Answer:
The thermodynamic parameter which is of significance in this case is the 'Reduction Potential' for molecular bromine which is ~ +1.1 v vs N.H.E. In other words, it is a strong oxidizing agent. The bromine will oxidize sulfur compounds in which the valence of sulfur is lower than six to sulfate.
There are many possible reactions. Here is one possible example:
Na2 S2O3 + 4Br2 + 5 H2O = 2NaHSO4 + 8 HBr
Answer:
2311.2 cal
Explanation:
540 cal / g * 4.28 g = 2311.2 cal
Answer:
She did no work while she posed because she did not move the torch or the book.
Explanation:
Answer:
% yield = 62.21 %
Explanation:
- C2H4(g) + H2O(l) → C2H6O(l)
∴ mass C2H4(g) = 4.6 g
∴ mass C2H6O(l) = 4.7 g
- % yield = ((real yield)/(theoretical yield))×100
theretical yield:
∴ molar mass C2H4(g) = 28.05 g/mol
⇒ mol C2H4(g) = (4.6 g)*(mol/28.05 g) = 0.164 mol
⇒ mol C2H6O(l) = (0.164 mol C2H4)*(mol C2H6O(l)/molC2H4(g))
⇒ mol C2H6O(l) = 0.164 mol
∴ molar mass C2H6O(l) = 46.07 g/mol
⇒ mass C2H6O(l) = (0.164 mol)*(46.07 g/mol) = 7.55 g
⇒ theoretical yield = 7.55 g
⇒ % yield = (4.7 g)/(7.55 g))*100
⇒ % yield = 62.21 %
