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DedPeter [7]
3 years ago
14

Pipettes are used when measuring the volumes of liquids with high degree of precision. The following volumes were obtained durin

g the calibration of a 10.00 mL pipette: 10.15 mL, 9.95 mL, 9.99 mL, and 10.02 mL. Calculate the standard deviation for these measurements.
Chemistry
1 answer:
never [62]3 years ago
6 0

Answer:

0.075

Explanation:

First obtain the mean of the measurement;

Mean = 10.15 + 9.95 + 9.99 + 10.02/4 = 10.03

Then obtain d^2= (mean-score)^2 for each score;

(10.15-10.03)^2 = 0.0144

(9.95-10.03)^2 = 0.0064

(9.99-10.03)^2 = 0.0016

(10.02-10.03)^2= 0.0001

∑d^2= 0.0144 + 0.0064 + 0.0016 + 0.0001

∑d^2= 0.0225

Variance = ∑d^2/N = 0.0225/4 = 0.005625

Standard deviation= √0.005625

Standard deviation= 0.075

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Explanation:

Step 1: Given and required data

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We will use the conversion factor 1 m = 10⁹ nm.

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E = 4.59 × 10⁻³³ J = 4.59 × 10⁻³⁶ kJ

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