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Question 2 1 pts A student determined the average molarity of their sample of acetic acid...
Question 19 2 pts What is the molarity of an acetic acid solution if 25.00 ml...
Question 19 2 pts What is the molarity of an acetic acid solution if 25.00 ml of HC2H30, is required to neutralize 0.424 g of sodium carbonate (105.99 g/mol)? 2 HC2H2O2(aq) + Na2CO3(s) — 2 NaC2H302 (aq) + H2O() + CO2(g) D 0.320 M O 0.100 M 0.160M O 0.0800 M O 0.200 M
Question 1 1 pts HC2H2O2 (aq) + NaOH(aq) - NaC2H302 (aq) + H2O(1) Trial V;(mL) V...
Question 1 1 pts HC2H2O2 (aq) + NaOH(aq) - NaC2H302 (aq) + H2O(1) Trial V;(mL) V (mL) 13.82 1 3.54 2 3 For their first trial, a student pipettes 5.00 mL of a vinegar solution into an Erlenmeyer flask. The student titrates the vinegar with standardized NaOH with a molarity of 0.229 M. The volume of titrant used was recorded in the table above. Calculate the molarity of the vinegar solution for this trial. Enter a number without units.
5. Determine the average value for the molarity of the acetic acid in vinegar. Show your...
5. Determine the average value for the molarity of the acetic acid in vinegar. Show your calculation 6. Determine the deviation between each individual molarity for acetic acid and the average value from question 4, above. Notice that the formula below uses absolute values: you only care about how big the difference is not whether cach trial value was higher or lower than the average. Deviation = (individual trial value)-(average value from #5) Trial Ideviation Trial 2 deviation Trial 3...
molarity of acid: 0.1435M volume:25ml 1. Use the molar mass and moles of acetic acid to...
molarity of acid: 0.1435M
volume:25ml
1. Use the molar mass and moles of acetic acid to determine the mass of the acetic acid in the 25.00-ml sample. 2. Calculate the mass/volume percentage (m/v %) by dividing the mass of acetic acid in grams by the volume of the vinegar sample in mL and multiplying by 100. 3. For the reaction of sodium hydroxide solution with a solution of hydrochloric acid: a. Write a balanced molecular equation for the reaction, including...
Calculate the % error. Compare the result with the average molarity of acetic vinegar as determined...
Calculate the % error. Compare the result with the average molarity of acetic vinegar as determined by previous students. .8488 Molarity of acetic acid in my experiment .8701 Molarity of previous students Is my result of -2.448% correct? Is my sigfigs correct? Is it supposed to be negative?
Analysis of Acetic Acid Unknown M Acetic Acid Average M Unknown Average % Acetic Acid Analysis...
Analysis of Acetic Acid Unknown
M Acetic Acid
Average M Unknown
Average % Acetic Acid
Analysis of Acetic Acid Unknown ID of Unknown used Trial 1 Trial 2 Vol. Unknown used Vol. NaOH used Mmol NaOH used M Acetic Acid, calc. Average M Unknown ml mmol 7. mmol Average % Acetic Acid HoAc Audlcnstn 17.4 (assume density - 1 g/mL) Show the calculations used for Sample 1 for the determination of the molarity of unknown Acetic acid sample and for...
Question 3 2 pts A solution is made by mixing 1.42 g of methanol, CH2OH(1), with...
Question 3 2 pts A solution is made by mixing 1.42 g of methanol, CH2OH(1), with 592.mL of acetic acid, HC2H2O2(1). Calculate the concentration of the solute in the solution using the units given below. (Density of methanol = 0.792 g/mL. Density of acetic acid - 1.05 g/mL. Assume volumes mix additively.) Enter your answers in decimal form to 3 significant figures. Do not enter units since they are already provided. Molar: M molal: m mole fraction: mass percent: %(m/m)...
6.31 M 7. Molarity of acetic acid, HC2H30in vinegar sample (show your calculations) Volume of vinegar...
6.31 M 7. Molarity of acetic acid, HC2H30in vinegar sample (show your calculations) Volume of vinegar sample = 0.003L 0.003L Vinegar contain 0.01893 moles IL 0.01893 = 6.3 IM 0.003 Vinegar contain 8. Grams of Acetic Acid, HC,H,O2 (show your calculations) 1.1368 g HC2H5O2 Grams acid = 12 *MW = 0.01893 mot x 60.052 g/ mot = 1-1368 g 31.90 % L 3 100 9. Percent (m/v) of HC,H,O, in vinegar sample (show your calculations) 3.0 mL vinegar contain 0.01893...
2.) Vinegar is an aqueous solution (homogenous mixture) of acetic acid and water laqueous aqua" -...
