* here is the lab experiment The entire report should be typed (handwritten port
ID: 484678 • Letter: #
Question
* here is the lab experiment
The entire report should be typed (handwritten portions will NOT be graded0. Please organize your report as shown below. You and your lab partner may not turn in the same work. Be sure to report all values to the correct number of significant figures and with appropriate units. 1. Station 1 a. List the two measurements made at this station with correct labels, units and significant figures. Which was more accurate measurement and why? Briefly explain. b. What is the name of the piece of glassware used to make these measurements? 2. Station 2 a. Calculate the mass of the metal sphere. Report the final answer to the correct number ofsignificant figures with units. Show the typed calculation to receive full credit. b. The known mass of the sphere is 13-878 g. Calculate the percent error in your measured mass. Show the typed calculation to receive full credit. 3. Station 3 a. Calculate the perimeter (A+B+C perimeter) of the triangle in units of meters. Report the final answer to the correct number of significant figures with units. Show the typed calculation to receive full credit. b. How would you change the measurement process to obtain a perimeter value with more significant figures? 4. Station 4 a. Report the label on the block you measured in lab. Calculate the density of the block measured in units of g/mm .Report the final answer to the correct number of significant figures with units. Show the typed calculation to receive full credit. b. If the mass of the block had been measured on a balance that reads to four decimal places, would that have changed the density value reported above? Briefly explain why or why not.Explanation / Answer
1(a)
Volume of colored water Meaurement 1 : 49mL
Measurement 2 : 6.4 mL
Second measumement is more accurate as it reports the volume to one place after the decimal place.
(b)
The glassware used is called measuring cylinder
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2.(a)
Mass of empty beaker = 4.873g
Mass of beaker + sphere =18.719 g
mass of the sphere = 18.719g-4.873g = 13.846 g
(b)
Error in measurement = actual value-calculated value =13.878g - 13.846 g = 0.032 g
% error = error/actual value * 100 = 0.032 *100/13.878 = 0.23 %
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