What mass of NH3 can be made from 35.0 grams of N2
1 answer:
42.5g
Explanation:
Given parameters:
Mass of N₂= 35g
Unknown:
Mass of NH₃ = ?
Solution:
Equation of the reaction:
N₂ + 3H₂ → 2NH₃
To solve this problem, we work from the known species to the unknown. The known here is the reacting mass of the N₂. From this we can find the number of moles of the N₂.
Number of moles of N₂ =
molar mass of N₂ = 28g/mol
Number of moles = = 1.25moles
From the equation of the reaction:
1 mole of N₂ produced 2 moles of NH₃
1.25 moles of N₂ will produce 2 x 1.25 = 2.5moles of NH₃
Mass of NH₃ = Number of moles of NH₃ x molar mass of NH₃
= 2.5 x (14 + 3)
= 42.5g
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Answer:
a) 75.5 degree relative to the North in north-west direction
b) 309.84 km/h
Explanation:
a)If the pilot wants to fly due west while there's wind of 80km/h due south. The north-component of the airplane velocity relative to the air must be equal to the wind speed to the south, 80km/h in order to counter balance it
So the pilot should head to the West-North direction at an angle of
relative to the North-bound.
b) As the North component of the airplane velocity cancel out the wind south-bound speed. The speed of the plane over the ground would be the West component of the airplane velocity, which is
Answer:
Explanation:
First, denote our known values;
Mass is impulse divided by change in velocity:
Hence, the mass of the ball is 141.30grams
Answer:
h = 10.4 m
R = 22.48 m
v= 16,2 m/s , α = 61.7°, below the horizontal
v = (7.7)i + (-14.3)j in meters per second (i horizontal, j downward)
Explanation:
The ball describes a parabolic path, and the equations of the movement are:
Equation of the uniform rectilinear motion (horizontal ) :
x = vx*t :
Equations of the uniformly accelerated rectilinear motion of upward (vertical ).
y = (v₀y)*t - (1/2)*g*t² Equation (2)
vfy² = v₀y² -2gy Equation (3)
vfy = v₀y -gt Equation (4)
Where:
x: horizontal position in meters (m)
t : time (s)
vx: horizontal velocity in m
y: vertical position in meters (m)
v₀y: initial vertical velocity in m/s
vfy: final vertical velocity in m/s
g: acceleration due to gravity in m/s²
Known data
y= 8.8 m
v = ( (7.7)i + (5.7)j ) m/s : vx= 7.7 m/s , vy= 5.7 m/s
g = 9.8 m/s²
Calculation of the initial vertical velocity ( v₀y)
We apply Equation (3) with the known data
(vfy)² = (v₀y)² -2*g*y
(5.7)² = (v₀y)²- (2)*(9.8)*(8.8)
(5.7)²+ 172.48 = (v₀y)²
v₀y = 14.3 m/s
Calculation of the maximum height the ball rise (h)
In the maximum height vfy=0
We apply the Equation (3) :
(vfy)² = (v₀y)² -2*g*y
0 = (14.3)² - 2*98*h
h = (14.3)² / 19.6
h = 10.4 m
Calculation of the time it takes for the ball to the maximum height
We apply the Equation (4) :
vfy = v₀y -gt
0 = v₀y -gt
gt = v₀y
t = v₀y/g
t = 14.3/9.8
t= 1.46 s
Flight time = 2t = 2.92 s
Total horizontal distance traveled by the ball (R)
We replace data in the equation (1)
x =vx*t vx= 7.7 m/s , t =2.92 s (Flight time)
R = (7.7)* (2.92) = 22.48 m
Velocity of the ball (magnitude (v) and direction (α)) the instant before it hits the ground
vx = 7.7 m/s
vy = v₀y -gt = 14.3 - 9.8* (2.92) = -14.3 m/s
v= 16,2 m/s
α = -61.7°
α = 61.7°, below the horizontal
i- j components of the v
v = (7.7)i + (-14.3)j in meters per second (i horizontal, j downward)