eraines1714
13.01.2021 •
Chemistry
The accepted density of water at 25oC is 0.99707 g/mL. Calculate the percent error forbothBeakers A and B.
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Ответ:
Percent error for beaker A is 0.26%
Percent error for beaker B is 0.25%
Explanation:
The remaining part of the question is shown in the attachment below.
For beaker A
Mass of water = 9.997 g
Volume = 10mL
First, we will determine the experimental density
From
Density = Mass / Volume
∴ Density = 9.997g / 10mL
Density = 0.9997 g/mL
To calculate the percent error
Percent error = (Error/True value) × 100%
Error = Experimental value - True value
From the question, True value = 0.99707 g/mL
Experimental value = 0.9997 g/mL
∴ Error = 0.9997 g/mL - 0.99707 g/mL
Error = 0.00263 g/mL
Then,
Percent error = (0.00263/0.99707) × 100%
Percent error = 0.26377%
Percent errror ≅ 0.26%
Hence, percent error for beaker A is 0.26%
For beaker B
Mass of water = 9.9956 g
Volume = 10mL
Also, we will first determine the experimental density
From
Density = Mass / Volume
∴ Density = 9.9956g / 10mL
Density = 0.99956 g/mL
To calculate the percent error
Percent error = (Error/True value) × 100%
Error = Experimental value - True value
True value = 0.99707 g/mL
Experimental value = 0.99956 g/mL
∴ Error = 0.99956 g/mL - 0.99707 g/mL
Error = 0.00249 g/mL
Then,
Percent error = (0.00249/0.99707) × 100%
Percent error = 0.24973%
Percent errror ≅ 0.25%
Hence, percent error for beaker B is 0.25%
Ответ: