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I am stuck on question 4, I found everything

I am stuck on question 4, I found everything

Chemistry for Engineering Students
Chemistry for Engineering Students
3rd Edition
ISBN:9781285199023
Author:Lawrence S. Brown, Tom Holme
Publisher:Lawrence S. Brown, Tom Holme
Chapter10: Entropy And The Second Law Of Thermodynamics
Section: Chapter Questions
Problem 10.18PAE
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I am stuck on question 4, I found everything else and I am stuck on the last part
Calculased mass Mg 20486 Fireworks: 7. My Alternate Report Sheet P. 2 3. Calalate the amount of My in your calibration sample and % error. Calalated moles My 00.002.00 124854. .60200) (24,31) MacPherson mol -g =10486 % error = Calculated mass - inital mass .1007 Thital mass .0486 0455 00485 0.0001 0485 .21% Percent error of Calibration 2190 4 Calculate the weight in grams and weight% magnesium in your unknown: 12 equations, 2unknowns) total mass of unknown = Mass 2n + mass mg NA₂ = total mules H₂ = moles H₂ from Mg + moles H₂ from Zn nH₂ = moles Mg B n = mass ma + moles Zn Mass zn + Molar muss My 1) rearrange to sore for mass of 2n: molar mass of Zn 1) since 1:1. Definition of SNichiometric ratio molar mass ii) plug into and solve for mass of Mg: 0.09935 > Mass 2n + Mass My ii) Percent Mg= mass of Mg mass of unknown × 100% Mass of magnesium= Percent Magnesium= g. % Unknown #
Transcribed Image Text:Calculased mass Mg 20486 Fireworks: 7. My Alternate Report Sheet P. 2 3. Calalate the amount of My in your calibration sample and % error. Calalated moles My 00.002.00 124854. .60200) (24,31) MacPherson mol -g =10486 % error = Calculated mass - inital mass .1007 Thital mass .0486 0455 00485 0.0001 0485 .21% Percent error of Calibration 2190 4 Calculate the weight in grams and weight% magnesium in your unknown: 12 equations, 2unknowns) total mass of unknown = Mass 2n + mass mg NA₂ = total mules H₂ = moles H₂ from Mg + moles H₂ from Zn nH₂ = moles Mg B n = mass ma + moles Zn Mass zn + Molar muss My 1) rearrange to sore for mass of 2n: molar mass of Zn 1) since 1:1. Definition of SNichiometric ratio molar mass ii) plug into and solve for mass of Mg: 0.09935 > Mass 2n + Mass My ii) Percent Mg= mass of Mg mass of unknown × 100% Mass of magnesium= Percent Magnesium= g. % Unknown #
Fireworks: Analysis of a Mixture for %. My Alternate Report Sheet MacPherson Dath Mg Calibration Mass of Sample 0.0485 Volume of f gas 15 ML 25 37.87 Mm Height of war han 1702 Temperature of room 21,55 °C Barometric Pressure 755.85 mm Calculations 20.2°C 7 Calculations: mm Hy 60456 600 unknown sample Mass of Sample 0.073 volume of gas 60.45 mL Height of water edumn 310.5 mm Unknown # 13 1. First, convert the height to water column into mmHg, using The densities: 31.05 (H₂) 37590 Height of water column in 310.5m 1.05 g/ml X- 13.6 g/ml 10fs/00! Calibraton: Height in mm Hy >29.1570 mmHg unknown? x 13.65/201 23.97.24mg If the total pressure inside the tube were the same as the barometric pressure, the water level inside the tube would be exactly at the surface of the water in the beaker. Since the column of water in the tube is above the surface of the water in the beaker, the pressure inside the tube must be less than the outside pressure. So, the pressure inside the tube is the barometric pressure minus this amount. Also, water vapor contributes to the pressure inside the tube, and this amount must be subtracted to finally calculate the pressure of hydrogen gas collected. water in mm Hy vapor PH2 Barometric Pressure - PH20-mmHg water column = 70%,00 mmHg 755.85mm 755.85mathy. - 29,25-18. Pitz caliban 708.00 mmHg 23.9724-18. Pilk 713.2776ts 2. Next, use the Ideal Gas Law to calculate the total moles of hydrogen generated: (watch units!) PV = nRT P= 708.00 ms/760 2.9316 V = 51.25mL 7.05732 0.08206 L.am R T = 21.55 + 273 = 294,70 R=0.08206 Latm/milk P=213.2776mmy/760 = .9385 √360.45w4 R=0.08204 = 0.040452 Cat Indk T₂ 21.55 +273.15 = 294.7 .9316 (53) (0.08206) (244,70) =-24.4393 200k 24542933 n = 0.00 2000 Measure In millimeters Heigh ланг of column (+9337) (06045) (.08206) (294.70) te, calib0.00200 MHz, untwun -3 =002341957 ایت 2.35×10 mul -0.0023459 mol 2.3457 X153 m
Transcribed Image Text:Fireworks: Analysis of a Mixture for %. My Alternate Report Sheet MacPherson Dath Mg Calibration Mass of Sample 0.0485 Volume of f gas 15 ML 25 37.87 Mm Height of war han 1702 Temperature of room 21,55 °C Barometric Pressure 755.85 mm Calculations 20.2°C 7 Calculations: mm Hy 60456 600 unknown sample Mass of Sample 0.073 volume of gas 60.45 mL Height of water edumn 310.5 mm Unknown # 13 1. First, convert the height to water column into mmHg, using The densities: 31.05 (H₂) 37590 Height of water column in 310.5m 1.05 g/ml X- 13.6 g/ml 10fs/00! Calibraton: Height in mm Hy >29.1570 mmHg unknown? x 13.65/201 23.97.24mg If the total pressure inside the tube were the same as the barometric pressure, the water level inside the tube would be exactly at the surface of the water in the beaker. Since the column of water in the tube is above the surface of the water in the beaker, the pressure inside the tube must be less than the outside pressure. So, the pressure inside the tube is the barometric pressure minus this amount. Also, water vapor contributes to the pressure inside the tube, and this amount must be subtracted to finally calculate the pressure of hydrogen gas collected. water in mm Hy vapor PH2 Barometric Pressure - PH20-mmHg water column = 70%,00 mmHg 755.85mm 755.85mathy. - 29,25-18. Pitz caliban 708.00 mmHg 23.9724-18. Pilk 713.2776ts 2. Next, use the Ideal Gas Law to calculate the total moles of hydrogen generated: (watch units!) PV = nRT P= 708.00 ms/760 2.9316 V = 51.25mL 7.05732 0.08206 L.am R T = 21.55 + 273 = 294,70 R=0.08206 Latm/milk P=213.2776mmy/760 = .9385 √360.45w4 R=0.08204 = 0.040452 Cat Indk T₂ 21.55 +273.15 = 294.7 .9316 (53) (0.08206) (244,70) =-24.4393 200k 24542933 n = 0.00 2000 Measure In millimeters Heigh ланг of column (+9337) (06045) (.08206) (294.70) te, calib0.00200 MHz, untwun -3 =002341957 ایت 2.35×10 mul -0.0023459 mol 2.3457 X153 m
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