Asked by eddie
how do i figure out the percentage error , as well as the absolute error of
a) (10.00+/- .02)mL x .42 mol/L
b) (10.00 +/- .02)mL + (10.00 +/- .02)mL
thanks in advancee plz help
a) (10.00+/- .02)mL x .42 mol/L
b) (10.00 +/- .02)mL + (10.00 +/- .02)mL
thanks in advancee plz help
Answers
Answered by
drwls
a) The % error is 0.02/10 = 0.2%
The absolute error is 0.2% of the product. The product is 4.2*10^-3 mol Take 0.2% of that for the absolute error.
b) The errors, being supposedly random, add as the square root of the sum of the squares. The result is 20.00 +/- 0.028. That is 0.14% relative error.
The absolute error is 0.2% of the product. The product is 4.2*10^-3 mol Take 0.2% of that for the absolute error.
b) The errors, being supposedly random, add as the square root of the sum of the squares. The result is 20.00 +/- 0.028. That is 0.14% relative error.
Answered by
eddie
for b where do u get the =/e .028?
and what about the absolute error?
and what about the absolute error?
Answered by
drwls
I already stated the absolute error in (b).
As I said before, 0.028 is the square root of the sum of two 0.02's squared.
It is possible that your teacher wants you to add the absolute errors, and not take the square root of the sum of the squares. To do so would give you an overly conservative worst case error. Errors should be handled statistically.
Refer to
http://en.wikipedia.org/wiki/Mean_squared_error
As I said before, 0.028 is the square root of the sum of two 0.02's squared.
It is possible that your teacher wants you to add the absolute errors, and not take the square root of the sum of the squares. To do so would give you an overly conservative worst case error. Errors should be handled statistically.
Refer to
http://en.wikipedia.org/wiki/Mean_squared_error