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Darya [45]
2 years ago
12

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:

<u>The consequences of soil erosion</u> 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:

<u>SOIL EROSION -:</u> 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.

<u>MAJOR EFFECTS OF SOIL EROSION -:</u>

  • <u>Pollution and Low Water Quality -:</u> 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.
  • <u>Structural Issues and Mudslides -:</u> 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.
  • <u>Flooding and Deforestation -:</u> 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.
  • <u>The Deterioration of Soil -:</u> 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.

<u>PROTECTION OF COMMUNITY FROM SOIL EROSION -</u>

  • <u>Maintaining a good, perennial cover for plants  -:</u> 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.
  • <u>MULCHING -:</u> 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.
  • <u>PLANTING A  CROP FOR COVER -: </u> 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.
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When 1.04g of cyclopropane was burnt in excess oxygen in a bomb calorimeter, the temperature rose by 3.69K. The total heat capac
STatiana [176]

Answer:

\Delta _{comb}H=-2,093\frac{kJ}{mol}

Explanation:

Hello!

In this case, since these calorimetry problems are characterized by the fact that the calorimeter absorbs the heat released by the combustion of the substance, we can write:

Q_{rxn}+Q_{cal}=0

Thus, given the temperature change and the total heat capacity, we obtain the following total heat of reaction:

Q_{rxn}=-14.01kJ/K*3.69K\\\\Q_{rxn}=-51.70kJ

Now, by dividing by the moles in 1.04 g of cyclopropane (42.09 g/mol) we obtain the enthalpy of combustion of this fuel:

n=\frac{1.04g}{42.09g/mol}=0.0247mol\\\\\Delta _{comb}H=\frac{Q_{rxn}}{n}\\\\  \Delta _{comb}H=-2,093\frac{kJ}{mol}

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4 0
1 year ago
Using the data, which of the following is the rate constant for the rearrangement of methyl isonitrile at 320 ∘C? (HINT: the act
svp [43]

Explanation:

Below is an attachment containing the solution.

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1 year ago
How many grams of NH3 can be prepared from the synthesis of 77.3 grams of nitrogen and 14.2 grams of hydrogen gas?
lbvjy [14]

Answer:

80.41 g

Explanation:

Data Given:

Mass of Nitrogen (N₂) = 77.3 g

Mass of Hydrogen (H₂) = 14.2 g

many grams of NH₃ = ?

Solution:

First we look at the balanced synthesis reaction

              N₂   +    3 H₂  ------—> 2 NH₃

             1 mol      3 mol

As 1 mole of Nitrogen react with 3 mole of hydrogen

Convert moles to mass

molar mass of N₂ = 2(14) = 28 g/mol

molar mass of H₂ = 2(1) + 2 g/mol

Now

                     N₂             +           3 H₂        ------—>      2 NH₃

             1 mol (28 g/mol)     3 mol(2g/mol)

                    28 g                        6 g

28 grams of N₂ react with 6 g of H₂  

So

if 28 grams of N₂ produces 6 g of H₂  so how many grams of N₂ will react with 14.2 g of H₂.

Apply Unity Formula

                 28 g of N₂ ≅ 6 g of H₂

                 X g of N₂ ≅ 14.2 g of H₂

Do cross multiply

                X g of N₂ = 28 g x 14.2 g / 6 g

                X g of N₂ = 66.3 g

As we have given with 77. 3 g of N₂ but from this calculation we come to know that 66.3 g will react with 14.2 g of hydrogen and the remaining 10 g N₂ will be in excess

So, Hydrogen is limiting reactant in this reaction and the amount of NH₃ depends on the amount of hydrogen.

Now

To find mass of NH₃ we will do following calculation

Look at the reaction

As we Know

                     N₂             +           3 H₂        ------—>      2 NH₃

                                                   6 g                            2 mol

So, 6 g of hydrogen gives 2 moles of NH₃, then how many moles of NH₃ will be produce by 14.2 g

Apply Unity Formula

                 6 g of H₂ ≅ 2 mol of NH₃

                14.2 g of H₂ ≅ X mol of NH₃

Do cross multiply

               X mol of NH₃= 14.2 g x 2 mol / 6 g

                X mol of NH₃ = 4.73 mol

So, 14.2 g of hydrogen gives 4.73 moles of NH₃

Now

Convert moles of NH₃ to mass

Formula will be used

        mass in grams = no. of moles x molar mass . . . . . . (2)

Molar mass of  NH₃

Molar mass of  NH₃ = 14 + 3(1)

Molar mass of  NH₃ = 14 + 3 = 17 g/mol

Put values in equation 2

        mass in grams = 4.73 mole x 17 g/mol

        mass in grams =  80.41 g

mass of NH₃=  80.41 g

3 0
1 year ago
Classify the reaction that makes a firefly glow in terms of energy input and output
user100 [1]
It glow, so light energy go out of the system, exotermic
4 0
2 years ago
The decomposition of nitramide, O 2 NNH 2 , O2NNH2, in water has the chemical equation and rate law O 2 NNH 2 ( aq ) ⟶ N 2 O ( g
valkas [14]

Answer:

Explanation:

The given overall reaction is as follows:

O 2 N N H ₂( a q ) k → N ₂O ( g ) + H ₂ O( l )

The reaction mechanism for this reaction is as follows:

O ₂ N N H ₂ ⇌ k 1 k − 1  O ₂N N H ⁻ + H ⁺ ( f a s t  e q u i l i b r i u m )

O ₂ N N H − k ₂→ N ₂ O + O H ⁻ ( s l ow )

H ⁺ + O H − k ₃→ H ₂ O ( f a s t )

The rate law of the reaction is given as follows:

k = [ O ₂ N N H ₂ ]  / [ H ⁺ ]

The rate law can be determined by the slow step of the mechanism.

r a t e = k ₂ [ O ₂ N N H ⁻ ] . . . ( 1 )

Since, from the equilibrium reaction

k e q = [ O ₂ N N H ⁻ ] [ H ⁺ ] /[ O ₂ N N H ₂ ] = k ₁ /k − 1

[ O ₂ N N H ⁻] = k ₁ /k − 1  × [ O ₂ N N H ₂ ] /[ H ⁺ ]. . . . ( 2 )

Substitituting the value of equation (2) in equation (1) we get.

r a t e = k ₂ k ₁/ k − 1  × [ O ₂ N N H ₂ ] /[ H ⁺ ]

Therefore, the overall rate constant is

k = k₂k₁/k-1

5 0
1 year ago
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