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natulia [17]
2 years ago
12

How do you think density plays a role in the atmosphere?

Physics
1 answer:
Feliz [49]2 years ago
4 0
Answer:
Also, gases like air are easily compressed. The weight of all the air above a given point in the atmosphere squeezes air molecules closer together. This causes the density to increase. The more air above a level (and hence the more weight of air above a level), the greater the compression.

Step By Step Explanation:
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Percent Yield Lab Report
Vlada [557]

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
  • repeated measurements are required in order to improve accuracy

<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>

moles of Mg = mass/molar mass

  • molar mass of Mg = 24 g

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)

Mass of MgO, crucible, and lid = 27.361 (g)

Mass of metal = 27.099 - 26.691 = 0.408

Mass of MgO = 27.361 - 26.691 = 0.670

<h3>Moles of Mg used</h3>

moles of Mg = mass/molar mass

  • molar mass of Mg = 24 g

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

  • moles of MgO produced =  mass/molar mass

molar mass of MgO = 40 g/mol

moles of MgO produced = 0.670/40 = 0.01675

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

Percent yield = 0.01675/0.0170 * 100%

Percent yield of MgO = 98.5%

Average percent yield = (87.7 + 98.5)% / 2

Average percent yield = 89.0%

Based on the data obtained from the reaction, the following conclusion 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
  • repeated measurements are required in order to improve accuracy

Learn more about law of conservation of mass at: brainly.com/question/1824546

6 0
2 years ago
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