All categories

3 years ago

______is a mixture dead organic material that can be used as fertilizer

Chemistry

2 answers:

DiKsa [7]3 years ago

8 0

The answer to this is compost.

bonufazy [111]3 years ago

6 0

Answer:

Compost is a mixture dead organic material that can be used as fertilizer

Explanation:

Compost is a natural organic fertilizer that is prepared with different organic materials, providing nutrients and improving soil structure. Compost converts waste into fertilizer and valuable organic matter for soils.

The transformation of organic matter is carried out with composting, that is, the controlled decomposition of organic materials. Among them are fruits, pruning remains, vegetables, among others. The result of this procedure is a completely organic product that improves the soil structure and provides the necessary nutrients to the soil, all naturally.

Compost has a wide variety of advantages. Among them, it can be mentioned that urban waste is reduced considerably because it minimizes environmental problems related to its treatment and transportation. The soil structure will also be improved, helping with water retention, ventilation and the possible effects of frost. And the earth will eventually contain more organic matter and more nutrients; and it is possible to avoid the use of chemical fertilizers that can negatively affect the soil or the health of people. In addition, the lack of these substances produces an increase in the amount of microorganisms beneficial to the life cycle. Earthworms and other organisms will aerate the earth, prevent it from composting and favor the roots of plants.

So, "compost is a mixture dead organic material that can be used as fertilizer"

You might be interested in

_______ is the classic structural formula for all the addictive narcotics

oee [108]

The answer to this statement is codein. Codein, or 3-Methylmorphine by its IUPAC name, is an opiate used as pain reliever and suppressor of coughs. Its structural formula is shown in the picture. The patient's dosage of these narcotics should be strictly prescribed by the doctor. When patients take this, they feel euphoria, hence, they tend to crave for that feeling once it's gone. Too much dosage of this drug would lead to addiction. Examples of drugs with this structural formula are Cotabflu, Nalex AC, T-Koff and Pediatuss.

7 0

3 years ago

A trench is a. Cold and shallow

nexus9112 [7]

Answer:

Explanation:

i think

4 0

3 years ago

Read 2 more answers

Identify the following reactions as combination, decomposition, or combustion reactions

kkurt [141]

Answer:

1. combustion

2. decomposition

3. decomposition

4. combination

5. decomposition

6. decomposition

Explanation:

combination of a reaction is when 2 more or elements/ compounds combine to form a compound. example: (A+B→AB)

decomposition of a reaction is when a chemical breakdown. example: (AB→A+B)

a combustion reaction is when an element/ compound reacts with oxygen to form the product of water and carbon dioxide.

example: ( $C_{x}$ $H_{y}$ + $O_{2}$ → $H_{2}$ O + $CO_{2}$ )

7 0

2 years ago

The Haber Process synthesizes ammonia at elevated temperatures and pressures. Suppose you combine 1580 L of nitrogen gas and 351

ikadub [295]

Answer : The volume of reactant measured at STP left over is 409.9 L

Explanation :

First we have to calculate the moles of $N_2$ and $H_2$ by using ideal gas equation.

For $N_2$ :

$PV_{N_2}=n_{N_2}RT$

where,

P = Pressure of gas at STP = 1 atm

V = Volume of $N_2$ gas = 1580 L

n = number of moles $N_2$ = ?

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of gas at STP = 273 K

Putting values in above equation, we get:

$1atm\times 1580L=n_{N_2}\times (0.0821L.atm/mol.K)\times 273K$

$n_{N_2}=70.49mole$

For $H_2$ :

$PV_{H_2}=n_{H_2}RT$

where,

P = Pressure of gas at STP = 1 atm

V = Volume of $H_2$ gas = 3510 L

n = number of moles $H_2$ = ?

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of gas at STP = 273 K

Putting values in above equation, we get:

$1atm\times 3510L=n_{H_2}\times (0.0821L.atm/mol.K)\times 273K$

$n_{H_2}=156.6mole$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$

From the balanced reaction we conclude that

As, 3 mole of $H_2$ react with 1 mole of $N_2$

So, 156.6 moles of $H_2$ react with $\frac{156.6}{3}\times 1=52.2$ moles of $N_2$

From this we conclude that, $N_2$ is an excess reagent because the given moles are greater than the required moles and $H_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the excess moles of $N_2$ reactant (unreacted gas).

Excess moles of $N_2$ reactant = 70.49 - 52.2 = 18.29 moles

Now we have to calculate the volume of reactant, measured at STP, is left over.

$PV=nRT$

where,

P = Pressure of gas at STP = 1 atm

V = Volume of gas = ?

n = number of moles of unreacted gas = 18.29 moles

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of gas at STP = 273 K

Putting values in above equation, we get:

$1atm\times V=18.29mole\times (0.0821L.atm/mol.K)\times 273K$

$V=409.9L$

Therefore, the volume of reactant measured at STP left over is 409.9 L

8 0

3 years ago

An atom has a diameter of 2.50 Å and the nucleus of that atom has a diameter of 9.00×10−5 Å . Determine the fraction of the volu

chubhunter [2.5K]

Answer:

The fraction of the volume of the atom that is taken up by the nucleus is $4.6656\times 10^{-14}$ .

The density of a proton is $6.278\times 10^{14} g/cm^3$ .

Explanation:

Diameter of the atom ,d = 2.50 Å

Radius of the atom ,r = 0.5 d=0.5 × 2.50 Å = 1.25Å

Volume of the sphere= $\frac{4}{3}\pi r^3$

Volume of atom = V

$V=\frac{4}{3}\pi r^3$ ..[1]

Diameter of the nucleus ,d' = $9.00\times 10^{-5}\AA$

Radius of the nucleus ,r' = 0.5 d'= $0.5\times 9.00\times 10^{-5}\AA=4.5\times 10^{-5}\AA$

Volume of nucleus = V'

$V=\frac{4}{3}\pi r'^3$ ..[2]

Dividing [2] by [1]

$\frac{V'}{V}=\frac{\frac{4}{3}\pi r'^3}{\frac{4}{3}\pi r^3}$

$=\frac{r'^3}{r^3}=\frac{(4.5\times 10^{-5}\AA)^3}{(1.25 \AA)^3}$

$\frac{V'}{V}=4.6656\times 10^{-14}$

The fraction of the volume of the atom that is taken up by the nucleus is $4.6656\times 10^{-14}$ .

Diameter of the proton ,d = $1.72\times 10^{-15} m = 1.72\times 10^{-13} cm$

1 m = 100 cm

Radius of the proton,r = 0.5 d= $0.5\times 1.72\times 10^{-13} cm=8.6\times 10^{-14} cm$

Volume of the sphere= $\frac{4}{3}\pi r^3$

Volume of atom = V

$V=\frac{4}{3}\times 3.14\times (8.6\times 10^{-14} cm)^3=2.664\times 10^{-39}cm^3$

Mass of proton, m = 1.0073 amu = $1.0073\times 1.66054\times 10^{-24} g$

$1 amu = 1.66054\times 10^{-24} g$

Density of the proton : d

$d=\frac{m}{V}=\frac{1.0073\times 1.66054\times 10^{-24} g}{2.664\times 10^{-39}cm^3}=6.278\times 10^{14} g/cm^3$

The density of a proton is $6.278\times 10^{14} g/cm^3$ .

5 0

3 years ago