Homework Answers
11.9
Molar concentration of the solution is number of moles of the solute present in one liter of the solution (moles/ltr).
Given that 0.297 moles of HCl is present in 25 mL of the solution.
We can find out the number of moles for 1 liter of this solution by simple cross multiplication. 1 liter = 1000 mL
25 mL of the solution contains 0.297 moles of HCl then
1000 mL of this solution contains
1000 mL x 0.297 moles / 25 mL = 11.88 moles
Thus, we get for the same 25 mL solution containing 0.297 moles, 1 liter of this solution contains 11.88 moles of HCl.
Thus the molarity of the solution is 11.88 moles/ltr or 11.88 M.
Therefore, the required answer is 11.9 (to 3 significant figures, without including units)
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A solution of hydrochloric acid in an erlenmeyer flask contains 0.297 moles HCI. If the volume...
Concentration of Standardized HCI Solution = 0.39 Flask Mass Mg (g) 0.0845 0.0815 Volume HCI (mL)...
Concentration of Standardized HCI Solution = 0.39 Flask Mass Mg (g) 0.0845 0.0815 Volume HCI (mL) T 10mL 10mL 12.15 m2 Initial volume NaOH (mL) 21.15ML Final volume NaOH (ml) 21. 15m2 31.5mL Volume used NaOH (mL) 19 mL 10.35 ml CALCULATIONS **Show work for each trial.** . Moles of Mg used • Initial moles of HCl (total moles of HCl placed into each Erlenmeyer flask) • Moles of NaOH Moles of HCl titrated NaOH + HCI -----> NaCl +...
Concentration of Standardized HCI Solution = 0.934 M Flask 2 Mass Mg (8) 10.0849 0.0815 Volume...
Concentration of Standardized HCI Solution = 0.934 M Flask 2 Mass Mg (8) 10.0849 0.0815 Volume HCI (mL) I 10mL 10mL Initial volume NaOH (mL) 12.15ml Final volume NaOH (mL) 21.154L 9 m 21. 15ML 31.5mL 10.35 mL Volume used NaOH (mL) CALCULATIONS **Show work for each trial.** • Moles of Mg used Trial 0.084 Hy loola 10.001g Mg = 0.00 0.003375 mol My 0.0035 molto • Initial moles of HCl (total moles of HCl placed into each Erlenmeyer flask)...
3. A student pipetted 25.00 mL of a 0.2531 M solution of HCl into an Erlenmeyer...
3. A student pipetted 25.00 mL of a 0.2531 M solution of HCl into an Erlenmeyer flask. After adding 3 drops of phenolphthalein indicator to the flask, the student started adding NaOH from the burette, until the color in the Erlenmeyer flask turned light pink, The student calculated that 21.40 mL NaOH was transferred in the flask to neutralize the acid. a) Calculate the number of moles of HCl initially present (Reaction: NaOH(aq) + HCl(aq) -NaCl(aq) + H2O() b) Calculate...
QUESTION 3 A 0.06 M HCl solution is used to simulate the acid concentration of the...
QUESTION 3 A 0.06 M HCl solution is used to simulate the acid concentration of the stomach. How many liters of "stomach acid" react with a tablet containing 0.37 g of magnesium hydroxide (58.33 g/mol)? Round answer to two sig figs and do not include units QUESTION 4 A 16.9 ml solution is 0.318 M HCI. What is the concentration after 56.8 ml of water is added to the solution? Round answer to 3 decimal places and do not include...
(10 pts) A solution is prepared by dissolving 1.239 grams of hydrochloric acid, HCI, in enough...
(10 pts) A solution is prepared by dissolving 1.239 grams of hydrochloric acid, HCI, in enough water to make 250.0 mL of solution. A 25.00 mL sample of this solution is diluted with water to a final volume of 100.0 mL. Now, you take 10.00 mL of that solution and dilute it to 100.00 mL What is the molarity of the final solution? (HCI 36.45 g/mol)
Concentration of NaOH = 1600 x10- Molar Sodium hydroxide Nach Trial 1 Trial 2 Trial Trial...
Concentration of NaOH = 1600 x10- Molar Sodium hydroxide Nach Trial 1 Trial 2 Trial Trial 4 2.60/1.20 1.40/1.30 28.20 29.10.26.1026.50 Volume of NaOH added Initial buret reading Final buret reading Moles of NaOH added Moles of HCl reacted Molarity of diluted HCI Molarity of undiluted HCI Average molarity of undiluted HCI Precision PROCEDURE 1. Obtain approximately 40 mL of an unknown acid in an Erlenmeyer flask from your teaching assistant. 2. Clean your 250-ml volumetric flask (contains exactly 250.00...
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...
5. During the titration of an acid with a base, the sides of the Erlenmeyer flask...
5. During the titration of an acid with a base, the sides of the Erlenmeyer flask are washed with distilled water. Do you think this rinsing will affect the outcome of the titration? Why or why not? 6. The approximate concentration of the NaOH solution you will prepare in Part A of this ex- periment is 0.12 M. Calculate the number of moles and the number of grams of oxalic acid (H,C,0, 2H,O) that will be required to neutralize 25...
help;( Data Sheet Titration and Buffers Name Date Part One: Buffer Solution (record the results from...
help;(
Data Sheet Titration and Buffers Name Date Part One: Buffer Solution (record the results from the video) Lab Section Number of Drops of HCl added to "water": Color changed from to Number of Drops of HCl added to "buffer": Color changed from to to Number of Drops of NaOH added to "water": Number of Drops of NaOH added to "buffer": _Color changed from Color changed from to Part Two: Titration Results for Acid-Base Neutralization (record the results from the...
