Answer:
Yes once a room is not rented for a couple of hours in the night then it is perishable.
Answer:
(c) Covalent bonding with some van der Waals bonding
Explanation:
Bonding present in the rubber is basically , covalent bonding with some van der waal bonding in between the individual chains .
Since ,
Rubber is mainly composed of the carbon and the hydrogen atoms ,
Hence , form the options , option (c) Covalent bonding with some van der Waals bonding is the most appropriate option .
Answer:
N_A=1.5*10^-8 kmol/s.m^2
Explanation:
<u>KNOWN: </u>
Molar concentration of helium at the inner and outer surfaces of a plastic membrane. Diffusion coefficient and membrane thickness.
<u>FIND:</u>
Molar diffusion flux.
<u>ASSUMPTIONS:</u>
(1) Steady-state conditions, (2) One-dimensional diffusion in a plane wall, (3) Stationary medium, (4) Uniform C = C_A + C_B.
<u>ANALYSIS:</u> The molar flux may be obtained from
N_A=D_AB/L(C_A,1-C_A,2)
=10^-9 m^2/s/0.001 m(0.02-0.005)kmol/m^3
N_A=1.5*10^-8 kmol/s.m^2
<u>COMMENTS:</u> The mass flux is:
n_A,x=M_a*N_A,x
n_A,x=6*10^-8 kg/s m^2
Answer:
Explanation:
a) for shifting reactions,
Kps = ph2 pco2/pcoph20
=[h2] [co2]/[co] [h2o]
h2 + co2 + h2O + co + c3H8 = 1
it implies that
H2 + 0.09 + H2O + 0.08 + 0.05 = 1
solving the system of equation yields
H2 = 0.5308,
H2O = 0.2942
B) according to Le chatelain's principle for a slightly exothermic reaction, an increase in temperature favors the reverse reaction producing less hydrogen. As a result, concentration of hydrogen in the reformation decreases with an increasing temperature.
c) to calculate the maximum hydrogen yield , both reaction must be complete
C3H8 + 3H2O ⇒ 3CO + 7H2( REFORMING)
CO + H2O ⇒ CO2 + H2 ( SHIFTING)
C3H8 + 6H2O ⇒ 3CO2 + 10 H2 ( OVER ALL)
SO,
Maximum hydrogen yield
= 10mol h2/3 molco2 + 10molh2
= 0.77
⇒ 77%