Step 2. Heat test tube until all of the O2 has evolved Step 3. Record the amount of O2 produced Observations Mass of O2 produced: 3.915576 grams or .1224 moles Mass of KCl produced: 6.083363 grams or .0816 moles Calculations Chemical Reaction 2KClO3-> 2KCl + 3O2 Theoretical Yield (96/245.1)*10 grams= 3.916 grams of O2 Percent Yield 3.915576/3.916= .99989 *100= 99.998 percent Data Actual amount of Oxygen Produced |3.915576 grams | |Theoretical amount of oxygen Produced |3.916 grams of O2 | |The Percent Yield |99.998 percent | | Results The amount of oxygen produced was almost exactly what was expected with a percent yield of 99.998 percent. The Law of conservation of mass was also up held with there being a combined mass of 10 grams from the resulting O2 and KCl. Conclusion 10 Grams of Potassium chlorate when decomposed produces 3.915576 grams oxygen gas and 6.083363 grams potassium chloride Atomic Weight of Magnesium Introduction In this lab we will determine the atomic weight of magnesium by measuring the amount of hydrogen gas evolved when hydrochloric acid reacts with magnesium.
Name that gas. CO2 Results Table 3. Effect of Hypercapnia on Pulmonary Ventilation 0.04% carbon dioxide (normal air) MV pCO2 pO2 pH (L/min) (mmHg) (mmHg) 6.2 39.1 102.7 7.41 6.1 39.3 107.8 7.38 6.1 40.1 102.4 7.4 6.1 39.5 104.3 7.4 4% carbon dioxide pCO2 pO2 (mmHg) (mmHg) 43 102.3 44.4 101.6 42.7 98.1 43.4 100.7 MV (L/min) 10.36 8.74 10.12 9.7 2% carbon dioxide pCO2 pO2 (mmHg) (mmHg) 41.6 100.3 41.6 105.5 42.2 104 41.8 103.3 6% carbon dioxide pCO2 pO2 (mmHg) (mmHg) 47.6 101.2 46.5 99.7 47.3 93.7 47.1 98.2 pH 7.38 7.4 7.38 7.39 Subj. 1 Subj. 2 Subj.
| Observations of Chemicals | Zinc Sulfate | Powder of a white solid | Barium Iodide | Powder of a white solid. | Deionized water | Liquid, transparent. | Trial # | BaI2 | ZnSO4 | Theoretical Yield of ZnI2 | Actual Yield | Percent Yield | 1 | .67g | .45g | .499820g | .52g | 104% | 2 | .67g | .45g | .499820g | .52g | 104% | 3 | .66g | .46g | .493117g | .48g | 97% | Calculations for Cost | Double Replacement | Synthesis | 0.48 grams of Zinc Sulfate - $0.02 | 1.00 gram Granular Zinc - $62.50 | 0.67grams of Barium Iodine Dihydrate - $0.886 | 2.00 gram Iodine - ($74.90 × 2) - $149.80 | 0.52 grams of Zinc Iodide - $0.906 | 1.00 gram zinc - $0.212 | 1000 grams of Zinc Iodide = $1,923.00 | 1000 grams of Zinc Ioidide = $212.30 | Focus Question Should chemists prepare Zinc Iodide, from its Elements or from a Double Replacement Reaction between Barium Iodide and Zinc Sulfate?
Acid test= solid + 3M Acetic acid cloudy + 6 ZnO ZnO + Heat yellow Zn Cooled White + ZnCO3 + heat yellow solid. Yellow solid cooled white solid + QUANTITATIVE DATA AND CALUCATIONS Desired Results Calculations Temperature of Water Bath 22.8oC Vapor Pressure (as
Determine the percent yield of this reaction, showing all steps of your calculation. (3 points) heoretical yield of H2 gas: (1.156 x 10^-3 moles)(1 mole H2/ 1 mole Mg) = 1.156 x 10^-3 moles Theoretical mass: (1.156 x 10^-3 moles)(2.02 g/mole) = 2.335 x 10^-3 g Using ideal gas law: P = (1.1 atm)(760 torr / 1 atm) - 19.8 torr = 816.2 torr V = 0.026 L T = 295 K Solve for n: n = PV/(RT) n = (816.2 torr)(0.026
There will have some error. 2) A volatile liquid was allowed to evaporate in a 43.298 g flask that has a total volume of 252 ml. the temperature of the water bath was 100˚C at the atmospheric pressure of 776 torr. The mass of the flask and condensed vapor was 44.173 g. calculate the molar mass of the liquid. T = 273 + 100 = 373 V = 252 mL = 1 L / 1000 mL = 0.252 L P = 776 Torr R= 0.0821 mass of 44.173 - 43.298 g = 0.875g moles of gas = PV / RT = 776 x .252 / 62.363 x (273+100) =0.00841 moles molar mass = 0.875g / 0.00841 moles = 104.1 g/
