B.) 3.0 x 10^2 g halothane. Please help me how to solve it. Thank you Properties
ID: 1043540 • Letter: B
Question
B.) 3.0 x 10^2 g halothane. Please help me how to solve it. Thank you Properties 44. (mole %): 4.5% HS, 3.0% CO2, balance N,"ntion pressure gauge attached to the cylinder reads 46 Calculate the partial pressure of each gas, in atmo- spheres, in the cylinder. and 38. A cylinder of compressed gas is labeled"Com 45. ure 39) A halothane-oxygen mixture (GHBrClF3 + 02) can be 46. used as an anesthetic. A tank containing such a mixture has the following partial pressures: P (halothane) - 170 mm Hg and P (02) 570 mm Hg. Hg at obile (a) What is the ratio of the number of moles of halo- thane to the number of moles of O2? (b) If the tank contains 160 g of O2, what mass of Diffus (See S 47 CHBrCIFs is present? e the 40. A collapsed balloon is filled with He to a volume of 12.5 L at a pressure of 1.00 atm. Oxygen, 02, is then added so that the final volume of the balloon is 26 L with a total pressure of 1.00 atm. The temperature, which remains constant throughout, is 21.5 °C. (a) What mass of He does the balloon contain? (b) What is the final partial pressure of He in the sure of O2 and 2O(g) in 02 75-L lete 22°C, balloon? 48. (c) What is the partial pressure of Oz in the balloon? (d) What is the mole fraction of each gas? 49 Kinetic-Molecular Theory (See Section 10-6 and Example 10.11.) You have two flasks of equal volume. Flask A H2 at 0 C and 1 atm pressure. Flask B contains gas at 25 °C and 2 atm pressure. Compare thes gases with respect to each of the following (a) average kinetic energy per molecule (b) root mean square speed (c) number of molecules (d) mass of gas wing contains 50 50-L be in the of Nonid See S us 42. Equal masses ofExplanation / Answer
39)
if halothane is denoted by 1
and O2 by 2
a)
partial pressure = (mole fraction)*(total pressure)
use,
ratio of mole fraction = ratio of number of mol
so,
ratio of number of mol = ratio of partial pressure
n1/n2 = p1/p2
= 170/570
= 0.30
b)
molar mass of O2 = 32 g/mol
number of mol of O2(n2) = (given mass of O2)/(molar mass of O2 )
= 160/32
= 5.0
use,
n1/n2 = 0.30
n1/5.0 = 0.30
n1 = 1.5 mol
molar mass of C2HBrClF3 = 197.4 g/mol
mass of C2HBrClF3 = (n1)*(molar mass of C2HBrClF3)
= 1.5*197.4
= (3.0*10^2) g
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.