Question

3 points

18) A student determines the density of gold to be 20.9g/L. The true
density of gold is 19.30g/L. What is the student's percent error?round
answer to 2 significant figures *

Answer 1

The answer is 8.29 %

Step-by-step explanation:

The percentage error of a certain measurement can be found by using the formula

`P(\%) = (error)/(actual \: \: number) * 100\% \\`

From the question

actual density = 19.30g/L

error = 20.9 - 19.3 = 1.6

We have

`p(\%) = (1.6)/(19.3) * 100 \\ = 8.290155440...`

We have the final answer as

8.29 %

Hope this helps you