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
The conversion of ethanol to ethanoic acid is an oxidation reaction.
Answer:
0.32M
Explanation:
<u>Step 1:</u> Balance the reaction
K2CO3 + Ba(NO3)2 ⇔ KNO3 + BaCO3
We have a 20 mL 0.2 M K2CO3 and a 30mL 0.4M Ba(NO3)2 solution
SinceK2CO3 is the limiting reactant, there will remain Ba(NO3)2 after it's consumed and produced KNO3 + BaCO3
<u>Step 2: </u>Calculate concentration
To find the concentration of the barium cation we use the following equation:
Concentration = moles of the <u>solute</u> / volumen of the <u>solution</u>
<u />
<u>[Ba2+] </u> = (20 * 10^-3 * 0.2M + 30 * 10^-3 * 0.4M) / ( 20 + 30mL) *10^-3
[Ba2+] = 0.32 M
The concentration of Barium ion in solution is 0.32 M
