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Artyom0805 [142]
3 years ago
6

By means of a schematic diagram show how a bacteria cell applied to the region of a cowpea root can end up becoming a nitrate io

n which can be absorbed by a subsequent crop​

Chemistry
1 answer:
GenaCL600 [577]3 years ago
4 0

Answer:

Nitrifying Bacteria are a group of aerobic bacteria important in the nitrogen cycle as converters of soil ammonia to nitrates, compounds usable by plants. An example is nitrosomonas or nitrobacter and species in that family.

The schematic diagram is attached below, which summarises the oxidation of ammonia or free nitrogen in the soil to nitrates for the cowpea plant's utilisation.

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How might you think it is possible to convert from moles to particles or grams to particles?
nevsk [136]

Explanation:

To convert moles to particles or grams to particles, let us have a firm understanding of what a mole is.

A mole is the unit of measuring quantity of particles.

It is the amount of substance that contains the Avogadro's number of particles.

The particle can be atoms, molecules, formula units, electrons, protons, neutrons, etc.

 So, to convert from moles to particles;

    1 mole of a substance  contains 6.02 x 10²³ particles

To convert from grams to particles;

  First convert to moles;

        number of moles  = \frac{mass}{molar mass}  

  So,    1 mole of a substance  contains 6.02 x 10²³ particles

4 0
2 years ago
Suppose 550.mmol of electrons must be transported from one side of an electrochemical cell to another in 49.0 minutes. Calculate
nekit [7.7K]

Answer:

18.0 Ampere is the size of electric current that must flow.

Explanation:

Moles of electron , n = 550 mmol = 0.550 mol

1 mmol = 0.001 mol

Number of electrons = N

N=N_A\times n

Charge on N electrons : Q

Q = N\times 1.602\times 10^{-19} C

Duration of time charge allowed to pass = T = 49.0 min = 49.0 × 60 seconds

1 min = 60 seconds

Size of current : I

I=\frac{Q}{T}=\frac{N\times 1.602\times 10^{-19} C}{49.0\times 60 seconds}

=\frac{n\times N_A\times 1.602\times 10^{-19} C}{49.0\times 60 seconds}

I=\frac{0.550 mol\times 6.022\times 10^{23} mol^{-1}\times 1.602\times 10^{-19} C}{49.0\times 60 seconds}=18.047 A\approx 18.0 A

18.0 Ampere is the size of electric current that must flow.

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3 years ago
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Answer:

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Explanation:

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