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tensa zangetsu [6.8K]
3 years ago
15

At construction sites, workers often reduce erosion by ____.

Chemistry
1 answer:
Naddik [55]3 years ago
5 0
<h3><u>Answer;</u></h3>

<em>All the above</em>

Workers at construction sites often reduce erosion by;

  • <em>Moving excess sediment back to its original location </em>
  • <em>Planting trees </em>
  • <em>Spraying water on bare soil</em>
<h3><u>Explanation;</u></h3>
  • Soil erosion is a naturally occurring process which involves the wearing away of the topsoil by natural forces such as wind, water or other forces associated with farming.
  • <em><u>Construction of roads and buildings results to large amounts of soil erosion around the world. It is therefore important to put measures that would help reduce soil erosion at construction sites</u></em>. These measures uses principals of soil control such as implementing sediment control, limiting soil exposure, reducing the runoff velocity, and modifying topography among others.
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Calcium Oxide will react with ammonium chloride to produce ammonia gas, water vapor, and calcium chloride. If only 16.3 g of amm
Vaselesa [24]

Answer:

                     %age Yield  =  22.72 %

Explanation:

                    The balance chemical equation is as follow;

                           2 NH₄Cl + CaO → 2 NH₃ + CaCl₂ + H₂O

Step 1: <u>Calculate Moles of CaO, NH₄Cl and NH₃;</u>

CaO:

        Moles  =  Mass / M.Mass

        Moles  =  112 g / 56.07 g/mol

        Moles  =  1.99 moles of CaO

NH₄Cl:

        Moles  =  Mass / M.Mass

        Moles  =  224 g / 53.49 g/mol

        Moles  =  4.18 moles of NH₄Cl

NH₃:

        Moles  =  Mass / M.Mass

        Moles  =  16.3 g / 17.03 g/mol

        Moles  =  0.95 moles of NH₃

Step 2: <u>Calculate Limiting reagent as:</u>

According to equation.

                   2 moles of NH₄Cl reacts with  =  1 mole of CaO

So,

              4.18 moles of NH₄Cl will react with  =  X moles of CaO

Solving for X,

                     X =  4.18 mol × 1 mol / 2 mol

                     X  =  2.09 mol of CaO

This means that none of the given reagent is limiting reagent. They both are almost equal in number of moles.

Step 3: <u>Calculate Theoretical Yield of NH₃ as;</u>

According to equation.

                   2 moles of NH₄Cl produced  =  2 moles of NH₃

So,

              4.18 moles of NH₄Cl will produce  =  X moles of NH₃

Solving for X,

                     X =  4.18 mol × 2 mol / 2 mol

                     X  =  4.18 mol of NH₃

Step 4: <u>Calculate Percentage Yield as;</u>

             %age Yield  =  Actual Yield / Theoretical Yield × 100

             %age Yield  =  0.95 mol / 4.18 × 100

             %age Yield  =  22.72 %

5 0
3 years ago
What is the pOH of a solution of 0.01 M HCI
Taya2010 [7]

Answer:

The answer is 2

Explanation:

5 0
3 years ago
What volume (in mL ) of 0.200   M NaOH do we need to titrate 40.00 mL of 0.140   M HBr to the equivalence point?
Ugo [173]
NaOH + HBr =⇒ NaBr + H2O

35.0 ml HBr x 1 liter/1000 mL x 0.140 moles HBr/ liter = 0.0049 moles HBr

0.0049 moles HBr x 1mole NaOH/1mole HBr = 0.0049 moles HBr

0.0049 moles HBr x 1 liter NaOH/0.200 moles NaOH x 1000 mL/1liter= 24.5 mL NaOH
7 0
3 years ago
Read 2 more answers
What is the formula for the compound?
4vir4ik [10]

Answer:

the correct answer is CO2 (the last one)

7 0
3 years ago
Read 2 more answers
Given the propane molecule C3H8(g)! a) What is the balanced equation for the combustion of propane in O2?(20 pts.) b) What is th
lubasha [3.4K]

Answer:

The answer to your question is below

Explanation:

Propane = C₃H₈

Process

1.- Write the chemical reaction

                     C₃H₈  +  O₂   ⇒   CO₂  +  H₂O

balanced chemical reaction

                      C₃H₈  +  5O₂   ⇒  3CO₂  +  4H₂O

               Reactants      Elements     Products

                      3                    C                  3

                      8                    H                  8

                     10                    O                 10

b) Standard enthalpy

Propane                  -104.7 kJ/mol

Oxygen                         0   kJ/mol

Carbon dioxide      -393.5 kJ/mol

Water                      -241.8 kJ/mol

ΔH° = ∑ΔH° products - ∑H° reactants

ΔH° = 3(-393.5) + 4(-241.8) - [(-104.7) + 0]

-Simplification

ΔH° = -1180.5 kJ/mol - 967.2 kJ/mol + 104.7 kJ/mol

ΔH° = - 2042.5 kJ/mol

This reaction is exothermic because ΔH° is negative.

4 0
3 years ago
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