iv. Data Tables. Changing mass of washer vs. acceleration Mass (g) | Acc #1 (m/s^2) | Acc #2(m/s^2) | Acc #3(m/s^2) | Average EXPERIMENTAL Acceleration | 60 | .77 | .70` | .76 | .743 | 80 | 1.28 | 1.26 | 1.25 | 1.263 | 100 | 1.71 | 1.74 | 1.76 | 1.736 | Changing mass of cart vs. acceleration Mass of Cart (g) | Acc #1(m/s^2) | Acc #2(m/s^2) | Acc #3(m/s^2) | Average EXPERIMENTAL Acceleration | 0.500 kg | 2.24 | 2.23 | 2.25 | 2.24 | 1.000 kg | .83 | .84 | .85 | .84 | 1.500 kg | .27 | .27 | .23 | .2566 | Changing mass of washer’s acceleration vs. Calculate acceleration Mass(g) | Experimental acceleration | Calculated Acceleration | Percent Error | 60 | .743 | 1.02 | 27.15 | 80 | 1.263 | 1.53 | 17.45 | 100 | 1.736 | 1.99 | 12.75 | Changing mass of cart’s acceleration vs. Calculate acceleration Mass(kg) | Experimental acceleration | Calculated Acceleration | Percent Error | .5 | 2.24 | 2.58 | 13.17 | 1 | 0.84 | 1.14 | 26.315
If the reaction is not spontaneous under standard conditions at 298K, at what temperature (if any) would the reaction become spontaneous? a) 2 PbS (s) + 3 O2 (g) → 2 PbO (s) + 2SO2 (g) ; ΔH° = -844 kJ; ΔS°= -165 J/K b) 2 POCl3 (g) → 2 PCl3 (g) + O2 (g) ; ΔH° = 572 kJ; ΔS°= 179 J/K 5. Consider the reaction H2 (g) + F2 (g) → 2 HF (g). a) Using data in your Appendix B, calculate ΔG° at 25°C b) Calculate ΔG at 298K if the reaction mixture consists of 8.0 bar of H2, 4.5 bar of F2 and 0.36 bar of HF. 6.
Why is it dangerous to use and acid containing toilet bowl cleaner with a chlorine liquid bleach? Lab 3: The Estimation of Avogadro’s Number (Expt 10) Understand the calculations in the lab report. Lab 4: Emission Spectra and Electronic Structure of Atoms (Expt 17) Equations! a. i. h = Planck’s constant 6.626 x 10-34J·s; = frequency (s-1) ii. c = speed of light 3.00 x 108m; = wavelength (m) iii.
Purpose Measure the enthalpy for three reactions and to confirm the value of Hess’ law in predicting the ΔH of a reaction A) NaOH(s) NaOH(aq) B) NaOH(aq) + HCl(aq) NaCl(aq) + H2(l) C) NaOH(s) + HCl(aq) NaCl(aq) + H2O(l) Materials Styrofoam Cup Calorimeter | Forceps | Thermometer | NaOH Pellets | Graduated Cylinder | 0.50M HCL | Beaker | Magnetic Stirrer | Stirring Hot Plate | | | | Procedure Thermometers were used instead of Lob pro. (Refer to lab handout) Observations Table One: Results for Part A: Trial | Mass m / g | Temperature of H2O (±0.1°C) | | Cups | Cups, H2O, Pellets | Cups, Pellets | Initial Ti / °C | Final Tf / °C | Change ΔT / °C | 1 | 7.14g | 56.15g | 8.15g | 19.3°C | 23.0°C | 3.7°C | 2 | 7.49g | 55.17g | 8.48g | 19.4°C | 23.5°C | 4.1°C | 3 | 8.02g | 54.15g | 9.03g | 19.4°C | 23.2°C | 3.8°C | Table Two: Results for Part B: Trial | Mass m / g | Temperature of H2O (±0.05°C) | Temperature of HCl (±0.1°C) | | Cups, H2O, Pellets | Initial Ti / °C | Final Tf / °C | Final Tf / °C | 1 | 104.66g | 21.2°C | 21.1°C | 24.9°C | 2 | 103.40g | 20.2°C | 22.3°C | 24.6°C | 3 | 102.56g | 20.0°C | 22.1°C | 24.7°C | Table Three: Results for Part C: Trial | Mass m / g | Temperature of H2O and HCl (±0.1°C) | | Cups | Cups, H2O, Pellets | Cups, H2O, Pellets | Initial Ti / °C | Final Tf / °C | Change ΔT / °C | 1 | 7.41g | 104.41g | 8.41g | 20.4°C | 23.9°C | 3.5°C | 2 | 7.85g | 103.12g | 8.84g | 20.1°C | 25.0°C | 4.9°C | 3 | 8.33g | 103.01g | 7.61g | 20.2°C | 24.2°C | 4.0°C | Analysis Sample Calculations: Trial One: Part A * ΔH=mH2OcH2OΔtnNaOH * ΔH=48.00g4.18Jg-1K-13.7K0.025mol * ΔH=29694.72Jmol-1=29.69kJmol-1 Sample Calculation: Trial One: Part A: Error * Error=mH2OcH2OΔtnNaOH * Error=0.010.20.01 *