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Lina20 [59]
3 years ago
6

4 kmol Methane (CH4) is burned completely with the stoichiometric amount of air during a steady-flow combustion process. If both

the reactants and the products are maintained at 25°C and 1 atm and the water in the products exists in the liquid form, determine the heat transfer from the combustion chamber during this process. What would your answer be if combustion were achieved with 100 percent excess air?
Chemistry
1 answer:
Svetach [21]3 years ago
7 0

Answer:

Heat transfer = 3564 Jolues

The same value

Explanation:

The heat of combustion is the heat released per 1 mole of substunce experimenting the combustion at standard conditions of pressure and temperature ( 1 atm, 298 K):

Qtransfer = - mol x ΔHºc Qtransfer

So look up in appropiate reference table ΔHºc  and solve the problem:

ΔHºc  = - 891 kJ/mol

Qtransfer = - (4 x 10³ mol x  -891 kJ/mol ) = 3564 J

if the combustion were achieved with 100 % excess air, the result will still be the same. As long as the standard conditions are maintained, the heat of combustion remains constant. In fact in many cases the combustion is performed under excess oxygen to ensure complete combustion.

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a balloon contains 30.0 L of helium gas at 103 kPA. what is the volume of the helium when the balloon rises to an altitude where
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Answer:
                V₂  =  123.6 L

Explanation:

According to Boyle's law pressure and volume of a gas are inversely related if amount and temperature are kept constant. For the initial and final states the gas law is given as,

                                                P₁ V₁  =  P₂ V₂     ----- (1)
Data Given;
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                   V₁  =  30 L

                   P₂  =  25 kPa

                   V₂  =  ?

Solution:

Solving equation 1 for V₂,

                                        V₂  =  P₁ V₁ / P₂

Putting values,
                                        V₂  =  (103 kPa × 30 L) ÷ 25 kPa

                                        V₂  =  123.6 L

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NH_{3} is a polar protic molecule. Hence it can form hydrogen bonding with another polar protic molecule CH_{2}NH. Also dipole-dipole attraction force along with london dispersion force exists between CH_{2}NH and NH_{3}.

CH_{3}I is a polar molecule. Hence only dipole-dipole attraction force along with london dispersion force operates between CH_{3}I and NH_{3}.

C_{2}H_{6} is a non-polar molecule. Hence only london dispersion force exists between C_{2}H_{6} and NH_{3}.

So, order of miscibility: HF>CH_{2}NH>CH_{3}I>C_{2}H_{6}

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