Volume of Na2S2O3(aq) required?
final burette reading
trial1= 45.00
trial2= 44.27
trial3= 45.28
initial burette reading
trial1= 0.52
trial2= 1.22
trial3= 2.33
volume of Na2S2O3 used in each trial?
average of the two volumes agreeing within 0.10mL?
can anyone help me with this question? i have no idea how to even approach it. can you please show your workings and if possible explain? Please and thank you!
a saturated solution of barium chloride at 30 c contains 150g water. How much additional barium chloride can be dissolved by heating this solution to 50c?
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To determine the volume of Na2S2O3(aq) used in each trial, you need to calculate the difference between the final and initial burette readings for each trial.
For trial 1:
Volume used = final reading - initial reading
= 45.00 mL - 0.52 mL
= 44.48 mL
For trial 2:
Volume used = final reading - initial reading
= 44.27 mL - 1.22 mL
= 43.05 mL
For trial 3:
Volume used = final reading - initial reading
= 45.28 mL - 2.33 mL
= 42.95 mL
Now, to find the average of the two volumes that agree within 0.10 mL, we can analyze the differences between the three trials:
|Trial 1 - Trial 2| = |44.48 mL - 43.05 mL| = 1.43 mL
|Trial 1 - Trial 3| = |44.48 mL - 42.95 mL| = 1.53 mL
|Trial 2 - Trial 3| = |43.05 mL - 42.95 mL| = 0.10 mL
From these differences, we can observe that the difference between trial 2 and trial 3 is the smallest (0.10 mL), meaning they agree within the required tolerance of 0.10 mL.
Therefore, the average volume of Na2S2O3(aq) used in these two trials is:
Average volume = (Volume of trial 2 + Volume of trial 3) / 2
= (43.05 mL + 42.95 mL) / 2
= 43.00 mL
So, the volume of Na2S2O3(aq) used in each trial is approximately 44.48 mL, 43.05 mL, and 42.95 mL, respectively. The average volume of the two trials that agree within 0.10 mL is 43.00 mL.