Trial 1 Trial 2 Trial 3 Initial burette reading (mL) 1.15 2.05 1.14 Molarity of NaOH (M) 0.100 0.100 0.100 Volume of vinegar sample (mL) 5.00 5.00 5.00 Final burette reading (mL) 48.70 48.98 48.36 Choose.. Choose.. Choose. Expected color at end point Choose... - Choose... - Choose... - Volume of NaoH used (mL) Average Volume of NaOH used in liters | Average moles of NaOH used (mol) Average moles of acetic acid (mol) Average molarity of acetic acid (M) Average mass of acetic acid (g) Average mass of vinegar (g) (assume the density of vinegar is 1.00 g/mL) ) Average mass % of acetic acid in vinegar Known mass % of acetic acid in vinegar is 5.45%
Trial 1 Trial 2 Trial 3 Initial burette reading (mL) 1.15 2.05 1.14 Molarity of NaOH (M) 0.100 0.100 0.100 Volume of vinegar sample (mL) 5.00 5.00 5.00 Final burette reading (mL) 48.70 48.98 48.36 Choose.. Choose.. Choose. Expected color at end point Choose... - Choose... - Choose... - Volume of NaoH used (mL) Average Volume of NaOH used in liters | Average moles of NaOH used (mol) Average moles of acetic acid (mol) Average molarity of acetic acid (M) Average mass of acetic acid (g) Average mass of vinegar (g) (assume the density of vinegar is 1.00 g/mL) ) Average mass % of acetic acid in vinegar Known mass % of acetic acid in vinegar is 5.45%
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...
Related questions
Question
![Trial 1
Trial 2
Trial 3
Initial burette reading
(mL)
1.15
2.05
1.14
Molarity of NaOH (M)
0.100
0.100
0.100
Volume of vinegar
sample (mL)
5.00
5.00
5.00
Final burette reading
(mL)
48.70
48.98
48.36
Choose.
Choose.
Choose..
Expected color at end
point
Choose... -
Choose...
Choose...
Volume of NaOH used
(mL)
Average Volume of NaOH used in liters
| Average moles of NaOH used (mol)
Average moles of acetic acid (mol)
Average molarity of acetic acid (M)
Average mass of acetic acid (g)
Average mass of vinegar (g) (assume the density of vinegar is
1.00 g/mL)
) Average mass % of acetic acid in vinegar
Known mass % of acetic acid in vinegar is 5.45%](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F889e0bc1-2d30-44db-bbe7-cf6686980e24%2F8b908182-26a6-4734-bd45-f12d0b60fcb6%2Ffblgw0j_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Trial 1
Trial 2
Trial 3
Initial burette reading
(mL)
1.15
2.05
1.14
Molarity of NaOH (M)
0.100
0.100
0.100
Volume of vinegar
sample (mL)
5.00
5.00
5.00
Final burette reading
(mL)
48.70
48.98
48.36
Choose.
Choose.
Choose..
Expected color at end
point
Choose... -
Choose...
Choose...
Volume of NaOH used
(mL)
Average Volume of NaOH used in liters
| Average moles of NaOH used (mol)
Average moles of acetic acid (mol)
Average molarity of acetic acid (M)
Average mass of acetic acid (g)
Average mass of vinegar (g) (assume the density of vinegar is
1.00 g/mL)
) Average mass % of acetic acid in vinegar
Known mass % of acetic acid in vinegar is 5.45%
![Answer the following questions.
) Suppose the concentration of the NaOH solution was 0.5 M
instead of 0.1 M. Would this titration have required more, less,
or the same amount of NaOH solution for a complete
reaction?
Choose...
:) Suppose you used 0.5 M NaOH to titrate your vinegar sample
instead of 0.1 M. What effect does the concentration of base
added have on the reliability of the results of your titration?
Choose...
Consider a 0.238 M aqueous solution of sodium hydroxide,
NaOH.
;)
Consider a 0.238 M aqueous solution of sodium hydroxide,
NaOH.
a. How many grams of NaOH are dissolved in 23.46 mL?
b. How many individual hydroxide ions (OH") are found in
23.46 mL?
c. How many moles of sulfuric acid, H2SO4, are neutralized
by 23.46 mL of 0.238 M NaOH(aq)? [Hint: begin by
writing a balanced equation for this neutralization
reaction.]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F889e0bc1-2d30-44db-bbe7-cf6686980e24%2F8b908182-26a6-4734-bd45-f12d0b60fcb6%2F20mq09_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Answer the following questions.
) Suppose the concentration of the NaOH solution was 0.5 M
instead of 0.1 M. Would this titration have required more, less,
or the same amount of NaOH solution for a complete
reaction?
Choose...
:) Suppose you used 0.5 M NaOH to titrate your vinegar sample
instead of 0.1 M. What effect does the concentration of base
added have on the reliability of the results of your titration?
Choose...
Consider a 0.238 M aqueous solution of sodium hydroxide,
NaOH.
;)
Consider a 0.238 M aqueous solution of sodium hydroxide,
NaOH.
a. How many grams of NaOH are dissolved in 23.46 mL?
b. How many individual hydroxide ions (OH") are found in
23.46 mL?
c. How many moles of sulfuric acid, H2SO4, are neutralized
by 23.46 mL of 0.238 M NaOH(aq)? [Hint: begin by
writing a balanced equation for this neutralization
reaction.]
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