Answer: the equilibrium conversion is 0.15 or 15mole %
K=1.33
Explanation:
Given the mole percent of the gases to be as follows
SO2(g) =15% = 0.15mol
O2(g) =20% = 0.20mol
N2(g) = 65% = 0.65mol
The equilibrium conversion reaction is given by
2SO2(g) + O2(g) <------> 2SO3(g)
2 1. 2
According to the reaction 2 moles of SO2 is converted to get 2 mole of SO3
Therefore 0.15 mole will produce 0.15 mole of SO3
So at equilibrium:
2SO2(g) + O2(g) ------> 2SO3(g)
0.15. 0.075. 0.15
Now the equilibrium constant K is given by:
K = [SO3]²/[SO2]²[O2]
Since the container is constant for all the reactants we can neglect the container capacity and use the mole as percent concentration.
K = [0.15]²/[0.15]²[0.075]
K = 1/0.075
K= 1.33
Answer:
mechanical efficiency = 54%
Explanation:
given data
power We = 44 kW
density = 860 kg/m³
rate of flow = 0.07 m³/s
inlet diameter d1 = 8 cm
outlet diameter d2 = 12 cm
pressure = 500 kPa
efficiency η = 90%
to find out
mechanical efficiency
solution
we get here shaft power is here
shaft power = efficiency × electric power
shaft power = 0.9 × 44 = 39.6
and
now we get DEmech that is
DE(mech)= m g h +
(v² - u ²) .....................1
that is we can write as
DEmech = ρVgh +
= VDP +
×
Plugging in values we now have DEmech that is
DEmech = (0.07m^3/s × 500kPa) +
solve we get
DEmech = 21342 W
DEmech = 21.3 kW
mechanical efficiency is
mechanical efficiency = 
mechanical efficiency = 0.54
mechanical efficiency = 54%
Answer: you need 4567 km but also mph Kikuyus
Explanation:
Answer:
Δr=20.45 %
Explanation:
Given that
Rake angle α = 15°
coefficient of friction ,μ = 0.15
The friction angle β
tanβ = μ
tanβ = 0.15
β=8.83°
2φ + β - α = 90°
φ=Shear angle
2φ + 8.833° - 15° = 90°
φ = 48.08°
Chip thickness r given as


r=0.88
New coefficient of friction ,μ' = 0.3
tanβ' = μ'
tanβ' = 0.3
β'=16.69°
2φ' + β' - α = 90°
φ'=Shear angle
2φ' + 16.69° - 25° = 90°
φ' = 49.15°
Chip thickness r' given as


r'=0.70
Percentage change


Δr=20.45 %
Answer:
Explanation:
To obtain 240 V from 12 V batteries they must be connected in series. The number needed is ...
240/12 = 20 . . . batteries needed
__
The current draw will be ...
(3000 W)/(240 V) = 12.5 A
Then the time available from the battery stack is ...
(120 Ah)/(12.5 A) = 9.6 h
The motor can run 9.6 hours from the series connection.