5 Determine the standard enthalpy of formation of Fe2O3(s) given the thermochemical equations below. 2Fe(s) + 3/2 O2(g) → Fe2O3(s) AH(Fe2O3) = ? Fe(s) + 3 H2O() Fe(OH)3(s) + 3/2 H2(g) AH° = +160.9 kJ/mol-rxn H2(g) + 1/2O2(g) → H2O(1) A. H° = -285.8 kJ/mol-rxn Fe2O3(s) + 3 H2O(1) ► Fe(OH),(s) A, Hº +288.6 kJ/mol-rxn

answer; 4 electrons;

(newtons first law)

conservation of momentum law states :

momentum before collision = momentum after collision

momentum p (kg-m/s)= mass * velocity

say moving car=1 000 kg, velocity = 10 m/s then p = 1 000*10= 10 000 kg-m/s

say still car = 2 000 kg, velocity = 0 m/s then p = 2 000* 0 = 0 kg-m/s

total momentum prior to collision = 10 000 + 0 = 10 000 kg-m/s

momentum after collision = 10 000 kg-m/s

mass now = 1 000 kg+2 000 kg = 3 000 kg

10 000 = 3 000 kg* velocity m/s

10 000/3 000 = velocity

3.33 m/s = velocity after collision

an impulse is a force applied over time , momentum is exclusive of external forces and a perfect collision is implied