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Harman [31]
3 years ago
12

Which organisms in soil helps relieve element so they can be recycled

Chemistry
1 answer:
Rudiy273 years ago
3 0

Answer:

Bacteria are also involved in many processes that are indispensable for our life on Earth. Many of these processes have to do with the recycling (reuse) of chemical elements that have been here since the formation of the planet

Explanation:

I hope it will help you

You might be interested in
1. The solubility of lead(II) chloride at some high temperature is 3.1 x 10-2 M. Find the Ksp of PbCl2 at this temperature.
solniwko [45]

Answer:

1) The solubility product of the lead(II) chloride is 1.2\times 10^{-4}.

2) The solubility of the aluminium hydroxide is 1.6\times 10^{-10} M.

3)The given statement is false.

Explanation:

1)

Solubility of lead chloride = S=3.1\times 10^-2M

PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)

                            S     2S

The solubility product of the lead(II) chloride = K_{sp}

K_{sp}=[Pb^{2+}][Cl^-]^2

K_{sp}=S\times (2S)^2=4S^3=4\times (3.1\times 10^{-2})^3=1.2\times 10^{-4}

The solubility product of the lead(II) chloride is 1.2\times 10^{-4}.

2)

Concentration of aluminium nitrate = 0.000010 M

Concentration of aluminum ion =1\timed 0.000010 M=0.000010 M

Solubility of aluminium hydroxide in aluminum nitrate solution = S

Al(OH)_3(aq)\rightleftharpoons Al^{3+}(aq)+3OH^-(aq)

                            S     3S

The solubility product of the aluminium nitrate = K_{sp}=1.0\times 10^{-33}

K_{sp}=[Al^{3+}][OH^-]^3

1.0\times 10^{-33}=(0.000010+S)\times (3S)^3

S=1.6\times 10^{-10} M

The solubility of the aluminium hydroxide is 1.6\times 10^{-10} M.

3.

Molarity=\frac{Moles}{Volume (L)}

Mass of NaCl= 3.5 mg = 0.0035 g

1 mg = 0.001 g

Moles of NaCl = \frac{0.0035 g}{58.5 g/mol}=6.0\times 10^{-5} mol

Volume of the solution = 0.250 L

[NaCl]=\frac{6.0\times 10^{-5} mol}{0.250 L}=0.00024 M

1 mole of NaCl gives 1 mole of sodium ion and 1 mole of chloride ions.

[Cl^-]=[NaCl]=0.00024 M

Moles of lead (II) nitrate = n

Volume of the solution = 0.250 L

Molarity lead(II) nitrate = 0.12 M

n=0.12 M]\times 0.250 L=0.030 mol

1 mole of lead nitrate gives 1 mole of lead (II) ion and 2 moles of nitrate ions.

[Pb^{2+}]=[Pb(NO_2)_3]=0.030 M

PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)

Solubility of lead(II) chloride = K_{sp}=1.2\times 10^{-4}

Ionic product of the lead chloride in solution :

Q_i=[Pb^{2+}][Cl^-]^2=0.030 M\times (0.00024 M)^2=1.7\times 10^{-9}

Q_i ( no precipitation)

The given statement is false.

3 0
3 years ago
Suppose that the C=O group in a peptide bond can be regarded as isolated from the rest of the molecule. Given that the force con
Stella [2.4K]

Answer:

a) v = 1497.2 cm^-1

b) v = 1465 cm^-1

Explanation:

In the attached image is the procedure explained to reach the answer.

4 0
3 years ago
A vehicle is moving with a velocity of 20 m/s.how far does it moves in one hrs
Airida [17]
Distance= Speed x Time

1hr=60mins

60mins=3600 secs

20 x 3600=72000

72000 is your answer.

5 0
3 years ago
What is the mass in grams of one mole of many of any substance known as
OLEGan [10]
The molecular weight of the substance.
7 0
2 years ago
A compound is found to contain 73.23% xenon name 26.77% oxygen by mass. What is the empirical formula for this compound ?
Luba_88 [7]

The empirical formula is XeO₃.

<u>Explanation:</u>

Assume 100 g of the compound is present. This changes the percents to grams:

Given mass in g:

Xenon = 73.23 g

Oxygen = 26.77 g

We have to convert it to moles.

Xe = 73.23/   131.293 = 0.56 moles

O = 26.77/ 16 = 1.67 moles

Divide by the lowest value, seeking the smallest whole-number ratio:

Xe = 0.56/ 0.56 = 1

O = 1.67/ 0.56 = 2.9 ≈3

So the empirical formula is XeO₃.

6 0
3 years ago
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