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GuDViN [60]
3 years ago
15

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.

<u>Explanation:</u>

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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Asparagine is a polar amino acid shown here at ph 7 what is the maximum
KonstantinChe [14]
Maximum number of water molcules which can take part in forming hydrogen bonding interaction with asparagine are 13. Following structure shows the interactions. Lone pair of electrons act as hydrogen bond acceptors and hydrogen atoms attached to heteroatoms acts as hydrogen bond donors.

3 0
3 years ago
Fuels cell automobiles use__ gas as a fuel
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Answer:

hydrogen gas

Explanation:

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3 years ago
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For the reaction A (g) → 3 B (g), Kp = 80100 at 298 K. When ∆G = -14.2 kJ/mol, what is the partial pressure of A when the partia
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5 0
3 years ago
5. Compare the threats faced by the Everglades and the Louisiana wetan
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5 0
3 years ago
How many grams of ethylene glycol (c2h6o2 must be added to 1.25 kg of water to produce a solution that freezes at -5.88 ∘c?
e-lub [12.9K]
The freezing point depression is calculated through the equation,
                                    ΔT = (kf)  x m 
where ΔT is the difference in temperature, kf is the freezing point depression constant (1.86°C/m), and m is the molality. Substituting the known values,
                                   5.88 = (1.86)(m)
m is equal to 3.16m

Recall that molality is calculated through the equation,
                                  molality = number of mols / kg of solvent
                                       number of mols = (3.16)(1.25) = 3.95 moles
Then, we multiply the calculated amount in moles with the molar mass of ethylene glycol and the answer would be 244.9 g.

6 0
3 years ago
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