A-D answers are correct. I found the equation steps in the textbook and answered

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A-D answers are correct. I found the equation steps in the textbook and answered the posted question before anyone was able to post an expert answer. I am leaving the posted question in hopes it may help a future or current student.

A method of analysis yields masses of gold that are low by 0.3 mg. This method is used for analysis of ores that assay about 1 3% gold. What minimum sample mass should be taken if the relative error resulting from a 0.3-mg loss is not to exceed a. -0.1%? Mass = g b. -0.4%? Mass = g c. -0.8%? Mass= g d. -1.1%? Mass= g Use the following formula to calculate the percent relative error: (experimental value (x_i) - true value(x_t)|/true value (x_t) times 100% a) (-0.3/-0.1)(110) = 300 rightarrow (300/1.30(100) = 23076.9 mg convert = 23 g B) (-0.3/-0.4)(100) = 75 rightarrow (75/1.3)(100) = 5769 mg convert = 5.77 g c) (-0.3/-0.8)(100 = 37.5 rightarrow (37.5/1.3)(100) = 2884.6 mg convert = 2.88 g D) (-0.3/-1.1)(100) = 27.27 rightarrow (27.27/1.3)(100) = 20927.9 mg convert = 2.098 g

Explanation / Answer

Error Percentage = (measured - accepted) / accepted * 100)

all your answer correct

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