Learning Task 1. Read each item carefully. Choose the letter of your answer.1. Which type of soil is characterized as having the finest particles holding greater amount of water?
A. Loam
B. Clay
C. Sand
2. Which type of soil is best for planting?
A. Loam
B. Clay
C. Sand
3. How does each soil types differ?
A. Texture
B. Color
C. Both A & B
4. Which type of soil do you usually expect if the community is along the seashore?
A. Loam
B. Clay
C. Sand
5. Why is soil important to living things?
A. Forms part of the earth where animals live
B. Provides the necessary nutrients needed by plants
C. Serves as a place where people live
D. All of the above​


Answer 1


1. B

2. A

3. C

4. C

5. D


Soil is regarded as the solid unconsolidated material of the earth crust. Soil is of three different types namely: Sandy soil, clay soil and loamy soil. These three different soil types possess different properties that distinguish them. Some of them are:

- CLAY soil is characterized as having the finest particles and can hold greater amount of water i.e. have a high water holding capacity.

- LOAMY SOIL is the best soil type for planting agricultural crops because it has the highest concentration of nutrients that suited for plant growth.

- loamy, Sandy and clay differ in how we feel when touched i.e. texture, and colour.

- SANDY soils are the kind of soils that are found in Sea shores and beaches.

- Soil is important to living things as it forms part of the earth where animals live, provides the necessary nutrients needed by plants, serves as a place where people live.

Related Questions

What experimental evidence led Rutherford to conclude that an atom is mostly space?
A binary feed mixture contains 40 mol% hexane (A) and 60 mol% toluene (B) is to be separated continuously into two products D (distillate) and B (bottoms) in a distillation unit. Distillate D is 90 mol% hexane and the bottoms B is 90 mol% toluene. Using a feed flow rate of 100 lbmoh as basis, compute the flow rates of products B and D in: (a) lbmol/h, and (b) kmol/h.
Be sure to answer all parts. Industrially, hydrogen gas can be prepared by combining propane gas (c3h8) with steam at about 400°c. The products are carbon monoxide (co) and hydrogen gas (h2). (a) write a balanced equation for the reaction. Include phase abbreviations. (b) how many kilograms of h2 can be obtained from 8.31 × 103 kg of propane
Consider the following chemical equilibrium: C(s) + 2H2 (g) <------> CH4 (g) Now write an equation below that shows how to calculate Kp from Kc for this reaction at an absolute temperature T. You can assume T is comfortably above room temperature. If you include any common physical constants in your equation be sure you use their standard symbols, found in the ALEKS Calculator.
A sample of solid calcium hdroxide, Ca(OH)2 is allowed to stand in water until a saturated solution is formed. A titration of 75.00mL of this solution with 5.00 x 10-2 M HCl 36.6 mL of the acid to reach the end pointCa(OH)2 + 2HCl ? CaCl + 2H2O What is the molarity?

Which particles are equal in number for an atom with a neutral charge? ​


electrons and protons

Metal vapor deposition is a process used to deposit a very thin layer of metal on a target substrate. A researcher puts a glass slide in the metal vapor deposition chamber and coats the slide with copper. After deposition, the glass slide had increased in mass by 2.26 milligrams. Approximately how many copper atoms were deposited on the glass slide



2.14x10¹⁹ atoms of Cu were deposited


The increased in mass of the glass slide is due the deposition of copper.

That means the mass of copper deposited is 2.26mg = 2.26x10⁻³g Cu

To know the copper atoms we need to convert this mass to moles of Cu using molar mass of copper (63.546g/mol), and these moles are converted to atoms using Avogadro's number (6.022x10²³ atoms = 1 mole)

Moles Cu:

2.26x10⁻³g Cu * (1 mol / 63.546g) = 3.556x10⁻⁵ moles Cu

Atoms Cu:

3.556x10⁻⁵ moles Cu * (6.022x10²³ atoms / 1 mole) =

2.14x10¹⁹ atoms of Cu were deposited

1. Based on the appearance of your reaction in the beaker, which reagent do you think was consumed and which reagent had some left over? The aluminum was consumed, and copper was left over as seen by the reddish particles. 2. If 5.0 g of iron metal is reacted with 15.0 g of Cl2 gas, how many grams of ferric chloride will form? About 14.52 grams will form. 3. For the reaction above the amount of ferric chloride obtained in the lab was 9.15 grams. Calculate the percent yield. The percent yield would be around 63.02%. 4. What are some reasons for obtaining a percent yield of less than 100 percent? Factors such as the reactants not reacting completely, human error in the experiment, the reactants might have too large of a surface area for reaction, multiple reactions occurring within an experiment, temperature, etc.



