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Dmitriy789 [7]
3 years ago
8

N2 + 3H2 mc033-1.jpg 2NH3

Physics
1 answer:
Svetradugi [14.3K]3 years ago
3 0

N2+3H2->2NH3

When 3 moles of H2 react, they produce 2 moles of NH3

3 moles of H2 have a mass of 2.02 g

2 moles of NH2 have a mass of 17.0 g

So when 2.02 g H2 react, they produce 17.0 g NH3

If 26.3g H2 react with a yield of 100%, we expect…

2.02g H2_____17.0gNH3

26.3g H2_____x=221gNH3

So now let’s calculate the percentage:

221gNH3­­­­­­­­_________100%

79.0gNH3_________x=79.0*100/221=35.7%

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Find the weight of an astronaut whose mass is 75 kg on the moon
gladu [14]
The formula for weight is always weight=mass X gravitational field strength.
We already know the mass is 75kg.
The gravitational field strength on the moon is 1.6N. To find out the weight, we can substitute these values in to the formula.
Weight=75 X 1.6
Weight= 120N
Weight is measured on Neutons as it is a force.
8 0
3 years ago
An object accelerates to a velocity of 230 m/s over a time of 2.5 s. The acceleration it experienced was 42 m/s2. What was its i
zmey [24]

Answer:

230 = x + 105

x= 125

Explanation:

v = v0 + at

5 0
3 years ago
Karen claps her hand and hears the echo from
Alinara [238K]

Answer:

89 m

Explanation:

Applying

v = 2d/t................... Equation 1

Where v = velocity of sound in air, d = distance of the wall from Karen, t = time taken to hear the echo.

make d the subject of equation 1

d = vt/2..................... Equation 2

From the question,

Given: v = 343 m/s, t = 0.519 s

Substitute these values into equation 2

d = (343×0.519)/2

d = 89.01 m.

d ≈ 89 m

3 0
2 years ago
A ball is thrown downward at 12 m/s from a windowsill 35 m above the ground. At the same time, another ball is thrown upward at
wariber [46]

Answer:

The second ball lands 1.5 s after the first ball.

Explanation:

Given;

initial velocity of the ball, u = 12 m/s

height of fall, h = 35 m

initial velocity of the second, v = 12 m/s

Time taken for the first ball to land;

t = \sqrt{\frac{2h}{g} }\\\\t =\sqrt{ \frac{2*35}{9.8}}\\\\t = 2.67 \ s

determine the maximum height reached by the second ball;

v² = u² -2gh

at maximum height, the final velocity, v = 0

0 = 12² - (2 x 9.8)h

19.6h = 144

h = 144 / 19.6

h = 7.35 m

time to reach this height;

t_1 = \sqrt{\frac{2h}{g} }\\\\t_1 =  \sqrt{\frac{2*7.35}{9.8}}\\\\t_1 = 1.23 \ s

Total height above the ground to be traveled by the second ball is given as;

= 7.35 m + 35m

= 42.35 m

Time taken for the second ball to fall from this height;

t_2 = \sqrt{\frac{2h}{g} }\\\\t_2 = \sqrt{\frac{2*42.35}{9.8} }\\\\t_2 = 2.94 \ s

total time spent in air by the second ball;

T = t₁ + t₂

T = 1.23 s + 2.94 s

T = 4.17 s

Time taken for the second ball to land after the first ball is given by;

t = 4.17 s -  2.67 s

T = 1.5 s

Therefore, the second ball lands 1.5 s after the first ball.

4 0
3 years ago
URGENT BY THE WAY!
Nastasia [14]

Answer:

So Nessa went so fast it made her pass the tile in a second. Lets take a look at this problem, It says "the" tile so we should assume that it means 1 tile. Then draw a diagram representing that tile then you should have your problem finished. Hope that helped and I'm willing to help if you have anymore questions!

Explanation:

6 0
3 years ago
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