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3 years ago

The process where bacteria takes nitrogen out of the air and makes it usable for other organisms is called

Chemistry

1 answer:

Likurg_2 [28]3 years ago

7 0

Answer:

The process is called Nitrogen fixation

Explanation:

The nitrogen fixation is a process carried out by some prokaryotic microorganisms (bacteria), specifically those have the presence of the nitrogenase enzyme. The bacteria absorb the atmospheric nitrogen (N2) from the roots of plants, and the nitrogenase enzyme, with the help of two proteins that act as electron donors and acceptors (nitrogenase complex) reduce the nitrogen to ammonia (NH3), then the ammonia is ionized to NH4+ (ammonium). Followed by that, the ammonia is oxidated to nitrates and nitrites, which are finally absorbed again by plants.

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Answer:

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Explanation:

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3 years ago

At 25 Â°c and 785 torr, carbon dioxide has a solubility of 0.0351 m in water. what is its solubility at 25 Â°c and 1510 torr?

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3 years ago

5. Didderent betwen group and period.

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3 years ago

How many moles of KOH are required to produce 4.79 g K3PO4 according to the following reaction? 3KOH + H3PO4 -----> K3PO4 + 3

8_murik_8 [283]

Answer:

0.677 moles

Explanation:

Take the atomic mass of K = 39.1, O =16.0, P = 31.0

no. of moles = mass / molar mass

no. of moles of K3PO4 used = 4.79 / (39.1x3 + 31 + 16x4)

= 0.02256 mol

From the equation, the mole ratio of KOH : K3PO4 = 3 :1,

meaning every 3 moles of KOH used, produces 1 mole of K3PO4.

So, using this ratio, let the no. of moles of KOH required to be y.

$\frac{3}{1} =\frac{y}{0.02256} \\$

y = 0.02256 x3

y = 0.0677 mol

If you don't find exactly 0.677 moles as one of the options, go for the closest one. A very slight error may occur because of taking different significant figures of atomic masses when calculating.

5 0

3 years ago

The half-life of cesium-137 is 30 years. Suppose we have a 200-mg sample. (a) Find the mass that remains after t years.

scZoUnD [109]

Given:

Half life(t^ 1/2) :30 years

A0( initial mass of the substance): 200 mg.

Now we know that

A= A0/ [2 ^ (t/√t)]

Where A is the mass that remains after t years.

A0 is the initial mass

t is the time

t^1/2 is the half life

Substituting the given values in the above equation we get

A= [200/ 2^(t/30) ] mg

Thus the mass remaining after t years is [200/ 2^(t/30) ] mg

5 0

4 years ago