Question

b. measurements for an unknown solution using a buret

solution # titrant	trial #1	trial #2	trial #3*
mass of beaker + solution	125.384	125.387
mass of solution	7.085	7.084
initial level in buret	0.04	0.06
final level in buret	10.06	10.07
volume of solution delivered	1.56ml	1.01ml
density of solution	4.543g/ml	7.028g/ml

|average density of solution|

percent difference

42.95%

*carry out a third trial on if the percent difference between trial #1 and trial #2 is greater than 5%.
use the following to calculate the percent different for trials #1 and #2:
percent difference = \\(\frac{|x_1 - x_2|}{average\\ of\\ x_1\\ and\\ x_2}\times100\\%\\) where x1 and x2 are the two calculated densities based on the experimental measurements
and show the calculation for the percent difference for each density below: (note: because percent difference is the absolute value of the difference of the two values divided by the average, it is always positive.)
average = \\(\frac{x_1 + x_2}{2}=\frac{4.543g/ml + 7.028g/ml}{2}=\frac{11.571g/ml}{2}=5.7855g/ml\\)
percent diff = \\(\frac{|x_1 - x_2|}{average}\times100\\%=\frac{2.485g/ml}{5.7855g/ml}\times100\\% = 42.95\\%\\)
volume measurements for various liquids

	volume of liquid
a	35ml
b	9.1ml
c	65ml
d	4.5ml

d. proper use of a bunsen burner
student was able to light the bunsen burner correctly:
student was able to adjust the bunsen burner to get a

Explanation:

Step1: Identify the density values for two trials

Let $x_1 = 4.543\ g/mL$ (density from Trial #1) and $x_2=7.028\ g/mL$ (density from Trial #2).

Step2: Calculate the average of the two density - values

The average $\bar{x}=\frac{x_1 + x_2}{2}=\frac{4.543+7.028}{2}=\frac{11.571}{2}=5.7855\ g/mL$.

Step3: Calculate the absolute difference between the two density - values

$|x_1 - x_2|=|4.543 - 7.028|=2.485\ g/mL$.

Step4: Calculate the percent difference

Percent difference $=\frac{|x_1 - x_2|}{\bar{x}}\times100\%=\frac{2.485}{5.7855}\times100\%\approx42.95\%$.

Answer:

The percent difference between the densities of Trial #1 and Trial #2 is approximately $42.95\%$.