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10. Working in the laboratory, a student finds the density of a piece of pure aluminum to be 2.850 g/cm3. The accepted value for the density of aluminum is 2.699 g/cm3. What is the student's percent error? ​

Respuesta :

babykutush2004

Answer: 5.6%

Step-by-step explanation:

Actual Value observed=2.850 g/cm3

Expected value=2.699 g/cm3

Error =Actual value - expected value

         = 2.850-2.699

Error  = 0.151

Error percent = [tex]\frac{Error}{Expected Value} 100%[/tex]%

                     =[tex]\frac{0.151}{2.699} *100\\0.056*100\\5.6[/tex]

The student's percent error ​is 5.6%

nathan2005
Put 2.85g and 2.69g in an open bracket over 2.69g. Multiply the answer by 100 to receive a percentage error of 5.95%.
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