And we see that the 5g electron is number 121,i.e.element 121 will be the lightest element containing this electron. C6.(a)A sample of a gas is at 50C in a 50.0 L container at 1.1 atm.The gas is expanded to 75 L and cooled to 0C.Calculate the new pressure. piV-PaVz TT Thus.P,-P:Vj-1.1atm(5OL)273 K)62atm TV2 (50+273)K(75L) (b)1.00 mol of He under certain conditions exerts a pressure of 5.00 atm.1.00 mol of Cl under the same conditions exerts a lower pressure.Why? Chlorine is a larger molecule than the very small He atoms.Larger molecules have larger attractive forces for one another.Hence the Cl2 molecules are pulled together,reducing the pressure they exert. (c)Calculate the density of Cl at 10.0 atm and 100C(ing/L) p=P(MW) 10 atm(0.070 kg mol-') RT 0.082LamK-m0P100+273K-2.29x10-2kgL'=2,9gL C7.Predict the shapes of the following molecules.Wrong name=zero marks. (a)AsC15-2 Electrons=5+(5x 7)+2=42.Making the five bonds uses 10 of these.Completing the octets on the Cl atoms uses another 30.This leaves two,which are placed as a lone pair on the As atom.Thus the shorthand notation is AXsE.This is a square pyramid. (b)PCIs Electrons=5+(5x 7)=40.Making the five bonds uses 10 of these.Completing the octets on the Cl atoms And we see that the 5g electron is number 121, i.e. element 121 will be the lightest element containing this electron. C6. (a) A sample of a gas is at 50o C in a 50.0 L container at 1.1 atm. The gas is expanded to 75 L and cooled to 0o C. Calculate the new pressure. pV pV T T pVT . atm( L)( K) Thus, p . atm T V ( ) K ( L) = = = = + 11 2 2 1 2 112 2 1 2 1 1 50 273 0 62 50 273 75 (b) 1.00 mol of He(g) under certain conditions exerts a pressure of 5.00 atm. 1.00 mol of Cl2(g) under the same conditions exerts a lower pressure. Why? Chlorine is a larger molecule than the very small He atoms. Larger molecules have larger attractive forces for one another. Hence the Cl2 molecules are pulled together, reducing the pressure they exert. (c) Calculate the density of Cl2(g) at 10.0 atm and 100o C (in g/L). p(MW) atm( . kg mol ) . kg L . g RT . L atm K mol ( )K − L − − − − − ρ= = = × = + 1 2 1 1 1 10 0 070 2 29 10 22 9 0 082 100 273 1 C7. Predict the shapes of the following molecules. Wrong name = zero marks. (a) AsCl5 −2 Electrons = 5 + (5 x 7) + 2 = 42. Making the five bonds uses 10 of these. Completing the octets on the Cl atoms uses another 30. This leaves two, which are placed as a lone pair on the As atom. Thus the shorthand notation is AX5E. This is a square pyramid. (b) PCl5 Electrons = 5 + (5 x 7) = 40. Making the five bonds uses 10 of these. Completing the octets on the Cl atoms 8