Homework Answers
I hope [CH3COOH] = 0.00466 M
And [H+] = 3.98*10-4 M
i.e. [CH3COO-] = 3.98*10-4 M
Formula: Ka = [CH3COO-][H+]/[CH3COOH]
i.e. Ka = 3.98*10-4 * 3.98*10-4 / 0.00466
Therefore, the Ka of acetic acid = 3.4*10-5
Add Answer to:
question 8 = 0.00466
question 11 = 3.98*10^-4
12. Using your answers from Questions 8 and...
Warner Chemical Equilibrium Page 6 of 11 Assuming the density of your initial acid solution is...
Warner Chemical Equilibrium Page 6 of 11 Assuming the density of your initial acid solution is 1.0 g/ml, what is the concentration (in molarity) of the diluted acetic acid solution? Show your work and include units. Hint: An initial 45% w/w solution (w/w means 4.5 g of acid per 100 g of solution) of CH3COOH (MW - 60.05 g/mol) was diluted by placing 0.5 mL into 80 mL of water to produce a final diluted solution with a total volume...
I MAINLY NEED HELP ON TABLES PLS HELP THANK YOU Materials: NaOH MW = 40 g/mL,...
I MAINLY NEED HELP ON TABLES
PLS HELP THANK YOU
Materials: NaOH MW = 40 g/mL, KHP - potassium hydrogen phthalate, KHC2H4O4, MW = 204.23 g/mol. Acetic acid, HC2H302, MW = 60.05 g/mol Part 1: Standardization of NaOH Assume 0.951 g of KHP is weighed and transferred to a 250 mL Erlenmeyer flask. Approximately 50 mL water is added to dissolve the KHP. Note that the exact volume of water is not important because you only need to know the...
Exp 9 titration of vinegar procedure b molarity of NaOH = .0859 ROCEDURE B. DETERMINATION OF...
Exp 9 titration of vinegar procedure b
molarity of NaOH = .0859
ROCEDURE B. DETERMINATION OF CONCENTRATION OF ACETIC ACID IN UNKNOWN SAMPLES Titration of vinegar solutions Trial 1 Trial 2 Trial 3 1. Volume of vinegar solution being titrated in milliliters (mL) 25.00 mL 25.00 mL 25.00 ml 2. Volume of vinegar solution being titrated in Liters (L) 0.0250 L 0.0250 L 0.0250 L 3. Final buret reading in ml 16. 20m 16.316L 16.25ml 4. Initial buret reading in...
Materials: NaOH MW = 40 g/mL, KHP - potassium hydrogen phthalate, KHC8H404, MW = 204.23 g/mol....
Materials: NaOH MW = 40 g/mL, KHP - potassium hydrogen phthalate, KHC8H404, MW = 204.23 g/mol. Acetic acid, HC2H302, MW = 60.05 g/mol Part 1: Standardization of NaOH Assume 0.951 g of KHP is weighed and transferred to a 250 mL Erlenmeyer flask. Approximately 50 ml water is added to dissolve the KHP. Note that the exact volume of water is not important because you only need to know the exact number of moles of KHP that will react with...
Following question: -Calulated molarity of diluted vinegar solution -Molarity of Commercial Vinegar (multiply by the dilution...
Following question:
-Calulated molarity of diluted vinegar solution
-Molarity of Commercial Vinegar (multiply by the dilution
factor
-Percent (mass/volume) of acetic acid in commefcial
vinegar
Data and Calculations Molarity of standardized sodium hydroxide solution Dilution factor for vinegar solution prepared in step 4 i 20 Volume of diluted vinegar titrated in each run Titrations: Run 3 Run 2 Run 1 G (o .00 ml Final Buret 22.20 Volume (mL) Initial Buret Volume (mL) 22.20 WL Volume NaOH 22.3 Delivered (mL)...
I need help calculating the mass percent and with the post lab question #2. Part 1...
I need help calculating the mass percent and with the post lab
question #2.
