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Science Chemistry Chemistry: An Introduction to General, Organic, and Biological Chemistry (13th Edition)

The volume percent of ethanol in the port wine (v/v) Solution: 18% (v/v) ethanol in the port wine Explanation: Given: The volume of a vintage port wine bottle is 750mL; it contains 135mL of ethanol Calculation: The volume percent = volume of ethanol x 100 / total volume of port wine Volume percent = 135 x 100 / 750 = 18% Conclusion: The volume percent of ethanol in the port wine (v/v) is 18% To determine: The molarity of ethanol in port wine Solution: 3.09 M ethanol Given: The volume of a vintage port wine bottle is 750mL, it contains 135mL of ethanol Density = 0.789g / mL Molecular formula of ethanol = C 2 H 6 O Formula Used: Molarity = moles of solute volume of solution ( L ) Moles of solute (ethanol) = mass of ethanol molar mass of ethanol Calculation: Mass of ethanol = Volume x Density = 135mL x 0 .789 g /mL = 106 .52 grams Molar mass of ethanol = 2 x atomic mass of carbon + 6 x atomic mass of hydrogen + 1 x atomic mass of oxygen = (2 x 12) + (6 x 1) + (1 x16) = 46g / mole Moles of ethanol = 106.52 46 = 2.32 m o l e s Molarity of ethanol = 2 .32 750 1000 = 2 .32 x1000 750 = 3 .09 M Conclusion: The molarity of ethanol is 3.09M To determine: The balanced chemical equation for fermentation of glucose Solution: C 6 H 12 O 6 → 2C 2 H 6 O ( l ) + 2CO 2 ( g ) Explanation: The balanced chemical equation is C 6 H 12 O 6 → 2C 2 H 6 O ( l ) + 2CO 2 ( g ) Here one mole of glucose will give two moles of ethanol and two moles of carbon dioxide. Conclusion: The balanced chemical equation for fermentation of glucose is C 6 H 12 O 6 → 2C 2 H 6 O ( l ) + 2CO 2 ( g ) To determine: The mass of glucose required for production of one bottle of port wine. Solution: 208.8 grams of glucose Explanation: Given: 2.32 moles of ethanol Formula used: Mass = moles x molar mass Calculation: As per balanced equation: C 6 H 12 O 6 → 2C 2 H 6 O ( l ) + 2CO 2 ( g ) For two moles of ethanol, one mole of glucose is required For one mole of ethanol, 0.5 moles of glucose are required For 2.32 moles of ethanol, 0.5 x 2.32 moles of glucose are required = 1.16 moles of glucose Molar mass of glucose (C 6 H 1 2 O 6 ) = 6 x atomic mass of carbon + 12 x atomic mass of hydrogen + 6 x atomic mass of oxygen = (6 x 12) + (6 x 1) + (6 x 16) = 180 g / mole Mass of glucose = moles of glucose x molar mass of glucose Mass of glucose required = 1.16 x 180 = 208.8 grams Conclusion: Mass of glucose required is 208.8 g To determine: The number of bottles of port wine that can be obtained from 1 ton of grapes. Given: 1 ton = 200lb Solution: 16 bottles of port wine Explanation: Given: Amount of grapes = 1 ton 1 ton = 200lb The mass of glucose in 1.5lb of grapes is 26 grams. Density = 0.789g / mL The volume of a vintage port wine bottle is 750mL, it contains 135mL of ethanol Molar mass of glucose = 180 g/mol Molar mass of ethanol = 46 g/mol Calculation: 1.5 lb of grapes will give 26grams of glucose So 1 lb of grapes will give 26 1.5 grams of glucose = 17.33 grams of glucose So 200 lb of grapes will give = 17.33 x 200 grams of glucose = 3466 grams of glucose Now, each mole of glucose will give two moles of ethanol Or 180grams of glucose will give 92 grams of ethanol So grams of ethanol from 3466 grams of glucose = 92 180 x 3466 = 1771 .5 grams of ethanol Volume of ethanol = Mass Density = 1771 .5 0.789 = 2245 .26 mL Each bottle of port wine contains = 135mL of ethanol So number of bottles of wine that can be obtained from 1 ton of grapes = Volume of ethanol that can be obtained from 1 ton of grapes Volume of ethanol in each bottle = 2245 .26 135 = 16 .63 bottles Conclusion: Number of bottles of wine that can be obtained from 1 ton of grapes is 16

