Answer:Just want to be carefree lately yeah
Explanation: Cradles Profanities
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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:
[HI] = 0.264M
Explanation:
Based on the equilibrium:
2HI(g) ⇄ H₂(g) + I₂(g)
It is possible to define Kc of the reaction as the ratio between concentration of products and reactants using coefficients of each compound, thus:
<em>Kc = 0.0156 = [H₂] [I₂] / [HI]²</em>
<em />
As initial concentration of HI is 0.660mol / 2.00L = <em>0.330M, </em>the equlibrium concentrations will be:
[HI] = 0.330M - 2X
[H₂] = X
[I₂] = X
<em>Where X is reaction coefficient.</em>
<em />
Replacing in Kc:
0.0156 = [X] [X] / [0.330M - 2X]²
0.0156 = X² / [0.1089 - 1.32X + 4X²
]
0.00169884 - 0.020592 X + 0.0624 X² = X²
0.00169884 - 0.020592 X - 0.9376 X² = 0
Solving for X:
X = - 0.055 → False solution, there is no negative concentrations
X = 0.0330 → Right solution.
Replacing in HI formula:
[HI] = 0.330M - 2×0.033M
<h3>[HI] = 0.264M</h3>
m = 7.84x107/(3x108)2kg = 7.84x107/9x1016kg = 0.871x10-9 kg = 8.71x10-10 kg
