In laboratory experiment, 10 g potassium chlorate sample on decomposition gives following data , The sample contains 3.8 g of oxygen and the actual mass of oxygen in the quantity of potassium chlorate is 3.92 g. Calculate absolute error and relative error.
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Explanation:
Solution
SolutionThe observed is 3.8 g and accepted value is 3.92 g
SolutionThe observed is 3.8 g and accepted value is 3.92 gAbsolute error = Observed value - True value = 3.8 - 3.92 = - 0.12 g
SolutionThe observed is 3.8 g and accepted value is 3.92 gAbsolute error = Observed value - True value = 3.8 - 3.92 = - 0.12 gThe negative sign indicates that your experimental result is lower than the true value.
SolutionThe observed is 3.8 g and accepted value is 3.92 gAbsolute error = Observed value - True value = 3.8 - 3.92 = - 0.12 gThe negative sign indicates that your experimental result is lower than the true value.The relative error =Absolute errorTrue value×100%
SolutionThe observed is 3.8 g and accepted value is 3.92 gAbsolute error = Observed value - True value = 3.8 - 3.92 = - 0.12 gThe negative sign indicates that your experimental result is lower than the true value.The relative error =Absolute errorTrue value×100%=(−0.12)/(3.92)×100%
SolutionThe observed is 3.8 g and accepted value is 3.92 gAbsolute error = Observed value - True value = 3.8 - 3.92 = - 0.12 gThe negative sign indicates that your experimental result is lower than the true value.The relative error =Absolute errorTrue value×100%=(−0.12)/(3.92)×100%=−3.06%
Answer:
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Explanation:
Absolute error = Observed value – True value = 3.8 – 3.92 = -0.12 g ii. Relative error = A b s o l u t e e r r o r T r u e v a l u e AbsoluteerrorTruevalue x 100% = − 0.12 3.92 −0.123.92 x 100% = - 3.06% [Note : The negative sign indicates that experimental result is lower than the true value.] ∴ i. Absolute error = - 0.12 g ii. Relative error = - 3.06%Read more on Sarthaks.com - https://www.sarthaks.com/2227004/laboratory-experiment-potassium-chlorate-sample-decomposition-following-contains-oxygen