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MatroZZZ [7]
3 years ago
6

Ammonia, NH3, can be made by reacting nitrogen and hydrogen and the equation is N2 + 3H2 --> 2NH3 How many moles of NH3 can b

e made if 19 moles of H2 react with enough N2? mol
Physics
2 answers:
horrorfan [7]3 years ago
8 0
Make a proportion
3 H2 - 2 NH3
19H2 - x
x = (19x2)/3= 12,666666
stira [4]3 years ago
4 0
You have a balanced equation that says one mole of N_2 plus three moles of H_2 react to produce two moles of NH_3. So if 2 moles of hydrogen are required for 3 moles of ammonia, then you can make the simplest of equations to solve this. That is,

3n_H=2n_A

Solving for the number of moles of hydrogen then

n_H=\frac{2}{3}n_A

You were given the number of moles of ammonia so plug it in and you are done!
You might be interested in
The amplitude of a lightly damped oscillator decreases by 3.42% during each cycle. What percentage of the mechanical energy of t
frozen [14]

Answer:

The percentage of the mechanical energy of the oscillator lost in each cycle is 6.72%

Explanation:

Mechanical energy (Potential energy, PE) of the oscillator is calculated as;

PE = ¹/₂KA²

During the first oscillation;

PE₁ = ¹/₂KA₁²

During the second oscillation;

A₂ = A₁ - 0.0342A₁ = 0.9658A₁

PE₂ = ¹/₂KA₂²

PE₂ = ¹/₂K (0.9658A₁)²

PE₂ = (0.9658²)¹/₂KA₁²

PE₂ = (0.9328)¹/₂KA₁²

PE₂ = 0.9328PE₁

Percentage of the mechanical energy of the oscillator lost in each cycle;

Change \ in \ percent= \frac{PE_2 - PE_1}{PE_1} \\\\Change \ in \ percent= \frac{0.9328PE_1 -PE_1}{PE_1} *100\\\\Change \ in \ percent= \frac{-0.0672PE_1}{PE_1}*100 \\\\Change \ in \ percent= -0.0672*100\\\\Change \ in \ percent= -6.72 \ \% \\\\Loss \ in \ percent= 6.72 \ \%

Therefore, the percentage of the mechanical energy of the oscillator lost in each cycle is 6.72%

4 0
3 years ago
Based on its orbit, which planet behaves the least like the others?
Gekata [30.6K]
I believe it's Mercury, because the only other option would be Pluto and it's not even considered a planet anymore
Hope this helps
3 0
3 years ago
Will the velocity of the book change as it moves across the surface with NO friction? Explain your answer.
MakcuM [25]

No velocity will not be changed

Why?

According to Newtons 1st law the velocity of a moving object remains unchanged unless a external force affect that.

6 0
2 years ago
Read 2 more answers
an object 24.0 cm from a concave mirror creates a virtual image at -33.5 cm. if the image is 7.25 cm tall, what is the height of
Hitman42 [59]

The height of the object will be -5.19 cm

A concave mirror's reflecting surface curves inward and away from the light source. Light is reflected inward to a single focus point via concave mirrors. Concave mirrors, in contrast to convex mirrors, produce a variety of images depending on the object's to the mirror.

Given an object 24.0 cm from a concave mirror creates a virtual image at -33.5 cm. if the image is 7.25 cm tall

So let,

v =  Image distance from the mirror = -33.5 cm

u = object distance from the mirror (concave) = 24 cm

hi = Image height = 7.25 cm

h = height of the object = ?

Using below formula to find height of the object

-v/u = hi/h

Putting all value in the formula we get

-(-33.5)/(-24) = 7.25/h

h = -5.19 cm

Therefore the height of the object will be -5.19 cm

Learn more about Concave mirror here:

brainly.com/question/3727024

#SPJ10

3 0
2 years ago
An electron in the n = 5 level of an h atom emits a photon of wavelength 1282.17 nm. to what energy level does the electron move
lions [1.4K]
This is an interesting (read tricky!) variation of Rydberg Eqn calculation.
Rydberg Eqn: 1/λ = R [1/n1^2 - 1/n2^2]
Where λ is the wavelength of the light; 1282.17 nm = 1282.17×10^-9 m
R is the Rydberg constant: R = 1.09737×10^7 m-1
n2 = 5 (emission)
Hence 1/(1282.17 ×10^-9) = 1.09737× 10^7 [1/n1^2 – 1/25^2]
Some rearranging and collecting up terms:
1 = (1282.17 ×10^-9) (1.09737× 10^7)[1/n2 -1/25]
1= 14.07[1/n^2 – 1/25]
1 =14.07/n^2 – (14.07/25)
14.07n^2 = 1 + 0.5628
n = √(14.07/1.5628) = 3
8 0
3 years ago
Read 2 more answers
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