The volume percent of ethanol in the port wine (v/v) Solution: 18% (v/v) ethanol in the port wine Explanation: Given: The volume of a vintage port wine bottle is 750mL; it contains 135mL of ethanol Calculation: The volume percent = volume of ethanol x 100 / total volume of port wine Volume percent = 135 x 100 / 750 = 18% Conclusion: The volume percent of ethanol in the port wine (v/v) is 18% To determine: The molarity of ethanol in port wine Solution: 3.09 M ethanol Given: The volume of a vintage port wine bottle is 750mL, it contains 135mL of ethanol Density = 0.789g / mL Molecular formula of ethanol = C 2 H 6 O Formula Used: Molarity = moles of solute volume of solution ( L ) Moles of solute (ethanol) = mass of ethanol molar mass of ethanol Calculation: Mass of ethanol = Volume x Density = 135mL x 0 .789 g /mL = 106 .52 grams Molar mass of ethanol = 2 x atomic mass of carbon + 6 x atomic mass of hydrogen + 1 x atomic mass of oxygen = (2 x 12) + (6 x 1) + (1 x16) = 46g / mole Moles of ethanol = 106.52 46 = 2.32 m o l e s Molarity of ethanol = 2 .32 750 1000 = 2 .32 x1000 750 = 3 .09 M Conclusion: The molarity of ethanol is 3.09M To determine: The balanced chemical equation for fermentation of glucose Solution: C 6 H 12 O 6 → 2C 2 H 6 O ( l ) + 2CO 2 ( g ) Explanation: The balanced chemical equation is C 6 H 12 O 6 → 2C 2 H 6 O ( l ) + 2CO 2 ( g ) Here one mole of glucose will give two moles of ethanol and two moles of carbon dioxide. Conclusion: The balanced chemical equation for fermentation of glucose is C 6 H 12 O 6 → 2C 2 H 6 O ( l ) + 2CO 2 ( g ) To determine: The mass of glucose required for production of one bottle of port wine. Solution: 208.8 grams of glucose Explanation: Given: 2.32 moles of ethanol Formula used: Mass = moles x molar mass Calculation: As per balanced equation: C 6 H 12 O 6 → 2C 2 H 6 O ( l ) + 2CO 2 ( g ) For two moles of ethanol, one mole of glucose is required For one mole of ethanol, 0.5 moles of glucose are required For 2.32 moles of ethanol, 0.5 x 2.32 moles of glucose are required = 1.16 moles of glucose Molar mass of glucose (C 6 H 1 2 O 6 ) = 6 x atomic mass of carbon + 12 x atomic mass of hydrogen + 6 x atomic mass of oxygen = (6 x 12) + (6 x 1) + (6 x 16) = 180 g / mole Mass of glucose = moles of glucose x molar mass of glucose Mass of glucose required = 1.16 x 180 = 208.8 grams Conclusion: Mass of glucose required is 208.8 g To determine: The number of bottles of port wine that can be obtained from 1 ton of grapes. Given: 1 ton = 200lb Solution: 16 bottles of port wine Explanation: Given: Amount of grapes = 1 ton 1 ton = 200lb The mass of glucose in 1.5lb of grapes is 26 grams. Density = 0.789g / mL The volume of a vintage port wine bottle is 750mL, it contains 135mL of ethanol Molar mass of glucose = 180 g/mol Molar mass of ethanol = 46 g/mol Calculation: 1.5 lb of grapes will give 26grams of glucose So 1 lb of grapes will give 26 1.5 grams of glucose = 17.33 grams of glucose So 200 lb of grapes will give = 17.33 x 200 grams of glucose = 3466 grams of glucose Now, each mole of glucose will give two moles of ethanol Or 180grams of glucose will give 92 grams of ethanol So grams of ethanol from 3466 grams of glucose = 92 180 x 3466 = 1771 .5 grams of ethanol Volume of ethanol = Mass Density = 1771 .5 0.789 = 2245 .26 mL Each bottle of port wine contains = 135mL of ethanol So number of bottles of wine that can be obtained from 1 ton of grapes = Volume of ethanol that can be obtained from 1 ton of grapes Volume of ethanol in each bottle = 2245 .26 135 = 16 .63 bottles Conclusion: Number of bottles of wine that can be obtained from 1 ton of grapes is 16

