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.
- 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.
This is an incomplete question, the given sketch is shown below.
Answer : The name of given unit cell is, FCC (face-centered cubic unit cell)
Explanation :
Unit cell : It is defined as the smallest 3-dimensional portion of a complete space lattice which when repeated over the and again in different directions produces the complete space lattice.
There are three types of unit cell.
- SCC (simple-centered cubic unit cell)
- BCC (body-centered cubic unit cell)
- FCC (face-centered cubic unit cell)
In SCC, the atoms are arranged at the corners.
The number of atoms of unit cell = Z = 1
In BCC, the atoms are arranged at the corners and the body center.
The number of atoms of unit cell = Z = 2
The given unit cell is, FCC because the atoms are arranged at the corners and the center of the 6 faces.
The number of atoms of unit cell = Z = 4
Thus, the name of given unit cell is, FCC (face-centered cubic unit cell)
Same as balancing a regular chemical reaction! Please see the related question to the bottom of this answer for how to balance a normal chemical reaction. This is for oxidation-reduction, or redox reactions ONLY! These instructions are for how to balance a reduction-oxidation, or redox reaction in aqueous solution, for both acidic and basic solution. Just follow these steps! I will illustrate each step with an example. The example will be the dissolution of copper(II) sulfide in aqueous nitric acid, shown in the following unbalanced reaction: CuS (s) + NO 3 - (aq) ---> Cu 2+ (aq) + SO 4 2- (aq) + NO (g) Step 1: Write two unbalanced half-reactions, one for the species that is being oxidized and its product, and one for the species that is reduced and its product. Here is the unbalanced half-reaction involving CuS: CuS (s) ---> Cu 2+ (aq) + SO 4 2- (aq) And the unbalanced half-reaction for NO 3 - is: NO 3 - (aq) --> NO (g) Step 2: Insert coefficients to make the numbers of atoms of all elements except oxygen and hydrogen equal on the two sides of each half-reaction. In this case, copper, sulfur, and nitrogen are already balanced in the two half-reaction, so this step is already done here. Step 3: Balance oxygen by adding H 2 O to one side of each half-reaction. CuS + 4 H 2 O ---> Cu 2+ + SO 4 2- NO 3 - --> NO + 2 H 2 O Step 4: Balance hydrogen atoms. This is done differently for acidic versus basic solutions. . For acidic solutions: Add H 3 O + to each side of each half-reaction that is "deficient" in hydrogen (the side that has fewer H's) and add an equal amount of H 2 O to the other side. For basic solutions: add H 2 O to the side of the half-reaction that is "deficient" in hydrogen and add an equal amount of OH - to the other side. Note that this step does not disrupt the oxygen balance from Step 3. In the example here, it is in acidic solution, and so we have: CuS + 12 H 2 O ---> Cu 2+ + SO 4 2- + 8 H 3 O + . NO 3 - + 4 H 3 O + --> NO + 6 H 2 O Step 5: Balance charge by inserting e - (electrons) as a reactant or product in each half-reaction. Oxidation: CuS + 12 H 2 O ---> Cu 2+ + SO 4 2- + 8 H 3 O + + 8 e - . Reduction: NO 3 - + 4 H 3 O + + 3 e - --> NO + 6 H 2 O . Step 6: Multiply the two half-reactions by numbers chosen to make the number of electrons given off by the oxidation step equal to the number taken up by the reduction step. Then add the two half-reactions. If done correctly, the electrons should cancel out (equal numbers on the reactant and product sides of the overall reaction). If H 3 O + , H 2 O, or OH - appears on both sides of the final equation, cancel out the duplication also. Here the oxidation half-reaction must be multiplied by 3 (so that 24 electrons are produced) and the reduction half-reaction must by multiplied by 8 (so that the same 24 electrons are consumed). 3 CuS + 36 H 2 O ---> 3 Cu 2+ + 3 SO 4 2- + 24 H 3 O + + 24 e - 8 NO 3 - + 32 H 3 O + + 24 e - ---> 8 NO + 48 H 2 O Adding these two together gives the following equation: 3 CuS + 36 H 2 O + 8 NO 3 - + 8 H 3 O + ---> 3 Cu 2+ + 3 SO 4 2- + 8 NO + 48 H 2 O Step 7: Finally balancing both sides for excess of H 2 O (On each side -36) This gives you the following overall balanced equation at last: 3 CuS (s) + 8 NO 3 - (aq) + 8 H 3 O + (aq) ---> 3 Cu 2+ (aq) + 3 SO 4 2- (aq) + 8 NO (g) + 12 H 2 O (l)
Displacement = √(3² + 4²)
Displacement = 5 meters north east
Velocity = displacement / time
Velocity = 5 / 35
Velocity = 0.14 m/s northeast
Answer:
Explanation:
Ketcher 01232019462D 1 1.00000 0.00000 0 5 4 0 0 0 999 V2000 -0.0330 2.2250 0.0000 C 0 0 0 0 0 0 0 0 0 0 0.8330 2.7250 0.0000 C 0 0 0 0 0 0 0 0 0 0 1.6990 2.2250 0.0000 C 0 0 0 0 0 0 0 0 0 0 0.8330 3.7250 0.0000 C 0 0 0 0 0 0 0 0 0 0 1.6990 1.2250 0.0000 C 0 0 0 0 0 0 0 0 0 0 1 2 1 0 0 0 2 3 1 0 0 0 2 4 1 0 0 0 3 5 1 0 0 0 M END