Concentration of NaOH = 1600 x10- Molar Sodium hydroxide Nach Trial 1 Trial 2 Trial Trial 4 2.60/1.20 1...
Concentration of NaOH = 1600 x10- Molar Sodium hydroxide Nach Trial 1 Trial 2 Trial Trial 4 2.60/1.20 11.40/1.30 28.20 29.10.26.1026.50 Initial buret reading Final buret reading Volume of NaOH added Moles of NaOH added Moles of HCl reacted Molarity of diluted HCI Molarity of undiluted HCI Average molarity of undiluted HCI Precision PROCEDURE 1. Obtain approximately 40 mL of an unknown acid in an Erlenmeyer flask from your teaching assistant. 2. Clean your 250-ml volumetric flask (contains exactly 250.00...
Concentration of Standardized HCI Solution = 0.39 Flask Mass Mg (g) 0.0845 0.0815 Volume HCI (mL) T 10mL 10mL 12.15 m2 Initial volume NaOH (mL) 21.15ML Final volume NaOH (ml) 21. 15m2 31.5mL Volume used NaOH (mL) 19 mL 10.35 ml CALCULATIONS **Show work for each trial.** . Moles of Mg used • Initial moles of HCl (total moles of HCl placed into each Erlenmeyer flask) • Moles of NaOH Moles of HCl titrated NaOH + HCI -----> NaCl +...
Concentration of Standardized HCI Solution = 0.934 M Flask 2 Mass Mg (8) 10.0849 0.0815 Volume HCI (mL) I 10mL 10mL Initial volume NaOH (mL) 12.15ml Final volume NaOH (mL) 21.154L 9 m 21. 15ML 31.5mL 10.35 mL Volume used NaOH (mL) CALCULATIONS **Show work for each trial.** • Moles of Mg used Trial 0.084 Hy loola 10.001g Mg = 0.00 0.003375 mol My 0.0035 molto • Initial moles of HCl (total moles of HCl placed into each Erlenmeyer flask)...
3. A student pipetted 25.00 mL of a 0.2531 M solution of HCl into an Erlenmeyer flask. After adding 3 drops of phenolphthalein indicator to the flask, the student started adding NaOH from the burette, until the color in the Erlenmeyer flask turned light pink, The student calculated that 21.40 mL NaOH was transferred in the flask to neutralize the acid. a) Calculate the number of moles of HCl initially present (Reaction: NaOH(aq) + HCl(aq) -NaCl(aq) + H2O() b) Calculate...
QUESTION 3 A 0.06 M HCl solution is used to simulate the acid concentration of the stomach. How many liters of "stomach acid" react with a tablet containing 0.37 g of magnesium hydroxide (58.33 g/mol)? Round answer to two sig figs and do not include units QUESTION 4 A 16.9 ml solution is 0.318 M HCI. What is the concentration after 56.8 ml of water is added to the solution? Round answer to 3 decimal places and do not include...
(10 pts) A solution is prepared by dissolving 1.239 grams of hydrochloric acid, HCI, in enough water to make 250.0 mL of solution. A 25.00 mL sample of this solution is diluted with water to a final volume of 100.0 mL. Now, you take 10.00 mL of that solution and dilute it to 100.00 mL What is the molarity of the final solution? (HCI 36.45 g/mol)
Concentration of NaOH = 1600 x10- Molar Sodium hydroxide Nach Trial 1 Trial 2 Trial Trial 4 2.60/1.20 1.40/1.30 28.20 29.10.26.1026.50 Volume of NaOH added Initial buret reading Final buret reading Moles of NaOH added Moles of HCl reacted Molarity of diluted HCI Molarity of undiluted HCI Average molarity of undiluted HCI Precision PROCEDURE 1. Obtain approximately 40 mL of an unknown acid in an Erlenmeyer flask from your teaching assistant. 2. Clean your 250-ml volumetric flask (contains exactly 250.00...
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...
5. During the titration of an acid with a base, the sides of the Erlenmeyer flask are washed with distilled water. Do you think this rinsing will affect the outcome of the titration? Why or why not? 6. The approximate concentration of the NaOH solution you will prepare in Part A of this ex- periment is 0.12 M. Calculate the number of moles and the number of grams of oxalic acid (H,C,0, 2H,O) that will be required to neutralize 25...
help;(
Data Sheet Titration and Buffers Name Date Part One: Buffer Solution (record the results from the video) Lab Section Number of Drops of HCl added to "water": Color changed from to Number of Drops of HCl added to "buffer": Color changed from to to Number of Drops of NaOH added to "water": Number of Drops of NaOH added to "buffer": _Color changed from Color changed from to Part Two: Titration Results for Acid-Base Neutralization (record the results from the...
Concentration of NaOH = 1600 x10- Molar Sodium hydroxide Nach Trial 1 Trial 2 Trial Trial 4 2.60/1.20 11.40/1.30 28.20 29.10.26.1026.50 Initial buret reading Final buret reading Volume of NaOH added Moles of NaOH added Moles of HCl reacted Molarity of diluted HCI Molarity of undiluted HCI Average molarity of undiluted HCI Precision PROCEDURE 1. Obtain approximately 40 mL of an unknown acid in an Erlenmeyer flask from your teaching assistant. 2. Clean your 250-ml volumetric flask (contains exactly 250.00...
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