14) Describe the Cloud Model.

Answer:

vague symptoms are characteristic of an acute toxin, because of the of the lack of well defined consistency that these symptoms have in relation to the course of the disease progress.

Answer:

9.09

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

Second step of the mechanism is slow step.

B. Only 2

Rate of reaction:

It is the speed at which a chemical reaction takes place, defined as proportional to the increase in the concentration of a product per unit time and to the decrease in the concentration of a reactant per unit time.

Chemical reaction:

The reaction mechanism  is as follows.

Second step is the slowest step, thus the rate determining step.

Therefore, rate of reaction can be represented as:

Thus, option B is correct.

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Answer:

Second step of the mechanism is slow step.

Explanation:

The given chemical reaction is as follows

The mechanism of the reaction is as follows.

The rate of the reaction.

Therefore, Step -2 must be slow.

Some of the effects of erosion on the environment includes :

• Washing away of soil nutrients
• Pollution of the waterways
• Blockage of drainages
• degradation of soil

Ways to protect the earth from soil erosion includes

• planting of cover crops
• Mulching
• use of crush rocks on certain areas

Soli erosion is the washing away of the top soil of the earth's crust which can be caused by the movement of wind , water or ice over the surface of the earth crust. this action leads to the degradation of the soil

Soli erosion leads to the washing away of soil nutrients and the pollution of waterways because of the deposition of soil particles into the waterways.  the washed away soil can also block the drainages leading to a bigger problem ( flooding ).

Some of the steps that would help protect the soil from the effects of soil erosion are planting of cover crops , mulching and use of crush rocks on areas that are used most frequently to prevent the washing away of the soil.

Hence we can conclude that the effects of soil erosion are Washing away of soil nutrients , Pollution of the waterways, Blockage of drainages while ways to protect the earth from erosion are ;planting of cover crops, Mulching, use of crush rocks on certain areas

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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.

Answer:

b. 3.35

Explanation:

To calculate the pH of a solution containing both acid and its salt (produced as a result of titration) we need to use Henderson’s equation i.e.

pH = pKa + log ([salt]/[acid])     (Eq. 01)

Where  

pKa = -log(Ka)        (Eq. 02)

[salt] = Molar concentration of salt produced as a result of titration

[acid] = Molar concentration of acid left in the solution after titration

Let’s now calculate the molar concentration of HNO2 and KOH considering following chemical reaction:

HNO2 + KOH ⇆ H2O + KNO2    (Eq. 03)

This shows that 01 mole of HNO2 and 01 mole of KOH are required to produce 01 mole of KNO2 (salt). And if any one of them (HNO2 and KOH) is present in lower amount then that will be considered the limiting reactant and amount of salt produced will be in accordance to that reactant.

Moles of HNO2 in 50 mL of 0.01 M HNO2 solution = 50/1000x0.01 = 0.005 Moles

Moles of KOH in 25 mL of 0.01 M KOH solution = 25/1000x0.01 = 0.0025 Moles

As it can be seen that we have 0.0025 Moles of KOH therefore considering Eq. 03 we can see that 0.0025 Moles of KOH will react with only 0.0025 Moles of HNO2 and will produce 0.0025 Moles of KNO2.

Therefore

Amount of salt produced i.e [salt] = 0.0025 moles       (Eq. 04)

Amount of acid left in the solution [acid] = 0.005 - 0.0025 = 0.0025 moles (Eq.05)

Putting the values in (Eq. 01) from (Eq.02), (Eq. 04) and (Eq. 05) we will get the following expression:

pH= -log(4.5x10 -4) + log (0.0025/0.0025)

Solving above we get  

pH = 3.35

Final answer:

The pH value in the titration flask after 25.00 mL of the 0.10 M KOH solution is added to 50.00 mL of 0.10 M HNO2 solution is 3.35.

Explanation:

The subject of this question is titration, which is a method used in chemistry to measure the concentration of an unknown solution. Given 50.00 mL of 0.10 M HNO2 (nitrous acid, Ka = 4.5 × 10-4), titrated with 0.10 M KOH (potassium hydroxide), we need to calculate the pH after 25.00 mL of the KOH solution is added.

First, we need to find the moles of the HNO2 and the KOH. Moles equals Molarity times Volume. So, for HNO2, it is 0.10 M * 0.050 L which equals 0.005 moles. For KOH, it is 0.10 M * 0.025 L which equals 0.0025 moles.

Then, subtract the moles of OH- from the moles of HNO2 to determine the concentration of HNO2 left, which is 0.005 moles - 0.0025 moles = 0.0025 moles. Divide this by the total volume of the solution (50.00 mL + 25.00 mL = 75.00 mL or 0.075 L to determine the new concentration of HNO2, 0.0025 moles / 0.075 L = 0.033 M. Then use the given Ka value with the equation [H+] = sqrt(Ka * [HNO2]) to get [H+].

To find acids' pH, we use the formula pH = -log[H+]. Use the calculated [H+] to find the pH.

Upon performing these calculations, the resulting pH value should be approximately 3.35 after 25.00 mL of the KOH solution is added, so the answer is (b) 3.35.

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