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julia-pushkina [17]
2 years ago
6

HELP PLEASE!!! Which of the following best describes the nitrogen fixation process?(1 point) Producers convert nitrogen gas into

nitrates in the soil. Producers can then absorb nitrates through their roots. Producers convert nitrogen gas into nitrates in the soil. Producers can then absorb nitrates through their roots. Producers use the energy from sunlight to fix nitrogen gas in the atmosphere. Producers can then absorb nitrogen through their leaves. Producers use the energy from sunlight to fix nitrogen gas in the atmosphere. Producers can then absorb nitrogen through their leaves. Soil bacteria convert nitrogen gas into ammonia, nitrites, and then nitrates. Nitrates are then absorbed by producers. Soil bacteria convert nitrogen gas into ammonia, nitrites, and then nitrates. Nitrates are then absorbed by producers. Consumers convert nitrogen gas into ammonia, nitrites, and then nitrates. Waste from consumers adds nitrogen to the soil, which can be used by producers.
Chemistry
1 answer:
BaLLatris [955]2 years ago
3 0

Soil bacteria convert atmospheric nitrogen into ammonia, nitrites, and then nitrates. This nitrates are then absorbed by producers.

<h3>What is nitrogen fixation?</h3>

Nitrogen fixation is a process that converts atmospheric nitrogen (N2), into available forms such as ammonia, nitrates, or nitrites to plants so that the plant can absorb it and make amino acids so we can conclude that Soil bacteria convert atmospheric nitrogen into ammonia, nitrites, and then nitrates. This nitrates are then absorbed by producers.

Learn more about fixation here: brainly.com/question/1380063

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After performing a dilution calculation, you determine you need 25.0 milliliters of an aqueous stock solution to make 100.0 mill
IRINA_888 [86]
A is obviously out because it leads to a volume of 125.0 milliliters of the new solution and gives you a lower concentration than you were aiming for.

D is out because you are adding 75 milliliters of the stock solution, so your concentration would be too high. You only need 25.0 milometers of stock solution per 100 milliliters of the new solution.

C is also out because it leads to 50.0 milliliters stock solution per 100 milliliters of the new solution and hence the wrong concentration.

B is by default the correct answer. It also details the correct technique. First you add the stock solution (This you know from your calculations to be 25 milliliters.) then you add the water up to the volume you needed. (Because the calculations only tell you the total volume of water not what you need to add) You also add the water last so you can rinse the neck of the flask to make sure you also get all the stock solution residue into the stock solution.

I would add the final step of stirring, but B is the only answer that can be correct.
8 0
3 years ago
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What will happen to the flask?
denis-greek [22]
B is the correct answer to select

7 0
3 years ago
(image attached)
Naily [24]

Answer:

D) He did not multiply the chlorine and oxygen atoms by the coefficient 4.

Explanation:

The coefficient 4 at the beginning of the chemical formula indicates that there are four Ca(ClO3)2 molecules. Think of this as Ca(ClO3)2 × 4. This means that he had to multiply the number of atoms for each element by 4 as well, so he should've ended up with 4 total calcium atoms (which is correct), 8 total chlorine atoms, and and 24 total oxygen atoms. He did not get all these answers because he didn't multiply the chlorine and oxygen atoms by the coefficient 4.

8 0
3 years ago
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Calculate the volume of an object that has a density of 4 g/mL and has a mass of 128 grams. Show your work
SOVA2 [1]
The first one is 32mL and the second one is 2.62 and I think it’s grams/mL I’m not for sure about the letters on the second one

4 0
3 years ago
Urea, CH4N2O (s), is manufactured from NH3 (g) and CO2 (g). H2O (l) is another product of this reaction. An experiment is starte
Katarina [22]

Answer:

a. 4.41 g of Urea

b. 1.5 g of Urea

Explanation:

To start the problem, we define the reaction:

2NH₃ (g) +  CO₂ (g) → CH₄N₂O (s)  +  H₂O(l)

We only have mass of ammonia, so we assume the carbon dioxide is in excess and ammonia is the limiting reactant:

2.6 g . 1mol / 17g = 0.153 moles of ammonia

Ratio is 2:1. 2 moles of ammonia can produce 1 mol of urea

0.153 moles ammonia may produce, the half of moles

0153 /2 = 0.076 moles of urea

To state the theoretical yield we convert moles to mass:

0.076 mol . 58 g/mol = 4.41 g

That's the 100 % yield reaction

If the percent yield, was 34%:

4.41 g . 0.34 = 1.50 g of urea were produced.

Formula is (Yield produced / Theoretical yield) . 100 → Percent yield

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