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lutik1710 [3]
3 years ago
15

A group of students perform three trials in a lab and determine the density of aluminum to be 2.1, 2.12, and 2.09 g/cm³ if the t

rue density of aluminum is 2.7 g/cm³ what is the percent error of their lab
Chemistry
1 answer:
jenyasd209 [6]3 years ago
4 0

Answer:

Percent error = -22%

Explanation:

Given data:

Density of aluminium = 2.1,2.12 and 2.09 g/cm³

Accepted value of density of aluminium = 2.7 g.cm³

Percent error = ?

Solution:

First of all we will calculate the average of data measured by students.

Average value = 2.1 g/cm³ + 2.12 g/cm³+ 2.09 g/cm³ / 3

Average value = 6.31 g/cm³/ 3

Average value = 2.1 g/cm³

Now we will calculate the percent error.

Percent error = [ Measured value - accepted value / accepted value ]× 100

Percent error = [ 2.1 g/cm³  - 2.70 g/cm³ / 2.70 g/cm³ ]× 100

Percent error = [-0.6/2.70 g/cm³ ] × 100

Percent error = -0.22 × 100

Percent error = -22%

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Answer:

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Explanation:

If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy (s_{2} - s_{1}), in joules per gram-Kelvin, by the following model:

s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }

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m - Mass, in kilograms.

c_{w} - Specific heat of water, in joules per kilogram-Kelvin.

T_{1}, T_{2} - Initial and final temperatures of water, in Kelvin.

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s_{2} - s_{1} = (1\,kg) \cdot \left(4190\,\frac{J}{kg\cdot K} \right)\cdot \ln \frac{288.15\,K}{373.15\,K}

s_{2} - s_{1} = -1083.112\,\frac{J}{kg\cdot K}

The change in entropy is -1083.112 joules per kilogram-Kelvin.

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