What is the volume of 0.24 mol oxygen (O2) gas at 345 K and a pressure of
1 answer:
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a. t=0.553 s
b. vox(horizontal speed) = 3.62 m/s
<h3>Further explanation</h3>Given
h = 1.5 m
x = 2 m
Required
a. time
b. vo=initial speed
Solution
Free fall motion
a. h = 1/2 gt²(vertical motion=h=voyt+1/2gt²⇒voy = 0)
t = √2h/g
t = √2.1.5/9.8
t=0.553 s
b. x=vox.t(horizontal motion)
vox=x/t
vox=2/0.553
vox=3.62 m/s
Answer:
The rate at which ammonia is being produced is 0.41 kg/sec.
Explanation:
Haber reaction
Volume of dinitrogen consumed in a second = 505 L
Temperature at which reaction is carried out,T= 172°C = 445.15 K
Pressure at which reaction is carried out, P = 0.88 atm
Let the moles of dinitrogen be n.
Using an Ideal gas equation:
According to reaction , 1 mol of ditnitrogen gas produces 2 moles of ammonia.
Then 12.1597 mol of dinitrogen will produce :
of ammonia
Mass of 24.3194 moles of ammonia =24.3194 mol × 17 g/mol
=413.43 g=0.41343 kg ≈ 0.41 kg
505 L of dinitrogen are consumed in 1 second to produce 0.41 kg of ammonia in 1 second. So the rate at which ammonia is being produced is 0.41 kg/sec.