Part 1 Mass Initial | Final Vol Nall Moles concen. KPHCG (ML) I (m2) Used (m2) KHP (mol)/NaOH(M) 0.28470.00 13.60r 13.600 10.00- 0.1025 2 UZ + ZANT 13.600 24.45 1 10.85 0.001128 10.40 Aug NaOH Concentration (M):0.1032 Part2 Vinegar identification # 4 Vinegar sample date: (011/ Vol Vinegat Initial Final Yol NaOH Mol Acotic Mass Acctic Mass % Titr ULST I OML Used (ml) Acid (mol)...
Materials: NaOH MW = 40 g/mL, KHP - potassium hydrogen phthalate, KHC8H404, MW = 204.23 g/mol....
Materials: NaOH MW = 40 g/mL, KHP - potassium hydrogen phthalate, KHC8H404, MW = 204.23 g/mol. Acetic acid, HC2H302, MW = 60.05 g/mol Part 1: Standardization of NaOH Assume 0.951 g of KHP is weighed and transferred to a 250 mL Erlenmeyer flask. Approximately 50 ml water is added to dissolve the KHP. Note that the exact volume of water is not important because you only need to know the exact number of moles of KHP that will react with...
in the following questions write your answers with detailed steps. In the following questions write your...
in the following questions write
your answers with detailed steps.
In the following questions write your answers with the detailed steps. Show all necessary equations or conversion factors. a. 4.50 g sample of glucose (C6H120) is dissolved in enough water to make 10.0 mL of solution. Calculate the molarity. b. 424g of solute is dissolved in enough water to give 1.00 L of solution. The density of the resulting solution is 1.18 g/mL. What is the %(w/w)? c. Write a...
2559b3823f31 > 8 of 28 A Review Constants Periodic Table ✓ Correct Part B What is...
2559b3823f31 > 8 of 28 A Review Constants Periodic Table ✓ Correct Part B What is the final volume V2 in milliliters when 0.572 L of a 43.8 % (m/v) solution is diluted to 21.2% (m/v)? Express your answer numerically in milliliters. View Available Hint(s) ? ml IVO AXO V₂V_2= Submit 8 of 28 M Review | Constants Periodic Table Part C A 554 ml NaCl solution is diluted to a volume of 1.08 L and a concentration of 2.00...
What information then do you need? This is Chem 2 and not Org Chem. Titration of...
What information then do you need? This is Chem 2 and not Org
Chem.
Titration of Vinegar Lab Report 1 /25 pts) Name: Part 1: Data and Calculations mass KHC.H.O. - 4.0744 g Molecular Weight potassium hydrogen phthalate KHC.H. - [1 pt) #moles KHP = [0.5 pts) Volume of solution = 250 ml_ L10.5 pts] Concentration of KHP = Mop = _M (1 pt) Volume of KHP Solution to be titrated mL = [1 pt) #mols of KHP to be...
Warner Chemical Equilibrium Page 6 of 11 Assuming the density of your initial acid solution is 1.0 g/ml, what is the concentration (in molarity) of the diluted acetic acid solution? Show your work and include units. Hint: An initial 45% w/w solution (w/w means 4.5 g of acid per 100 g of solution) of CH3COOH (MW - 60.05 g/mol) was diluted by placing 0.5 mL into 80 mL of water to produce a final diluted solution with a total volume...
I MAINLY NEED HELP ON TABLES
PLS HELP THANK YOU
Materials: NaOH MW = 40 g/mL, KHP - potassium hydrogen phthalate, KHC2H4O4, MW = 204.23 g/mol. Acetic acid, HC2H302, MW = 60.05 g/mol Part 1: Standardization of NaOH Assume 0.951 g of KHP is weighed and transferred to a 250 mL Erlenmeyer flask. Approximately 50 mL water is added to dissolve the KHP. Note that the exact volume of water is not important because you only need to know the...
