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omeli [17]
3 years ago
11

Shelby measured the volume of a cylinder and determined it to be 54.5cm3. The teacher told her that she was 4.25% too high in he

r determination of the volume. What is the actual volume of the cylinder ?
Chemistry
1 answer:
FinnZ [79.3K]3 years ago
6 0

Answer:

56.8cm³

Explanation:

Given parameters:

Measured volume = 54.5cm³

Percentage error  = 4.25%

Unknown:

Actual volume of the cylinder = ?

Solution:

The percentage error shows the amount of error introduced into a measurement.

We need to find out this amount of error from the data given.

   Error = Percentage error x measured volume

                       Error = \frac{4.25}{100} x 54.5  = ±2.32cm³

Since the error introduced = 2.32cm³

  Actual volume of cylinder = measured volume ± Error

  Since the percentage error quoted as higher 4.25%;

 Actual volume  = 54.5cm³ + 2.32cm³ = 56.8cm³

       

 

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Limiting Reactants—————-
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Answer:

21.8 grams.

Explanation:

Molar mass data from a modern periodic table:

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  • O: 15.999.

How many moles of MgO will be produced if Mg is the limiting reactant?

Number of moles of Mg:

\displaystyle n = \frac{m}{M} = \frac{16.3}{24.301} = 0.670644\;\text{mol}.

The ratio between the coefficient of Mg and that of MgO is 2:2. Two moles of Mg will make two moles of MgO. 0.670644 moles of MgO will be produced if Mg is the limiting reactant.

How many moles of MgO will be produced if O₂ is the limiting reactant?

Number of moles of O₂:

\displaystyle n = \frac{m}{M} = \frac{4.33}{15.999} = 0.270642\;\text{mol}.

The ratio between the coefficient of O₂ and that of MgO is 1:2. One mole of O₂ will make two moles of MgO. 2\times 0.270642 = 0.541284\;\text{mol} of MgO will be produced if O₂ is in excess.

How many moles of MgO will be produced?

0.541284 is smaller than 0.670644. Only 0.541284 moles of MgO will be produced since O₂ will run out before all 16.3 grams of Mg is consumed.

What's the mass of 0.541284 moles of MgO?

Formula mass of MgO:

24.301 + 15.999 = 40.300\;\text{g}\cdot\text{mol}^{-1}.

Mass of 0.541284 moles of MgO:

m = n \cdot M = 0.541284\times 40.300 = 21.8\;\text{g}.

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3 years ago
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