About 300 grams of glucose was fermented with 20 grams of yeast inside a fermenting vessel for 2 weeks. Calculate the percent yie alcohol in mL or %(v/v) if the actual yield of the fermentation set-up is at 84.50 mL alcohol. Assume the density of ethanol is at 0.789 C6H1206 + H20 → 2C2H5OH + 2CO2 Molar Masses: Glucose: 180.156 g/mole; Ethanol: 46.07 g/mole Water: 18.00 g/mole Report your answers in two decimal places. Please include the unit. For example: 25.78%

About 300 grams of glucose was fermented with 20 grams of yeast inside a fermenting vessel for 2 weeks. Calculate the percent yie alcohol in mL or %(v/v) if the actual yield of the fermentation set-up is at 84.50 mL alcohol. Assume the density of ethanol is at 0.789 C6H1206 + H20 → 2C2H5OH + 2CO2 Molar Masses: Glucose: 180.156 g/mole; Ethanol: 46.07 g/mole Water: 18.00 g/mole Report your answers in two decimal places. Please include the unit. For example: 25.78%

Chemistry: The Molecular Science

5th Edition

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

Publisher:John W. Moore, Conrad L. Stanitski

Chapter13: The Chemistry Of Solutes And Solutions

Section: Chapter Questions

Problem 16QRT

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About 200 grams of glucose was fermented with 20 grams of yeast inside a fermenting vessel for 2 weeks. Calculate the percent yield of ethyl alcohol in mL or %(v/v) if the actual yield of the fermentation set-up is at 84.50 mL alcohol. Assume the density of ethanol is at 0.789 g/mL. C6H1206 + H20 → 2C2H5OH + 2CO2 Molar Masses: Glucose: 180.156 g/mole; Ethanol: 46.07 g/mole Water: 18.00 g/mole
(d) A sample contains a mixture of ammonium carbonate (NH4)2CO3, sodium carbonate Na2CO3, and sodium chloride NaCl. Thermogravimetric analysis indicates that a 0.0965-g portion of this sample loses 0.0574 g over a temperature range of 50-75 oC and then loses another 0.0124 g at 800 oC.(i) What causes the weight loss at 50-75 oC? (ii) What species will remain after being heated to 800 oC? (iii) What is the composition of the original sample? Given: Atomic masses: H = 1.008; C = 12.011; N = 14.007; Na = 22.99; O = 15.999;Cl = 35.45.
1)The production of authentic tequila is restricted to the blue agave and a defined geographical area in west-central Mexico. Two basic categories of tequila can be distinguished: “100% agave” and “mixed” tequila. High-quality “100% agave” tequila uses pure agave juice for fermentation and Mexican law states that these authentic tequilas must be bottled in Mexico. A mixed tequila is manufactured by adding up to 49% (w/v) of sugar from sugar cane for fermentation. The low-quality tequila is usually shipped in bulk containers for diluting and bottling in the importing countries. Due to frequent fraud in the labeled quality of tequila, there is a need to verify the authenticity of a tequila using chemical analysis. Figure 1 shows an ion chromatogram for an authentic 100% agave tequila and a commercial mixed tequila. The difference in anion content is due to the different sources of water used in final dilution and bottling steps. a) In reversed-phase liquid chromatography, a…
Oral rehydration salts are stated to contain the following components: Sodium Chloride 3.5g Potassium Chloride 1.5g Sodium Citrate 2.9g Anhydrous Glucose 20.0g 8.342 g of oral rehydration salts are dissolved in 500 ml of water. 5 ml of the solution is diluted to 100 ml and then 5 ml is taken from the diluted sample and is diluted to 100 ml. The sodium content of the sample is then determined by flame photometry. The sodium salts used to prepare the mixture were: Trisodium citrate hydrate (C6H5Na3O7, 2H2O) MW 294.1 and sodium chloride (NaCl) NW 58.5. Atomic weight of Na = 23. The content of Na in the diluted sample was determined to be 0.3210 mg/100 ml. Determine the % of stated content of Na in the sample. The stated should be 104.5, how??
