Question

Flexible container is charged with 25.00 L of gas at 371 K. under conditions of constant pressure and moles of gas what is the temperature of the gas when the volume is decreased by a factor of three

of the following molecules does not have a d. iodine trichloride, ICs sulfur dioxide, SO2 C. nitrogen trifluoride, NF3 d sulfur hexafluoride, SFe b. e. sulfur tetrafluoride, SF4 49. Which of the underlined atoms (Ci, C2. N, and O) are sp hybridized? NH2 a.Ci and C2 b. c1, C2, N, and O c. N and O d. Ci, C2, and N e. none

Answer 1

In these compounds option d sulfur hexafluoride SF_6 does not have a dipole moment because SF_6 have a octahedral of F atom is greater than s sulfur atom That the dipole moment of bonds cancel/polar bonds cancel Underlined atom (C_1,C_2,N,O) are not sp hybrid option e is correct C_1 and C_2 is sp^2 hybridized N and O is sp^3 hybridized According to problem constant pressure and moles from ideal gas PV = nRT so V directlyproportional T volume of gas V_1 = 25. L Temperature of gas T_1 = 373 K when volume decreases by a factor of three the V_2 = 25/3 V_1/T_1 = V_2/T_2 put the value in the equation 25L/373K = 25/3L/T_2 T_2 = 25/3L times 373K/25L 124.3 = 123K Temperature of gas (T_2) = 123K mass of He Helium = 988g molar mass of helium = 4 g/mole so moles of helium = 988/4 g/mole = 24.7 mole volume of gas = 45.0L temperature = 23 Degree C or 296 K according to ideal gas law PV = nRT P = 247 mole times 0.08206L atm mole^-1 k^-1 times 296K/45.0L = 133.33 = 133 atm pressure inside the cylinder = 133 atm