If the sun shining through windows hits the air in a sealed room what happens to the air pressure in that room

In the Haber process for ammonia synthesis, K " 0.036 for N 2 (g) ! 3 H 2 (g) ∆ 2 NH 3 (g) at 500. K. If a 2.0-L reactor is char

lisabon 2012 [21]

Answer : The partial pressure of $N_2,H_2\text{ and }NH_3$ at equilibrium are, 1.133, 2.009, 0.574 bar respectively. The total pressure at equilibrium is, 3.716 bar

Solution : Given,

Initial pressure of $N_2$ = 1.42 bar

Initial pressure of $H_2$ = 2.87 bar

$K_p$ = 0.036

The given equilibrium reaction is,

$N_2(g)+H_2(g)\rightleftharpoons 2NH_3(g)$

Initially 1.42 2.87 0

At equilibrium (1.42-x) (2.87-3x) 2x

The expression of $K_p$ will be,

$K_p=\frac{(p_{NH_3})^2}{(p_{N_2})(p_{H_2})^3}$

Now put all the values of partial pressure, we get

$0.036=\frac{(2x)^2}{(1.42-x)\times (2.87-3x)^3}$

By solving the term x, we get

$x=0.287\text{ and }3.889$

From the values of 'x' we conclude that, x = 3.889 can not more than initial partial pressures. So, the value of 'x' which is equal to 3.889 is not consider.

Thus, the partial pressure of $NH_3$ at equilibrium = 2x = 2 × 0.287 = 0.574 bar

The partial pressure of $N_2$ at equilibrium = (1.42-x) = (1.42-0.287) = 1.133 bar

The partial pressure of $H_2$ at equilibrium = (2.87-3x) = [2.87-3(0.287)] = 2.009 bar

The total pressure at equilibrium = Partial pressure of $N_2$ + Partial pressure of $H_2$ + Partial pressure of $NH_3$

The total pressure at equilibrium = 1.133 + 2.009 + 0.574 = 3.716 bar

6 0

3 years ago

When is a roman numeral most likely needed in the name of an ionic compound?

Sauron [17]

The answer is I’m guessing c

5 0

3 years ago

Read 2 more answers

8. How do you know when kinetic energy is not changing?

Nostrana [21]

Answer:

Kinetic energy is the energy of motion so to figure out that it’s not changing is if the object is still moving. If it’s staying still or is at rest, it is presenting potential energy, which is when energy is being stored inside the object.

5 0

3 years ago

Acetic acid, ethanol, acetaldehyde and ethane form a series of 2-carbon molecules which differ in their extent of oxidation. Con

yanalaym [24]

Explanation:

Ethanol can be oxidized to ethanal or acetaldehyde which is further oxidized to acid that is acetic acid.

$CH_3CH_2OH$ → $CH_3CHO$ [oxidation by loss of hydrogen]

An oxidizing agent potassium dichromate(VI) solution is used to remove the hydrogen from the ethanol.

An oxidizing agent used along with dilute sulphuric acid for acidification.

Acetaldehyde can also be reduced back to ethanol again by adding hydrogen to it by using a reducing agent that is sodium tetrahydro borate, NaBH4.

The oxidation of aldehydes to carboxylic acids can be done by the two-step process.

In the first step, one molecule of water is added in the presence of a catalyst that is acidic.

There is a generation of a hydrate. (geminal 1,1-diol).

$CH_3CHO$ → $CH_3CH_2COOH$ [reduction by the gain of electrons]

Here, the oxidizing agent used is $Cr_3O$ in the presence of acetone.

7 0

3 years ago

Oh no! Igor stepped on a piece of glass on the sidewalk in front of

Helen [10]

Answer: The correct sequence for the series of event would be.

1. Igor's toe is being cut by the glass.

2. The wound surrounding the injury becomes infected with bacteria from Igor's foot.

3. Antibodies and circulating white blood cells stick to the bacteria creating a large complex in the lymph vessel.

4. The bacteria enter his lymph system and travel towards a lymph node.

5. The complex becomes trapped in a lymph node and is engulfed by a phagocyte.

6. The bacteria is destroyed.

Explanation:

Whenever there is any cut or wound in the body and body encounters invasion of the foreign materials it considers it as harmful pathogen.

These pathogens when enters the body it is considered as antigen, it then travels to the lymphatic system.

These bacterial complex is then killed by the phagocytes and digested by the body.

In this way the complex is killed and the bacteria is destroyed.

8 0

3 years ago