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A certain analytical method for the determination of lead yields masses for lead that are low by 0.5 g. Calculate the percent relative error caused by this deviation for each measured mass of lead. Report the percent relative error with the correct number of significant figures.

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MrRoyal

Question Continuation

if the measured weight of lead in the sample is

a.) 764.9g lead

b.)226.3g lead

c.) 53.5g lead

Answer:

a.

Relative Error = 0.065

b.

Relative Error = 0.221

c.

Relative Error = 0.935

Step-by-step explanation:

Given

Absolute Error = 0.5g

Relative error = absolute error/magnitude of measurement.

Relative error % = Relative error * 100

a.

Relative Error = 0.5/764.9 * 100

Relative Error = 50/764.9

Relative Error = 0.065

b.

Relative Error = 0.5/226.3 * 100

Relative Error = 50/226.3

Relative Error = 0.221

c.

Relative Error = 0.5/53.5 * 100

Relative Error = 50/53.5

Relative Error = 0.935

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