2. What affect, in any, would this error have on the molarity of acid in part two? Would you calculate a molarity that is higher than/lower than/the same as the actual molarity? Explain your reasoning. 3. Can you confidently determine if your molarities are accurate? Explain why or why not. 4. Can you confidently determine if your molarities are precise? Explain why or why not.

Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Using the experiment results, answer the following.

 

Part 2:
Use the volume of NaOH used and the molarity of NaOH to calculate moles of NaOH reacted
Use stoichiometry to determine moles of the acid reacted.
●
●
Divide moles acid by volume of the sample used to get molarity (mol/L),
Average the molarity for all trials.
0.1137
Data
Volume of
HCI
Initial
Burette
Reading
Final
Burette
Reading
Net Volume
NaOH
Moles NaOH
Used
Moles HCI
Used
Molarity of
HCI
Average
Molarity of
HCI
0.0084.
Average Molaity
Trial 1
Trial 2
10AL
0.15 Me
8.58
8.43
B5
0.000958
BULA
1200009%
34 $80.000958
60.874
Males Hel
Net volun
0.000 9/8.43
0.01 0.958
1.284
01137
lome
8.5PML
15.88 ML
7.3m2
0.000830
2007
0.00083.
0.083
Na OH
Trial 3
10 ML
15.88ML
23.4512
7.3ML
7.7712
23.
0.000983
468
0.02973
0.0883
10ML = 0.01L
Trial 4
10ML
23.65ML
30.75ML
7.1072
0-000883
0.000807
0.0001
0.000807
0807
Trial 5
10 ML
30:35A2
37.28m2
4.53m2
0.000742
0.000742
0.0742
Transcribed Image Text:Part 2: Use the volume of NaOH used and the molarity of NaOH to calculate moles of NaOH reacted Use stoichiometry to determine moles of the acid reacted. ● ● Divide moles acid by volume of the sample used to get molarity (mol/L), Average the molarity for all trials. 0.1137 Data Volume of HCI Initial Burette Reading Final Burette Reading Net Volume NaOH Moles NaOH Used Moles HCI Used Molarity of HCI Average Molarity of HCI 0.0084. Average Molaity Trial 1 Trial 2 10AL 0.15 Me 8.58 8.43 B5 0.000958 BULA 1200009% 34 $80.000958 60.874 Males Hel Net volun 0.000 9/8.43 0.01 0.958 1.284 01137 lome 8.5PML 15.88 ML 7.3m2 0.000830 2007 0.00083. 0.083 Na OH Trial 3 10 ML 15.88ML 23.4512 7.3ML 7.7712 23. 0.000983 468 0.02973 0.0883 10ML = 0.01L Trial 4 10ML 23.65ML 30.75ML 7.1072 0-000883 0.000807 0.0001 0.000807 0807 Trial 5 10 ML 30:35A2 37.28m2 4.53m2 0.000742 0.000742 0.0742
2. What affect, in any, would this error have on the molarity of acid in part two? Would you calculate a
molarity that is higher than/lower than/the same as the actual molarity? Explain your reasoning.
3. Can you confidently determine if your molarities are accurate? Explain why or why not.
4. Can you confidently determine if your molarities are precise? Explain why or why not.
Transcribed Image Text:2. What affect, in any, would this error have on the molarity of acid in part two? Would you calculate a molarity that is higher than/lower than/the same as the actual molarity? Explain your reasoning. 3. Can you confidently determine if your molarities are accurate? Explain why or why not. 4. Can you confidently determine if your molarities are precise? Explain why or why not.
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