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LUCKY_DIMON [66]
3 years ago
12

Why is it difficult to pull nitrogen from the atmosphere into the nitrogen cycle of the biosphere?

Chemistry
1 answer:
Jlenok [28]3 years ago
5 0

Answer:

Because only a few bacterias can "fix" the atmosphere nitrogen.

Explanation:

The nitrogen at the atmosphere is in the form of N₂ and represents 78% of the atmosphere composition. The element is part of the constitution of nucleic acids and proteins, so the living beings needed them.

However, the animals and the plants can't catch the N₂. Some bacterias that live in mutualism with plants have this ability, and they "fix" the atmosphere nitrogen, transforming the N₂ in the ions nitrite (NO₃⁻) or ammonia (NH₃), which can be caught by the plants.

Them, when the primary consumers eat the plants they catch the nitrogen, which will be passed through the food chain.

So, it's difficult to pull nitrogen from the atmosphere into the nitrogen cycle of the biosphere because only a few bacterias can do it.

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What makes calcium and nitrogen different list two differences
liberstina [14]

Answer:

the difference in electronegativity is so large (2.04) that the bonding electrons spend almost all their time on the nitrogen atom.

Explanation:

Because calcium loses 2 electrons to become Ca2+, and nitrogen gains 3 electrons to become N3−, you need two calcium atoms and three nitrogen atoms in order to form a neutral compound.

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Write the following elements in increasing order of atomic radius.<br> Ca, Mg, Be, Sr
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In increasing order:

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Please help,<br> How could you tell a Ca(NO3)2 solution from a Zn(NO3)2 solution?
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3 years ago
Read 2 more answers
A 1.59 mol sample of Kr has a volume of 641 mL. How many moles of Kr are in a 4.41 L sample at the same temperature and pressure
Marina86 [1]

Answer:

The correct answer is 10.939 mol ≅ 10.94 mol

Explanation:

According to Avogadro's gases law, the number of moles of an ideal gas (n) at constant pressure and temperature, is directly proportional to the volume (V).

For the initial gas (1), we have:

n₁= 1.59 mol

V₁= 641 mL= 0.641 L

For the final gas (2), we have:

V₂: 4.41 L

The relation between 1 and 2 is given by:

n₁/V₁ = n₂/V₂

We calculate n₂ as follows:

n₂= (n₁/V₁) x V₂ = (1.59 mol/0.641 L) x 4.41 L = 10.939 mol ≅ 10.94 mol

5 0
2 years ago
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