I need help with parts a-e on this homework problem Question 4-Propane gas (C3H8) is used as a propellant in aerosol cans.a. Calculate the density of propane gas in a 455 mL container at a pressure of 1.75 atm at25.0°C.b. If the can in part a is cooled to 5.0 °C, what is the new pressure?c. If an additional 10.0 grams of propane is added to the can described in part b, what isthe new pressure of the propane gas?d. If all the propane described in part c were transferred to a balloon at STP, whatvolume would the propane now occupy? Using the postulates of Kinetic MolecularTheory, explain why there is a new pressure?e. 1.682 g ofa different gas is placed in a 500.0 mL can at 20.0 °C and is found to exert apressure of 2.50 atm. What is the molar mass of the gas? Identify the gas-is ithydrogen gas, oxygen gas, chlorine gas or radon gas?

Note: As per our guidelines, we are only allowed to answer first three subparts. To get the solutions for the rest of the subparts, kindly re-submit the whole question again with the mention of the other two subparts to be specifically answered. part a) Given: Volume occupied by propane, V = 455 mL = 0.000455 m3 Pressure exerted by the gas, P = 1.75 atm = 177318.8 Pa Temperature of the gas, T = 250C = 298.15 K molar mass of the gas, M = 44.1 g/mol by Ideal gas law in terms of density, PM=ρRTρ=PMRT=1.75×44.10.0821×298.15=3.1528 gL=3.1528kgm3 Part b) If the can is cooled to 50C, new pressure? P1 = 177318.8 Pa ; T1 = 298.15 K ; T2 = 278.15 K As we know, Pressure is directly proportional to temperature, P1P2=T1T2P2=T2T1P1=278.15×177318.8298.15=165424.19 Pa Part c) Additional 10 g of propane is added to the can, what is the new pressure? As we know, Pressure is directly proportional to the number of moles, To find n1: n1=(PVRT)1=177318.8×0.0004558.314×298.15=0.0325 mol To find n2, number of moles in 10 g of propane = 1044=0.2272 mol n2 = 0.2272+0.0325=0.2597 mol Now, P1P2=n1n2P2=n2n1P1=0.2597×177318.80.0325=1417310.4 Pa