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3 years ago

Microorganisms and humus have little impact on soil health. Please select the best answer from the choices provided T F

Chemistry

2 answers:

egoroff_w [7]3 years ago

7 0

First of all what the heck to the other guy that answered. And the correct answer is FALSE :))

hram777 [196]3 years ago

6 0

The right answer is false, they have great impact on soil health.

In general, microorganisms play a fundamental role in the fertility of the soil (production of the minerals necessary for the plant: N, P, K, etc.) and the study of their activity is therefore important. In addition, certain bacteria participate in the release and transformation of metals in the soil and thus contribute to increasing or reducing the toxicity of heavy metals (nickel, cobalt, etc.).

Humus is essential in the function of soil production.

Humus is used primarily as a "pantry" for plants, and also plays a vital role in structuring the soil.

You might be interested in

Will something that is water soluble dissolve in oil

lawyer [7]

Something like that wouldn't dissolve in oil

4 0

3 years ago

Sodium hydroxide solution reacts with copper (II) nitrate solution to produce a

dsp73

Answer:

brown precipitate.. d.. option

7 0

2 years ago

Write the balanced half-reaction equation for when H 2 O 2 ( aq ) acts as an oxidizing agent in an acidic solution. Phases are o

Radda [10]

Explanation:

An oxidizing agent is a reactant that removes electrons from other reactants during a redox reaction. The oxidizing agent typically takes these electrons for itself, thus gaining electrons and being reduced. An oxidizing agent is thus an electron acceptor.

Examples of Oxidizing Agents include the following; Hydrogen peroxide, ozone, oxygen.

A half-equation shows you what happens at one of the electrodes during electrolysis . Electrons are shown as e -. A half-equation is balanced by adding, or taking away, a number of electrons equal to the total number of charges on the ions in the equation.

The balanced half reaction equation for H2O2 as an oxidizing agent is given as;

H2O2 + 2e^- + 2H^+ ==> 2H2O

3 0

3 years ago

3) Lithium metal (Li) react with hydrosulfuric acid (HS) to produce hydrogen gas and magnesium chloride (Li 2 S) . How many gram

Vaselesa [24]

Answer:

3.066g

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2Li + H2S —> Li2S + H2

Step 2:

Data obtained from the question.

Volume (V) of H2 = 4.5L

Temperature (T) = 315K

Pressure (P) = 1.258 atm

Note:

Gas constant (R) = 0.0821atm.L/Kmol

Number of mole (n) of H2 =..?

Step 3:

Determination of the number of mole of H2 produced.

This can be obtained by using the ideal gas equation as follow

PV = nRT

Divide both side by RT

n = PV /RT

n = 1.258 x 4.5 / 0.0821 x 315

n = 0.219 mole

Therefore, 0.219 mole of H2 is produced.

Step 4:

Determination of the number of mole of Li that will produce 0.219 mole of H2.

This is shown below:

2Li + H2S —> Li2S + H2

From the balanced equation above,

2 moles of Li reacted to produce 1 mole of H2.

Therefore, Xmol of Li will react to produce 0.219 mole of H2 i.e

Xmol of Li = 2 x 0.219

Xmol of Li = 0.438 mole

Step 5:

Conversion of 0.438 mole of Li to grams.

Number of mole of Li = 0.438 mole

Molar Mass of Li = 7g/mol

Mass = number of mole x molar Mass

Mass of Li = 0.438 x 7

Mass of Li = 3.066g

Therefore, 3.066g if Li is needed for the reaction.

6 0

3 years ago

Calculate the amount of heat energy in KJ required to convert 45.0 g of ice at -15.5'C to steam at 124.0°C. (Cwater 118 Jig'c, G

dangina [55]

Answer : The enthalpy change or heat required is, 139.28775 KJ

Solution :

The conversions involved in this process are :

$(1):H_2O(s)(-15.5^oC)\rightarrow H_2O(s)(0^oC)\\\\(2):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(3):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)\\\\(4):H_2O(l)(100^oC)\rightarrow H_2O(g)(100^oC)\\\\(5):H_2O(g)(100^oC)\rightarrow H_2O(g)(124^oC)$

Now we have to calculate the enthalpy change.

$\Delta H=[m\times c_{p,s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]+n\times \Delta H_{vap}+[m\times c_{p,g}\times (T_{final}-T_{initial})]$

where,

$\Delta H$ = enthalpy change or heat required = ?

m = mass of water = 45 g

$c_{p,s}$ = specific heat of solid water = $2.09J/g^oC$

$c_{p,l}$ = specific heat of liquid water = $4.18J/g^oC$

$c_{p,g}$ = specific heat of liquid water = $1.84J/g^oC$

n = number of moles of water = $\frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{45g}{18g/mole}=2.5mole$

$\Delta H_{fusion}$ = enthalpy change for fusion = 6.01 KJ/mole = 6010 J/mole

$\Delta H_{vap}$ = enthalpy change for vaporization = 40.67 KJ/mole = 40670 J/mole

Now put all the given values in the above expression, we get

$\Delta H=[45g\times 4.18J/gK\times (0-(-15.5))^oC]+2.5mole\times 6010J/mole+[45g\times 2.09J/gK\times (100-0)^oC]+2.5mole\times 40670J/mole+[45g\times 1.84J/gK\times (124-100)^oC]$

$\Delta H=139287.75J=139.28775KJ$ (1 KJ = 1000 J)

Therefore, the enthalpy change is, 139.28775 KJ

3 0

3 years ago