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Nat2105 [25]
3 years ago
5

What is the most likely fate of a protein with an N-terminal hydrophobic sorting signal and an additional internal hydrophobic d

omain of 22 amino acids
Chemistry
1 answer:
scoundrel [369]3 years ago
4 0

Explanation:

The  fate of a protein with an N-terminal hydrophobic sorting signal and an additional internal hydrophobic domain of 22 amino acids.When it is in the translocation channel and emitted laterally into the membrane, the hydrophobic region is known as a trans-membrane domain.

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What is the potential energy of a 2-kg book sitting on a shelf 2 meters above<br> the ground?
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<u><em>What is the potential energy of a 2-kg book sitting on a shelf 2 meters above the ground?</em></u>

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

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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

<u>Molarity of CH4</u>:

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<u>Molarity of H20</u>:

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At final temperature:

<u>Molarity of H2</u>:

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.

<u>Molarity of CO</u>:

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:

<u>Molarity of CH4</u>:

CH4 = 0.16 - 0.048 = 0.112M

<u>Molarity of H2O</u>:

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}

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The concentration equilibrium constant for the process is K_{c} = 2.10^{-2}.

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