What is the symbol of iron

Nitrogen and oxygen do not react appreciably at room temperature, as illustrated by our atmosphere. But at high temperatures, th

Gre4nikov [31]

Answer : The equilibrium concentration of NO is, 0.0092 M.

Solution :

First we have to calculate the concentration of NO.

$\text{Concentration of NO}=\frac{\text{Moles of }NO}{\text{Volume of solution}}=\frac{0.3152mol}{2.0L}=0.1576M$

The given equilibrium reaction is,

$N_2(g)+O_2(g)\rightleftharpoons 2NO(g)$

Initially conc. 0 0 0.1576

At eqm. (x) (x) (0.1576-2x)

The expression of $K_c$ will be,

$K_c=\frac{[NO]^2}{[N_2][O_2]}$

$0.0153=\frac{(0.1576-2x)^2}{(x)\times (x)}$

By solving the term, we get:

$x=0.0742,0.0839$

Neglecting the 0.0839 value of x because it can not be more than initial value.

Thus, the value of 'x' will be, 0.0742 M

Now we have to calculate the equilibrium concentration of NO.

Equilibrium concentration of NO = (0.1576-2x) = [0.1576-2(0.0742)] = 0.0092 M

Therefore, the equilibrium concentration of NO is, 0.0092 M.

5 0

3 years ago

What are the different areas of an electron cloud called?

max2010maxim [7]

I believe they are called Energy Levels or Energy Orbitals.

5 0

3 years ago

Naming chlorine as an Ion

Katarina [22]

The name of the ion is chloride.

To name a monatomic anion

• Drop the ending of the element name (chlorine → chlor)

• Add the ending ide (chlor + ide → chloride)

4 0

3 years ago

g Consider an ideal atomic gas in a cylinder. The upper part of the cylinder is a moveable piston of negligible weight. The heig

kumpel [21]

A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.

A cylindrical weight with a mass (m) of 3 kg is dropped, that is, its initial velocity (u) is 0 m/s and travels 10 m (s). Assuming the acceleration (a) is that of gravity (9.8 m/s²). We can calculate the velocity (v) of the weight in the instant prior to the collision with the piston using the following kinematic equation.

$v^{2} = u^{2} + 2as = 2 (9.8 m/s^{2} ) (10m) \\\\v = 14 m/s$

The object with a mass of 3 kg collides with the piston at 14 m/s, The kinetic energy (K) of the object at that moment is:

$K = \frac{1}{2} m v^{2} = \frac{1}{2} (3kg) (14m/s)^{2} = 294 J$

The kinetic energy of the weight is completely converted into heat transferred into the gas cylinder. Thus, Q = 294 J.

Given all the process is at 250 K (T), we can calculate the change of entropy of the gas using the following expression.

$\Delta S_{gas} = \frac{Q}{T} = \frac{294 J}{250K} = 1.18 J/K$

The change in the entropy of the environment, has the same value but opposite sign than the change in the entropy of the gas. Thus, $\Delta S_{env} = -1.18 J/K$

A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.

Learn more: brainly.com/question/22655760

6 0

2 years ago

Ulfur dioxide gas can react in the presence of oxygen and vanadium(V) oxide to form sulfur trioxide. Sulfur trioxide is the only

LenaWriter [7]

According to the illustration, the vanadium (V) oxide would be a catalyst.

<h3>What are catalysts?</h3>

Catalysts are substances that are utilized in reactions that are not themselves consumed in reactions but only speed up the rate of the reactions.

Catalysts speed up the rate of reactions by lowering the activation energy of the reactants.

Sulfur dioxide reacts with oxygen to produce sulfur trioxide. The vanadium (v) oxide is not consumed in the reaction. Thus it only serves as a catalyst.

More on catalysts can be found here: brainly.com/question/12260131

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4 0

2 years ago