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raketka [301]
3 years ago
10

Vapour-liquid equilibrium of a two-component ideal solution of trichloroethene (C2HCl3) and trichloromethane (CHCl3) is establis

hed at 25 °C. The mole fraction of CHCl3 in the vapour phase is 0.59. What is the mass fraction of C2HCl3 in the liquid phase? Round your answer to two significant figures.
The vapour pressures of trichloroethene and trichloromethane at 25 °C are:

Pvap,C2HCl3 = 73.0 mmHg

Pvap,CHCl3 = 199.1 mm Hg
Chemistry
1 answer:
riadik2000 [5.3K]3 years ago
4 0

Answer:

dajdajkwdanx

Explanation:

i need points

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There are two steps in the usual industrial preparation of acrylic acid, the immediate precursor of several useful plastics. In
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The net change in enthalpy for the formation of one mole of acrylic acid from calcium carbide, water and carbon dioxide is 523.2 kJ.

Explanation:

Step 1:

CaC_2(s) + 2H_2O(g)\rightarrow C_2H_2(g) + Ca(OH)_2(s),\Delta H_1=414.0 kJ...[1]

Step 2 :

6C_2H_2(g) + 3CO_2(g) + 4H_2O(g)\rightarrow 5CH_2CHCO_2H(g) \Delta H_2=132.0kJ..[2]

Adding 6 × [1] and [2]:

6CaC_2(s) + 12H_2O(g)\rightarrow 6C_2H_2(g) + 6Ca(OH)_2(s)

6C_2H_2(g)+3CO_2(g)+16H_2O(g)\rightarrow 5CH_2CHCO_2H(g)

we get :

6CaC_2(s) + 8H_2O(g)+3CO_2(g)\rightarrow 5CH_2CHCO_2H(g)+ 6Ca(OH)_2(s),\Delta H'=?

\Delta H'=6\times \Delta H_1+\Delta H_2

\Delta H'=6\times 414.0 kJ+132.0kJ

\Delta H'=2,626 kJ

Energy released on formation of 5 moles of acrylic acid = 2,626 kJ

Energy released on formation of 1 mole of acrylic acid:

\frac{ 2,626 kJ}{5 } = 523.2 kJ

7 0
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