Volume of Na2S2O3(aq) required
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Volume of Na2S2O3(aq) required

[From: ] [author: ] [Date: 11-11-17] [Hit: ]
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?......
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!

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1) To get the volume of Na2S2O3 used in each trial -> final reading - initial reading = VOLUME (in mL)

Trial 1: 45.00 mL - 0.52 mL = 44.48 mL

Trial 2: 44.27 mL - 1.22 mL = 43.05 mL

Trial 3: 45.28 mL - 2.33 mL = 42.95 mL

2) To get the average of the two volumes agreeing within 0.10 mL, you would add the volumes of the two trials together and then divide by 2. Unfortunately, none of the trials here are within .10 mL of each other -> the digit in the tenth place must be within one digit of each other. For example:

44.48 and 44.35 ---> both are 44 mL with a .10 mL difference in the digit after the decimal (.4 and .3).

Sorry I couldn't fully help. Hope this makes sense! :)

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Final burette reading - initial burette reading = volume used.


Choose the two volumes that are the closest. E.g. if you get 44, 45 and 44.2, then you take the first and third volume.


To find the average of any numbers, add all the numbers together, then divide that number by how many numbers you added together. E.g. if you had 10, 10, 12, 20, 30, then you would go (10 + 10 + 12 + 20 + 30)/5
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