2.) Vinegar is an aqueous solution (homogenous mixture) of acetic acid and water laqueous aqua" - Latin for water). To simplify the calculations let us assume that all of the acetic acid in the vinegar reacts and there is some left over "excess" baking soda. The vinegar solution is a 5% acetic acid solution. If you look at the bottle of vinegar somewhere on there it probably says "5% acidity" or similar notation. Let us estimate that about 25.0 g...
Calculate the moles of acetic acid, molarity of the vinegar solution, and mass %. of acetic...
Calculate the moles of acetic acid, molarity of the vinegar solution, and mass %. of acetic acid in the vinegar using the average of three good trials used to titrate 5.00 mL of the vinegar. You may assume that the density of your vinegar sample is 1.01 g mL^-1. Drain off unused NaOH solution into a clean container and place it into a designated container. Wash the buret two times with regular tap water, then rinse with deionized water, and...
Question 19 2 pts What is the molarity of an acetic acid solution if 25.00 ml of HC2H30, is required to neutralize 0.424 g of sodium carbonate (105.99 g/mol)? 2 HC2H2O2(aq) + Na2CO3(s) — 2 NaC2H302 (aq) + H2O() + CO2(g) D 0.320 M O 0.100 M 0.160M O 0.0800 M O 0.200 M
Question 1 1 pts HC2H2O2 (aq) + NaOH(aq) - NaC2H302 (aq) + H2O(1) Trial V;(mL) V (mL) 13.82 1 3.54 2 3 For their first trial, a student pipettes 5.00 mL of a vinegar solution into an Erlenmeyer flask. The student titrates the vinegar with standardized NaOH with a molarity of 0.229 M. The volume of titrant used was recorded in the table above. Calculate the molarity of the vinegar solution for this trial. Enter a number without units.
5. Determine the average value for the molarity of the acetic acid in vinegar. Show your calculation 6. Determine the deviation between each individual molarity for acetic acid and the average value from question 4, above. Notice that the formula below uses absolute values: you only care about how big the difference is not whether cach trial value was higher or lower than the average. Deviation = (individual trial value)-(average value from #5) Trial Ideviation Trial 2 deviation Trial 3...
molarity of acid: 0.1435M
volume:25ml
1. Use the molar mass and moles of acetic acid to determine the mass of the acetic acid in the 25.00-ml sample. 2. Calculate the mass/volume percentage (m/v %) by dividing the mass of acetic acid in grams by the volume of the vinegar sample in mL and multiplying by 100. 3. For the reaction of sodium hydroxide solution with a solution of hydrochloric acid: a. Write a balanced molecular equation for the reaction, including...
Analysis of Acetic Acid Unknown
M Acetic Acid
Average M Unknown
Average % Acetic Acid
Analysis of Acetic Acid Unknown ID of Unknown used Trial 1 Trial 2 Vol. Unknown used Vol. NaOH used Mmol NaOH used M Acetic Acid, calc. Average M Unknown ml mmol 7. mmol Average % Acetic Acid HoAc Audlcnstn 17.4 (assume density - 1 g/mL) Show the calculations used for Sample 1 for the determination of the molarity of unknown Acetic acid sample and for...
Question 3 2 pts A solution is made by mixing 1.42 g of methanol, CH2OH(1), with 592.mL of acetic acid, HC2H2O2(1). Calculate the concentration of the solute in the solution using the units given below. (Density of methanol = 0.792 g/mL. Density of acetic acid - 1.05 g/mL. Assume volumes mix additively.) Enter your answers in decimal form to 3 significant figures. Do not enter units since they are already provided. Molar: M molal: m mole fraction: mass percent: %(m/m)...
6.31 M 7. Molarity of acetic acid, HC2H30in vinegar sample (show your calculations) Volume of vinegar sample = 0.003L 0.003L Vinegar contain 0.01893 moles IL 0.01893 = 6.3 IM 0.003 Vinegar contain 8. Grams of Acetic Acid, HC,H,O2 (show your calculations) 1.1368 g HC2H5O2 Grams acid = 12 *MW = 0.01893 mot x 60.052 g/ mot = 1-1368 g 31.90 % L 3 100 9. Percent (m/v) of HC,H,O, in vinegar sample (show your calculations) 3.0 mL vinegar contain 0.01893...
2.) Vinegar is an aqueous solution (homogenous mixture) of acetic acid and water laqueous aqua" - Latin for water). To simplify the calculations let us assume that all of the acetic acid in the vinegar reacts and there is some left over "excess" baking soda. The vinegar solution is a 5% acetic acid solution. If you look at the bottle of vinegar somewhere on there it probably says "5% acidity" or similar notation. Let us estimate that about 25.0 g...
Calculate the moles of acetic acid, molarity of the vinegar solution, and mass %. of acetic acid in the vinegar using the average of three good trials used to titrate 5.00 mL of the vinegar. You may assume that the density of your vinegar sample is 1.01 g mL^-1. Drain off unused NaOH solution into a clean container and place it into a designated container. Wash the buret two times with regular tap water, then rinse with deionized water, and...
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