Propane : ).............................................
Explanation:
What are the two oxidation states of sulphur in sodium thiosulphate ? In thiosulphate, two sulphurs have oxidation state of -2 and +6. (as per suggestions). Sulphur bonded to three oxygen is considered to have +6 (Sulphur A) and other sulphur has -2 (Sulphur B).
Data:
Q = 402.7 J → releases → Q = - 402.7 J
m = 16.25 g
T initial = 54 ºC
adopting: c = 4.184J/g/°C
ΔT (T final - T initial) = ?
Solving:
Q = m*c*ΔT
-402.7 = 16.25*4.184*ΔT
-402.7 = 67.99*ΔT
If: ΔT (T final - T initial) = ?