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20. Use the following information and following graph to calculate the mass of sodium chloride in...
which is correct? please explain Identify the predominant reaction taking place in the solution at point...
which is correct? please explain
Identify the predominant reaction taking place in the solution at point A, 0 mL NaOH added. Acetic Acid Titration Curve 14.00 13.00 12.00 11.00 10.00 9.00 7.00 6.00 5.00 4.00 3.00 2.00 1.00 0.00 0.00 25.00 30.00 35.00 5.00 10.00 15.00 20.00 Volume NaOH (mL) HC2H302(aq) + OH - (aq)=C2H302-(aq) + H2O(1) HC2H302(aq) + H20(1) = C2H302 (aq) + H30+(aq) o C2H302 (aq) + H2O(1) = HC2H302(aq) + OH- (aq)
Given that the pKa for the half equivalence point is = 5.5, and the pKa of...
Given that the pKa for the half equivalence point is =
5.5, and the pKa of Acetic acid is 4.76, what is the percent
error?
Is 4.76 the accepted value?
and 5.5 the experimental?
and so...
|4.76-5.5| = -0.74
--------- ----- = 0.15546 ×100% = 15.55%
|4.76|
|4.76|
Please verify I am correct.
Vinegar Titration Curve / / 14.00 13.00 12.00 11.00 10.00 9.00 8.00 E 7.00 6.00 5.00 4.00 3.00 2.00 1.00 0.00 0.00 5.00 10.00 25.00 30.00...
1. The following graph is provided for the titration of 10.00 mL benzoic acid solution with...
1. The following graph is provided for the titration of 10.00 mL benzoic acid solution with a strong base. The concentration of NaOH is 0.100 M. A. Based on this graph given below, determine the Ka. B. Determine the molarity of the acid. Benzoic acid titration with 0.1 M NaOH 12.00 10.00 8.00 6.00 4.00 2.00 0.00 0.00 5.00 10.00 15.00 Volume of NaOH mL
1. The following graph is provided for the titration of 10.00 mL benzoic acid solution with...
1. The following graph is provided for the titration of 10.00 mL benzoic acid solution with a strong base. The concentration of NaOH is 0.100 M. A. Based on this graph given below, determine the Ka. B. Determine the molarity of the acid. Benzoic acid titration with 0.1 M NaOH 12.00 10.00 8.00 pH 6.00 4.00 2.00 0.00 0.00 5.00 10.00 15.00 Volume of NaOH mL
1. The following graph is provided for the titration of 10.00 mL benzoic acid solution with...
1. The following graph is provided for the titration of 10.00 mL benzoic acid solution with a strong base. The concentration of NaOH is 0.100 M. A. Based on this graph given below, determine the Ka. B. Determine the molarity of the acid. Benzoic acid titration with 0.1 M NaOH 12.00 10.00 8.00 6.00 4.00 2.00 0.00 0.00 5.00 10.00 15.00 Volume of NaOH mL
1. The following graph is provided for the titration of 10.00 mL benzoic acid solution with...
1. The following graph is provided for the titration of 10.00 mL benzoic acid solution with a strong base. The concentration of NaOH is 0.100 M. A. Based on this graph given below, determine the Ka. B. Determine the molarity of the acid. Benzoic acid titration with 0.1 M NaOH 12.00 10.00 8.00 6.00 4.00 2.00 0.00 0.00 5.00 10.00 15.00 Volume of NaOH mL
Im still not understanding how to calculate the molarity of Fe(SCN)^2+ I have the absorbance for...
Im
still not understanding how to calculate the molarity of Fe(SCN)^2+
I have the absorbance for them but I dont know what to do from
there.
I would really appreciate the help. please and thank
you.
H115 lab manual_rev2016_MOD19AUG18_JSchesser.pdf (2.82 MB) Page < 65 > of 130 Tortas experiment unless you can measure their absorbance values within one hour 0 - ZOOM + Label 6 screw top test tubes A-F Using pipets, measure the solutions as outlined in the table...
Enter the following data into the calculator and graph pH versus mL of NaOH added. At...
Enter the following data into the calculator and graph pH versus mL of NaOH added. At the endpoint you should see a rapid change in pH. Identify the two volumes on each side of the endpoint, and determine the color change of the indicator for the end point. Data Analysis Table Volume of NaOH Solution Added pH Color 0.00 mL 2.79 Clear 1.00 mL 3.99 Magenta then clear 2.00 mL 4.49 Magenta then clear 3.00 mL 4.99 Magenta then clear...
Following the Procedure of this experiment, a student titrated 0.653 g of an unknown weak, monoprotic...