Exp 9 titration of vinegar procedure b
molarity of NaOH = .0859
ROCEDURE B. DETERMINATION OF CONCENTRATION OF ACETIC ACID IN UNKNOWN SAMPLES Titration of vinegar solutions Trial 1 Trial 2 Trial 3 1. Volume of vinegar solution being titrated in milliliters (mL) 25.00 mL 25.00 mL 25.00 ml 2. Volume of vinegar solution being titrated in Liters (L) 0.0250 L 0.0250 L 0.0250 L 3. Final buret reading in ml 16. 20m 16.316L 16.25ml 4. Initial buret reading in...
Materials: NaOH MW = 40 g/mL, KHP - potassium hydrogen phthalate, KHC8H404, MW = 204.23 g/mol. Acetic acid, HC2H302, MW = 60.05 g/mol Part 1: Standardization of NaOH Assume 0.951 g of KHP is weighed and transferred to a 250 mL Erlenmeyer flask. Approximately 50 ml water is added to dissolve the KHP. Note that the exact volume of water is not important because you only need to know the exact number of moles of KHP that will react with...
Following question:
-Calulated molarity of diluted vinegar solution
-Molarity of Commercial Vinegar (multiply by the dilution
factor
-Percent (mass/volume) of acetic acid in commefcial
vinegar
Data and Calculations Molarity of standardized sodium hydroxide solution Dilution factor for vinegar solution prepared in step 4 i 20 Volume of diluted vinegar titrated in each run Titrations: Run 3 Run 2 Run 1 G (o .00 ml Final Buret 22.20 Volume (mL) Initial Buret Volume (mL) 22.20 WL Volume NaOH 22.3 Delivered (mL)...
I need help calculating the mass percent and with the post lab
question #2.
Part 1 Mass Initial | Final Vol Nall Moles concen. KPHCG (ML) I (m2) Used (m2) KHP (mol)/NaOH(M) 0.28470.00 13.60r 13.600 10.00- 0.1025 2 UZ + ZANT 13.600 24.45 1 10.85 0.001128 10.40 Aug NaOH Concentration (M):0.1032 Part2 Vinegar identification # 4 Vinegar sample date: (011/ Vol Vinegat Initial Final Yol NaOH Mol Acotic Mass Acctic Mass % Titr ULST I OML Used (ml) Acid (mol)...
Materials: NaOH MW = 40 g/mL, KHP - potassium hydrogen phthalate, KHC8H404, MW = 204.23 g/mol. Acetic acid, HC2H302, MW = 60.05 g/mol Part 1: Standardization of NaOH Assume 0.951 g of KHP is weighed and transferred to a 250 mL Erlenmeyer flask. Approximately 50 ml water is added to dissolve the KHP. Note that the exact volume of water is not important because you only need to know the exact number of moles of KHP that will react with...
in the following questions write
your answers with detailed steps.
In the following questions write your answers with the detailed steps. Show all necessary equations or conversion factors. a. 4.50 g sample of glucose (C6H120) is dissolved in enough water to make 10.0 mL of solution. Calculate the molarity. b. 424g of solute is dissolved in enough water to give 1.00 L of solution. The density of the resulting solution is 1.18 g/mL. What is the %(w/w)? c. Write a...
2559b3823f31 > 8 of 28 A Review Constants Periodic Table ✓ Correct Part B What is the final volume V2 in milliliters when 0.572 L of a 43.8 % (m/v) solution is diluted to 21.2% (m/v)? Express your answer numerically in milliliters. View Available Hint(s) ? ml IVO AXO V₂V_2= Submit 8 of 28 M Review | Constants Periodic Table Part C A 554 ml NaCl solution is diluted to a volume of 1.08 L and a concentration of 2.00...
What information then do you need? This is Chem 2 and not Org
Chem.
Titration of Vinegar Lab Report 1 /25 pts) Name: Part 1: Data and Calculations mass KHC.H.O. - 4.0744 g Molecular Weight potassium hydrogen phthalate KHC.H. - [1 pt) #moles KHP = [0.5 pts) Volume of solution = 250 ml_ L10.5 pts] Concentration of KHP = Mop = _M (1 pt) Volume of KHP Solution to be titrated mL = [1 pt) #mols of KHP to be...
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