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.
Part I: Density of Unknown Liquid | | Trial 1 | Trial 2 | Trial 3 | Mass of Empty 10 mL graduated cylinder (grams) | 25.5g | 25g | 25g | Volume of liquid (milliliters) | 8.6mL | 8.7mL | 8.4mL | Mass of graduated cylinder and liquid (grams) | 36g | 36g | 35.5g | Part II: Density of Irregular-Shaped Solid | Mass of solid (grams) | 38.74g | 39.002g | 42.489g | Volume of water (milliliters) | 50mL | 49mL | 51mL | Volume of water and solid (milliliters) | 54mL | 53mL | 56mL | Part III: Density of Regular-Shaped Solid | Mass of solid (grams) | 26g | 27g | 26g | Length of solid (centimeters) | 5.2cm | 5cm | 4.5cm | Width of solid (centimeters) | 3cm | 4cm | 3.5cm | Height of solid (centimeters) | 2.5cm | 3cm | 2cm | Calculations Show all of your work for each of the following calculations and be careful to follow significant figure rules in each calculation. Part I: Density of Unknown Liquid 1. Calculate the mass of the liquid for each trial. (Subtract the mass of the empty graduated cylinder from the mass of the graduated cylinder with liquid.) * Trial 1 36-25.5=10g * Trial 2 36-25=11g * Trial 3 35.5-25=10g 2.
Using the same electronic balance, the average mass of five copper slugs, in grams, will be determined. Lastly, by using the electronic balance again, the weight of two different unknown weights, in grams, will be determined by the weighing by difference method. Using both the direct weight and weighing by difference techniques, the weight of the copper slug (2.98 g) and the two unknown weights can be fairly accurately determined using the centigram balance. However, since the electronic balance can determine mass out to three decimal places, the electronic balance was more accurate weighing the copper slug (3.022 g) than the centigram balance using the direct weight and weighing by difference methods. Determining the mass of the two unknown weights (unknown weight #1 and #2) was determined using only the centigram balance using the weighing by difference method.
How many moles of gas does it take to fill a 1.0 L flask at a pressure of 1.5 atm at 100 celsius? (.049 mol) 2. What is the atmospheric pressure if the partial pressures of nitrogen, oxygen and argon are 604.5 mm Hg, 162.8 mm and .5 mm respectively? (767.8 mm Hg) 3. Describe the
How many kilograms of pure HCl would be used to make this hydrochloric acid? (Assume that 30% has two significant figures. There are 2000 lb/ton.) (Obj 15) 30 ton HCl 2000 lb 1 kg ? kg HCl = 6.0 × 105 ton HCl soln 100 ton HCl soln 1 ton 2.205 lb 30 ton HCl 2000 lb 453.6 g 1 kg ?
8) See column 8 and back for work. 9) My accuracy was not really bad but it wasn’t that good, some of the guesses I made were almost accurate but others were way off. Atom | Average mass of one atom | Mass relative to carbon (#:1) | Atomic mass (from periodic table) (g) | Number of atoms in a relative mass (column 4/ column 2) | Carbon | 2.00E-23 | 1 | 12.01 (g) | 6.005 x 1023 | Iron | 9.30E-23 | 4.65:1 | 55.85 (g) | -65.15 | Aluminum | 4.49E-23 | 2.245 | 26.98 (g) | 6.010 x 1023 | Zinc | 1.08E-22 | 5.4 | 55.85 (g) | 5.17 x 1023 | Lead | 3.44E-22 | 17.2 | 207.2 (g) | 6.024 x 1023 | Copper | 1.05E-22 | 5.25 | 63.55 (g) | 6.052 x 1023 | PART:2 Question 1) I think carbon’s role is something to do with global warming I think I’m not sure, this doesn’t make sense because why does global warming have anything to do with the mole or even chemistry. | 2) In column 5 all the numbers are very close to Avogadro’s number 6.02 x 1023. 3) One gram = 6.02 x 1023 amu.