1. Al is consumed first and CuSO₄ remains left.

2. The grams of ferric chloride that forms is 14.5 g.

3. The percent yield is 63.1%

4. Trace of impurities present in the reagents, bad manipulations when preparing solutions, etc.


1. The reaction is:

2Al + 3CuSO₄ = Al₂(SO₄)₃ + 3Cu

The number of moles of Al is less than the number of moles of CuSO₄. Therefore, Al is the limiting reagent and CuSO₄ is the excess reagent. This means that Al is consumed first and CuSO₄ remains left.

2. The reaction is:

2Fe + 3Cl₂ = 2FeCl₃

The number of moles of Fe is:

n_(Fe) =(m_(Fe) )/(MW_(Fe) ) =(5)/(55.85) =0.0895moles

The number of moles of Cl₂ is:

n_(Cl2) =(15)/(70.9) =0.211moles

We know that 2 moles of Fe react with 3 moles of Cl₂, thus:

2 moles Fe---------------3 moles Cl₂

0.0895 moles Fe-------X moles Cl₂

Clearing X:

Xmoles_(Cl2) =(3*0.0895)/(2) =0.134moles

It needs 0.134 moles of Cl₂ but it only has 0.211 moles, thus, Cl₂ is the excess reagent. Fe is the limiting reagent.

2 moles Fe-----------2 moles FeCl₃

0.0895 moles Fe------X moles FeCl₃

Clearing X:

Xmoles_(FeCl3) =(2*0.0895)/(2) =0.0895moles

m_(FeCl3) =0.0895molesFeCl3*(162.2gFeCl3)/(1molFeCl3) =14.5g

3. The actual yield of FeCl₃ is 9.15 g, the theoritical yield is 14.5 g, thus, ther percent yield is:

Percent-yield=(Actual-yield)/(Theoritical-yield) *100=(9.15)/(14.5) *100=63.1%

4. Trace of impurities present in the reagents, bad manipulations when preparing solutions, etc.

The salinity of sea water represents the amount of NaCl (salt) dissolved in the water.True


True because the sea water is mostly salt

In a lab experiment 80.0 g of ammonia [NH3] and 120 g of oxygen are placed in a reaction vessel. At the end of the reaction 72.2 g of water are obtained. Determine the percent yield of the reaction.


The percent yield of the reaction : 89.14%

Further explanation

Reaction of Ammonia and Oxygen in a lab :

4 NH₃ (g) + 5 O₂ (g) ⇒ 4 NO(g)+ 6 H₂O(g)

mass NH₃ = 80 g

mol NH₃ (MW=17 g/mol):


mass O₂ = 120 g

mol O₂(MW=32 g/mol) :

\tt (120)/(32)=3.75

Mol ratio of reactants(to find limiting reatants) :

\tt (4.706)/(4)/ (3.75)/(5)=1.1765/ 0.75\rightarrow O_2~limiting~reactant(smaller~ratio)

mol of H₂O based on O₂ as limiting reactants :

mol H₂O :

\tt (6)/(5)* 3.75=4.5

mass H₂O :

4.5 x 18 g/mol = 81 g

The percent yield :

\tt \%yield=(actual)/(theoretical)* 100\%\n\n\%yield=(72.2)/(81)* 100\%=89.14\%

Radioactive gold-198 is used in the diagnosis of liver problems. the half-life of this isotope is 2.7 days. if you begin with a sample of 8.1 mg of the isotope, how much of this sample remains after 2.6 days?



See explanation below


To solve this problem, we need to use the expression of half life decay of concentration (or mass) which is the following:

m = m₀e^-kt  (1)

In this case, k will be the constant rate of this element. This is calculated using the following expression:

k = ln2/t₁/₂  (2)

Let's calculate the value of k first:

k = ln2/2.7 = 0.2567 d⁻¹

Now, we can use the expression (1) to calculate the remaining mass:

m = 8.1 * e^(-0.2567 * 2.6)

m = 8.1 * e^(-0.6674)

m = 8.1 * 0.51303

m = 4.16 mg remaining

Final answer:

The half-life of gold-198 is the time it takes for half of it to decay. Given that the half-life is 2.7 days, and the period in consideration is 2.6 days, approximately half of the original amount of 8.1 mg, which is 4.05 mg, will remain.


This problem is related to the concept of half-life in radioactive decay. The half-life of a substance is the time it takes for half of it to decay. As the half-life of gold-198 is 2.7 days and we are considering a period of 2.6 days, which is almost one half-life, therefore, approximately half the substance should have decayed.

So, if you start with 8.1 mg of gold-198, at the end of one half-life (or close to it at 2.6 days), you should have approximately half of this amount remaining. Half of 8.1 mg is 4.05 mg, thus, approximately 4.05 mg remains after 2.6 days.

Learn more about half-life here: