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lilavasa [31]
3 years ago
14

The equation shows the formation of ammonia from nitrogen and hydrogen. How will increasing the temperature affect the equilibri

um? A. more energy is released B. more ammonia forms C. the system is unaffected by the change D. more hydrogen and nitrogen form

Physics
2 answers:
bogdanovich [222]3 years ago
8 0

Your answer is D.

Hope this helps:)

Serjik [45]3 years ago
3 0

Answer: D. more hydrogen and nitrogen form

Explanation:

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

For the given equation:

2N_2(g)+3H_2(g)\leftrightharpoons 2NH_3+energy

This is a type of exothermic reaction because heat is released in the reaction.

If the temperature is increased, so according to the Le-Chatlier's principle , the equilibrium will shift in the direction where decrease in temperature occurs. As forward reaction is exothermic reaction, it will increase the temperature and backward reaction being endothermic will decrease the temperature. Hence, the equilibrium will shift in the left or backward direction.

Thus more hydrogen and nitrogen will be formed.

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A solid non-conducting sphere of radius R carries a charge Q1 distributed uniformly. The sphere is surrounded by a concentric sp
STALIN [3.7K]

Answer:

E = k Q₁ / r²

Explanation:

For this exercise that asks us for the electric field between the sphere and the spherical shell, we can use Gauss's law

           Ф = ∫ E .dA = q_{int} / ε₀

where Ф the electric flow, qint is the charge inside the surface

To solve these problems we must create a Gaussian surface that takes advantage of the symmetry of the problem, in this almost our surface is a sphere of radius r, that this is the sphere of and the shell, bone

           R <r <R_a

for this surface the electric field lines are radial and the radius of the sphere are also, therefore the two are parallel, which reduces the scalar product to the algebraic product.

         E A = q_{int} /ε₀

The charge inside the surface is Q₁, since the other charge Q₂ is outside the Gaussian surface, therefore it does not contribute to the electric field

          q_{int} = Q₁

The surface area is

          A = 4π r²

we substitute

          E 4π r² = Q₁ /ε₀

          E = 1 / 4πε₀ Q₁ / r²

          k = 1/4πε₀

 

          E = k Q₁ / r²

6 0
3 years ago
What kind of energy involves the flow of positive charges?
solniwko [45]
Electrical

Electrical would be the energy that involves the flow of positive changes
7 0
3 years ago
Read 2 more answers
A solid sphere of radius 40.0cm has a total positive charge of 26.0μC uniformly distributed throughout its volume. Calculate the
Rudiy27

The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C

R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m

Q\;(\text{total charge of the solid sphere})=(26\;\mathrm{\mu C})\left(\dfrac{1\;\mathrm{C}}{10^6\;\mathrm{\mu C}} \right)={26\times 10^{-6}\;\mathrm{C}}

Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:

E=\dfrac{Q}{4\pi\epsilon_0 r^2}

Substitute numerical values:

E&=\dfrac{24\times 10^{-6}}{4\pi (8.8542\times 10^{-12})(0.6)}\\ &={6.49\times 10^5\;\mathrm{N/C}\;\text{directed radially outward}}}

The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.

As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).

Learn more about Gaussian sphere here:

brainly.com/question/2004529

#SPJ4

6 0
2 years ago
Velocity of an object of mass 5 kg increases from 3 m/s to 7 m/s on applying a force continuously for 2 s. Find out the force ap
Alenkasestr [34]

Answer:

Force=10N

Velocity=13m/s

Explanation:

Given,

Mass = 5kg

initial velocity=3m/s

Final velocity=7m/s

Time=2s

Now acceleration= v-u/t

=7-3/2

=4/2

=2m/s²

So, acceleration=2m/s²

Now,

F=ma

=5*2

=10N

7 0
2 years ago
A main sequence star cools and expands. It must be entering which stage?
aev [14]
If a main sequence star is cooling and expanding, it is entering the red giant stage. This means that the star has burned up all of it's hydrogen and is now starting to burn its helium making it cooler.
4 0
4 years ago
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