20. mol H2 reacts with 8.0 mol O2 to produce H2O. Determine the number of grams reactant in excess and number of grams H2O produced. Identify the limiting reactant. 8.1 g H2 , 2.9 x 102 g H2O 17. How many litres of O2 gas are required to produce 100. g Al2O3?
Record the readings of the three instruments at eight different speed settings of the tunnel: 15, 20, 25, 30, 35, 40, 45, & 50. 4. Plot two calibration curves with pressure transducer reading as the abscissa and micromanometer reading as the ordinate for the first, and micromanometer versus scannivalve as the second. Convert micromanometer data to read as total pressure in SI units [Pa]. 5.
If 0.100 mol of hydrogen iodide is placed in a 1.0 L container and allowed to reach equilibrium, find the concentrations of all reactants and products at equilibrium. 2 HI (g) === H2 (g) + I2 (g) Ke = 1.84(10-2 [H2]=[I2]= 1.07(10-2 mol/L, [HI]=7.86(10-2 mol/L 6. A 1.00 L reaction vessel initially contains 9.28(10-3 moles of H2S. At equilibrium, the concentration of H2S of 7.06(10-3 mol/L. Calculate the value of Ke for this system.
5) Information about Clearwater Company's direct materials cost follows: Standard price per materials ounce $ 100 Actual quantity used 8,700 grams Standard quantity allowed for production 9,100 grams Price variance $ 76,125 F ________________________________________ Required: What was the actual purchase price per gram? (Round your answer to 2 decimal places. Omit the "$" sign in your response.) Actual purchase price $ 91.25 Total grade: 0.0×1/1 = 0% Feedback: Actual Costs = AP × 8,700 Actual Inputs at Standard Price = $100 × 8,700 =$870,000 Price Variance = $76,125 F 8,700 × AP = $870,000 – $76,125 AP = $91.25 ________________________________________ Question 3: Score
What is the volume of both tanks if the radius of the tank #1 is 15 feet and the height if tank #2 is 120 feet. You must explain your answer using words, and you must show your work to receive credit. - Since the formula for cylinders volume is V=π r² h, you would just plug in the feet and the height. V=3.14*9^2*72 1) 3.14 * 15² *120 2) 3.14 * 225* 120 3) 706.5 * 120 4) 84,780 ft The volume is 84,780ft for one whole tank, but the tanks are half of a cylinder, so that means that the volume would be half of what the whole would be. One tank volume would be 42,390ft.
29.4 atm B. 4.89 atm C. 25.1 atm D. 36.0 atm _____ 5. The vapor pressure of pure ethanol at 60 °C is 349 mm Hg. Calculate the vapor pressure at 60 °C of a solution prepared by dissolving 10.0 mol of naphthalene (nonvolatile) in 90.0 mol of ethanol. A.
Repeat the titration until there are two titres within 0.1cm3 of each other. Record results in a suitable table. Results: Rough Titre: 7.653 First Run: 6.553 Second Run: 6.453 Third Run: 6.553 Calculations: During the titration, iron(II) ions are oxidised to iron(III) ions and manganate(VII) ions are reduced to manganese(II) ions. The equation is as follows: 5Fe2+(aq) + MnO4-(aq) + 8H+(aq) ? 5Fe3+(aq) + Mn2+(aq) + 4H2O(l) The above equation shows that one mole of manganate(VII) ions reacts with 5 moles of iron(II) ions in acid solution.
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/
Comparing and Contrasting: Compare the ratio of moles of iron to moles of copper from the balanced chemical equation to the mole ratio calculated using your data. 8. Evaluating Results: Use the balanced chemical equation to calculate the mass of copper that should have been produced from the sample of iron you used. Use this number and the mass of copper you actually obtained to calculate the percent yield. 9.
2 The Design of the Sprue Bushing Traditionally, the molten material is injected into the mold structure through the sprue bushing. Then, the runner carries out the molten material through gate into the mold cavity. Hot loss and turbulence are minimized by the cross- section of the runner. To balance the material flow through sprue bushing, proper design and dimensions are critical. The “O” diameter of the sprue must be roughly 20% larger than the nozzle diameter.