Oral rehydration salts are stated to contain the following components: Sodium Chloride 3.5g Potassium Chloride 1.5g Sodium Citrate 2.9g Anhydrous Glucose 20.0g 8.342 g of oral rehydration salts are dissolved in 500 ml of water. 5 ml of the solution is diluted to 100 ml and then 5 ml is taken from the diluted sample and is diluted to 100 ml. The sodium content of the sample is then determined by flame photometry. The sodium salts used to prepare the mixture were: Trisodium citrate hydrate (C6H5Na3O7, 2H2O) MW 294.1 and sodium chloride (NaCl) NW 58.5. Atomic weight of Na = 23. The content of Na in the diluted sample was determined to be 0.3210 mg/100 ml. Determine the % of stated content of Na in the sample.
Table 5, Dissolved oxygen of distilled water at different temperatures Temperature (C) Volume of aliquot (ml.) 5.00 Volume of sodium thiosulfate used (mL) Dissolved oxygen (ppm) 10.3 4 13.60 25 5.00 12.20 8.23 73 5.00 5.40 4.03 1. From your data, calculate the dissolved oxygen content in distilled water at 73 °C (data in Table 5) without using the DO formula. Detailed calculation and formulas are required in this answer.
The hydration of a hydrocarbon, but-2-ene, C4H$ produces butan-2-ol, C4H100 as follows: Penghidratan terhadap satu hidrokarbon, but-2-ena, C4H3 menghasilkan- butan-2-ol, C4H100 seperti berikut: C4H8(g)+ H2O (g)→ C4H10O (g) The enthalpy values for the following reactions are: Nilai entalpi untuk tindak balas berikut ialah: 4C(s)+ 4H2(g)→C4Hg(g) AH = -7.1 kJ 4C(s)+ 5H2(g)+ ¿O,¬C,Hj„O(g) AH = -292.8 kJ 2H2(g)+ O2(g)→2H2O(g) AH = -483.6 kJ By using the data above, calculate the enthalpy change, AH (in kJ) for the production of butan-2-ol. Dengan menggunakan data di atas, hitungkan perubahan entalpi, AH (dalam kJ) bagi menghasilkan butan-2-ol.
= STATES OF MATTER Applying like dissolves like For each solute, click the button under the better solvent. H H HO U C solute H I 2 CIH Η 2=0 CaC1₂ H H CH₂ - CH₂-OH Which is the better solvent? CH₂(CH₂) CH3 H₂O CH₂ CH₂OH :0: || CH₂- - CH3 I CH CC14 CH₂OH - CH3 0/5
When 20 grams of potassium chlorate, KCIO3, is dissolved in 100 grams of water at 80 °C, the solution can be correctly described as: 100 90 80 NANO, 70 Pb(NO)2 KNOS 60 CaC 50 O 40 NaCl KCI 30 20 KCIO, 10 Ce,(SO 10 20 30 40 50 60 70 80 90 100 Temperature ("C) supersaturated saturated unsaturated Solubility (g of salt in 100 g H,O) CONY 00 0 K,Cr,O,
Consider the Solubility Curve for this Problem: Analysis for KClO3 & NH4Cl MW (KClO3) = 122.55 g/mol , MW (NH4Cl) = 53.49 g/mol i) What is the temperature at which KClO3 & NH4Cl both have the same solubility? Round to nearest degree. +/- 1° T = °C ii) What is the solubility (g per 100g H2O) at this temperature. Round to nearest gram. +/- 1 g s (KClO3) = g/100g H2O s (NH4Cl) = g/100g H2O iii) What is the molar solubility for KClO3 at this temperature? Assume the density of the solution is 1.21 g/cc. Round off to 2 sig. figs. |s| KClO3 = M iv) Calculate the Ksp for each of these chemicals at this temperature. Ksp = v) What is the unit of Ksp for KClO3? Unit for Ksp =
Diluc asks Connor, the winemaker on Dawn Winery, about the status of the new batch of dandelion wine. Connorreports that it is 2.65 M of ethanol (CH3CH2OH) content. What is the concentration of ethanol in % w/v?
5.20 g of tea leaves were used for the extraction of caffeine in a laboratory experiment, If the mass of crude caffeine obtained and the mass of pure caffeine recovered after recrystallization are 0.166 g and 0.082 g respectively, what is the percentage yield of pure caffeine from the tea leaves? O 2.10% O 2.30% O 1.58% O 3.62%