Answer:
The enthalpy of vaporization of water at 273 K and 1 bar = 44.9 KJ/mol
Explanation:
Enthalpy of vaporization of water at 273 K, ΔHvap(T₂) is given as;
ΔHvap(T₂) = ΔHvap(T₁) + ΔCp * (T₂ - T₁)
where ΔCp = molar heat capacity of gas - molar heat capacity of liquid
Therefore, ΔCp = (33.6 - 75.3) = -41.70 J/(mol K) = 0.0417 kJ/(molK)
substituting ΔCp = 0.0417 kJ/(mol K) in the initial formula
;
ΔHvap(T) = ΔHvap(T1) + ΔCp * (T₂ - T₁)
ΔHvap(T₂)= 40.7 kJ/mol + {-0.0417 kJ/(mol K) * (272 - 373 K)}
ΔHvap(T₂) = 44.9 kJ/mol
Therefore, enthalpy of vaporization of water at 273 K and 1 bar = 44.9kJ/mol
Answer:
The rate of evaporation decreases, or slows down
Explanation:
Yes yes you are totally right
The balanced equation for the neutralisation reaction is as follows
Ca(OH)₂ + H₂SO₄ ---> CaSO₄ + 2H₂O
stoichiometry of Ca(OH)₂ to H₂SO₄ is 1:1
equivalent number of acid reacts with base
number of H₂SO₄ mol reacting - 2 mol
according to molar ratio of 1:1
number of Ca(OH)₂ mol = number of H₂SO₄ moles
therefore number of Ca(OH)₂ moles required - 2 mol
If we consider a combustion reaction of Methane:
The balanced equation is:
CH4 + 2O2 ---> 2H2O + CO2
The rate of appearance of H2O is rH2O, rate of disappearance of O2 is -rO2
(rH2O)^2 = (-rO2)^2
rH2O = -rO2