Chemistry: An Introduction to General, Organic, and Biological Chemistry (13th Edition)

Chemistry: An Introduction to General, Organic, and Biological Chemistry (13th Edition)

13th Edition

ISBN: 9780134421353

Author: Karen C. Timberlake

Publisher: PEARSON

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Chemistry: An Introduction to General, Organic, and Biological Chemistry (13th Edition)

Ch. 9.1 - Prob. 9.1PP Ch. 9.1 - Prob. 9.2PP Ch. 9.1 - Describe the formation of an aqueous KI solution,...Ch. 9.1 - Describe the formation of an aqueous LiBr...Ch. 9.1 - Prob. 9.5PP Ch. 9.1 - Water is a polar solvent and hexane (C6H12)is a...Ch. 9.2 - Prob. 9.7PP Ch. 9.2 - NaOHis a strong electrolyte, and CH3OH is a...Ch. 9.2 - Write a balanced equation for the dissociation of...Ch. 9.2 - Write the balanced equation for the dissociation...

Ch. 9.2 - Indicate whether aqueous solutions of each of the...Ch. 9.2 - Prob. 9.12PP Ch. 9.2 - Prob. 9.13PP Ch. 9.2 - Prob. 9.14PP Ch. 9.2 - Calculate the number of equivalents in each of the...Ch. 9.2 - Calculate the number of equivalents in each of the...Ch. 9.2 - An intravenous saline solution contains 154 mEq/L...Ch. 9.2 - An intravenous solution to replace potassium loss...Ch. 9.2 - Prob. 9.19PP Ch. 9.2 - A Ringer's solutioncontains the following...Ch. 9.2 - When Michael's blood was tested, the chloride...Ch. 9.2 - After dialysis, the level of magnesium in...Ch. 9.3 - State whether each of the following refers to a...Ch. 9.3 - State whether each of the following refers to a...Ch. 9.3 - Determine whether each of the following solutions...Ch. 9.3 - Determine whether each of the following solutions...Ch. 9.3 - A solution containing 80. g of KClin 200 g of H2O...Ch. 9.3 - A solution containing 80. g of NaNO3 in 75 g of...Ch. 9.3 - Explain the following observations More sugar...Ch. 9.3 - Explain the following observations: And open can...Ch. 9.3 - Predict whether each of the following organic...Ch. 9.3 - Predict whether each of the following organic...Ch. 9.4 - Calculate the mass percent (m/m) for the solute in...Ch. 9.4 - Calculate the mass percent (m/m) for the solute in...Ch. 9.4 - A mouthwash contains 22.5% (v/v) alcohol.If the...Ch. 9.4 - A bottle of champagne is 11% (v/v) alcohol. If...Ch. 9.4 - What is the difference between a 5.0% (m/m)...Ch. 9.4 - What is the difference between a 10.0% (v/v)...Ch. 9.4 - Calculate the mass/volume(m/v) percent for the...Ch. 9.4 - Calculate the mass/volume (m/v) percent for the...Ch. 9.4 - Calculate the grams or milliliters of solute...Ch. 9.4 - Calculate the grams or ml of solute needed to...Ch. 9.4 - Prob. 9.43PP Ch. 9.4 - For each of the following solutions, calculate...Ch. 9.4 - Prob. 9.45PP Ch. 9.4 - Prob. 9.46PP Ch. 9.4 - Calculate the gram of solely needed to prepare...Ch. 9.4 - Calculate the gram of solute needed to prepare...Ch. 