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ANS_EX_8 - Swieter p 1 of 5 CHM1040 Answer to Exercise Set...

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Swieter; p 1 of 5 X:\~Natsci\ADJUNCT\bhaskar\pdf\1041\ANS_EX#8.DOC CHM1040 Answer to Exercise Set 8 QUESTIONS ABOUT GASES 1. a. atm 0.714 = torr 760 atm 1 torr x 543 b. torr 10 x 1.33 = atm 1 torr 760 x atm 1.75 3 c. mL 10 x 4.32 = L 1 mL 10 x L 43.2 4 3 d. g/L 10 x 1.004 = L 1 mL 10 x mL g 1.004 3 3 e. f. K 293 = 273.15 + C 20 ° C 173 _ = 273.15 _ K 100 ° 2. Since a torr is defined as a mm Hg at 0 ° C and sea level, the mercury would be more dense (it would contract at the sub-zero temperatures at the north pole) and thus one would need to add mm Hg to the reading to get the correct pressure. Conversely, at the equator (here the temperature is much higher than zero and the mercury would expand) one would need to subtract mm Hg from the reading to get the correct pressure. 3. a. L 1.12 = torr 800 torr 760 x L 1.18 = P P V = V or V P = V P 2 1 1 2 2 2 1 1 b. L 2.24 = torr 760 atm 1 x atm 0.526 torr 760 x L 1.18 = P P V = V or V P = V P 2 1 1 2 2 2 1 1 c. 2 P = V 2 V P = V V P = P or V P = V P 1 1 1 1 2 1 1 2 2 2 1 1 4. a. mL 133 = K . 250 K . 333 x mL 100 = T T V = V or T V = T V 15 15 1 2 1 2 2 2 1 1 b. C _ 112 or K . 385 = mL 100 K . 250 x mL 154 = V T V = T or T V = T V 23 15 1 1 2 2 2 2 1 1 c. V 2 = T T 2 x V = T T V = V or T V = T V 1 1 1 1 1 2 1 2 2 2 1 1 5. a. mL 10 x 2.00 = mol 0.250 mol 1.25 x mL 400 = n n V = V or n V = n V 3 1 2 1 2 2 2 1 1 b. mol 0.200 = mL 400 mL 320 x mol 0.250 = V V n = n or n V = n V 1 2 1 2 2 2 1 1 c. V 2 = n n 2 x 1 Vsub = n n V = V or n V = n V 1 1 1 1 2 1 2 2 2 1 1 6. 3 P 2 = T n ) V x (3 ) T x (4 ) n x (‰ V P = T n V T n V P = P or T n V P = T n V P 1 1 1 1 1 1 1 1 1 1 2 2 2 1 1 2 2 2 2 2 1 1 1 1
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