1. PV = nRT P = 534 torr = 534/760 = 0.7 atm V = 345ml = 0.345L T = 25+273 = 298
ID: 699962 • Letter: 1
Question
1.
PV = nRT
P = 534 torr = 534/760 = 0.7 atm
V = 345ml = 0.345L
T = 25+273 = 298K
PV = nRT
n = PV/RT
= 0.7*0.345/0.0821*298 = 0.00987moles
2.
Initial Final
V1 = 736ml V2 =
T1 = 45+273 = 318K T2 = 105+273 = 378K
P1 = 719 torr P2 = 972 torr
P1V1/T1 = P2V2/T2
V2 = P1V1T2/T1P2
= 719*736*378/318*972 = 647.15ml >>>>answer
3. PV = nRT
n = W/M
PV = WRT/M
M = WRT/PV
P = 496torr = 496/760 = 0.652atm
V = 250ml = 0.25L
T = 37+273 = 310K
W = 2.43g
M = WRT/PV
= 2.43*0.0821*310/0.652*0.25 = 379g/mole
Explanation / Answer
If an ideal gas sample at 534 torr pressure occupies 345 mL at 25°C, how many moles of that gas are present? Report your answer to 3 significant figures.
If If an ideal gas occupies 736 mL at 45°C and 719 torr, what volume will it occupy at 105°C and 972 torr? Report your answer in mL to 3 significant figures.
A 2.43 g sample of a gaseous compound has a pressure of 496 torr in a volume of 250 mL at 37°C. What is its molar mass? Report your answer to 3 significant figures
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