9.4 - For each of the following solutions, calculate...Ch. 9.4 - Prob. 9.50PP Ch. 9.4 - A patient received 100 mL of a 20.0% (m/v)...Ch. 9.4 - A patient received 250 mL of a 4.0% (m/v) amino...Ch. 9.4 - A patient needs 100. g of glucose in the next 12...Ch. 9.4 - A patient received 2.0 g of NaCl in 8 h. How many...Ch. 9.5 - Prob. 9.55PP Ch. 9.5 - Prob. 9.56PP Ch. 9.5 - Determine the final volume, in milliliters, of...Ch. 9.5 - Prob. 9.58PP Ch. 9.5 - Prob. 9.59PP Ch. 9.5 - Prob. 9.60PP Ch. 9.5 - You need 500. mL of a 5.0% (m/v) glucose solution....Ch. 9.5 - A doctor ordered 100. mL of 2.0% (m/v)...Ch. 9.6 - Prob. 9.63PP Ch. 9.6 - Identify each of the following as characteristics...Ch. 9.6 - A 10% (m/v) starch solution is separated form a 1%...Ch. 9.6 - A 0.1% (m/v) albumin solution is separated form a...Ch. 9.6 - Indicate the compartment (A or B) that will...Ch. 9.6 - Indicate the compartment (A or B) that will...Ch. 9.6 - Will a red blood cell undergo creation, hemolysis,...Ch. 9.6 - Will a red blood cell undergo creation, hemolysis,...Ch. 9.6 - Each of the following mixtures is placed in a...Ch. 9.6 - Prob. 9.72PP Ch. 9.6 - 9.73 After her latest dialysis treatment, Michael...Ch. 9.6 - Prob. 9.74PP Ch. 9.6 - 9.75 A CaCl2solution is given to increase blood...Ch. 9.6 - 9.76 10 intravenous solution of mannitol is used...Ch. 9 - Prob. 9.77UTC Ch. 9 - Prob. 9.78UTC Ch. 9 - Prob. 9.79UTC Ch. 9 - Prob. 9.80UTC Ch. 9 - Prob. 9.81UTC Ch. 9 - Whydo lettuces leaves in a salad with after a...Ch. 9 - Prob. 9.83UTC Ch. 9 - Prob. 9.84UTC Ch. 9 - Prob. 9.85APP Ch. 9 - Prob. 9.86APP Ch. 9 - Prob. 9.87APP Ch. 9 - Prob. 9.88APP Ch. 9 - Prob. 9.89APP Ch. 9 - Prob. 9.90APP Ch. 9 - Prob. 9.91APP Ch. 9 - Prob. 9.92APP Ch. 9 - Prob. 9.93APP Ch. 9 - Prob. 9.94APP Ch. 9 - Prob. 9.95APP Ch. 9 - Prob. 9.96APP Ch. 9 - Prob. 9.97APP Ch. 9 - Prob. 9.98APP Ch. 9 - Prob. 9.99APP Ch. 9 - Prob. 9.100APP Ch. 9 - Prob. 9.101APP Ch. 9 - Prob. 9.102APP Ch. 9 - Prob. 9.103APP Ch. 9 - Prob. 9.104APP Ch. 9 - Prob. 9.105APP Ch. 9 - Prob. 9.106APP Ch. 9 - Calculate the final concentration of the solution...Ch. 9 - Calculate the final concentration of the solution...Ch. 9 - Prob. 9.109APP Ch. 9 - Prob. 9.110APP Ch. 9 - Prob. 9.111APP Ch. 9 - Prob. 9.112APP Ch. 9 - Prob. 9.113CP Ch. 9 - 9.114. In a laboratory experiment, a 15.0-sample...Ch. 9 - Prob. 9.115CP Ch. 9 - Prob. 9.116CP Ch. 9 - Prob. 9.117CP Ch. 9 - Prob. 9.118CP Ch. 9 - Prob. 13CI Ch. 9 - Automobile exhaust is a major cause of air...Ch. 9 - Bleach is often added to a wash to remove stains...Ch. 9 - The compound butyric acid gives rancid butter its...Ch. 9 - Methane is a major component of purified natural...Ch. 9 - The active ingredient in Turns is calcium...Ch. 9 - Tamiflu (oseltamivir), C16H28N2O4, is an antiviral...Ch. 9 - Prob. 20CI

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