Question

The Gas Laws h. Measured final volume of gas in the flask (V2 measured): mL 26laml 1. Calculated final volume of gas in the flask (V2 calculated): mL 3247610 371.65 295.75 j. Percent of error: % 3% Calculations (include all units): 0.872 = 0.899 2. Review Questions: a. Suppose a student performed this experiment but did NOT close off the rubber tubing during the transfer to the cold water bath. Would their V2 measured be larger or smaller than it should be? EXPLAIN CLEARLY AND COMPLETELY! b. Why is it necessary to level the water in the flask with the water in the cold water bath? c. A 525.0 mL sample of a gas at 25.20 C and 717.8 mm Hg is pumped into a 115.0 mL container. If the temperature were maintained, what would be the new pressure of the gas? (Show calculations.) d. Nitrogen gas is stored in a 200 mL container at 0.93 atm and 27.40 C. How many grams of gas are present in the container? (Show calculations.) a. I think that V2 (volume of gas) would be less because of you failed to clamp the tubing when you submerged the flash then. water would rush in before the al temperature had a chance to cool. If water had a chance to enter the flash before cooling to the terup of the cold water then the volume would o be incorrect. b. it is necessary to raise the flash to the water level because this action oquializas pressure. If it is not done properly then. pressure is not equalized within and Charles Law would not apply. 5.525.m. - 1.16 15 -0.39 (25.2 +22915 T2 = 298.35 -0.39 D.P'= 0.93 atm T' 27.4" (300.55k)

part c and d please ?

Answer 1

C. Temperature is maintained. Pq V = P2 V2 (Boyle's law) 717.8. atm X 525.0mL = P2 X 11 s.Om P2 = 4:312 atm = 4.312x 760 mmng 213276.9 mm Hg do PV=MRT n = PV -0.93 atm x 0.2L RT 0.0821 Latm x 300.55K - Knol z 0.0075 38 mol Mass = (14 g/mol) x2 xo.007538 mal = 0.21 g