Ethylene () is the starting point for a wide array of industrial chemical syntheses. For example, worldwide about of polyethylene are made from ethylene each year, for use in everything from household plumbing to artificial joints. Natural sources of ethylene are entirely inadequate to meet world demand, so ethane () from natural gas is "cracked" in refineries at high temperature in a kinetically complex reaction that produces ethylene gas and hydrogen gas. Suppose an engineer studying ethane cracking fills a reaction tank with of ethane gas and raises the temperature to . He believes at this temperature. Calculate the percent by mass of ethylene the engineer expects to find in the equilibrium gas mixture. Round your answer to significant digits. Note for advanced students: the engineer may be mistaken about the correct value of , and the mass percent of ethylene you calculate may not be what he actually observes.

The question is incomplete, here is the complete question:

Ethylene is the starting point for a wide array of industrial chemical syntheses. For example, worldwide about 8.0 x 10¹⁰ of polyethylene are made from ethylene each year, for use in everything from household plumbing to artificial joints. Natural sources of ethylene are entirely inadequate to meet world demand, so ethane from natural gas is "cracked" in refineries at high temperature in a kinetically complex reaction that produces ethylene gas and hydrogen gas.

Suppose an engineer studying ethane cracking fills a 30.0 L reaction tank with 38.0 atm of ethane gas and raises the temperature to 400°C. He believes Kp = 0.4 at this temperature. Calculate the percent by mass of ethylene the engineer expects to find in the equilibrium gas mixture.

The mass percent of ethylene gas is 9.20 %

Explanation:

We are given:

Initial partial pressure of ethane gas = 38.0 atm

The chemical equation for the dehydrogenation of ethane follows:

Initial:             38

At eqllm:      38-x         x       x

The expression of for above equation follows:

We are given:

Putting values in above equation, we get:

Neglecting the negative value of 'x' because partial pressure cannot be negative

So, equilibrium partial pressure of ethane = 38 - x = 38 - 3.70 = 34.30 atm

Equilibrium partial pressure of ethene = x = 3.70 atm

To calculate the number of moles, we use the equation given by ideal gas which follows:

        ..........(1)

To calculate the number of moles, we use the equation:

    .....(2)

We are given:

Putting values in equation 1, we get:

Molar mass of ethane gas = 30 g/mol

Moles of ethane gas = 18.62 mol

Putting values in equation 2, we get:

We are given:

Putting values in equation 1, we get:

Molar mass of ethylene gas = 28 g/mol

Moles of ethylene gas = 2.01 mol

Putting values in equation 2, we get:

Mass of ethylene = 56.28 g

Mass of mixture = [558.6 + 56.28] g = 641.88 g

Putting values in above equation, we get:

Hence, the mass percent of ethylene gas is 9.20 %

answer; i believe the correct answer is (c) /// is a homogeneous mixture of gaseous substance nitrogen,oxygen and smaller amounts of other substancesmixtures can be either ( homogeneous or heterogeneous)plz check first good luck

a rain shadow is a dry region of land on the side that is protected against the prevailing winds of a mountain range. the prevailing winds are the winds that occur most of the time in a particular place of the earth

prevailing winds carry air toward the mountain range. as the air rises up over a mountain range, the air cools, water vapor condenses, and clouds form. on this side of the mountains, called the windward side, precipitation falls in the form of  rain  or snow. this dry air produces a  rain shadow.