Using the equation for boiling point elevation Δt
Δt = i Kb m
we can find the new boiling point T for the solution:
Δt = T - 100∘C
since we know that pure water boils at 100 °C.
We know that the van't Hoff Factor i is equal to 1 because sugar does not dissociate in water.
Also, the value of Ebullioscopic constant Kb for water is listed as 0.512 °C·kg/mol.
The molality m of the solution of 6 moles of sugar dissolved in a kilogram of water can be calculated as
m = 6 moles / 1 kg
= 6 mol/kg
Therefore the new boiling point T would be
T - 100 °C = i Kb m
T = i Kb m + 100 °C.
= (1) (0.512 °C·kg/mol) (6 mol/kg) + 100 °C
= 3.072 °C + 100 °C
= 103.072 °C
Answer:
d
Explanation:
H3O =1×10_12=12 ooh is 12
Answer:
maybe speak in english than we can understand and help you sorryy
Explanation:
Combustion of a compound is the reaction with oxygen , hence , the process of combustion is an oxidation reaction.
The carbohydrates contain more amount of oxygen as compared to the fats ,
Hence ,
carbohydrates , have a lot of oxygen contents , are are already partially oxidized , but fats have lower oxygen content .
Therefore ,
The partially oxidized carbohydrates are very difficult to oxidized in comparison to fats .
The original results have not been replicated consistently and reliably.
