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Lecture

lab4 chem.docx

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Department
Psychology
Course
PSY100Y5
Professor
Dax Urbszat
Semester
Winter

Description
NAME: YVONNE EIFEDIYI LAB PARTNER:ANICK KARERANGABO STUDENT NUMBER: 1001045102 LAB SECTION NO: PRA101 SECT.A TIME: 2PM -5PM BEHAVIOUR OF GASES PARTA. DETERMINATION OF THE ATOMIC MASS OF METALLIC ELEMENT The atomic mass of a metal can be determined in different ways. Two of these methods are based on the reaction of the metal with aqueous acid: H + + +n H M(s) + n 3 O  M (aq) + n/2 2 (g) + n H2O The atomic mass can be determined by either (1) determining the number of moles of H gas produced by a known 2 weight of metal, or (2) determining the number of moles of H O that are consumed by a known weight of metal. 3 In this experiment we’ll use the first method. PROCEDURE: 1. The apparatus used in this experiment consists of an 800/1000mL gas burette set up . 2. 350mL of distilled was mixed with 150mL of 1M HCl solution in an 800/1000mL beaker . Fill the gas burette with this solution and, wearing a glove, close the open end with your finger and invert the burette in the solution remaining in the beaker. Clamp the burette vertically, with the end immersed just below the surface. Take care that air bubbles are not trapped in the gas burette. 3. The unknown metal was obtained from my TA.An approx. 2 cm strip of a 3 mm wide metal ribbon was cleaned by rubbing with a small piece of sand paper until the metal was bright and no black spots was left on the surface. The clean metal was wiped with tissue or filter paper and then weighed, accurately to 0.1 mg, without handling. 4. Fold the metal two or three times into a fairly compact mass and press it into a 3-in. test tube. It should fit snugly against the walls. Fill this tube with distilled water; then, wearing a glove, insert it, open end up, into the end of the gas burette and lower the gas burette to hold the tube captive. The acid will diffuse into the test tube and react with the metal. The H evolved was then collected in the gas burette and the dilute acid was displaced. Ensure your 2 were washed hands after. 5. When all the metal had reacted, the volume (V, mL) of the gas and the temperature of the solution in the beaker were measured. It can be assumed that the temperature of the gas in the burette (T, °K) is the same as that of the solution. Also the difference in height (h, mm) of the solution levels in the burette and in the beaker were measured, and the atmospheric pressureatm , torr) was recorded. OBSERVATION: During the reaction we can see that the volume of the solution reduces while hydrogen gas is evolved in the burette. CONCLUSION: From the experiment we can conclude that the atomic mass of an element can be determined using this method(reacting the unknown metal with HCl) CALCULATIONS 1) To calculate the hydrostatic pressure we’ll use the equation; p h = p h and 1 torr = 1 mm Hg, where the density of the solution is soln soln Hg Hg 1.00g/mL , change in height of the solution from the experiment is 20.10 cm(201.0mm) and density of Hg is 13.6g/mL .  1.00 g/mL ×201.0mm= 13.6g/mL × h Hg make h Hg the subject of formula h Hg = 1.00g/mL×201.0mm⁄13.6g/mL = 14.779mmHg to convert to torr, 1mmHg=1torr. 14.77mmHg/1mmHg×1torr. = 14.78torr. 2) To calculate the total total pressure of gas trapped ; PH 2PH O=p2essureinsideburette PH +P H O=P 2 2 atm PH +PH O=P - P 2 2 atm hydrostatic P atm747.2torr ¿experiment∧P hydrostatic78torr. total pressure of trapped gas =747.2 torr.-14.78 torr. =732.40 torr The graph of P as a function of T if the given data is in the next page T(°C) 16 18 20 22 24 26 28 30 P(torr.) 13.6 15.5 17.5 19.8 22.4 25.2 28.3 31.8 3) To calculate the pressure of trapped hydrogen gas ; PH 2PH O=P2 atm PH 2P H O=2 −P atn hydrostatic PH 2 +25 torr(from graph)= 747.2 torr -14.78 torr make PH 2 the subject of formula = PH = 747.2torr-14.78torr- 25 torr 2 = 707.42 torr 4) To calculate the number of moles of hydrogen gas , using the ideal gas relation we use PV=nRT P of hydrogen gas = 707.42 torr = 0.9308 atm V=42.10mL = 0.0421L −1 −1 R= 0.08206 L·atm·mol ·K T= 296.26 °K 0.9308atm 0.0421L= n(0.08206 L·atm·mol ·K 296.26K) n= 0.03918/24.
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