All categories

2 years ago

Soil is an essential component of the earth's crust. It enabled life to exist and

Chemistry

1 answer:

goblinko [34]2 years ago

7 0

Answer:

The consequences of soil erosion go beyond the loss of fertile land. It has contributed to increased runoff and sedimentation in streams and rivers, clogging these waters and causing declines in fish and other animals.

We can protect the community from soil erosion by -:

Maintaining a good, perennial cover for plants.

From mulching.

Planting a crop for cover

Explanation:

SOIL EROSION -: The soil erosion mechanism is both natural and man-made. In nature, this refers to the removal of the top layer of soil caused by wind and water, while human activity may increase exposure to these elements.

MAJOR EFFECTS OF SOIL EROSION -:

Pollution and Low Water Quality -: Sedimentation is created by gradual soil erosion, a process by which rocks and minerals in the soil are separated from the soil and deposited elsewhere, often in streams and rivers. Soil contaminants, such as fertilizers and pest control agents, often settle in the streams and rivers to protect crops. Water contaminants contribute to low water quality, including drinking water quality, if the contaminants are not removed prior to ingestion. As sunlight can get through the sediment, sedimentation also leads to the excessive growth of algae. According to the World Wildlife Fund, high levels of algae drain too much oxygen from the water, resulting in the mortality of marine species and reduced fish stocks.
Structural Issues and Mudslides -: Soil erosion contributes to mudslides, impacting the stability of buildings and roadways and their structural integrity. Mudslides affect not only soil-supported structures, but also buildings and roads that are in the path of slides. Mudslides occur when, as a result of the intensity and energy of heavy rainfall, fine sand , clay, silt, organic matter and soil spill off the sides of hills and slopes. According to Envirothon, a program of the National Conservation Foundation and North America's largest high school environmental education competition, this runoff happens rapidly, because there is not enough time for the surface to reabsorb or catch the eroding soil.
Flooding and Deforestation -: Deforestation erodes soil — the removal of trees to create space for towns and agriculture. Trees help to maintain soil in place, so winds and rains drive the loose soil and rocks to streams and rivers when they are uprooted, resulting again in unnecessary sedimentation. The thick layers of sediment keep streams and rivers from flowing smoothly, ultimately contributing to flooding. Excess water, especially during rainy seasons and when the snow melts, gets trapped by the sediment and has nowhere to go except back on land.

The Deterioration of Soil -: Soil nutrient depletion is often the result of poorly performed cultivation and cultivation practices that contribute to soil erosion. For natural vegetation and agricultural purposes, excessive irrigation and obsolete tilling practices decrease the amount of nutrients in the soil and make it less fertile.

PROTECTION OF COMMUNITY FROM SOIL EROSION -

Maintaining a good, perennial cover for plants -: Your perennial garden's care and upkeep need not be difficult or overwhelming. A blend of certain simple horticultural values with common sense and a good eye is a great part of good gardening.
MULCHING -: The amount of water that evaporates from your soil will be reduced by mulch, greatly reducing the need to water the plants. By breaking up clay and permitting better movement of water and air through the soil. Mulch supplements sandy soil with nutrients and enhances its ability to retain water.

PLANTING A CROP FOR COVER -: Winter rye in vegetable gardens, for instance. This includes annual grasses, small grains , legumes and other forms of vegetation that have been planted to provide temporary vegetative cover. Cover crops are also often tilled as a 'green manure' crop under serving.

You might be interested in

A container holds 15.0 g of phosphorous gas at a pressure of 2.0 atm and a temperature of 20.0 Celsius. What is the density of t

erik [133]

Density is a value for mass, such as kg, divided by a value for volume, such as m3. Density is a physical property of a substance that represents the mass of that substance per unit volume. We calculate as follows:

PV = nRT
PV = mRT/ Molar mass
m/V = P(molar mass)/RT
Density = P(molar mass)/RT
Density = 2.0 ( 30.97 ) / 0.08206 ( 20 + 273.15) = 2.57 g/L <----First option

5 0

2 years ago

Read 2 more answers

Ammonia gas is compressed from 21°C and 200 kPa to 1000 kPa in an adiabatic compressor with an efficiency of 0.82. Estimate the

Evgen [1.6K]

Explanation:

It is known that efficiency is denoted by $\eta$ .

The given data is as follows.

$\eta$ = 0.82, $T_{1}$ = (21 + 273) K = 294 K

$P_{1}$ = 200 kPa, $P_{2}$ = 1000 kPa

Therefore, calculate the final temperature as follows.

