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irinina [24]
4 years ago
9

Sonar is the detection of sound waves reflected off boundaries in water. a region of warm water in a cold lake can produce a ref

lection, as can the bottom of the lake. which would you expect to produce the stronger echo? explain.
Physics
2 answers:
kirill [66]4 years ago
5 0
The stronger echo will originate from the bottom and between the lake since the sound will bounce back harder as the bottom of the lake is solid. Unlike the boundary between the cold and warm water.
svetlana [45]4 years ago
4 0

The echo will be stronger when originated from the bottom of the lake.

Explanation

Echo is created when sound waves get reflected due to difference in the densities of the medium.

As we know that SONAR technique works on the detection of the echo strength to navigate, to determine the thickness of any object in water and also to measure the depth of the water bodies.

A region of warm water and cold water will create very negligible echo as there is no object which can resist the penetration of sound waves in this region.

But the sound waves will get reflected when hits the solid object or the bottom of the lake.

As the bottom of the lake is made up of solid, the sound waves will not be able to penetrate the bottom layer.

This will lead to reflection of sound waves creating echo of stronger energy.

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Based on the data obtained from the reaction, the following conclusions can be made;

  • according to the law of conservation of mass, the mass of the product was higher than the reactant because of the mass of oxygen added during the combustion.
  • the percent yield is less than 100% because of loss in mass of either reactants or products.
  • the errors could have occurred during the weighing and transfer of reactants and products
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<h3>What is the percent yield of the reaction?</h3>

Equation of the reaction is given below:

  • 2 Mg + O₂ ----> 2 MgO

Trial 1

Mass of empty crucible with lid = 26.698 (g)

Mass of Mg metal, crucible, and lid  = 27.040 (g)

Mass of MgO, crucible, and lid = 27.198 (g)

Mass of metal = 27.040 - 26.698 = 0.342

Mass of MgO = 27.198 - 26.698 = 0.500

<h3>Moles of Mg used</h3>

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moles of Mg = 0.342/24 = 0.01425

<h3>Moles of MgO expected</h3>

Based on the equation of reaction;

moles of MgO expected = 0.01425

moles of MgO produced =  mass/molar mass

  • molar mass of MgO = 40 g/mol

moles of MgO produced = 0.500/40 = 0.0125

<h3>Percent yield</h3>
  • Percent yield = (moles of MgO produced/moles of MgO expected) * 100%

Percentage yield = 0.0125/0.01425 * 100%

Percent yield of MgO = 87.7%

Trial 2

Mass of empty crucible with lid = 26.691 (g)

Mass of Mg metal, crucible, and lid = 27.099 (g)

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Mass of metal = 27.099 - 26.691 = 0.408

Mass of MgO = 27.361 - 26.691 = 0.670

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moles of Mg = 0.408/24 = 0.0170

<h3>Moles of MgO expected</h3>

Based on the equation of reaction;

moles of MgO expected = 0.0170

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Percent yield = 0.01675/0.0170 * 100%

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