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3 years ago

Select the correct answer.

Chemistry

1 answer:

ankoles [38]3 years ago

4 0

It would still have oceans but no atmospheric water in Earth if no icy debris had arrived.

A. It would still have oceans but no atmospheric water.

Explanation:

Seas characterize our home planet, covering most of the Earth's surface and driving the water cycle that commands our territory and climate. However, progressively significant still, the narrative of our seas wraps our home in a far bigger setting that ventures profound into the universe and spots us in a rich group of sea universes that range our nearby planetary group and past.

It would in any case have seas yet no air water on Earth if no frigid flotsam and jetsam had shown up. For a long time, it was accepted that the frosty moons were only that - solidified husks, strong to their center. However, lately that thought has steadily been supplanted by a fresher, additionally energizing worldview.

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I think the answer is B

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2 years ago

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At a certain concentration of H2 and NH3, the initial rate of reaction is 0.120 M / s. What would the initial rate of the reacti

mel-nik [20]

The question is incomplete, here is the complete question:

The rate of certain reaction is given by the following rate law:

$rate=k[H_2]^2[NH_3]$

At a certain concentration of $H_2 and [tex]I_2, the initial rate of reaction is 0.120 M/s. What would the initial rate of the reaction be if the concentration of [tex]H_2 were halved.Answer : The initial rate of the reaction will be, 0.03 M/sExplanation :Rate law expression for the reaction:[tex]rate=k[H_2]^2[NH_3]$

As we are given that:

Initial rate = 0.120 M/s

Expression for rate law for first observation:

$0.120=k[H_2]^2[NH_3]$ ....(1)

Expression for rate law for second observation:

$R=k(\frac{[H_2]}{2})^2[NH_3]$ ....(2)

Dividing 2 by 1, we get:

$\frac{R}{0.120}=\frac{k(\frac{[H_2]}{2})^2[NH_3]}{k[H_2]^2[NH_3]}$

$\frac{R}{0.120}=\frac{1}{4}$

$R=0.03M/s$

Therefore, the initial rate of the reaction will be, 0.03 M/s

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3 years ago

What are weak bonds that allow flexibility in an enzyme

34kurt

Enzymes are characterized to have weak bonds because their tertiary structure could easily bend and break because it will have to adjust to the shape of the substrate. It could be done via induced fitting or lock-and-key theory. These weak bonds are intermolecular forces like the London forces, electrostatic interactions and hydrogen bonding.

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Answer:

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Explanation:

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