PatsDynastyidn
Answered

IDNLearn.com provides a collaborative platform for sharing and gaining knowledge. Our Q&A platform offers reliable and thorough answers to ensure you have the information you need to succeed in any situation.

The true value was 0.120. What was your percent error?

Sagot :

Fichohidn

The question is incomplete as the measured value isn't given. However, an hypothesized measured value will be used to explain how to calculate the percentage error.

Answer:

2.5%

Explanation:

Percentage error :

[(measured value - True value)/ True value] * 100%

The measured value is the value obtained during an while using a measuring device.

If the measured value is = 0.123

Then, the percent error will be :

[(0.123 - 0.120) / 0.120] * 100%

(0.003 / 0.120) * 100%

0.025 * 100%

= 2.5%

Answer:

For each trial, compute the mol of titrant; (molarity x L) and keep the number of significant figures to 4.

Trial 1: 12.49 mL =

0.0025

mol NaOH

Trial 2: 12.32 mL =  

0.0025

mol NaOH

Trial 3: 11.87 mL =  

0.0024

mol NaOH

Because the moles of titrant should equal the moles of analyte at the equilibrium point, copy these values into the row of the data table marked “mol of analyte.”

Compute the concentration of HCl for all three trials using the formula molarity = mol / L.

Trial 1 (20.61 mL HCl):

0.121

M

Trial 2 (20.06 mL HCl):

0.125

M

Trial 3 (19.67 mL HCl):

0.122

M

Compute the average concentration (molarity) of HCl for all three trials:

0.123

M.

The true value was 0.120. What was your percent error?

2.5

%

Explanation:

Thank you for being part of this discussion. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Your questions deserve reliable answers. Thanks for visiting IDNLearn.com, and see you again soon for more helpful information.