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HACTEHA [7]
3 years ago
14

Gas chromatography separates compounds depending on their__________ . Benzene, m-xylene, and toluene have similar_________ , the

refore, the main basis for separation is_______ . The more volatile a component the_______ , hence the more time it spends in the________ , giving it a________ retention time. Therefore, components of a liquid mixture will elute in order of_________ .
Engineering
1 answer:
amm18123 years ago
3 0

Answer and Explanation:

Gas chromatography separates compounds depending on their **polarity and volatility**. Benzene, m-xylene, and toluene have similar **polarities**, therefore, the main basis for separation is **volatility**. The more volatile a component the ** higher its vapor pressure**, hence the more time it spends in the **gaseous mobile phase**, giving it a **shorter** retention time. Therefore, components of a liquid mixture will elute in order of **increasing boiling points/decreasing volatilities/increasing polarities with the stationary phase**.

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Ethylene (C2H4) gas enters a well-insulated reactor and reacts completely with 400% of theoretical air, each at 25°C, 2 atm. The
GuDViN [60]

Answer:see explanation.

Explanation:

(a).The equation of reaction is;

C2H4 + 3(O2 + 3.76N2) -----> 2CO2 + 2H2O + 11.28 N2.

Theoretical air is the amount of air that will allow the complete combustion of the fuel. Air contains 21% of oxygen,79% of Nitrogen. Thus, the molecular mass of air = 29 [Kg/Kmol] . The number of oxygen needed to oxidize the hydrocarbon(C2H4) is 79/21= 3.76 moles of Nitrogen.

(b). N(total) =2+2+11.28 =15.28 kmoles of product

15.28[h(T) - h°] of air

= 15.28(Mass of air) (Cp,1000k)

T(adiabatic) = 802,310 kJ/k. Mole.

When mass of air= 29 Kg/ k.mol

Cp,1000k = 1.142 kJ.k which is the specific heat capacity of air.

T(adiabatic) = 1275k -----> assuming all the products are air.

(C). 2{∆h} co2 + 2{∆h} H20 +11.28 {∆h} N2

After series of calculations and checking of tables:

S(o2) = 320.173 - 8.314 ln 0.12= 337.46 kJ/ K.mol.k

S(H2O)= 273.986- 8.314 ln 0.1406 = 290.30 KJ/ kmol. K

S(N2) = 258.503- 8.314 ln 0.7344= 261.07 kJ/kmol.k

= 2(337.46) + 2(290.30)+11.28(261.07)-360.79-3(218.01)-11.28(193.46)

=1003.3378 kJ/kmol. K

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