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STALIN [3.7K]
4 years ago
7

in Analytical chemistry laboratory. Nickle(ppm)in water sample estimated by two different students was as ,student 1 (25 24 24 2

4 24 28 26 ) and student 2 (24 25 24 24 25 25 ), if the true valude on Nickle in water was 20 ppm estimate the precision and accuracy of both students
Chemistry
1 answer:
frosja888 [35]4 years ago
6 0

Answer:

Answers are in the explanation

Explanation:

Precision and accuracy are both statistical concepts. Precision is about how close are the measured values to each other. Accuracy is closeness of the measured values to a specific value or reference value.

There are many ways to calculate precision and accuracy, one is stimating precision as <em>Relative Uncertainty </em>and accuracy as <em>Relative Error.</em>

Formulas are:

Relative Uncertainty: Standard desviation /  Measured value.

Relative Error: Measured Value - Expected Value / Expected value

<em>Measured value is the aveage of the values.</em>

<em />

For student 1:

Mean = 25

Standard desviation = 1.5275

Relative Uncertainty: 1.5275 /  25

Relative Uncertainty: 0.061

Relative Error: 25 - 20 / 20

Relative Error: 0.25

For student 2:

Mean = 24.5

Standard desviation = 0.5477

Relative Uncertainty: 0.5477 / 24.5

Relative Uncertainty: 0.02

Relative Error: 24.5 - 20 / 20

Relative Error: 0.225

<em>As the student 2 has a low Relative error and uncertainty, you can say student 2 is more accurate and precise</em>

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Answer:

see explanation below

Explanation:

The question is incomplete. Here's the complete question:

<em>Consider the following reaction: </em>

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This is a problem of equilibrium, therefore, we need to solve this using the expression of equilibrium constant. To do that, we need to wirte an ICE chart and solve from there:

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Replacing the values from the chart:

2.99x10⁻⁷ = x² / 0.168 - x

However, Kc is a very very small value, therefore, we can assume that the value of "x" would be very small too, and we can neglect the 0.168-x and just round it to 0.168:

2.99x10⁻⁷ = x²/0.168

2.99x10⁻⁷ * 0.168 = x²

√5.02x10⁻⁸ = x

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This means then, that the concentration of Cl2 in equilibrium would be:

<em>[Cl₂] = 2.24x10⁻⁴ M</em>

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