Sucrose parafilm sulfur after sting heating what kind of of physical or chemical

While ethanol is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by reacting e

sp2606 [1]

The question is incomplete, here is the complete question:

While ethanol is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by reacting ethylene with water vapor at elevated temperatures.

A chemical engineer studying this reaction fills a 1.5 L flask at 12°C with 1.8 atm of ethylene gas and 4.7 atm of water vapor. When the mixture has come to equilibrium she determines that it contains 1.16 atm of ethylene gas and 4.06 atm of water vapor.

The engineer then adds another 1.2 atm of ethylene, and allows the mixture to come to equilibrium again. Calculate the pressure of ethanol after equilibrium is reached the second time. Round your answer to 2 significant digits.

<u>Answer:</u> The partial pressure of ethanol after equilibrium is reached the second time is 1.0 atm

<u>Explanation:</u>

We are given:

Initial partial pressure of ethylene gas = 1.8 atm

Initial partial pressure of water vapor = 4.7 atm

Equilibrium partial pressure of ethylene gas = 1.16 atm

Equilibrium partial pressure of water vapor = 4.06 atm

The chemical equation for the reaction of ethylene gas and water vapor follows:

$CH_2CH_2(g)+H_2O(g)\rightleftharpoons CH_3CH_2OH(g)$

<u>Initial:</u> 1.8 4.7

<u>At eqllm:</u> 1.8-x 4.7-x

Evaluating the value of 'x'

$\Rightarrow (1.8-x)=1.16\\\\x=0.64$

The expression of $K_p$ for above equation follows:

$K_p=\frac{p_{CH_3CH_2OH}}{p_{CH_2CH_2}\times p_{H_2O}}$

$p_{CH_2CH_2}=1.16atm\\p_{H_2O}=4.06atm\\p_{CH_3CH_2OH}=0.64atm$

Putting values in above expression, we get:

$K_p=\frac{0.64}{1.16\times 4.06}\\\\K_p=0.136$

When more ethylene is added, the equilibrium gets re-established.

Partial pressure of ethylene added = 1.2 atm

$CH_2CH_2(g)+H_2O(g)\rightleftharpoons CH_3CH_2OH(g)$

<u>Initial:</u> 2.36 4.06 0.64

<u>At eqllm:</u> 2.36-x 4.06-x 0.64+x

Putting value in the equilibrium constant expression, we get:

$0.136=\frac{(0.64+x)}{(2.36-x)\times (4.06-x)}\\\\x=0.363,13.41$

Neglecting the value of x = 13.41 because equilibrium partial pressure of ethylene and water vapor will become negative, which is not possible.

So, equilibrium partial pressure of ethanol = (0.64 + x) = (0.64 + 0.363) = 1.003 atm

Hence, the partial pressure of ethanol after equilibrium is reached the second time is 1.0 atm

3 0

3 years ago

How do you think the temperatures of these two cooking appliances compare?

asambeis [7]

What is the two coking appliances

6 0

3 years ago

Positive Deviation from Raoult's Law occurs when the vapour pressure of component is greater than what is expected in Raoult's L

8090 [49]

Answer:

All of the above.

Explanation:

In positive deviation from Raoult's Law occur when the vapour pressure of components is greater than what is expected value in Raoult's law.

When a solution is non ideal then it shows positive or negative deviation.

Let two solutions A and B to form non- ideal solutions.let the vapour pressure of component A is $P_A$ and vapour pressure of component B is $P_B$ .

$P^0_A$ = Vapour pressure of component A in pure form

$P^0_B$ = Vapour pressure of component B in pure form

$x_A$ =Mole fraction of component A

$x_B=$ =Mole fraction of component B

The interaction between A- B is less than the interaction A- A and B-B interaction.Therefore, the escaping tendency of liquid molecules in mixture is greater than the escaping tendency in pure form.Hence, the vapour pressure of a mixture is greater than the initial value of vapour pressure.

$P_A >P^0_A\cdot x_A$ , $P_B>P^0_B\cdot x_B$

Therefore, $P_A+P_B >P^0_A\cdot x_A+P^0_B \cdot x_B$

Therefore, the enthalpy of mixing is greater than zero and change in volume is greater than zero.

Hence, option a,b,c and d are true.

6 0

3 years ago

. the density (d) of a substance is an intensive property that is defined as the ratio of its mass (m) to its volume (v). densit

ale4655 [162]

Density is intensive because it is the ratio of two extensive properties that is mass to volume.

<h3>What is Density ?</h3>

Density is defined as mass per unit volume. S.I unit if density is kg/m³.

It is expressed as

Density = $\frac{\text{Mass}}{\text{Volume}}$ or $d = \frac{m}{V}$

<h3>What is Intensive Property ? </h3>

Intensive property is the property which does not depend on the size of the system. Intensive property can be easily identified. Intensive property is the independent property. The size does not change in this.

Example of intensive property are Density, Freezing point, melting point, colour, Lustre, etc.

<h3>What is Extensive property ?</h3>

Extensive property is the property which depend on the substance. Extensive property cannot be easily identified.

Example of extensive property length, weight, volume, mass etc.

Thus from the above conclusion we can say that Density is intensive because it is the ratio of two extensive properties that is mass to volume.

Learn more about Density here: brainly.com/question/1354972

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Disclaimer: The given question is incomplete on the portal. Here is the complete question.

Question: The density (d) of a substance is an intensive property that is defined as the ratio of its mass (m) to its volume (v)

density = $\frac{\text{Mass}}{\text{Volume}}$ or $d = \frac{m}{V}$

Considering that mass and volume are both extensive properties, explain why their ratio, density is intensive.

5 0

2 years ago

"You have a solution of glucose in water that has a concentration of 2.50 M and a volume of 0.442 liters. You dilute this soluti

taurus [48]

Answer:

0.737M

Explanation:

C1V1=C2V2

2.50 x 0.442 = C x 1.5

1.105 = C x 1.5

C = 1.105/1.5

C = 0.737M

7 0

3 years ago