$\eta = \frac{T_{2} - T_{1}}{T_{2}}$

0.82 = $\frac{T_{2} - 294 K}{T_{2}}$

$T_{2}$ = 1633 K

Final temperature in degree celsius = $(1633 - 273)^{o}C$

= $1360^{o}C$

Now, we will calculate the entropy as follows.

$\Delta S = nC_{v} ln \frac{T_{2}}{T_{1}} + nR ln \frac{P_{1}}{P_{2}}$

For 1 mole, $\Delta S = C_{v} ln \frac{T_{2}}{T_{1}} + R ln \frac{P_{1}}{P_{2}}$

It is known that for $NH_{3}$ the value of $C_{v}$ = 0.028 kJ/mol.

Therefore, putting the given values into the above formula as follows.

$\Delta S = C_{v} ln \frac{T_{2}}{T_{1}} + R ln \frac{P_{1}}{P_{2}}$

= $0.028 kJ/mol \times ln \frac{1633}{294} + 8.314 \times 10^{-3} kJ \times ln \frac{200}{1000}$

= 0.0346 kJ/mol

or, = 34.6 J/mol (as 1 kJ = 1000 J)

Therefore, entropy change of ammonia is 34.6 J/mol.

3 0

2 years ago

For the reaction of oxygen and nitrogen to form nitric oxide, consider the following thermodynamic data (Due to variations in th

CaHeK987 [17]

Answer:

a. 7278 K

b. 4.542 × 10⁻³¹

Explanation:

Let´s consider the following reaction.

N₂(g) + O₂(g) ⇄ 2 NO(g)

The reaction is spontaneous when:

ΔG° < 0 [1]

Let's consider a second relation:

ΔG° = ΔH° - T × ΔS° [2]

Combining [1] and [2],

ΔH° - T × ΔS° < 0

ΔH° < T × ΔS°

T > ΔH°/ΔS°

T > (180.5 × 10³ J/mol)/(24.80 J/mol.K)

T > 7278 K

First, we will calculate ΔG° at 25°C + 273.15 = 298 K

ΔG° = ΔH° - T × ΔS°

ΔG° = 180.5 kJ/mol - 298 K × 24.80 × 10⁻³ kJ/mol.K

ΔG° = 173.1 kJ/mol

We can calculate the equilibrium constant using the following expression.

ΔG° = - R × T × lnK

lnK = - ΔG° / R × T

lnK = - 173.1 × 10³ J/mol / (8.314 J/mol.K) × 298 K

K = 4.542 × 10⁻³¹

7 0

2 years ago

A converging lens with a focal length of 70.0cm forms an image of a 3.20-cm-tall real object that is to the left of the lens. Th

Setler [38]

Answer: (i)The object is at a distance of 120cm from the lens while the image is at a distance of 160cm from the lens. (ii)The image is Real

Note: This is the complete question; A converging lens with a focal length of 70.0cm forms an image of a 3.20-cm-tall real object that is to the left of the lens. The image is 4.50cm tall and inverted. Where are the object and image located in relation to the lens? Is the image real or virtual?

Explanation:

The required formulas are;

1. lens formula given by, 1/s' + 1/s = 1/f

2. magnification = h'/h = s'/s

where h' is image height, h is object height, s' is image distance, s is object distance, f is focal length of lens.

h' = 4.50cm, h = 3.20cm, f = 70cm ( f of a converging lens is positive)

solving for s' and s in eqn (2)

4.50/3.20=s'/s

1.4=s'/s

s'=1.4s

to find the values of s' and s, we use equation (1) and substitute s' = 1.4s

1/1.4s + 1/s = 1/70

2.4/1.4s =1/70

s = 2.4*70/1.4 = 120cm

s' = 1.4*120 = 160cm

Therefore, the object is on the left, 120cm from the lens while the image is 160cm on the right of the lens

Since the object is at a distance between f and 2f from the lens, the image formed is real, inverted and magnified

3 0

2 years ago

Marie and Calvin dissolved 10 grams of KNO3 in 100 grams of water at 25oC. Next they added 5 grams more. Calvin told Marie that

PSYCHO15rus [73]

Hello!

Calvin told Marie that they could continue to add solute until the reached 40 grams because the solution was still unsaturated.

Unsaturated solutions are those in which the solvent (in this case water) can still dissolve more solute (in this case KNO₃) at the given pressure and temperature. This can be seen visually when adding more solute doesn't result in the presence of grains of solids that settle in the bottom of the flask. That happens because the rate of dissolving is higher than the rate of crystallization.

Have a nice day!

8 0

2 years ago