1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
stira [4]
2 years ago
7

In what ways will the addition of earth worms most likely affect the terrarium?

Chemistry
2 answers:
Gre4nikov [31]2 years ago
4 0

Answer: The soil biota benefits soil productivity and contributes to the sustainable function of all ecosystems. The cycling of nutrients is a critical function that is essential to life on earth. Earthworms (EWs) are a major component of soil fauna communities in most ecosystems and comprise a large proportion of macrofauna biomass. Their activity is beneficial because it can enhance soil nutrient cycling through the rapid incorporation of detritus into mineral soils. In addition to this mixing effect, mucus production associated with water excretion in earthworm guts also enhances the activity of other beneficial soil microorganisms. This is followed by the production of organic matter. So, in the short term, a more significant effect is the concentration of large quantities of nutrients (N, P, K, and Ca) that are easily assimilable by plants in fresh cast depositions. In addition, earthworms seem to accelerate the mineralization as well as the turnover of soil organic matter. Earthworms are known also to increase nitrogen mineralization, through direct and indirect effects on the microbial community. The increased transfer of organic C and N into soil aggregates indicates the potential for earthworms to facilitate soil organic matter stabilization and accumulation in agricultural systems, and that their influence depends greatly on differences in land management practices. This paper summarises information on published data on the described subjects.

Explanation: Protection of the soil habitat is the first step towards sustainable management of its biological properties that determine long-term quality and productivity. It is generally accepted that soil biota benefits soil productivity but very little is known about the organisms that live in the soil and the functioning of the soil ecosystem. The role of earthworms (EWs) in soil fertility is known since 1881, when Darwin (1809–1882) published his last scientific book entitled “The formation of vegetable mould through the action of worms with observations on their habits.’’ Since then, several studies have been undertaken to highlight the soil organisms contribution to the sustainable function of all ecosystems [1]. Soil macrofauna, such as EWs, modify the soil and litter environment indirectly by the accumulation of their biogenic structures (casts, pellets, galleries, etc.) (Table 1). The cycling of nutrients is a critical ecosystem function that is essential to life on earth. Studies in the recent years have shown increasing interest in the development of productive farming systems with a high efficiency of internal resource use and thus lower input requirement and cost [2, 3]. At present, there is increasing evidence that soil macroinvertebrates play a key role in SOM transformations and nutrient dynamics at different spatial and temporal scales through perturbation and the production of biogenic structures for the improvement of soil fertility and land productivity [4, 5]. EWs are a major component of soil fauna communities in most natural ecosystems of the humid tropics and comprise a large proportion of macrofauna biomass [6]. In cultivated tropical soils, where organic matter is frequently related to fertility and productivity, the communities of invertebrates—especially EWs—could play an important role in (SOM) dynamics by the regulation of the mineralization and humification processes [7–9]. The effects of EWs on soil biological processes and fertility level differ in ecological categories [12]. Anecic species build permanent burrows into the deep mineral layers of the soil; they drag organic matter from the soil surface into their burrows for food. Endogeic species live exclusively and build extensive nonpermanent burrows in the upper mineral layer of soil, mainly ingested mineral soil matter, and are known as “ecological engineers,’’ or “ecosystem engineers.’’ They produce physical structures through which they can modify the availability or accessibility of a resource for other organisms [13]. Epigeic species live on the soil surface, form no permanent burrows, and mainly ingest litter and humus, as well as on decaying organic matter, and do not mix organic and inorganic matter [14]. In the majority of habitats and ecosystems (Table 2), it is usually a combination of these ecological categories which together or individually are responsible for maintaining the fertility of soils [15–17]. EWs influence the supply of nutrients through their tissues but largely through their burrowing activities; they produce aggregates and pores (i.e., biostructures) in the

OleMash [197]2 years ago
3 0
Significantly modify the physical, chemical and biological properties of the soil profile.
You might be interested in
A student assisting with the experiment would observe all of the following about the electron transport chain EXCEPT:A. Electron
zheka24 [161]

Answer:

C. All electron carriers are mobile and hydrophobic

Explanation:

Hello,

In this case, it is widely known that the electron carriers move inside the inner mitochondrial membrane and consequently move electrons from one to another. In such a way, they are mobile, therefore they are largely hydrophobic as long as they are inside the membrane.

For instance, the cytochrome c is a water-soluble protein in a large range, therefore, the answer is: C. All electron carriers are mobile and hydrophobic.

Best regards.

5 0
3 years ago
In a nuclear reaction, the energy released is equal to 8.1 x 1016 joules. Calculate the mass lost in this reaction. (1 J = 1 kg
makvit [3.9K]
Use the formula E=mc^2
energy given=<span>8.1 x 10^16 joules
</span>speed of <span>light = 3.00 × 10^8 m/s
</span>
plug in the values we'll get mass=<span>9.0 x 10-1 kg</span>

5 0
3 years ago
Read 2 more answers
What action leads to crystal formation in minerals?
ArbitrLikvidat [17]
I believe c is the right answer.
4 0
3 years ago
Read 2 more answers
Please help!!!! ⚠️⚠️⚠️⚠️⚠️⚠️⚠️⚠️
Bumek [7]

Polyatomic ions: CH_3COO^-, NO_3^-, NH^+, CN^-, Li^+, and OH^-

Monatomic ions: Ca^{2+, O^{2-, and Fe^{3+

<h3>Monoatomic vs Polyatomic Ions</h3>

In chemistry, monoatomic ions are ions that consist of only a single type of atom. They are usually positive or negatively charged and are otherwise known as simple ions. Examples include  Ca^{2+, O^{2-, and Fe^{3+

Polyatomic ions, on the other hand, are ions that consist of more than one atom, unlike monoatomic ions. The two or more atoms are covalently bonded and the entire structure behaves like a single chemical entity in reactions. Polyatomic ions are otherwise known as molecular ions.

Examples of polyatomic ions are  CH_3COO^-, NO_3^-, NH^+, CN^-, Li^+, and OH^-

Thus, from the diagram:

  • Polyatomic ions: CH_3COO^-, NO_3^-, NH^+, CN^-, Li^+, and OH^-
  • Monatomic ions: Ca^{2+, O^{2-, and Fe^{3+

More on ions can be found here: brainly.com/question/14982375

#SPJ1

6 0
1 year ago
Glucose (C6H12O6) can be fermented to yield ethanol (CH3CH2OH) and carbon dioxide (CO2). C6H12O6⟶2CH3CH2OH+2CO2 The molar mass o
jekas [21]

Answer:

The % yield is 74.45 %

Explanation:

<u>Step 1:</u> The balanced equation

C6H12O6⟶2CH3CH2OH+2CO2

<u>Step 2</u>: Data given

Molar mass glucose = 180.15 g/mol

Molar mass of ethanol = 46.08 g/mol

Molar mass of carbon dioxide = 44.01 g/mol

Mass of glucose = 61.5 grams

Mass of ethanol = 23.4 grams

<u>Step 3:</u> Calculate moles of glucose

Moles glucose = Mass glucose  / Molar mass of glucose

Moles glucose = 61.5 grams / 180.15 g/mol

Moles glucose = 0.341 moles

<u>Step 4:</u> Calculate moles of ethanol

1 mole of glucose consumed, produces 2 moles of ethanol and 2 moles of CO2

0.341 moles of glucose, will produce 2*0.341 = 0.682 moles of ethanol

<u>Step 5:</u> Calculate mass of ethanol

Mass ethanol = moles ethanol * Molar mass ethanol

Mass ethanol = 0.682 moles * 46.08 g/mol

Mass ethanol = 31.43 grams = theoretical mass

<u>Step 6:</u> Calculate % yield

% yield = actual mass / theoretical mass

% yield = (23.4 grams / 31.43 grams) * 100%

% yield = 74.45 %

The % yield is 74.45 %

7 0
3 years ago
Other questions:
  • This process of heat transfer by conduction would NOT work __________
    5·2 answers
  • Match the mixture with its description.
    9·2 answers
  • If the smallest markings on a ruler are one tenth (1/10) of a centimeter, which of the following measurements could be recorded
    5·2 answers
  • What is a compound?
    8·2 answers
  • Which of the following is a characteristic of the antinides?
    14·1 answer
  • A small windmill is to be built at a science competition. A large floor fan is used to power each windmill The task
    9·1 answer
  • 12. Draw the lewis structure for the element to help you answer the question *
    11·1 answer
  • which wavelength of the band on the hydrogen emission line spectrum that has the most energy ? a. 410nm b.434nm c.486nm d.656nm
    6·1 answer
  • How many molecules are in 0.73 mol of C12H22011? *
    5·1 answer
  • Why does water have a very high boiling point?
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!