The accepted density of water at 25oC is 0.99707 g/mL. Calculate the percent error forbothBeakers A and B.

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%

a) the particles of the chair are strongly held together.

the answer is b. dilution