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Answer:
- Reduced biodiversity
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Answer:
72.8 % (But verify explanation).
Explanation:
Hello,
In this case, with the following obtained results, the percent error is computed as follows:
Volume of vinegar= 7.0 mL
Volume of NaOH= (7+6.6+6.4)/3= 6.7 mL
Used concentration of NaOH= 1.5M
Concentration of acetic acid= (concentration of NaOH*volume of NaOH)/volume of vinegar= (6.7mL*1.5M)/7.0M= 1.44M
Assuming we have 100 mL (0.100L) of vinegar, moles of acetic acid in vinegar = 1.44M x 0.100 L= 0.144 mol
Mass of acetic acid in 100g of vinegar = 0.144 mol x 60.0g/mol= 8.64 g
% of acetic acid in vinegar=8.64 %
% error in percentage of acetic acid = [(8.64% - 5.0%)/5.0] x 100=72.8 %
Clearly, this result depend on your own measurements, anyway, you can change any value wherever you need it.
Regards.
Answer:
Acetic acid is a weak monoprotic acid.
In aqueous solution it has a pKa value of 4.76.
It conjugate base is acetate (CH3COO-) A 1.0m
solution ( about the concentration of domestic vinegar) has pH of 2.4, *indicating that merely
0.4% of the acetic acid molecules are dissociated.
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