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AysviL [449]
3 years ago
6

How does bacteria help soil?

Chemistry
2 answers:
horsena [70]3 years ago
5 0
Hi hi hello hi hey hey do khan know if I have any pictures I have a picture of you
bulgar [2K]3 years ago
3 0

Answer:

They are decomposers, eating dead plant material and organic waste. By doing this, the bacteria release nutrients that other organisms could not access. The bacteria do this by changing the nutrients from inaccessible to usable forms. The process is essential in the nitrogen cycle.

Explanation:

Most are decomposers that consume simple carbon compounds, such as root exudates and fresh plant litter. By this process, bacteria convert energy in soil organic matter into forms useful to the rest of the organisms in the soil food web. A number of decomposers can break down pesticides and pollutants in soil.

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Write two sentences explaining how the illustration and student observations support the cell theory?
SSSSS [86.1K]

Answer:

The classical cell theory was proposed by Theodor Schwann in 1839. There are three parts to this theory.

Explanation:

5 0
3 years ago
WILL MARK BRAINLIEST
levacccp [35]
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3 years ago
Based on this graph what was the oxygen content of the earths atmosphere about 500million years ago
igomit [66]

12 percent is your answer :))

4 0
4 years ago
How many moles are in 4.38 of AgNO2
Gwar [14]

Answer: 0.02 moles

Explanation:

4 0
3 years ago
1.24 grams of magnesium phosphate tribasic dissolved in 1 L of lemon juice. What is the Ksp of the magnesium phosphate tribasic
vivado [14]

<u>Answer:</u> The solubility product of magnesium phosphate tribasic is 2.52\times 10^{-10}

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

Given mass of magnesium phosphate = 1.24 g

Molar mass of magnesium phosphate = 262.85 g/mol

Volume of solution = 1 L

Putting values in above equation, we get:

\text{Molarity of solution}=\frac{1.24g}{262.85g/mol\times 1}\\\\\text{Molarity of solution}=4.72\times 10^{-3}M

The equation for the ionization of the magnesium phosphate is given as:

Mg_3(PO_4)_2\leftrightharpoons 3Mg^{2+}+2PO_4^{3-}

Expression for the solubility product of Mg_3(PO_4)_2 will be:

K_{sp}=[Mg^{2+}]^3[PO_4^{3-}]^2

We are given:

[Mg^{2+}]=(3\times 4.72\times 10^{-3})=1.416\times 10^{-2}M

[PO_4^{3-}]=(2\times 4.72\times 10^{-3})=9.44\times 10^{-3}M

Putting values in above expression, we get:

K_{sp}=(1.416\times 10^{-2})^3\times (9.44\times 10^{-3})^2=2.52\times 10^{-10}

Hence, the solubility product of magnesium phosphate tribasic is 2.52\times 10^{-10}

4 0
3 years ago
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