How many moles of Al2O3 are produced by the reaction 200. g Al? 4Al + 302 → 2Al2O3 3.70 mole 12. How many moles Al are required to produce 300. g Al2O3? 4Al + 302 → 2Al2O3 5.88 mole 13. 100. g Al reacts with excess O2 to produce 150. g Al2O3 according to Calculate the theoretical and percentage yield.
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.
0.00512g ZnI2/mL of solution 0.00512g/319.18 g/mol=1.6*10-5 mol 1.6*10-5 mol/(1*10-3L)=0.016M c. 0.00806 moles of ZnI2/500 mL of solution 0.00806mol/(500*10-3)L=0.016M d. 0.0161 moles of ZnI2/L of solution 0.0161mol/1L=0.016M Exercise 4: a. The moles of ZnI2: 0.25M*(250*10-3)L=0.0625mol b. 0.25M*(250*10-3)L=0.0625mol The mass of ZnI2: 0.0625mol*319.18 g/mol=19.95g c. 0.25M*(500*10-3)L=0.125mol 0.125mol*319.18 g/mol=39.9g ZnI2 d. 0.0125mol/0.25M=0.05L Exercise 5: a. 0.125M*(100*10-3)L=0.0125mol b. 0.0625mol/0.125M=0.5L=500mL Calculation for preparing the EDTA solution Exercise 6 a.
25 cm3 of a solution of sodium hydroxide reacts with 15 cm3 of 0.1 mol/dm3 HCl. What is the molar concentration of the sodium hydroxide solution? 4. Succinic acid has the formula (CH2)n(COOH)2 and reacts with dilute sodium hydroxide as follows: (CH2)n(COOH)2 + 2NaOH → (CH2)n(COONa)2 + 2H2O 2.0 g of succinic acid were dissolved in water and the solution made up to 250 cm3. This solution was placed in a burette and 18.4 cm3 was required to neutralise 25 cm3 of 0.1 moldm-3 NaOH.
Empirical formula: CH5N Steps for molecular formula: 1- Calculate the molar mass of the empirical formula. 2- Divide the known (given) molar mass by the calculated empirical formula molar mass to get a whole number 3- Multiply that whole number through subscripts of the empirical formula to obtain the molecular formula. Example CH5N 12.01 g C x 1 C= 12.01 g/mol 1.008 g H x 5 H = 5.040
| 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?
Warm Up Questions: 1. Exercise 1: a. 2.56 ZnI2=x moles of ZnI2 b. 500 ml=0.500L c. The flask would be labeled as 0.0161 ZnI2/L Solution 2. Exercise 2: a.
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. The reaction is as follows: Mg + 2HCl -> H2 + Mg2+ (aq) + 2Cl- (aq) There is a one to one relationship between the number of moles of hydrogen gas evolved and the
The chemical reaction used to find this constant is as follows: MgC2O4 (s) ↔Mg(aq)2++ C2O4 (aq)2- Kc= Mg2+[C2O42-][MgC2O4] Ksp=Mg2+[C2O42-] The solid salt magnesium oxalate is prepared through the following precipitation reaction: Mg(SO4)(aq)+NaC2O4 (aq) → MgC2O4 (s)+NaSO4 (aq) Next, the concentration of the Mg2+ and C2O42- ions is found through a redox titration. This redox titration uses a standardized potassium permanganate solution. The potassium permanganate solution is standardized by titrating it with samples of iron(II)ammonium sulfate hexahydrate . The end point is reached when the solution has turned light purple which is a result of excess amounts of MNO4-. This reaction can be summed up using the following formula: 5Fe2++8H++MnO4- →5Fe3++Mn2++4H2O After standardization, the potassium permanganate solution is then titrated with 3 different magnesium oxalate solutions.
See Table 3: Effect of Sucrose Concentration on Sucrase Activity See Graph: Effect of Sucrose Concentration on Sucrase Activity 4. Was sucrase activity higher at 25 ºC or 55 ºC? 5. State how sucrase activity changes with increasing sucrose concentration. Sucrase activity steadily increases with increasing sucrose concentration until a plateau is reached.