_2^4)He → (_0^1)n + 126C


(_27^59)Co + ______ → (_25^56)Mn + (_2^4)He

(_7^14)N + (_0^1)n → (_6^14)C + ______

(_11^23)Na + (_1^2)H → ______ + (_1^1)H

(_5^11)B + ______ → (_6^13)C + (_1^1)H

(_92^238)U + (_0^1)n → ______ + (_-1^0)e

(_94^239)Pu + (_0^1)n → ______ (_95^240)Pu+ (_-1^0)e

(_92^238)U + (_6^12)C → ______ + 4 (_0^1)n

(_3^7)Li + (_1^1)H → ______ (_2^4)He + _____

(_3^7)Li + (_1^1)H → (_2^4)He + _____

Answer 1

Solution.

$Be^94 + He^42 = n^10 + C^{12}6$

$Co^{59}27 + 4n^10=Mn^{56}25 + He^42$

$N^{14}7 + n^10 = C^{14}6 + H^11$

$Na^{23}11 + H^21 = Na^{24}11 + H^11$

$B^{11}5 \ + He^32 = C^{13}6 + H^11$

$U^{238}92 \ + n^10 = Np^{239}93 + e^0(-1)$

$Pu^{239}94 \ + n^10 = Pu^{240}95 \ + e^0(-1)$

$U^{238}92 \ + C^{12}6 = Cf^{246}98 + 4n^10$

$Li^73 + H^11 = He^42 + He^42$

$Li^73 + H^11 = He^42 + He^42$

Answer:

$Be^94 + He^42 = n^10 + C^{12}6$

$Co^{59}27 + 4n^10=Mn^{56}25 + He^42$

$N^{14}7 + n^10 = C^{14}6 + H^11$

$Na^{23}11 + H^21 = Na^{24}11 + H^11$

$B^{11}5 \ + He^32 = C^{13}6 + H^11$

$U^{238}92 \ + n^10 = Np^{239}93 + e^0(-1)$

$Pu^{239}94 \ + n^10 = Pu^{240}95 \ + e^0(-1)$

$U^{238}92 \ + C^{12}6 = Cf^{246}98 + 4n^10$

$Li^73 + H^11 = He^42 + He^42$

$Li^73 + H^11 = He^42 + He^42$