a)
nr.moli = m/M = 355/71 = 5 moli Cl2
1 mol Cl2 ......................... 2x6,022x10la23 atomi Cl
5 moli Cl2 .................... x = 3,011x10la24 atomi Cl
b)
PV = nRT => P = nRT/V
= 5x0,082x(273+17)/20
= 5,95 atm
c)
n = V/Vm => V = nxVm = 10x22,4 = 224 L
d)
PVM = mRT
=> m = 2PVM/RT = 2x5,95x20x71/0,082(273+17)
= 710,6 g Cl2