(b) Calculate the volume of 0.2M UO3- needed to react with 20.00 cm3 of 0.1M Cr2O72-. 3. 24.40 g of hydrated iron(II) sulphate, FeSO4.xH2O was dissolved and made up to 1.0 dm3 of aqueous solution, acidified with sulphuric acid. 25.00 cm3 of the solution was titrated with 20.00 cm3 of 0.022M potassium manganate(VII) solution for complete oxidation. a) Write the equation for the reaction.
* Trial 1 36-25.5=10g * Trial 2 36-25=11g * Trial 3 35.5-25=10g 2. Calculate the density of the unknown liquid for each trial. (Divide the mass of the liquid calculated above by the volume of the liquid.) * Trial 1: 10.5/50=0.20g/mL * Trial 2: 11/49=0.20 g/mL
Find the volume of 2.40 mol of gas whose temperature is 50.0 °C and whose pressure is 202 kPa. 6. How many moles of gas are contained in a 50.0 L cylinder at a pressure of 10100 kPa and a temperature of 35.0 °C? 7. Determine the number of grams of carbon dioxide in a 450.6 mL tank at 1.80 atm and minus 50.5 °C.
The average bond enthalpies for O—O and O==O are 146 and 496 kJ mol−1 respectively. What is the enthalpy change, in kJ, for the reaction below? H—O—O—H(g) ® H—O—H(g) + ½O==O(g) A. – 102 B. + 102 C. + 350 D. + 394 (1) 7.
Initial Concentration Chemical Mass of Graduated Cylinder (g) Mass of Sugar (g) Molar Mass (g) Moles in Graduated Cylinder Total Volume (L) Molarity (mol/L) Sugar (C12H22O11) 16.6 10 342.29 .29 .1 .29 Data Table 9. Dilution Series Dilution Volume (mL) Mass (g) Density (g/mL) Initial Concentration (M) Volume Transferred (mL) Final Concentration (M) 0 25.0 mL 25 25 0 mL 0ml 1 25.0 mL 10 10 .29 2.5 mL 5.4M 2 25.0 mL 15 40 .435 4.5 mL 6.8M 3 25.0 mL 20 45 .58 3.0 mL 9.4M 4 25.0 mL 25 50 7.25 6.0 mL 11.2M Data Table 10. Molarity vs. Density Molarity vs. Density molarity is the concentration of a solution given in gram moles of solute per liter of solution. Density is the degree of compactness of a substance. Questions: How would you prepare 10 mL of a 0.25M HCl solution if 1M HCl was available?
0.15 M NaCl solution = 0.15 moles of Na+ atoms + 0.15 moles of Cl- atoms = 0.30 Osmoles In other words, the solution is said to have an osmolarity of 0.30 Osm (or 300 mOsm) Assume the osmolarity of the ICF of body cells to be 0.300 Osm (300 mOsm) 2nd, determine if the solute is a PENETRATING particle or is NON-PENETRATING. 3rd, determine whether a concentration gradient of NON-PENETRATING particles exist or not. If a gradient exists, determine where the higher concentration of non-penetrating particles exist; in the ICF or the ECF? (Assume the 300 mOsm concentration of particles in the ICF of body cells is composed of non-penetrating particles) 4th, repeat your definition of osmosis in order to determine if water will move: a. into the cell b. out of the cell c. not move into or out of the cell Osmosis is the diffusion of water across a membrane from a region of lower concentration of non-penetrating particles into a region of higher concentration of non-penetrating
Part I: Density of Unknown Liquid | | Trial 1 | Trial 2 | Trial 3 | Mass of Empty 10 mL graduated cylinder (grams) | 26.10 | 26.15 | 26.05 | Volume of liquid (milliliters) | 8.69 | 8.50 | 8.31 | Mass of graduated cylinder and liquid (grams) | 37.00 | 36.70 | 36.10 | Part II: Density of Irregular-Shaped Solid | Mass of solid (grams) | 38.954 | 39.068 | 42.885 | Volume of water (milliliters) | 49.9 | 49.9 | 50.0 | Volume of water and solid (milliliters) | 54.1 | 54.1 | 55.0 | Part III: Density of Regular-Shaped Solid | Mass of solid (grams) | 27.50 | 26.70 | 27.40 | Length of solid (centimeters) | 5.25 | 5.00 | 4.50 | Width of solid (centimeters) | 3.00 | 4.00 | 3.50 | Height of solid (centimeters) | 2.50 | 3.00 | 2.00 | 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 37.00(g) – 26.10(g) = 10.90(g) * Trial 2 36.70(g) – 26.15(g) = 10.55(g) * Trial 3 36.10(g) – 26.05(g) = 10.05(g) 2.
734 812 + 50 + 1 062 328 = A 1 977 180 C 1 797 180 B 1 797 190 ( D 1 779 190 9. 2 450 000 − 36 000 − 300 500 = A 1 789 500 C 2 113 500 ( B 1 798 500 D 2 383 950 10. 158 780 ÷ 25 = 6 351 remainder . What is the number need to be written in the above? A 1 C 10 B 5 ( D 15 11.
Through these calculations, the concentration of hydrochloric acid was found to be 8.3 mol•L-1. Purpose Refer to handout “Standardize an Approximately 1 mol•L-1 Hydrochloric Acid” Material and Method Refer to handout “Standardize an Approximately 1 mol•L-1 Hydrochloric Acid” Results Quantitative Results Trial Initial Final Reading (in mL) Volume HCl Used (in mL) 1 0.2 12.4 12.2 2 12.4 23.9 11.5 3 23.9 36.3 12.4 Calculations Average volume of HCl
Calculate the exact normality of Na2S2O3 knowing that in this chemical reaction 1 gram-equivalent of K2Cr2O7 react with 1 gram-equivalent of Na2S2O3 (1 mole K2Cr2O7 react with 6 moles Na2S2O3). Determination of peroxide value. Weigh 3.00 g oil (with precision of 0.001 g) into a 250 ml Erlenmeyer flask. Add 10 ml chloroform and swirl to dissolve oil. Add 15 ml acetic acid,