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3 years ago

What happens to the particles during condensation

Chemistry

1 answer:

Paraphin [41]3 years ago

8 0

Answer:

molecus in a gas cool down

Explanation:

the molecules lose heat and energy do they slow down they move closer to other molecules and form a liquid

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3 years ago

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Do chemical equations with no reaction still have to be balanced

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3 years ago

Why do you think it is a bad idea to use taste, touch, or smell to identify an unknown chemical?

miv72 [106K]

Answer:

Because it can cause health problems or injuries to our sense organs.

Explanation:

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3 years ago

Problem PageQuestionSteam reforming of methane ( ) produces "synthesis gas," a mixture of carbon monoxide gas and hydrogen gas,

Serhud [2]

The question is incomplete. Her eis the complete question.

Steam reforming methane (CH4) produces "synthesis gas", a mixture of carbon monoxide gas and hydrogen gas, which is the starting point for many important industrial chemical syntheses. An industrial chemist studying this reaction fills a 125L tank with 20 mol of methane gas and 10 mol of water vapor at 38°C. He then raises the temperature, and when the mixture has come to equilibrium measures the amount of gas hydrogen to be 18 mol. Calculate the concentration equilibrium constant for the steam reforming of methane at the final temperature of the mixture. Round your answer to significant digits.

Answer: $K_{c}$ = 2. $10^{-2}$

Explanation: The reaction for steam reforming methane is:

$CH_{4} + H_{2}O$ ⇒ $CO_{} + 3H_{2}$

To calculate the concentration equilibrium constant, first calculate the molarity ( $\frac{mol}{L}$ ) of each molecule of the reaction.

At 38°C: At the initial temperature, there no products yet

Molarity of CH4:

CH4 = $\frac{20}{125}$ = 0.16M

Molarity of H20:

H2O = $\frac{10}{125}$ = 0.08M

At final temperature:

Molarity of H2:

H2 = $\frac{18}{125}$ = 0.144M

According to the chemical reaction, the combination of 1 mol of each reagents produces 1 mol of CO and 3 mols of H2, so, for the products, the ratio is 1:3.

Molarity of CO:

CO = $\frac{0.144}{3}$ = 0.048M

For the reagents, the proportion is 1:1, but they had an initial concentration, so, when in equilibrium, the concentration will be:

Molarity of CH4:

CH4 = 0.16 - 0.048 = 0.112M

Molarity of H2O:

H20 = 0.08 - 0.048 = 0.032M

The equilibrium constant is given by:

$K_{c} = \frac{[CO][H_{2}]^{3} }{[CH_{4}][H_{2}O ] }$

$K_{c}$ = $\frac{0.048.0.144^{3} }{0.112.0.032}$

$K_{c}$ = 2. $10^{-2}$

The concentration equilibrium constant for the process is $K_{c}$ = 2. $10^{-2}$ .

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3 years ago