Following the Procedure of this experiment, a student titrated 0.653 g of an unknown weak, monoprotic acid with 0.100 M NaOH and monitored the titration with a pH meter. His titration data were: Volume of NaOH solution added, mL /// pH 0.00 | 3.30 2.00 | 4.22 4.00 | 4.55 6.00 | 4.76 8.00 | 4.92 10.00 | 5.06 12.00 | 5.18 14.00 | 5.29 16.00 | 5.40 18.00 | 5.51 20.00 | 5.62 22.00 | 5.74 24.00 | 5.88...
please explain and show work рон 0.000398 TOH Acid/Base/Neutral 2.1 110 4.21 2.- In the following...
please explain and show work
рон 0.000398 TOH Acid/Base/Neutral 2.1 110 4.21 2.- In the following reaction identify the acid and its conjugate base and the base and its conjugate acid: Acid 3.- Determine the hydrogen and hydroxide concentration, when we dissolve 0.50 moles of the AH in 250 mL of solution. (Hint consider that [H+] is small) pKaa. 6.5 4.25 c. 5.6 x=[H*) = a. 2.12 * 104 b. 1.06 * 102 c. 3.86 * 10-5 [OH--a. 1.08 *1010...
which is correct? please explain
Identify the predominant reaction taking place in the solution at point A, 0 mL NaOH added. Acetic Acid Titration Curve 14.00 13.00 12.00 11.00 10.00 9.00 7.00 6.00 5.00 4.00 3.00 2.00 1.00 0.00 0.00 25.00 30.00 35.00 5.00 10.00 15.00 20.00 Volume NaOH (mL) HC2H302(aq) + OH - (aq)=C2H302-(aq) + H2O(1) HC2H302(aq) + H20(1) = C2H302 (aq) + H30+(aq) o C2H302 (aq) + H2O(1) = HC2H302(aq) + OH- (aq)
Given that the pKa for the half equivalence point is =
5.5, and the pKa of Acetic acid is 4.76, what is the percent
error?
Is 4.76 the accepted value?
and 5.5 the experimental?
and so...
|4.76-5.5| = -0.74
--------- ----- = 0.15546 ×100% = 15.55%
|4.76|
|4.76|
Please verify I am correct.
Vinegar Titration Curve / / 14.00 13.00 12.00 11.00 10.00 9.00 8.00 E 7.00 6.00 5.00 4.00 3.00 2.00 1.00 0.00 0.00 5.00 10.00 25.00 30.00...
1. The following graph is provided for the titration of 10.00 mL benzoic acid solution with a strong base. The concentration of NaOH is 0.100 M. A. Based on this graph given below, determine the Ka. B. Determine the molarity of the acid. Benzoic acid titration with 0.1 M NaOH 12.00 10.00 8.00 6.00 4.00 2.00 0.00 0.00 5.00 10.00 15.00 Volume of NaOH mL
1. The following graph is provided for the titration of 10.00 mL benzoic acid solution with a strong base. The concentration of NaOH is 0.100 M. A. Based on this graph given below, determine the Ka. B. Determine the molarity of the acid. Benzoic acid titration with 0.1 M NaOH 12.00 10.00 8.00 pH 6.00 4.00 2.00 0.00 0.00 5.00 10.00 15.00 Volume of NaOH mL
1. The following graph is provided for the titration of 10.00 mL benzoic acid solution with a strong base. The concentration of NaOH is 0.100 M. A. Based on this graph given below, determine the Ka. B. Determine the molarity of the acid. Benzoic acid titration with 0.1 M NaOH 12.00 10.00 8.00 6.00 4.00 2.00 0.00 0.00 5.00 10.00 15.00 Volume of NaOH mL
1. The following graph is provided for the titration of 10.00 mL benzoic acid solution with a strong base. The concentration of NaOH is 0.100 M. A. Based on this graph given below, determine the Ka. B. Determine the molarity of the acid. Benzoic acid titration with 0.1 M NaOH 12.00 10.00 8.00 6.00 4.00 2.00 0.00 0.00 5.00 10.00 15.00 Volume of NaOH mL
Im
still not understanding how to calculate the molarity of Fe(SCN)^2+
I have the absorbance for them but I dont know what to do from
there.
I would really appreciate the help. please and thank
you.
H115 lab manual_rev2016_MOD19AUG18_JSchesser.pdf (2.82 MB) Page < 65 > of 130 Tortas experiment unless you can measure their absorbance values within one hour 0 - ZOOM + Label 6 screw top test tubes A-F Using pipets, measure the solutions as outlined in the table...
please explain and show work
рон 0.000398 TOH Acid/Base/Neutral 2.1 110 4.21 2.- In the following reaction identify the acid and its conjugate base and the base and its conjugate acid: Acid 3.- Determine the hydrogen and hydroxide concentration, when we dissolve 0.50 moles of the AH in 250 mL of solution. (Hint consider that [H+] is small) pKaa. 6.5 4.25 c. 5.6 x=[H*) = a. 2.12 * 104 b. 1.06 * 102 c. 3.86 * 10-5 [OH--a. 1.08 *1010...
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