Calculate the number of moles of ethylene (C_2H_2) in 950 mL of ethylene at STP.
ID: 1014858 • Letter: C
Question
Calculate the number of moles of ethylene (C_2H_2) in 950 mL of ethylene at STP. 42.4 moles 1.12 moles 2.24 moles 0.0424 moles Calculate the volumes at STP occupied by 4.5 times 10^23 molecules of helium gas. 0.0698 15.6 L 6.98 L 22.4 L Calculate the molecular mass of a gas where 2.50 liters at STP has a mass of 7.14 grams. A sample of gas has a volume of 1000.mL when measured at 30degree/c and 1.00 atm. What will be the pressure exerted by the gas if the volume is reduced to 350 mL while the temperature remains at 30 degree C? 1.00 atm 12.0 atm 2.86 atm 5.40 atm A gas sample occupies a volume of 5.00 L at 40.0 degree C and 1.00 atm. What will be the volume occupied by this gas at 75.0 degree C and 0.500 atm. 1.00 L 0.0900 L 50.0 L 11.1 L none of these A gas sample measures 2.00 L at STP. Calculate its temperature, in C, if the volume is changed to 1.00 L and the pressure is changed to 0.500 atm. 150 degree C 68.3 degree C 204.7 degree C 273 degree C -204.7 degree C What is the volume occupied by 13.7 g Cl_2(g) at 45 degree C and 745 mmHg? 1.00 L 5.14 L 12.0 L 2.86 L none of these How many molecules of N_2 remain in an ultrahigh vacuum system of 128-mL volume when the pressure is reduced to 5 times 10^-10 mm Hg at 25 degree C? 1 times 10^10 5 times 10^-10 6 times 10^9 2 times 10^9 none of these A mixture of gases has the following partial pressures for the component gases at 20 degree C in a volume of 1L: nitrogen, 112 mm Hg; argon, 252 mm Hg; helium, 385 mmHg; neon, 155 mm Hg. Calculate the total pressure (in mm Hg) and the total moles of the mixture, and the moles of helium gas. 904 mm Hg; 0.0495 mols(total); 0.0211moles(He) 904 mmHg; 0.495 mols(total); 0.00211 moles(He) 904 mmHg; 1.12 mols(total); 0.495 moles(He) 904 mm Hg; 2.11 mols(total); 1.15 moles(He) 904 mm Hg; 0.0434 mols(total); 0.0211 moles(He)Explanation / Answer
23. 22400ml ------------> 1 mole
950ml --------------> 1*950/22400 = 0.0424moles >>>>> answer d
24. 6.023*1023 molecules -------------> 22400ml
4.2*1023 molecules ---------> 22400*4.2*1023/6.023*1023 = 15620ml = 15.6L >>>> answer b
25. 22.4 L -------------> 1 mole
2.5L ---------------> 1mole*2.5/22.4 = 0.11 moles
molar mass = mass of substance/no of moles
= 7.14/0.11 = 64 g/mole >>>>> answer a
26. initial Final
P1 = 1atm P2 =
V1 = 1000ml V2 = 350ml
T1 = 30C0 = 30+273 = 303K T2= 30c0 = 30+273 = 303K
P1V1/T1 = P2V2/T2
P2 = P1V1T2/T1V2
= 1*1000*303/303*350 = 2.857atm
27. P1V1/T1 = P2V2/T2
V2 = P1V1T2/P2T1
= 1*5*348/0.5*313 = 11.1L >>>>> answer d
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